Partial Differential Equations — UPSC Maths Optional Notes

1. Understanding the Formation of a PDE by Eliminating Constants

Why we need it: many families of surfaces are given by one formula with adjustable numbers in it, such as 'all unit spheres whose centres lie in the xy-plane'. We want a single equation that every member of the family satisfies and that contains none of the adjustable numbers.

Such an equation involves z and its partial derivatives, so it is a partial differential equation (PDE). Solving PDEs later means running this process backwards: from the equation we recover the family.

In words: in z = f(x, y, a, b), x and y are the independent variables, z depends on them, and a, b are arbitrary constants: fixed on any one surface, but different from one surface to another.

Differentiating partially with respect to x keeps y and also a, b fixed; this gives p = ∂z/∂x, and similarly q = ∂z/∂y. We now have three equations (for z, p and q) containing the two unknown constants.

Eliminating two unknowns from three equations leaves exactly one relation F(x, y, z, p, q) = 0. Notice what it says: at every point of every member, the slopes p, q are tied to x, y, z in the same way.

First example, slowly: form the PDE of the spheres (x − a)² + (y − b)² + z² = 1.

Step 1: count. There are two constants a, b and two independent variables x, y, so first derivatives should be enough and the PDE should be of first order.

Step 2: differentiate with respect to x, remembering that z depends on x (chain rule on z²): 2(x − a) + 2zp = 0, so x − a = −zp.

Step 3: differentiate with respect to y in the same way: 2(y − b) + 2zq = 0, so y − b = −zq.

Step 4: substitute both into the equation of the family: z²p² + z²q² + z² = 1.

Step 5: the answer is z²(p² + q² + 1) = 1; read it once more to make sure no a or b is left.

Step 6: a sanity check: at the top of each sphere the tangent plane is horizontal, p = q = 0, and the PDE gives z² = 1, which is right since the top is at height 1.

Non-example: z = (a + b)x + y looks like a two-constant family, but only the sum c = a + b matters, so it is really a one-constant family. Then p = c and q = 1, and the counting breaks down: q = 1 and z = px + y are both valid PDEs. The condition that fails is the independence of the constants.

Building up: the general rule is 'one new equation for each constant to be removed'. With two independent variables the first derivatives supply two equations, so two constants give a first-order PDE.

A third constant forces the use of r, s or t and the PDE becomes of second order (for z = ax + by + cxy, s = c and the PDE is z = px + qy − xys, as in 'Choosing the Method' below).

Run backwards, the same count explains why a complete integral of a first-order PDE has exactly two independent constants (see 'Types of Solutions' below).

Where it is used: 10-mark 'form the PDE' questions, the tangent-plane families of quadrics that lead to Clairaut equations, and, in reverse, every complete integral found by Charpit's method in the next topics.

2. Understanding the Formation of a PDE by Eliminating Arbitrary Functions

Why we need it: some families cannot be described by a few numbers. 'All surfaces of revolution about the z-axis' contains a surface for every profile curve, so its description z = f(x² + y²) contains a whole unknown function f.

We want a PDE that every such surface satisfies whatever f is.

In words: in z = f(u) with u = u(x, y) known, f is an arbitrary differentiable function (it may be anything), while u is a fixed, known combination of x and y.

The chain rule gives p = f′(u)u_x and q = f′(u)u_y. At a given point f′(u) is one unknown number appearing in both equations, so it can be eliminated by division: u_yp − u_xq = 0.

First example, slowly: eliminate f from z = f(x² + y²).

Step 1: name the inner combination: u = x² + y², so u_x = 2x and u_y = 2y.

Step 2: differentiate with respect to x by the chain rule: p = f′(u) · 2x, where f′ is evaluated at u.

Step 3: differentiate with respect to y: q = f′(u) · 2y.

Step 4: multiply the first by y and the second by x: yp = 2xyf′(u) = xq.

Step 5: hence yp − xq = 0, which contains no f: this is the PDE.

Step 6: test it on two members. For z = x² + y², yp − xq = 2xy − 2xy = 0; for z = sin(x² + y²), p = 2x cos u and q = 2y cos u, and again yp − xq = 0.

Step 7: notice the geometry: yp − xq = 0 says the normal (p, q, −1) is perpendicular to (y, −x, 0), the direction of the horizontal circle about the z-axis, so the surface contains these circles, as a surface of revolution must.

Non-example: z = f(x) + g(y) contains two arbitrary functions. The first derivatives p = f′(x), q = g′(y) are two equations with two new unknowns, so nothing can be eliminated at first order; only the second derivative s = ∂²z/∂x∂y = 0 removes both. The rule 'one function gives a first-order PDE' does not apply.

Building up: for the more general relation φ(u, v) = 0, where u and v may contain z, differentiate with respect to x and y, remembering that ∂u/∂x on the surface is u_x + u_zp.

The two equations are linear in (φ_u, φ_v), which is not (0, 0), so their 2 × 2 determinant vanishes; expanding it gives Pp + Qq = R with Jacobians P, Q, R (worked out in 'Jacobian Elimination' below).

Whatever φ is, the result is linear in p and q. This is the seed of Lagrange's method in the next topic.

Where it is used: PDEs of cones, cylinders and surfaces of revolution; Lagrange's method, which reverses the calculation; and second-order cases such as f(x + ct) + g(x − ct), which give the wave equation.

3. Understanding Complete, General and Singular Integrals

Why we need it: a first-order PDE has a huge supply of solutions. To answer questions such as 'find the complete and singular integrals' we need to sort them into three kinds and know how each is obtained from the others.

In words: a complete integral z = f(x, y, a, b) is a solution for every choice of the two independent constants a, b.

A general integral is obtained by tying b to a by an arbitrary relation b = φ(a) and taking the envelope (the surface that touches every member) of the resulting one-parameter family.

A singular integral is the envelope of the whole two-parameter family: eliminate a and b from f, f_a = 0 and f_b = 0. It touches every member but is not itself a member.

First example, slowly: sort the solutions of the Clairaut equation z = px + qy + pq.

Step 1: try planes with constant slopes p = a, q = b: z = ax + by + ab. Check: p = a, q = b and px + qy + pq = ax + by + ab = z. This is a complete integral.

Step 2: for the singular integral, set f_a = x + b = 0 and f_b = y + a = 0, so a = −y and b = −x.

Step 3: substitute: z = (−y)x + (−x)y + (−y)(−x) = −xy.

Step 4: check it in the PDE: p = −y, q = −x, and px + qy + pq = −xy − xy + xy = −xy = z. It is a solution.

Step 5: notice that z = −xy is not a plane, so no choice of a, b produces it: it is genuinely new.

Step 6: for one general integral take b = a: z = a(x + y) + a²; the envelope condition x + y + 2a = 0 gives a = −(x + y)/2 and z = −(x + y)²/4.

Step 7: check: p = q = −(x + y)/2, so px + qy + pq = −(x + y)²/2 + (x + y)²/4 = −(x + y)²/4 = z. Another choice of φ gives another surface.

Non-example: for p = q the complete integral is z = a(x + y) + b. Here f_b = 1 can never be 0, so the system f_a = f_b = 0 has no solution and there is no singular integral. A constant that enters additively never produces one.

Building up: why is an envelope a solution? At each of its points it has the same z, p and q as the member it touches there, and that member satisfies F = 0 (proved in 'Derivation: Why Envelopes Are Solutions' below).

The singular integral can also be read off the PDE itself: here F = px + qy + pq − z, F_p = x + q = 0 and F_q = y + p = 0 give p = −y, q = −x, and F = 0 then gives z = −xy again.

Where it is used: Clairaut-type questions, the tangent planes of a quadric whose singular integral is the quadric itself, Charpit's method (which produces complete integrals), and Cauchy problems, which pick out one particular member.

4. Surfaces and Notation

Notation: z = z(x, y), p = ∂z/∂x, q = ∂z/∂y, r = ∂²z/∂x², s = ∂²z/∂x∂y, t = ∂²z/∂y².

A surface F(x, y, z) = 0 has normal direction (F_x, F_y, F_z); for z = z(x, y) the normal direction is (p, q, −1).

Idea: writing the surface as F = z − z(x, y) = 0 gives (F_x, F_y, F_z) = (−p, −q, 1), so p and q are simply the slopes of the surface in the x and y directions; a first-order PDE F(x, y, z, p, q) = 0 is therefore a condition on the tangent plane at each point.

One-parameter family f(x, y, z, a) = 0: its envelope is obtained by eliminating a from f = 0 and ∂f/∂a = 0; it touches each member along a curve (the characteristic of the family).

Idea: two neighbouring members f(x, y, z, a) = 0 and f(x, y, z, a + δa) = 0 meet in a curve; dividing their difference by δa and letting δa → 0 gives ∂f/∂a = 0 on that curve, which is why the envelope comes from f = 0, ∂f/∂a = 0. Picture: the unit spheres x² + y² + (z − a)² = 1 strung along the z-axis; ∂f/∂a = −2(z − a) = 0 gives a = z, and the envelope is the cylinder x² + y² = 1, touching each sphere along its equator.

Two-parameter family f(x, y, z, a, b) = 0: its envelope is obtained by eliminating a, b from f = 0, ∂f/∂a = 0, ∂f/∂b = 0.

Idea: a two-parameter envelope touches each member at a point, not along a curve. Example: the unit spheres (x − a)² + (y − b)² + z² = 1 with centres in the xy-plane; ∂f/∂a = 0 and ∂f/∂b = 0 give a = x, b = y, so z² = 1 and the envelope is the pair of planes z = ±1, touching each sphere at its top and bottom points.

5. Eliminating Arbitrary Constants

From z = f(x, y, a, b) find p and q, then eliminate a and b from the three equations to get F(x, y, z, p, q) = 0.

If there are more constants than independent variables, second (or higher) derivatives are needed and the PDE is of higher order.

Idea: the three equations for z, p, q can remove exactly two constants, so 'two constants, two independent variables' gives a first-order PDE; a third constant forces the use of r, s or t. With fewer constants than variables the PDE is not unique: z = ax + y gives both q = 1 and z = px + y, and either is a correct answer.

Example 1: z = (x − a)² + (y − b)². Then p = 2(x − a), q = 2(y − b), so p² + q² = 4z.

Example 2: z = ax + by + ab. Then p = a, q = b, so z = px + qy + pq (Clairaut form).

6. Eliminating Arbitrary Functions

z = f(u) with u = u(x, y): differentiate with respect to x and y and eliminate f′.

Example 3: z = f(x² + y²). Then p = 2xf′, q = 2yf′, so yp − xq = 0.

φ(u, v) = 0 with u, v functions of x, y, z: eliminating φ gives Pp + Qq = R, where P = ∂(u, v)/∂(y, z), Q = ∂(u, v)/∂(z, x), R = ∂(u, v)/∂(x, y).

Idea: φ(u, v) = 0 says only that u and v are functionally related on the surface; whatever φ is, the resulting PDE is linear in p and q (it is derived in Section 8 below). Lagrange's method in the next topic runs this calculation backwards: from Pp + Qq = R it recovers u and v.

Eliminating one arbitrary function gives a first-order quasilinear PDE; eliminating two arbitrary functions usually gives a second-order PDE.

Idea: an arbitrary function carries far more freedom than a constant, so it cannot be removed by counting equations; it is removed only because it enters through a known combination (x² + y², x + ct, ...). The word 'usually' matters: z = f(x) + g(y) gives the second-order PDE s = 0, but z = f(x + y) + g(x + y) is really one function of x + y and gives the first-order p = q.

Example 4: z = f(x + ct) + g(x − ct). Then z_tt = c²(f″ + g″) and z_xx = f″ + g″, so z_tt = c²z_xx (one-dimensional wave equation).

Example 5: z = f(x + iy) + g(x − iy) gives z_xx + z_yy = 0 (Laplace equation), since ∂²/∂y² brings a factor i² = −1.

7. Types of Solutions of F(x, y, z, p, q) = 0

Complete integral: a solution z = f(x, y, a, b) with two independent arbitrary constants, that is, the matrix [[f_a, f_xa, f_ya], [f_b, f_xb, f_yb]] has rank 2.

Idea: 'independent' means the two constants cannot be merged into one. z = (a + b)x + y looks like a two-constant family but is really z = cx + y; its matrix has rows (x, 1, 0) and (x, 1, 0), of rank 1, so it is not a complete integral of any first-order PDE.

General integral: put b = φ(a) in the complete integral and take the envelope of this one-parameter family; it involves an arbitrary function.

Idea: choosing b = φ(a) picks a one-parameter sub-family out of the complete integral, and its envelope is again a solution (proved in the derivation below). For a Clairaut equation the members are planes, so the general integral is made of developable surfaces built from those planes; different φ give different surfaces, which is how one complete integral produces infinitely many solutions.

Singular integral: the envelope of the whole two-parameter family, found by eliminating a, b from f = 0, f_a = 0, f_b = 0; it must be checked against the PDE.

Idea: the singular integral touches every member of the complete integral, so at each of its points it shares x, y, z, p, q with some member and therefore satisfies the PDE, yet no choice of a, b produces it. Not every PDE has one: if the constant enters additively, z = g(x, y, a) + b, then ∂f/∂b = 1 ≠ 0 and there is no singular integral (for example z = ax + ay + b for p = q).

The singular integral can also be found from the PDE by eliminating p, q from F = 0, F_p = 0, F_q = 0.

Idea: both eliminations can throw up extra factors, such as a locus of points where the members are not smooth; keep only the factor that actually satisfies the PDE, and say that you checked it.

Example 6: z = px + qy + p² + q² has complete integral z = ax + by + a² + b².

Then x + 2a = 0 and y + 2b = 0 give a = −x/2, b = −y/2, so the singular integral is 4z = −(x² + y²), a paraboloid.

Orthogonal surfaces: surfaces cutting every member of f(x, y, z) = c at right angles satisfy f_xp + f_yq = f_z, a Lagrange equation.

Why it works: the surface z = z(x, y) has normal (p, q, −1) and the member f = c has normal (f_x, f_y, f_z); the surfaces cut at right angles when these normals are perpendicular, f_xp + f_yq − f_z = 0, which is the stated equation.

Example 7: surfaces orthogonal to the spheres x² + y² + z² = c satisfy xp + yq = z, whose solution is φ(y/x, z/x) = 0 (cones with vertex at the origin).

8. Jacobian Elimination, Envelopes and Geometric Families

Why φ(u, v) = 0 gives Pp + Qq = R: differentiating with respect to x and y gives φ_u(u_x + u_zp) + φ_v(v_x + v_zp) = 0 and φ_u(u_y + u_zq) + φ_v(v_y + v_zq) = 0; a non-zero (φ_u, φ_v) needs the 2 × 2 determinant to vanish, and expanding it gives Pp + Qq = R with the Jacobians P, Q, R.

Two arbitrary functions z = f(u) + g(v): compute p, q, r, s, t; the five equations contain f′, g′, f″, g″, and eliminating these four leaves one second-order PDE (look first for a combination of r, s, t that removes f″, g″).

Tangent planes to a central quadric: the plane lx + my + nz = d touches x²/a² + y²/b² + z²/c² = 1 iff a²l² + b²m² + c²n² = d²; writing the plane as z = px + qy + k gives the Clairaut PDE (z − px − qy)² = a²p² + b²q² + c², whose singular integral is the quadric itself.

Standard geometric families: cones with vertex (a, b, c) satisfy (x − a)p + (y − b)q = z − c; cylinders with generators parallel to (l, m, n) satisfy lp + mq = n; surfaces of revolution about the z-axis, z = f(x² + y²), satisfy yp − xq = 0.

After forming a second-order PDE you may be asked its type: read off A, B, C (coefficients of r, s, t) and use the sign of B² − 4AC in the stated region.

9. Derivation: Why Envelopes Are Solutions

Let z = f(x, y, a, b) be a complete integral of F(x, y, z, p, q) = 0, so F(x, y, f, f_x, f_y) = 0 for all x, y, a, b.

On the envelope, a and b become functions a(x, y), b(x, y) fixed by f_a = 0, f_b = 0, and z = f(x, y, a(x, y), b(x, y)).

By the chain rule, ∂z/∂x = f_x + f_aa_x + f_bb_x = f_x and ∂z/∂y = f_y, because f_a = f_b = 0.

So at each point the envelope has the same z, p, q as the member with those values of a, b; that member satisfies F = 0, hence so does the envelope.

With b = φ(a) there is one condition f_a + f_bφ′(a) = 0, and the same calculation shows that the general integral is a solution.

Why F = F_p = F_q = 0 gives the singular integral: differentiating F(x, y, f, f_x, f_y) = 0 with respect to a gives F_zf_a + F_pf_xa + F_qf_ya = 0, and similarly with b.

On the singular integral f_a = f_b = 0, so F_pf_xa + F_qf_ya = 0 and F_pf_xb + F_qf_yb = 0; when f_xaf_yb − f_yaf_xb ≠ 0 this forces F_p = F_q = 0.

Check on Clairaut: z = px + qy + pq has complete integral z = ax + by + ab; f_a = x + b = 0, f_b = y + a = 0 give a = −y, b = −x and z = −xy. Then p = −y, q = −x and px + qy + pq = −xy, so the singular integral z = −xy (a hyperbolic paraboloid) does satisfy the PDE.

10. Choosing the Method

If the family has as many constants as independent variables → find p and q once and eliminate; the PDE is of first order.

If there are more constants than variables → bring in r, s, t and look for a combination that the constants cannot survive (for z = ax + by + cxy, s = c and the PDE is z = px + qy − xys).

If z = f(u) with u = u(x, y) known → p = f′u_x, q = f′u_y, so u_yp − u_xq = 0.

If the relation is φ(u, v) = 0 and u or v contains z → compute the Jacobians P, Q, R directly and write Pp + Qq = R.

If z = f(u) + g(v) → go to second derivatives; with linear arguments such as x ± ct the combination is immediate, otherwise solve the p, q equations for f′, g′ first.

If the family is described in words (cones, cylinders, tangent planes, surfaces of revolution, orthogonal surfaces) → write its equation with constants or a function first, or quote the standard PDE in Section 8 and justify it in one line.

11. Common Mistakes

Leaving a constant behind: the final PDE must contain no a, b or f; re-read it before writing 'Answer'.

Forgetting that u and v contain z: in φ(u, v) = 0, ∂u/∂x means u_x + u_zp, not u_x alone.

Unchecked singular integral: the eliminant of f = f_a = f_b = 0 may contain extraneous factors; substitute back into the PDE.

Mixing up the integrals: a complete integral has two constants, a general integral an arbitrary function, a singular integral neither.

Wrong sign in the orthogonality condition: with normal (p, q, −1) the condition is f_xp + f_yq = f_z, not f_xp + f_yq + f_z = 0.

Dividing by a coefficient that may vanish: writing a plane as z = px + qy + k assumes it is not perpendicular to the xy-plane; state this, as UPSC questions do.

Link: the singular integral of a Clairaut PDE is the two-variable form of the singular solution of Clairaut's ODE y = px + f(p) (Ordinary Differential Equations), found the same way from an envelope.

Link: the tangency condition a²l² + b²m² + c²n² = d² comes from Analytic Geometry (tangent planes to conicoids), and z = f(x + ct) + g(x − ct) is D'Alembert's solution of the wave equation read backwards (Vibrating String topic).

12. Past UPSC Questions Solved

Example 8 (UPSC 2018, 10 marks): Find the partial differential equation of the family of all tangent planes to the ellipsoid : x² + 4y² + 4z² = 4, which are not perpendicular to the xy plane.

The ellipsoid is x²/4 + y²/1 + z²/1 = 1, so a² = 4, b² = 1, c² = 1.

A plane not perpendicular to the xy-plane contains z, so it can be written z = px + qy + k, i.e. px + qy − z = −k, where p, q are its slopes ∂z/∂x, ∂z/∂y.

Tangency a²l² + b²m² + c²n² = d² with (l, m, n) = (p, q, −1), d = −k gives k² = 4p² + q² + 1.

Eliminating the constant k = z − px − qy gives the PDE.

Answer: (z − px − qy)² = 4p² + q² + 1, i.e. z = px + qy ± √(4p² + q² + 1), a Clairaut equation whose singular integral is the ellipsoid.

Example 9 (UPSC 2016, 10 marks): Find the general equation of surfaces orthogonal to the family of spheres given by x² + y² + z² = cz.

Write the family as f = (x² + y² + z²)/z = c: f_x = 2x/z, f_y = 2y/z, f_z = (z² − x² − y²)/z².

Orthogonal surfaces satisfy f_xp + f_yq = f_z; multiplying by z²: 2xz p + 2yz q = z² − x² − y².

Auxiliary equations dx/(2xz) = dy/(2yz) = dz/(z² − x² − y²); the first two give dx/x = dy/y, so y/x = c₁.

Multipliers x, y, z give the denominator z(x² + y² + z²), so d(x² + y² + z²)/(x² + y² + z²) = dx/x and (x² + y² + z²)/x = c₂.

Answer: φ(y/x, (x² + y² + z²)/x) = 0, φ arbitrary.

Example 10 (UPSC 2020, 10 marks): Form a partial differential equation by eliminating the arbitrary functions f(x) and g(y) from z = y f(x) + x g(y) and specify its nature (elliptic, hyperbolic or parabolic) in the region x > 0, y > 0.

p = yf′(x) + g(y), q = f(x) + xg′(y) and s = ∂²z/∂x∂y = f′(x) + g′(y).

Then xp + yq = xy(f′ + g′) + (xg + yf) = xys + z, which contains no arbitrary function.

Type: A = 0, B = xy, C = 0 (coefficients of r, s, t), so B² − 4AC = x²y² > 0 in x > 0, y > 0.

Answer: xy ∂²z/∂x∂y − x ∂z/∂x − y ∂z/∂y + z = 0, hyperbolic in the region x > 0, y > 0.

Example 11 (UPSC 2022, 10 marks): It is given that the equation of any cone with vertex at (a, b, c) is f((x − a)/(z − c), (y − b)/(z − c)) = 0. Find the differential equation of the cone.

Put u = (x − a)/(z − c), v = (y − b)/(z − c) and differentiate f(u, v) = 0; after multiplying by (z − c)²: f_u[(z − c) − (x − a)p] − f_v(y − b)p = 0 and −f_u(x − a)q + f_v[(z − c) − (y − b)q] = 0.

For non-zero (f_u, f_v) the determinant vanishes: [(z − c) − (x − a)p][(z − c) − (y − b)q] − (x − a)(y − b)pq = 0.

Expanding, (z − c)² − (z − c)[(x − a)p + (y − b)q] = 0, and z ≠ c away from the vertex.

Answer: (x − a)p + (y − b)q = z − c.

Example 12 (UPSC 2023, 10 marks): By eliminating the arbitrary functions f and g from z = f(x² − y) + g(x² + y), form partial differential equation.

p = 2x(f′ + g′), q = −f′ + g′, r = 2(f′ + g′) + 4x²(f″ + g″), t = f″ + g″ (f at x² − y, g at x² + y).

From p, f′ + g′ = p/(2x); substituting this and f″ + g″ = t in r gives r = p/x + 4x²t.

Answer: x ∂²z/∂x² − ∂z/∂x − 4x³ ∂²z/∂y² = 0.

Example 13 (UPSC 2024, 10 marks): Show that if f and g are arbitrary functions of their respective arguments, then u = f(x − kt + iαy) + g(x − kt − iαy), is a solution of
∂²u/∂x² + ∂²u/∂y² = (1/C²) ∂²u/∂t², where α² = 1 − k²/C².

Let ξ = x − kt + iαy and η = x − kt − iαy, with f″ = f″(ξ), g″ = g″(η).

u_xx = f″ + g″, u_yy = (iα)²f″ + (−iα)²g″ = −α²(f″ + g″), u_tt = (−k)²(f″ + g″) = k²(f″ + g″).

So u_xx + u_yy = (1 − α²)(f″ + g″) = (k²/C²)(f″ + g″) = (1/C²)u_tt, using α² = 1 − k²/C².

Answer: u satisfies ∂²u/∂x² + ∂²u/∂y² = (1/C²) ∂²u/∂t² for all twice-differentiable f, g. Hence proved.

13. Practice Exercises

Exercise 1 (basic): Form the PDE of the family z = ax + by.

Answer: p = a, q = b, so z = px + qy.

Exercise 2 (basic): Eliminate the arbitrary function f from z = f(y/x).

Answer: p = −(y/x²)f′, q = (1/x)f′, so xp + yq = 0.

Exercise 3 (basic): Is z = ax + by + a² + b² a complete integral of z = px + qy + p² + q²? Justify with the rank test.

Answer: yes: it satisfies the PDE, and the rows (f_a, f_xa, f_ya) = (x + 2a, 1, 0) and (f_b, f_xb, f_yb) = (y + 2b, 0, 1) have rank 2.

Exercise 4 (standard): Form the PDE by eliminating a and b from z = (x² + a)(y² + b).

Answer: p = 2x(y² + b), q = 2y(x² + a), so pq = 4xyz.

Exercise 5 (standard): Eliminate f from z = e^yf(x − y).

Answer: p = e^yf′ and q = e^yf − e^yf′ = z − p, so p + q = z.

Exercise 6 (standard): Eliminate φ from φ(x + y + z, x² + y² − z²) = 0.

Answer: the Jacobians give P = −2(y + z), Q = 2(x + z), R = 2(y − x), so (y + z)p − (x + z)q = x − y.

Exercise 7 (standard): Find a complete integral and the singular integral of z = px + qy + log(pq).

Answer: complete integral z = ax + by + log(ab) (ab > 0); x + 1/a = 0, y + 1/b = 0 give the singular integral z = −2 − log(xy), for xy > 0.

Exercise 8 (UPSC level): Find the PDE satisfied by all surfaces of revolution whose axis is the line x = y = z.

Answer: such surfaces are φ(x + y + z, x² + y² + z²) = 0 (circles in planes perpendicular to the axis, centred on it); eliminating φ gives (z − y)p + (x − z)q = y − x.

Exercise 9 (UPSC level): Eliminate the arbitrary functions from z = f(xy) + g(x/y).

Answer: with u = xy, v = x/y one finds xp = uf′ + vg′ and yq = uf′ − vg′; going to second derivatives and eliminating f′, g′, f″, g″ gives x²r − y²t + xp − yq = 0.

Exercise 10 (UPSC level): For z = px + qy + p² + q², take the complete integral z = ax + by + a² + b², put b = a and find the corresponding surface of the general integral; verify that it satisfies the PDE.

Answer: z = a(x + y) + 2a² and x + y + 4a = 0 give z = −(x + y)²/8; then p = q = −(x + y)/4 and px + qy + p² + q² = −(x + y)²/4 + (x + y)²/8 = z.

1. Understanding the Auxiliary Equations and Characteristic Curves

Why we need it: Pp + Qq = R is a single equation for an unknown surface z = z(x, y), and we cannot integrate it directly. We want to replace it by ordinary differential equations, which we know how to integrate.

In words: P, Q, R are given functions of x, y, z, so at every point of space they define an arrow (P, Q, R). The surface z = z(x, y) has normal (p, q, −1).

Pp + Qq − R = 0 is the dot product of these two vectors, so the PDE says exactly this: the arrow (P, Q, R) lies in the tangent plane of the surface at every point.

A curve that follows the arrows has dx : dy : dz = P : Q : R, which is written dx/P = dy/Q = dz/R (the auxiliary equations). These curves are the characteristic curves.

The auxiliary equations are two ODEs, so their solution is two relations u(x, y, z) = c₁, v(x, y, z) = c₂; each pair (c₁, c₂) labels one characteristic. A surface built from characteristics is a relation between the labels, φ(u, v) = 0.

First example, slowly: solve yzp + xzq = xy.

Step 1: read off P = yz, Q = xz, R = xy and write dx/(yz) = dy/(xz) = dz/(xy).

Step 2: take the first two fractions and cross-multiply: xz dx = yz dy; cancel z (z ≠ 0) to get x dx = y dy.

Step 3: integrate: x² − y² = c₁. Check that it is constant along characteristics: Pu_x + Qu_y + Ru_z = yz(2x) + xz(−2y) + 0 = 0.

Step 4: take the first and third fractions: xy dx = yz dz, so x dx = z dz and x² − z² = c₂.

Step 5: the two integrals are independent: the first contains y and not z, the second z and not y, so neither is a function of the other.

Step 6: general solution φ(x² − y², x² − z²) = 0, or x² − z² = f(x² − y²).

Step 7: test one member: f = 0 gives z = x, so p = 1, q = 0 and yzp + xzq = yz = xy, as required.

Non-example: p² + q = z is not of the form Pp + Qq = R, because p occurs squared. The PDE no longer fixes one direction (P, Q, R) in the tangent plane, so there are no auxiliary equations of Lagrange's type; such equations need Charpit's method (next topic).

Building up: along a characteristic u stays constant, so Pu_x + Qu_y + Ru_z = 0, and likewise for v. Solving these two equations gives P : Q : R as the Jacobians of u, v, and the Jacobian elimination of the previous topic shows that every φ(u, v) = 0 satisfies Pp + Qq = R.

Conversely, every solution surface contains the arrow (P, Q, R) in its tangent plane, so a characteristic starting on it stays on it; the surface is a union of characteristics. The full argument is in 'Derivation: Proof of Lagrange's Theorem' below.

Where it is used: every 'find the general solution of Pp + Qq = R' question, surfaces orthogonal to a given family, and the method of characteristics for Cauchy problems in the next topic.

2. Understanding Multipliers for the Auxiliary Equations

Why we need it: often only one pair of fractions separates into two variables, or none does. A second integral then has to be manufactured by combining all three fractions.

In words: call the common value of the fractions k, so dx = kP, dy = kQ, dz = kR. For any multipliers l, m, n (numbers or functions), l dx + m dy + n dz = k(lP + mQ + nR).

So if we choose l, m, n with lP + mQ + nR = 0, then l dx + m dy + n dz = 0 along every characteristic. If the left side is an exact differential dw, then w = constant is an integral.

The multipliers are guessed from the shape of P, Q, R: look for terms that cancel when the three products are added.

First example, slowly: solve z(x + y)p + z(x − y)q = x² + y².

Step 1: the auxiliary equations are dx/(z(x + y)) = dy/(z(x − y)) = dz/(x² + y²); no pair separates directly.

Step 2: R is quadratic in x, y, while P, Q are z times linear terms; multiplying dx, dy by x, y and dz by z makes every product of the form z × (quadratic). Try l, m, n = x, −y, −z.

Step 3: check: xz(x + y) − yz(x − y) − z(x² + y²) = z(x² + xy − xy + y² − x² − y²) = 0.

Step 4: so x dx − y dy − z dz = 0, which is exact: it is (1/2)d(x² − y² − z²). Hence x² − y² − z² = c₁.

Step 5: for a second integral try y, x, −z: yz(x + y) + xz(x − y) − z(x² + y²) = z(xy + y² + x² − xy − x² − y²) = 0.

Step 6: so y dx + x dy − z dz = 0, that is d(xy − z²/2) = 0, and 2xy − z² = c₂.

Step 7: the integrals are independent (one contains xy, the other x² − y²), so the general solution is φ(x² − y² − z², 2xy − z²) = 0.

Non-example: for dx/1 = dy/1 = dz/1 the multipliers y, −y, 0 satisfy lP + mQ + nR = y − y = 0, but y dx − y dy is not an exact differential (∂(y)/∂y = 1 while ∂(−y)/∂x = 0). The condition that fails is exactness; dividing by the integrating factor y repairs it and gives d(x − y) = 0.

Building up: the same identity gives the ratio rule: each fraction equals (l dx + m dy + n dz)/(lP + mQ + nR) even when the denominator is not zero.

Choosing the numerator to be the differential of the denominator, or a multiple of an existing fraction, produces a logarithmic integral, as in Example 5 below; the table 'Useful Multipliers' lists the standard patterns.

Where it is used: almost every Lagrange question in UPSC papers (cyclic systems, systems with factors x, y, z), and the integration of Charpit's auxiliary system in the next topic.

3. Understanding the Integral Surface through a Given Curve

Why we need it: the general solution φ(u, v) = 0 contains an arbitrary function, so it describes infinitely many surfaces. Data such as 'the surface passes through this curve' pick out one of them.

In words: through each point of the given curve passes exactly one characteristic, with labels c₁ = u and c₂ = v evaluated at that point.

As the point moves along the curve (parameter t), the labels trace a relation between c₁ and c₂; eliminating t gives F(c₁, c₂) = 0. The required surface consists of exactly these characteristics, so it is F(u, v) = 0.

First example, slowly: find the integral surface of yzp + xzq = xy (first lesson above) through the line x = 0, z = 2y.

Step 1: the integrals are x² − y² = c₁ and x² − z² = c₂.

Step 2: parametrise the line: x = 0, y = t, z = 2t.

Step 3: evaluate the labels on it: c₁ = −t² and c₂ = −4t².

Step 4: eliminate t: c₂ = 4c₁.

Step 5: replace the labels by the functions: x² − z² = 4(x² − y²), that is 3x² − 4y² + z² = 0.

Step 6: check the curve: 0 − 4t² + 4t² = 0. Check the PDE: implicit differentiation gives p = −3x/z and q = 4y/z, so yzp + xzq = −3xy + 4xy = xy.

Step 7: check transversality: on the line (P, Q) = (yz, xz) = (2t², 0), while the projected line has direction (x′, y′) = (0, 1); x′Q − y′P = −2t² ≠ 0 for t ≠ 0, so the surface is unique near the line (away from the origin).

Non-example: for the same PDE take the curve x² − y² = 1, x² − z² = 1. On it c₁ = 1 and c₂ = 1 for every point, so elimination gives no relation at all: the curve is itself a characteristic, and every surface φ(u, v) = 0 with φ(1, 1) = 0 contains it. The condition that fails is transversality.

Building up: the method is the Cauchy problem in Lagrange's language: the curve supplies one characteristic per value of t, and these sweep out a surface. When the projected curve crosses the projected characteristics (x′Q − y′P ≠ 0), each characteristic meets the curve once and the surface is unique; when the curve is a characteristic, uniqueness is lost (see 'Curve that is a characteristic' below).

Where it is used: the 15-mark 'integral surface through a curve' questions (Examples 9, 10 and 11 below) and Cauchy's method of characteristics in the next topic.

4. Lagrange's Theorem

Quasilinear first-order PDE: P(x, y, z)p + Q(x, y, z)q = R(x, y, z), linear in p and q; this is Lagrange's linear equation.

Idea: 'quasilinear' means the derivatives p, q occur only to the first power, although P, Q, R may depend on z itself; the equation is 'linear' in the strict sense only when P, Q do not involve z and R is linear in z. Lagrange's method handles both, but not equations such as p² + q² = 1 (these need Charpit's method, next topic).

Theorem: if u(x, y, z) = c₁ and v(x, y, z) = c₂ are two independent solutions of the auxiliary equations dx/P = dy/Q = dz/R, then φ(u, v) = 0, with φ an arbitrary differentiable function, is the general solution.

Idea: dx/P = dy/Q = dz/R is a system of two ODEs in space, so its solution curves form a two-parameter family labelled by (c₁, c₂). An integral surface is a one-parameter family of these curves, and picking a one-parameter family of pairs (c₁, c₂) is exactly a relation φ(c₁, c₂) = 0. Independence matters: if v were a function of u, φ(u, v) = 0 would only give the special surfaces u = constant.

Geometric meaning: the normal (p, q, −1) of an integral surface is perpendicular to (P, Q, R), so the surface contains the direction (P, Q, R) at each point and is generated by the characteristic curves u = c₁, v = c₂.

Idea: think of (P, Q, R) as a velocity field; the characteristics are its streamlines, and an integral surface is any surface woven from whole streamlines. A characteristic that touches an integral surface at one point lies on it entirely.

n variables: P₁p₁ + ... + P_np_n = R has auxiliary equations dx₁/P₁ = ... = dx_n/P_n = dz/R and general solution φ(u₁, ..., u_n) = 0.

5. Solving the Auxiliary Equations

Grouping: pick two fractions involving only two variables and integrate directly.

Using a known integral: use u = c₁ to eliminate one variable from another pair of fractions.

Multipliers: choose l, m, n (functions of x, y, z) with lP + mQ + nR = 0; then l dx + m dy + n dz = 0, which gives an integral when it is exact.

Why it works: if lP + mQ + nR = 0 while the fractions are finite, the combined fraction (l dx + m dy + n dz)/(lP + mQ + nR) can stay finite only if its numerator vanishes, so l dx + m dy + n dz = 0 along every characteristic.

Ratio rule: each fraction equals (l dx + m dy + n dz)/(lP + mQ + nR); choosing the numerator as the differential of the denominator gives a logarithmic integral.

Why it works: if dx/P = dy/Q = dz/R = k, then dx = kP, dy = kQ, dz = kR, so l dx + m dy + n dz = k(lP + mQ + nR); dividing gives the same k. This is the familiar rule a/b = c/d = (la + mc)/(lb + md).

6. Derivation: Proof of Lagrange's Theorem

Since u = c₁ along each characteristic, du = u_xdx + u_ydy + u_zdz = 0 with dx : dy : dz = P : Q : R, so Pu_x + Qu_y + Ru_z = 0; likewise Pv_x + Qv_y + Rv_z = 0.

Solving these two linear equations for P : Q : R (cross-multiplication) gives P : Q : R = ∂(u, v)/∂(y, z) : ∂(u, v)/∂(z, x) : ∂(u, v)/∂(x, y).

By the Jacobian elimination of the previous topic, φ(u, v) = 0 satisfies ∂(u, v)/∂(y, z) p + ∂(u, v)/∂(z, x) q = ∂(u, v)/∂(x, y), which is a non-zero multiple of Pp + Qq = R. So every φ(u, v) = 0 is a solution.

Conversely, on an integral surface z = z(x, y) let U = u(x, y, z(x, y)), V = v(x, y, z(x, y)); expanding, ∂(U, V)/∂(x, y) = ∂(u, v)/∂(x, y) − p ∂(u, v)/∂(y, z) − q ∂(u, v)/∂(z, x), which is proportional to R − Pp − Qq = 0.

A zero Jacobian means U and V are functionally dependent on the surface, that is φ(u, v) = 0 for some φ; hence φ(u, v) = 0 contains every solution, which justifies the word 'general'.

7. Choosing the Method

If two fractions involve only their own two variables (for example dx/P = dy/Q with P, Q free of z) → integrate that pair directly; this is the first thing to look for.

If one integral u = c₁ is known but every remaining pair has three variables → use u = c₁ to remove one variable, integrate with c₁ held constant, and only then replace c₁ by u(x, y, z).

If P, Q, R are cyclic, such as (y − z, z − x, x − y) or (mz − ny, nx − lz, ly − mx) → multipliers 1, 1, 1 (or l, m, n) and x, y, z.

If P, Q, R carry factors x, y, z → try 1/x, 1/y, 1/z (giving log xyz) or x, y, z (giving x² + y² + z²).

If P + Q or P − Q factorises neatly → use d(x + y)/(P + Q) and d(x − y)/(P − Q).

If a combination lP + mQ + nR equals a simple multiple of the denominator you already have → use the ratio rule to get a logarithmic integral, as in Example 5.

8. Common Mistakes

Stopping at two integrals: u = c₁, v = c₂ is a family of curves, not the solution; the answer is φ(u, v) = 0 (or v = f(u)).

Treating c₁ as a variable: after substituting u = c₁, c₁ is a constant during the integration; putting u back too early makes the integration wrong.

Multipliers that are not exact: lP + mQ + nR = 0 gives l dx + m dy + n dz = 0, but this yields an integral only if the left side is a total differential; check it.

Dependent integrals: y/x = c₁ and (x + y)/x = c₂ are not independent (the second is 1 + c₁); find a genuinely new combination.

Log constants: dx/x = dy/y gives y = c₁x, not y = x + c₁.

Integral surface through a curve: using the curve in only one of the two integrals; both u = c₁ and v = c₂ must be evaluated on the curve before eliminating x, y, z.

Link: the characteristics are the field lines of the vector field (P, Q, R) of Vector Analysis, and the same equations dx/u = dy/v = dz/w give streamlines in Fluid Dynamics.

Link: Cauchy's method of characteristics (next topic) is Lagrange's method with the data curve built in from the start; the orthogonal-surface problems of the previous topic always end in a Lagrange equation.

9. Worked Examples

Example 1: xp + yq = z. From dx/x = dy/y = dz/z: y/x = c₁, z/x = c₂; general solution φ(y/x, z/x) = 0.

Example 2: p + q = z. From dx = dy: x − y = c₁; from dx = dz/z: log z − x = c₂; so z = e^xf(x − y).

Check: p = e^xf + e^xf′, q = −e^xf′, so p + q = e^xf = z.

Example 3: (y − z)p + (z − x)q = x − y. Multipliers 1, 1, 1 give dx + dy + dz = 0, so x + y + z = c₁.

Multipliers x, y, z give x dx + y dy + z dz = 0, so x² + y² + z² = c₂.

General solution: φ(x + y + z, x² + y² + z²) = 0, surfaces of revolution about the line x = y = z.

Example 4: Solve x(y − z)p + y(z − x)q = z(x − y).

Multipliers 1, 1, 1: P + Q + R = xy − xz + yz − xy + xz − yz = 0, so dx + dy + dz = 0 and x + y + z = c₁.

Multipliers 1/x, 1/y, 1/z: (y − z) + (z − x) + (x − y) = 0, so dx/x + dy/y + dz/z = 0 and xyz = c₂.

General solution: φ(x + y + z, xyz) = 0.

Example 5: Solve (x² − y² − z²)p + 2xyq = 2xz.

From dy/(2xy) = dz/(2xz): dy/y = dz/z, so y/z = c₁.

Multipliers x, y, z: xP + yQ + zR = x(x² − y² − z²) + 2xy² + 2xz² = x(x² + y² + z²).

So (x dx + y dy + z dz)/(x(x² + y² + z²)) = dz/(2xz), that is d(x² + y² + z²)/(x² + y² + z²) = dz/z, giving (x² + y² + z²)/z = c₂.

General solution: φ(y/z, (x² + y² + z²)/z) = 0.

10. Integral Surface through a Given Curve

Step 1: Find two independent integrals u = c₁, v = c₂.

Step 2: Combine them with the equations of the given curve and eliminate x, y, z to get a relation F(c₁, c₂) = 0.

Step 3: Replace c₁ by u and c₂ by v: F(u, v) = 0 is the required surface.

Example 6: Find the integral surface of xp + yq = z through the curve x + y = 1, z = 1.

Here y = c₁x, z = c₂x; with z = 1, x = 1/c₂, y = c₁/c₂, and x + y = 1 gives 1 + c₁ = c₂.

So 1 + y/x = z/x, that is z = x + y (check: p = q = 1 and xp + yq = x + y = z).

Uniqueness remark: if the projection (x(t), y(t)) of the given curve on the xy-plane is nowhere tangent to the projected characteristic direction (P, Q), that is x′(t)Q − y′(t)P ≠ 0 along it, a unique integral surface contains it near the curve; if the curve is itself a characteristic, infinitely many integral surfaces contain it.

11. More Tricks for the Auxiliary Equations

Sum and difference: when P + Q and P − Q factor neatly, use d(x + y)/(P + Q) and d(x − y)/(P − Q) with dz/R; this is the standard route when P, Q involve x + y (or when P = xz − y, Q = yz − x type pairs appear).

Exact differential multipliers: if lP + mQ + nR = 0 and l dx + m dy + n dz = dw for some w, then w = c is an integral; try l, m, n equal to partial derivatives of a guessed w (such as x/y² + y/x², xy + z, x² + y² − z²).

Curve given parametrically (x = x(t), y = y(t), z = z(t)): substitute in u = c₁, v = c₂, eliminate t to get F(c₁, c₂) = 0, then put c₁ = u, c₂ = v and check that the curve lies on the surface.

Curve that is a characteristic: if u and v are both constant along the given curve, no relation F(c₁, c₂) = 0 is forced; every surface φ(u, v) = 0 with φ(c₁, c₂) = 0 contains it, so the integral surface is not unique.

12. Past UPSC Questions Solved

Example 7 (UPSC 2015, 15 marks): Solve for the general solution p cos(x + y) + q sin(x + y) = z, where p = ∂z/∂x and q = ∂z/∂y.

Auxiliary equations dx/cos(x + y) = dy/sin(x + y) = dz/z; adding and subtracting the first two, d(x + y)/(cos(x + y) + sin(x + y)) = d(x − y)/(cos(x + y) − sin(x + y)).

With w = x + y: d(x − y) = [(cos w − sin w)/(cos w + sin w)] dw, whose right side is d log(cos w + sin w); so x − y − log(cos w + sin w) = const, i.e. e^{y − x}[cos(x + y) + sin(x + y)] = c₁.

Also dz/z = dw/(cos w + sin w) = dw/(√2 sin(w + π/4)), so log z = (1/√2) log tan(w/2 + π/8) + const.

Hence z[tan((x + y)/2 + π/8)]^−1/√2 = c₂.

Answer: z = [tan((x + y)/2 + π/8)]^1/√2 f(e^{y − x}[cos(x + y) + sin(x + y)]), f arbitrary (checked by substitution).

Example 8 (UPSC 2016, 10 marks): Find the general integral of the partial differential equation
(y + zx)p − (x + yz)q = x² − y².

Auxiliary equations dx/(y + zx) = dy/(−(x + yz)) = dz/(x² − y²).

Multipliers x, y, −z: x(y + zx) − y(x + yz) − z(x² − y²) = 0, so x dx + y dy − z dz = 0 and x² + y² − z² = c₁.

Multipliers y, x, 1: y(y + zx) − x(x + yz) + (x² − y²) = 0, so y dx + x dy + dz = 0 and xy + z = c₂.

Answer: φ(x² + y² − z², xy + z) = 0.

Example 9 (UPSC 2018, 15 marks): Find the general solution of the partial differential equation :
(y³x − 2x⁴)p + (2y⁴ − x³y)q = 9z(x³ − y³),
where p = ∂z/∂x, q = ∂z/∂y, and find its integral surface that passes through the curve : x = t, y = t², z = 1.

Auxiliary equations dx/(x(y³ − 2x³)) = dy/(y(2y³ − x³)) = dz/(9z(x³ − y³)).

Multipliers 1/x, 1/y, 1/(3z): (y³ − 2x³) + (2y³ − x³) + 3(x³ − y³) = 0, so d log x + d log y + (1/3)d log z = 0 and x³y³z = c₁.

Multipliers 1/y² − 2y/x³, 1/x² − 2x/y³, 0 (the partial derivatives of x/y² + y/x²): the sum lP + mQ expands to xy − 2x⁴/y² − 2y⁴/x² + 4xy + 2y⁴/x² − xy − 4xy + 2x⁴/y² = 0, so (x³ + y³)/(x²y²) = c₂.

General solution φ(x³y³z, (x³ + y³)/(x²y²)) = 0.

On x = t, y = t², z = 1: c₁ = t⁹ and c₂ = (t³ + t⁶)/t⁶ = 1 + t⁻³, so c₁(c₂ − 1)³ = 1.

Substituting: x³y³z[(x³ + y³ − x²y²)/(x²y²)]³ = 1.

Answer: general solution φ(x³y³z, (x³ + y³)/(x²y²)) = 0; required surface z(x³ + y³ − x²y²)³ = x³y³ (on the curve x³ + y³ − x²y² = t³, giving z = 1).

Example 10 (UPSC 2020, 15 marks): Find the integral surface of the partial differential equation : (x − y)y² ∂z/∂x + (y − x)x² ∂z/∂y = (x² + y²)z that contains the curve : xz = a³, y = 0 on it.

Auxiliary equations dx/((x − y)y²) = dy/(−(x − y)x²) = dz/((x² + y²)z); the first two give x²dx + y²dy = 0, so x³ + y³ = c₁.

(dx − dy)/((x − y)(y² + x²)) = dz/((x² + y²)z) gives d(x − y)/(x − y) = dz/z, so z/(x − y) = c₂.

On y = 0, xz = a³: c₁ = x³ and c₂ = z/x = a³/x², so c₂³c₁² = a⁹.

Answer: z³(x³ + y³)² = a⁹(x − y)³ (at y = 0 it reduces to x³z³ = a⁹, i.e. xz = a³).

Example 11 (UPSC 2024, 15 marks): Find the integral surface of the following quasi-linear equation
(y − φ) ∂φ/∂x + (φ − x) ∂φ/∂y = x − y,
which passes through the curve φ = 0, xy = 1 and through the circle x + y + φ = 0, x² + y² + φ² = a².

Write z for φ: dx/(y − z) = dy/(z − x) = dz/(x − y), the same system as Example 3, so x + y + z = c₁ and x² + y² + z² = c₂.

(i) On z = 0, xy = 1: c₁ = x + y and c₂ = (x + y)² − 2xy = c₁² − 2; so x² + y² + z² = (x + y + z)² − 2, i.e. xy + yz + zx = 1.

(ii) On the circle x + y + z = 0, x² + y² + z² = a², both integrals are constant (c₁ = 0, c₂ = a²): the circle is itself a characteristic curve.

Hence every surface φ(x + y + z, x² + y² + z²) = 0 with φ(0, a²) = 0 passes through it, e.g. the sphere x² + y² + z² = a², the plane x + y + z = 0, or x² + y² + z² − a² = k(x + y + z).

Answer: through φ = 0, xy = 1 the surface is xy + yφ + φx = 1 (unique); through the circle there are infinitely many integral surfaces, x² + y² + φ² − a² = k(x + y + φ) for any constant k being one family.

Example 12 (UPSC 2026, 10 marks): Find the integral surface of the equations dx/(xz − y) = dy/(yz − x) = dz/(1 − z²).

Adding the first two: d(x + y)/((x + y)(z − 1)) = dz/(1 − z²), so d(x + y)/(x + y) = −dz/(1 + z) and (x + y)(1 + z) = c₁.

Subtracting: d(x − y)/((x − y)(z + 1)) = dz/((1 − z)(1 + z)), so d(x − y)/(x − y) = dz/(1 − z) and (x − y)(1 − z) = c₂.

The two integrals are independent, and the system is the auxiliary system of (xz − y)p + (yz − x)q = 1 − z².

Answer: φ((x + y)(1 + z), (x − y)(1 − z)) = 0, φ arbitrary.

13. Practice Exercises

Exercise 1 (basic): Write the auxiliary equations of x²p + y²q = z² and find its general solution.

Answer: dx/x² = dy/y² = dz/z²; integrals 1/x − 1/y = c₁, 1/x − 1/z = c₂; φ(1/x − 1/y, 1/x − 1/z) = 0.

Exercise 2 (basic): Show that u = xy and v = z/x are integrals of dx/x = dy/(−y) = dz/z, and write the PDE whose general solution is φ(xy, z/x) = 0.

Answer: x(y) + (−y)(x) = 0 and x(−z/x²) + z(1/x) = 0; the PDE is xp − yq = z.

Exercise 3 (basic): Solve p + q = 1.

Answer: dx = dy = dz gives x − y = c₁, z − x = c₂, so z = x + f(x − y).

Exercise 4 (standard): Solve xp + yq = 3z.

Answer: y/x = c₁ and z/x³ = c₂, so z = x³f(y/x) (homogeneous surfaces of degree 3).

Exercise 5 (standard): Solve x²(y − z)p + y²(z − x)q = z²(x − y).

Answer: multipliers 1/x², 1/y², 1/z² give 1/x + 1/y + 1/z = c₁; multipliers 1/x, 1/y, 1/z give xyz = c₂; φ(1/x + 1/y + 1/z, xyz) = 0.

Exercise 6 (standard): Solve (x² − yz)p + (y² − zx)q = z² − xy.

Answer: P − Q = (x − y)(x + y + z) and Q − R = (y − z)(x + y + z) give (x − y)/(y − z) = c₁; multipliers x, y, z and 1, 1, 1 give xy + yz + zx = c₂; φ((x − y)/(y − z), xy + yz + zx) = 0.

Exercise 7 (standard): Find the integral surface of xp + yq = 2z through the curve x = 1, z = y².

Answer: y/x = c₁, z/x² = c₂; on the curve c₂ = c₁², so z = y² (check: yq = 2y² = 2z).

Exercise 8 (UPSC level): Find the general solution of (y + z)p + (z + x)q = x + y.

Answer: d(x − y)/(y − x) = d(y − z)/(z − y) = d(x + y + z)/(2(x + y + z)) give (x − y)/(y − z) = c₁ and (x + y + z)(x − y)² = c₂; φ((x − y)/(y − z), (x + y + z)(x − y)²) = 0.

Exercise 9 (UPSC level): Find the integral surface of x(y² + z)p − y(x² + z)q = (x² − y²)z which contains the line x + y = 0, z = 1.

Answer: multipliers 1/x, 1/y, 1/z give xyz = c₁; multipliers x, y, −1 give x² + y² − 2z = c₂; on x = t, y = −t, z = 1, c₂ = −2c₁ − 2, so x² + y² + 2xyz − 2z + 2 = 0.

Exercise 10 (UPSC level): Show that the line x = t, y = 2t, z = 3t is a characteristic curve of xp + yq = z, and find two different integral surfaces that contain it.

Answer: on the line y/x = 2 and z/x = 3 are both constant, so it is a characteristic; any φ(y/x, z/x) = 0 with φ(2, 3) = 0 contains it, for example z = 3x and z = x + y.

1. Understanding the Method of Characteristics for a Cauchy Problem

Why we need it: in applications we rarely want 'all' solutions; we want the one solution that takes given values on a given curve, such as u(x, 0) = f(x) at time 0. The method of characteristics builds that solution directly, without first finding a general solution.

In words: a u_x + b u_y is the rate at which u changes when the point (x, y) moves with velocity (a, b). So the equation a u_x + b u_y = c says: 'travelling with velocity (a, b), u changes at rate c'.

The data curve Γ is described by a parameter s (which point of the curve) and the travel by a parameter t (how far along the characteristic). At t = 0 we stand on Γ, where x, y and u are known.

Solving dx/dt = a, dy/dt = b, du/dt = c gives x, y, u as functions of (s, t). The last step is to invert: express s and t through x and y, and substitute into u.

First example, slowly: solve x u_x + u_y = 0 with u(x, 0) = x².

Step 1: here a = x, b = 1, c = 0. Parametrise the data: x = s, y = 0, u = s² at t = 0.

Step 2: check transversality: J = x₀′(s)b − y₀′(s)a = 1 · 1 − 0 · s = 1 ≠ 0, so a unique solution exists near the x-axis.

Step 3: solve the ODEs: dx/dt = x gives x = se^t; dy/dt = 1 gives y = t; du/dt = 0 gives u = s² (u is carried unchanged).

Step 4: invert: t = y and s = xe^−t = xe^−y.

Step 5: substitute: u = s² = x²e^−2y.

Step 6: check: x u_x = 2x²e^−2y and u_y = −2x²e^−2y add to 0, and u(x, 0) = x².

Step 7: notice the picture: the characteristics x = se^y fan out from the x-axis, and u is constant along each of them.

Non-example: for the same equation put the data on the curve x = e^y (x = e^s, y = s). Then J = e^s · 1 − 1 · e^s = 0: the curve is itself a characteristic. Along it u must be constant, so data such as u = s have no solution, and constant data have infinitely many. Transversality is the condition that fails.

Building up: along any curve with dx/dt = a, dy/dt = b the chain rule gives du/dt = a u_x + b u_y = c, so the graph of the solution is woven from these curves.

The inversion (s, t) → (x, y) is possible near Γ when its Jacobian is non-zero, and at t = 0 that Jacobian is exactly J; this is the existence and uniqueness theorem (see 'Derivation: Why the Characteristic ODEs Solve the PDE' below).

Where it is used: 'solve by the method of characteristics' questions (Example 9 below), traffic and gas-dynamics equations such as u_y + uu_x = 0 where characteristics can cross, and the nonlinear strips of the third lesson.

2. Understanding Charpit's Method

Why we need it: equations such as pq = xy or p² + q² = x + y are not linear in p and q, so there is no single direction (P, Q, R) and Lagrange's method fails. We still want a complete integral, a two-constant family of solutions.

In words: F(x, y, z, p, q) = 0 is one equation between the two unknowns p and q. If we had a second equation G(x, y, z, p, q, a) = 0 containing a constant a, we could solve both for p and q and integrate dz = p dx + q dy.

Not every G works: dz = p dx + q dy must be integrable. Charpit showed that it is, exactly when G is an integral of the auxiliary system dx/F_p = dy/F_q = dz/(pF_p + qF_q) = −dp/(F_x + pF_z) = −dq/(F_y + qF_z).

So the method is: compute the five partial derivatives of F, write the fractions, and look for the simplest pair that integrates to a relation containing p or q.

First example, slowly: find a complete integral of pq = xy.

Step 1: F = pq − xy, so F_x = −y, F_y = −x, F_z = 0, F_p = q, F_q = p.

Step 2: Charpit's fractions become dx/q = dy/p = dz/(2pq) = dp/y = dq/x (the two minus signs cancel those in F_x, F_y).

Step 3: pair dx/q with dp/y, and use the PDE to replace q by xy/p: p dx/(xy) = dp/y.

Step 4: cancel y and separate: dp/p = dx/x, so p = ax, where a is the new constant.

Step 5: the PDE now gives q = xy/p = y/a.

Step 6: integrate dz = ax dx + (y/a) dy, which is exact: z = ax²/2 + y²/(2a) + b.

Step 7: check: p = ax, q = y/a and pq = xy. The two constants a, b are independent, so this is a complete integral.

Non-example: for the same PDE try the 'second relation' p = q, which is not an integral of Charpit's system. Then p = q = √(xy) and dz = √(xy)(dx + dy); but ∂(√(xy))/∂y = √x/(2√y) differs from ∂(√(xy))/∂x = √y/(2√x) unless x = y, so this dz is not exact and no surface results. Compatibility is the condition that fails.

Building up: integrability of dz = p dx + q dy is the condition p_y + qp_z = q_x + pq_z. Substituting p and q from F = 0, G = 0 turns it into [F, G] = 0, a linear first-order PDE for G in five variables.

Its Lagrange auxiliary equations are precisely Charpit's fractions ('Derivation: Charpit's Equations' below). The standard forms in the table are cases in which the fractions integrate at sight.

Where it is used: 15-mark 'find a complete integral by Charpit's method' questions, the standard forms f(p, q) = 0, Clairaut and separable equations, and, through characteristic strips, the nonlinear Cauchy problem.

3. Understanding Characteristic Strips for Nonlinear Cauchy Problems

Why we need it: for a nonlinear F, questions often ask for the integral surface through a given curve ('find the surface of z = p² − q² through the parabola ...'). Charpit gives a complete integral, but fitting a curve to it is awkward; Cauchy's strips build the required surface directly.

In words: for nonlinear F the direction in which information travels, (F_p, F_q), depends on the slopes p, q themselves. So we must carry the slopes along with the point: a strip is a curve together with a tangent plane at each of its points, described by x, y, z, p, q.

The five ODEs are dx/dt = F_p, dy/dt = F_q, dz/dt = pF_p + qF_q, dp/dt = −(F_x + pF_z), dq/dt = −(F_y + qF_z).

At t = 0 the strip starts on the data curve x₀(s), y₀(s), z₀(s); the initial slopes p₀, q₀ are fixed by two conditions: F(x₀, y₀, z₀, p₀, q₀) = 0 (the plane is allowed by the PDE) and z₀′ = p₀x₀′ + q₀y₀′ (the plane contains the curve).

First example, slowly: find the integral surface of pq = z through the parabola y = 0, z = x².

Step 1: F = pq − z, so F_p = q, F_q = p, F_x = F_y = 0, F_z = −1. Parametrise the data: x₀ = s, y₀ = 0, z₀ = s².

Step 2: the strip condition 2s = p₀ · 1 + q₀ · 0 gives p₀ = 2s; then F = 0 gives 2s · q₀ = s², so q₀ = s/2.

Step 3: the slope equations are dp/dt = p and dq/dt = q, so p = 2se^t and q = (s/2)e^t.

Step 4: dx/dt = q and dy/dt = p give x = s + (s/2)(e^t − 1) and y = 2s(e^t − 1); dz/dt = 2pq = 2s²e^2t gives z = s²e^2t.

Step 5: invert: s(e^t − 1) = y/2, so x = s + y/4, i.e. s = x − y/4, and se^t = s + y/2 = x + y/4.

Step 6: hence z = (se^t)² = (x + y/4)².

Step 7: check: p = 2(x + y/4), q = (x + y/4)/2, so pq = (x + y/4)² = z; and at y = 0, z = x².

Non-example: for p² + q² = 1 with data z = 2x on y = 0, the strip condition gives p₀ = 2, and then q₀² = 1 − 4 < 0: there is no real initial strip, so no real integral surface. The data curve is steeper than any surface with |∇z| = 1 can be; the condition that fails is the solvability of F = 0 with the strip condition.

Building up: along a curve with dx/dt = F_p, dy/dt = F_q, differentiating F = 0 and using p_y = q_x shows that p and q change exactly as the last two ODEs say; so every solution surface is a union of strips.

When F is linear in p and q, the first three ODEs no longer involve p, q and reduce to Lagrange's dx/P = dy/Q = dz/R. When F has several roots (p₀, q₀), each root gives its own surface, as in Examples 7 and 10 below.

Where it is used: UPSC questions of the form 'find the characteristics of F = 0 and the integral surface through ...', and, when F does not contain z, Hamilton's canonical equations of Mechanics.

4. The Cauchy Problem for Quasilinear Equations

Equation: a(x, y, u)u_x + b(x, y, u)u_y = c(x, y, u), with a, b, c continuously differentiable.

Idea: a u_x + b u_y is the rate of change of u when (x, y) moves with velocity (a, b), so the equation says 'along this direction u grows at rate c'. Knowing u at one point of such a path fixes it along the whole path, and the PDE collapses to an ODE.

Cauchy data: u is prescribed on an initial curve Γ: x = x₀(s), y = y₀(s), u = u₀(s).

Characteristic equations: dx/dt = a, dy/dt = b, du/dt = c, with x = x₀(s), y = y₀(s), u = u₀(s) at t = 0.

Solving gives x(s, t), y(s, t), u(s, t); invert the first two to express s, t in terms of x, y and substitute in u.

Theorem (local existence and uniqueness): if x₀, y₀, u₀ are continuously differentiable and J = x₀′(s) b − y₀′(s) a ≠ 0 along Γ (a, b evaluated on the data), then there is a unique solution in a neighbourhood of Γ.

Idea: J is the determinant of the matrix with rows (x₀′, y₀′) and (a, b); J ≠ 0 says Γ cuts across the characteristic directions instead of running along them, so each characteristic meets Γ once and carries exactly one data value away from it. Picture the characteristics as parallel threads and Γ as a cut across them.

If J = 0 along Γ, there is either no solution or infinitely many; infinitely many when Γ is itself a characteristic curve.

Idea: example: u_x + u_y = 0 has solutions u = F(x − y), constant on each line y = x + k. With data on y = x, J = 1 · 1 − 1 · 1 = 0: data u(x, x) = h(x) admit no solution unless h is constant, and then every F with F(0) = h gives one.

5. Derivation: Why the Characteristic ODEs Solve the PDE

Along a curve with dx/dt = a, dy/dt = b, the chain rule gives du/dt = u_x(dx/dt) + u_y(dy/dt) = a u_x + b u_y, which the PDE says equals c.

So the graph of any solution is a union of curves solving dx/dt = a, dy/dt = b, du/dt = c; starting one such curve from each point of Γ (parameter s) sweeps out the solution surface x(s, t), y(s, t), u(s, t).

To write u as a function of x, y we must invert (s, t) → (x, y); by the inverse function theorem this is possible near t = 0 when ∂(x, y)/∂(s, t) = x_sy_t − x_ty_s ≠ 0.

At t = 0, x_s = x₀′, y_s = y₀′, x_t = a, y_t = b, so the Jacobian is x₀′b − y₀′a = J: the transversality condition is exactly the invertibility condition.

Far from Γ the inversion can fail: in Example 2 the characteristics x = s(1 + y) all pass through (0, −1), so the solution u = x/(1 + y) cannot be continued to y = −1.

6. Steps and Examples

Step 1: Parametrise the initial data by s.

Step 2: Solve the characteristic ODEs in t with these initial values.

Step 3: Check J ≠ 0, eliminate s and t, and verify both the PDE and the data.

Example 1: u_x + u_y = 2, u(x, 0) = x². Γ: x = s, y = 0, u = s²; J = 1 · 1 − 0 · 1 = 1 ≠ 0.

dx/dt = 1, dy/dt = 1, du/dt = 2 give x = s + t, y = t, u = s² + 2t.

So t = y, s = x − y and u = (x − y)² + 2y (check: u_x + u_y = 2(x − y) − 2(x − y) + 2 = 2).

Example 2: u_y + uu_x = 0, u(x, 0) = x. Γ: x = s, y = 0, u = s; J = 1 ≠ 0.

dx/dt = u, dy/dt = 1, du/dt = 0 give u = s, y = t, x = s + st, so s = x/(1 + y).

Solution u = x/(1 + y) for y > −1 (check: u_y + uu_x = −x/(1 + y)² + x/(1 + y)² = 0).

Example 3: Solve xu_x + yu_y = 2u with u = 1 on the circle x² + y² = 1.

Γ: x = cos s, y = sin s, u = 1; J = (−sin s)(sin s) − (cos s)(cos s) = −1 ≠ 0, so the solution is unique near Γ.

dx/dt = x, dy/dt = y, du/dt = 2u give x = e^tcos s, y = e^tsin s, u = e^2t.

Since x² + y² = e^2t, u = x² + y² (check: x(2x) + y(2y) = 2u).

7. Nonlinear First-Order Equations

Charpit's method for F(x, y, z, p, q) = 0: from dx/F_p = dy/F_q = dz/(pF_p + qF_q) = −dp/(F_x + pF_z) = −dq/(F_y + qF_z) find a second relation containing p or q and a constant a.

Idea: one equation F = 0 cannot fix both p and q; Charpit looks for a second equation G(x, y, z, p, q, a) = 0 that is compatible with F, so that p and q can be solved and dz = p dx + q dy becomes exact. The Charpit fractions are just the Lagrange auxiliary equations of the condition that G must satisfy (derived below).

Solve the two relations for p and q, and integrate dz = p dx + q dy to get the complete integral.

Cauchy's characteristic strips: the same system with parameter t, dx/dt = F_p, dy/dt = F_q, dz/dt = pF_p + qF_q, dp/dt = −(F_x + pF_z), dq/dt = −(F_y + qF_z).

Its initial strip must satisfy F(x₀, y₀, z₀, p₀, q₀) = 0 and the strip condition z₀′(s) = p₀x₀′(s) + q₀y₀′(s).

Idea: a strip is a curve carrying a small tangent plane at each of its points; the strip condition says the plane really contains the curve (dz = p dx + q dy along it), and F = 0 says the plane is one the PDE allows. For a nonlinear F these two equations may have several roots (p₀, q₀), and each root gives its own solution surface.

Example 4 (Charpit): (p² + q²)y = qz. Here dp/(pq) = dq/(−p²), so p dp + q dq = 0 and p² + q² = a².

Then q = a²y/z and p = (a/z)√(z² − a²y²); dz = p dx + q dy becomes d(√(z² − a²y²)) = a dx.

Complete integral: z² = a²y² + (ax + b)².

Example 5 (standard form f(p, q) = 0): pq = 1 has complete integral z = ax + y/a + c.

Example 6 (Charpit): Find a complete integral of p²x + q²y = z.

F = p²x + q²y − z gives dx/dt = 2px, dp/dt = −(F_x + pF_z) = p − p², so d(p²x)/dt = 2px(p − p²) + p²(2px) = 2p²x; likewise d(q²y)/dt = 2q²y.

Hence p²x and q²y grow in the same ratio, giving the second relation p²x = aq²y.

With the PDE: q²y = z/(1 + a), p²x = az/(1 + a), so √(1 + a) dz/√z = √a dx/√x + dy/√y.

Complete integral: √((1 + a)z) = √(ax) + √y + b.

8. Cauchy's Method for Nonlinear Equations: Working Steps

Step 1: parametrise the initial curve x₀(s), y₀(s), z₀(s).

Step 2: find p₀(s), q₀(s) from F(x₀, y₀, z₀, p₀, q₀) = 0 and the strip condition z₀′ = p₀x₀′ + q₀y₀′; each admissible root gives a separate integral surface.

Step 3: solve the five characteristic ODEs; look first for combinations whose derivative vanishes (such as d(p − x)/dt = 0) or that are linear.

Step 4: eliminate s and t between x, y, z; check the PDE and the data.

Reducible forms: F(x^mp, yⁿq) = 0 becomes f(P, Q) = 0 with X = x^1−m/(1 − m), Y = y¹⁻ⁿ/(1 − n) (log x, log y when m = 1, n = 1); F(z^kp, z^kq, x, y) = 0 becomes an equation in P = Z_x, Q = Z_y with Z = z^k+1/(k + 1) (log z when k = −1).

Compatibility: Charpit's auxiliary relation G(x, y, z, p, q, a) = 0 is exactly a G compatible with F, i.e. [F, G] = 0; any integrable combination of the Charpit fractions (for example dy/F_q with dq/(F_y + qF_z)) can be used.

9. Derivation: Charpit's Equations

Suppose F = 0 and G(x, y, z, p, q) = 0 are solved for p = p(x, y, z), q = q(x, y, z). The Pfaffian dz = p dx + q dy is integrable iff (p, q, −1) · curl(p, q, −1) = 0, that is p_y + qp_z = q_x + pq_z.

Differentiating F = 0 with respect to x and z and combining: (F_x + pF_z) + F_p(p_x + pp_z) + F_q(q_x + pq_z) = 0; the same with G, and the analogous pair with y in place of x.

Solving these by Cramer's rule for q_x + pq_z and p_y + qp_z and equating them gives the compatibility condition [F, G] = F_p(G_x + pG_z) + F_q(G_y + qG_z) − G_p(F_x + pF_z) − G_q(F_y + qF_z) = 0.

Read as an equation for the unknown G of the five variables x, y, z, p, q, this is linear: F_pG_x + F_qG_y + (pF_p + qF_q)G_z − (F_x + pF_z)G_p − (F_y + qF_z)G_q = 0.

Its Lagrange auxiliary equations are dx/F_p = dy/F_q = dz/(pF_p + qF_q) = −dp/(F_x + pF_z) = −dq/(F_y + qF_z), the Charpit system; any integral of it containing p or q serves as G.

10. Choosing the Method

If the equation is linear or quasilinear and data are given on a curve → Cauchy's characteristics (or Lagrange's general solution, then fit the curve); check J ≠ 0 first.

If F is nonlinear and matches a standard form (table below) → use its trial solution; it is quicker than Charpit and examiners accept it.

If F contains x^mp, yⁿq or z^kp, z^kq → substitute (Section 8) to reach a standard form.

If no standard form fits → Charpit; scan the fractions for a denominator that vanishes (then dp = 0 or dq = 0, so p = a or q = a) or for a pair that integrates at once.

If a nonlinear equation comes with an initial curve and the question names Cauchy's method → characteristic strips (Section 8 steps), finding p₀, q₀ first.

11. Common Mistakes

Skipping the transversality check, or checking J at a single point; it must be non-zero along the whole data curve.

Ignoring the domain of validity: the inversion of x(s, t), y(s, t) may hold only in part of the plane (y > −1 in Example 2); state it.

Sign slips in Charpit: the p and q fractions carry minus signs, −dp/(F_x + pF_z) and −dq/(F_y + qF_z); dropping them gives a wrong second relation.

Stopping at p and q: the complete integral needs the final integration of dz = p dx + q dy and a second constant b.

Losing a branch: when p₀, q₀ come from a quadratic, both roots must be followed, or the omission must be justified by the data.

Link: when F does not contain z, the strip equations dx/dt = F_p, dp/dt = −F_x (and the same for y, q) are Hamilton's canonical equations of Mechanics with F as the Hamiltonian; this is the Hamilton–Jacobi connection.

Link: the crossing of characteristics in u_y + uu_x = 0 is how shock waves form in gas dynamics, and J ≠ 0 is the inverse function theorem of Calculus applied to (s, t) → (x, y).

12. Past UPSC Questions Solved

Example 7 (UPSC 2016, 15 marks): Determine the characteristics of the equation z = p² − q², and find the integral surface which passes through the parabola 4z + x² = 0, y = 0.

F = p² − q² − z: dx/dt = 2p, dy/dt = −2q, dz/dt = 2p² − 2q² = 2z, dp/dt = −(F_x + pF_z) = p, dq/dt = q.

Initial curve x₀ = s, y₀ = 0, z₀ = −s²/4; the strip condition gives p₀ = −s/2, and F = 0 gives s²/4 − q₀² = −s²/4, so q₀ = ±s/√2.

Characteristic strips: p = −(s/2)e^t, q = q₀e^t, z = −(s²/4)e^2t, x = s(2 − e^t), y = −2q₀(e^t − 1).

For q₀ = s/√2: s(e^t − 1) = −y/√2 and x = s − s(e^t − 1), so s = x − y/√2 and se^t = x − √2y; then z = −(se^t)²/4.

Check: z = −(x − √2y)²/4 gives p = −(x − √2y)/2, q = (x − √2y)/√2 and p² − q² = −(x − √2y)²/4 = z; at y = 0, 4z + x² = 0.

Answer: characteristics x = s(2 − e^t), y = ∓√2 s(e^t − 1), z = −(s²/4)e^2t; integral surfaces 4z + (x − √2y)² = 0 and 4z + (x + √2y)² = 0.

Example 8 (UPSC 2017, 15 marks): Find a complete integral of the partial differential equation
2(pq + yp + qx) + x² + y² = 0.

F = 2pq + 2yp + 2qx + x² + y²: F_x = 2(q + x), F_y = 2(p + y), F_z = 0, F_p = 2(q + y), F_q = 2(p + x).

Charpit: dx/(2(q + y)) = dy/(2(p + x)) = dp/(−2(q + x)) = dq/(−2(p + y)); then d(p + x) has denominator 2(y − x) and d(q + y) has 2(x − y), so d(p + x) = −d(q + y) and (p + x) + (q + y) = a.

The PDE is 2(p + x)(q + y) = 2xy − x² − y², so P = p + x and Q = q + y satisfy P + Q = a, PQ = −(x − y)²/2, giving P, Q = [a ± √(a² + 2(x − y)²)]/2.

dz = (P − x)dx + (Q − y)dy = (a/2)d(x + y) + (1/2)√(a² + 2v²) dv − x dx − y dy with v = x − y.

∫√(a² + 2v²) dv = (v/2)√(a² + 2v²) + (a²/(2√2)) log(√2v + √(a² + 2v²)).

Answer: z = a(x + y)/2 − (x² + y²)/2 + ((x − y)/4)√(a² + 2(x − y)²) + (a²/(4√2)) log[√2(x − y) + √(a² + 2(x − y)²)] + b (verified numerically).

Example 9 (UPSC 2019, 15 marks): Solve the first order quasilinear partial differential equation by the method of characteristics :
x ∂u/∂x + (u − x − y) ∂u/∂y = x + 2y in x > 0, −∞ < y < ∞ with u = 1 + y on x = 1.

Characteristic equations dx/dt = x, dy/dt = u − x − y, du/dt = x + 2y with x = 1, y = s, u = 1 + s at t = 0; J = 0 · b − 1 · a = −1 ≠ 0, so the solution is unique.

x = e^t; adding the last two, d(u + y)/dt = u + y, so u + y = (1 + 2s)e^t.

Then dy/dt = (u + y) − x − 2y = 2se^t − 2y, and with y(0) = s, y = (s/3)(2e^t + e^−2t).

With e^t = x: s = 3x²y/(2x³ + 1) and u = (1 + 2s)x − y = x + y(4x³ − 1)/(2x³ + 1).

Answer: u = x + y(4x³ − 1)/(2x³ + 1) for x > 0 (at x = 1 it gives 1 + y; substitution returns x + 2y).

Example 10 (UPSC 2020, 15 marks): Find the solution of the partial differential equation : z = (1/2)(p² + q²) + (p − x)(q − y); p ≡ ∂z/∂x, q ≡ ∂z/∂y which passes through the x-axis.

F = (p² + q²)/2 + (p − x)(q − y) − z: F_p = p + q − y, F_q = p + q − x, F_x = −(q − y), F_y = −(p − x), F_z = −1.

dx/dt = p + q − y = dp/dt and dy/dt = p + q − x = dq/dt, so p − x and q − y are constant on each strip.

x-axis: x₀ = s, y₀ = z₀ = 0; the strip condition gives p₀ = 0 and F = 0 gives q₀²/2 − sq₀ = 0, so q₀ = 2s or q₀ = 0.

q₀ = 2s: p − x = −s, q − y = 2s; dx/dt = x + s, dy/dt = y + s give x + s = 2se^t, y + s = se^t, so s = x − 2y, p = 2y, q = 2x − 3y and dz = 2y dx + (2x − 3y)dy, z = 2xy − 3y²/2.

q₀ = 0: p − x = −s, q = y and dx/dt = 0, so x = s, p = 0, q = y and z = y²/2.

Check: for z = y(4x − 3y)/2, (4y² + (2x − 3y)²)/2 + (2y − x)(2x − 4y) = 2xy − 3y²/2 = z; for z = y²/2, y²/2 + (−x)(0) = z.

Answer: z = y(4x − 3y)/2 (the root q₀ = 2s); the other root q₀ = 0 gives a second surface through the x-axis, 2z = y².

Example 11 (UPSC 2021, 15 marks): Find a complete integral of the partial differential equation p = (z + qy)² by using Charpit’s method.

F = (z + qy)² − p: F_y = 2q(z + qy), F_z = 2(z + qy), F_p = −1, F_q = 2y(z + qy).

Charpit: dy/F_q = −dq/(F_y + qF_z) gives dy/(2y(z + qy)) = −dq/(4q(z + qy)), so dy/y = −dq/(2q) and qy² is constant; write q = −a/y².

Then z + qy = z − a/y and p = (z − a/y)²; with w = z − a/y, dw = dz + (a/y²)dy = p dx = w² dx.

So −1/w = x + b, i.e. z − a/y = −1/(x + b).

Answer: complete integral z = a/y − 1/(x + b) (check: z + qy = −1/(x + b), whose square is p).

Example 12 (UPSC 2025, 15 marks): Find the complete integral of z(p² − q²) = x − y; p ≡ ∂z/∂x, q ≡ ∂z/∂y.

Write the equation as (√z p)² − (√z q)² = x − y and put Z = (2/3)z^3/2, so that P = Z_x = √z p and Q = Z_y = √z q.

Then P² − x = Q² − y = a (separable form), P = √(x + a), Q = √(y + a).

dZ = √(x + a) dx + √(y + a) dy, so Z = (2/3)(x + a)^3/2 + (2/3)(y + a)^3/2 + (2/3)b.

Answer: z^3/2 = (x + a)^3/2 + (y + a)^3/2 + b.

Example 13 (UPSC 2025, 15 marks): Find the characteristics of the partial differential equation
p² + q² = 2; p ≡ ∂z/∂x, q ≡ ∂z/∂y
and determine the integral surface which passes through x = 0, z = y.

F = p² + q² − 2: dx/dt = 2p, dy/dt = 2q, dz/dt = 2(p² + q²) = 4, dp/dt = dq/dt = 0; p, q are constant on each strip, so the characteristics are straight lines.

Initial curve x₀ = 0, y₀ = s, z₀ = s: the strip condition gives q₀ = 1 and F = 0 gives p₀ = ±1.

x = 2p₀t, y = s + 2t, z = s + 4t, so z − y = 2t = x/p₀.

Answer: characteristics x = ±2t, y = s + 2t, z = s + 4t (with p = ±1, q = 1); integral surfaces z = y + x and z = y − x.

13. Practice Exercises

Exercise 1 (basic): Solve 3u_x + 2u_y = 0 with u(x, 0) = sin x by the method of characteristics.

Answer: x = s + 3t, y = 2t, u = sin s, so u = sin(x − 3y/2).

Exercise 2 (basic): For u_x − u_y = 0 with u(x, x) = h(x), compute J and find u.

Answer: J = 1 · (−1) − 1 · 1 = −2 ≠ 0; u = F(x + y) with F(2x) = h(x), so u = h((x + y)/2).

Exercise 3 (basic): Find a complete integral of p + q = pq.

Answer: standard form f(p, q) = 0: p = a, q = a/(a − 1), so z = ax + ay/(a − 1) + b (a ≠ 1).

Exercise 4 (standard): Solve x u_x + y u_y = u with u = 1 + y on the line x = 1.

Answer: x = e^t, y = se^t, u = (1 + s)e^t (J = 1 ≠ 0), so u = x + y.

Exercise 5 (standard): Solve u_x + u_y = u² with u(x, 0) = 1, and state where the solution exists.

Answer: du/dt = u² with u(0) = 1 gives u = 1/(1 − t) and t = y, so u = 1/(1 − y), valid for y < 1 (it blows up as y → 1).

Exercise 6 (standard): Use Charpit's method to find a complete integral of p² + q² = x + y.

Answer: dx/(2p) = dp gives p² = x + a, so q² = y − a; z = (2/3)(x + a)^3/2 + (2/3)(y − a)^3/2 + b.

Exercise 7 (standard): Use Charpit's method to find a complete integral of 2xz − px² − 2qxy + pq = 0.

Answer: F_y + qF_z = −2qx + 2qx = 0, so dq = 0 and q = a; then p = 2x(z − ay)/(x² − a), and d(z − ay)/(z − ay) = 2x dx/(x² − a) gives z = ay + b(x² − a).

Exercise 8 (UPSC level): Find a complete integral of z² = pqxy.

Answer: with X = log x, Y = log y, Z = log z the equation becomes PQ = 1, so Z = aX + Y/a + log b and z = bx^ay^1/a.

Exercise 9 (UPSC level): Using characteristic strips, find the integral surface of pq = z that passes through the line x = 0, z = y.

Answer: q₀ = 1, p₀ = s; p = se^t, q = e^t, x = e^t − 1, y = se^t, z = se^2t; hence z = y(x + 1).

Exercise 10 (UPSC level): Solve (y + u)u_x + yu_y = x − y in y > 0, with u = 1 + x on y = 1.

Answer: y = e^t, x + u = (1 + 2s)e^t and x − u = e^t − 2e^−t (J = 1 ≠ 0), which give u = x − y + 2/y.

1. Understanding the Complementary Function of a Homogeneous Equation

Why we need it: as for linear ODEs, the general solution of F(D, D′)z = f(x, y) is 'all solutions of F(D, D′)z = 0' plus one particular solution. So we first need every solution of the equation with right side zero.

In words: D = ∂/∂x and D′ = ∂/∂y are operators: they take a function and return its partial derivative. 'Homogeneous' means every term of F has the same total order, as in D² − 3DD′ + 2D′².

The key fact is that D − mD′ kills every function of y + mx: for z = φ(y + mx), Dz = mφ′ and mD′z = mφ′. So once F is split into linear factors, each factor contributes one arbitrary function.

To find the factors, put D = m and D′ = 1; the resulting polynomial in m is the auxiliary equation, and each root m gives the factor D − mD′.

First example, slowly: find the complementary function (CF) of (D² − 3DD′ + 2D′²)z = 0.

Step 1: the auxiliary equation is m² − 3m + 2 = 0, with roots m = 1 and m = 2.

Step 2: so F = (D − D′)(D − 2D′); expanding gives D² − 2DD′ − DD′ + 2D′², which is correct.

Step 3: D − D′ kills φ₁(y + x), and because constant-coefficient operators commute, F = (D − 2D′)(D − D′) kills it too.

Step 4: in the same way D − 2D′ kills φ₂(y + 2x).

Step 5: by linearity CF = φ₁(y + x) + φ₂(y + 2x): two arbitrary functions for an equation of order 2.

Step 6: test one member: z = (y + x)² has z_xx = z_xy = z_yy = 2, and 2 − 3(2) + 2(2) = 0.

Non-example: the heat operator D² − D′ is not homogeneous. Trying z = φ(y + mx) gives m²φ″ − φ′, which cannot vanish for every φ. So its CF is not made of arbitrary functions of y + mx; it is a sum of exponentials Ae^{hx + h2y}. The condition that fails is homogeneity.

Building up: for z = φ(y + mx) every term D^rD′^s of order n gives m^rφ⁽ⁿ⁾, so F(D, D′)z = F(m, 1)φ⁽ⁿ⁾, which vanishes for all φ exactly when m is a root.

A double root m gives only one function this way; the second is xφ₂(y + mx), because (D − mD′)[xφ(y + mx)] = φ(y + mx), which one more factor D − mD′ kills (see 'Why it works' under the complementary function below).

Where it is used: the CF part of every 'solve F(D, D′)z = f' question, and the wave operator D² − c²D′² = (D − cD′)(D + cD′), whose CF is D'Alembert's f(y + cx) + g(y − cx).

2. Understanding Non-Homogeneous Operators and the Factor D − mD′ − c

Why we need it: operators such as D² − D′² + D − D′ contain terms of different orders. Functions of y + mx alone no longer solve them, and we need the right building block for each factor.

In words: (D − mD′ − c)z = 0 is the Lagrange equation p − mq = cz. Its auxiliary equations dx/1 = dy/(−m) = dz/(cz) give y + mx = c₁ and ze^−cx = c₂, so z = e^cxφ(y + mx).

So each linear factor, written in the exact shape D − mD′ − c, contributes e^cx times an arbitrary function of y + mx. Read m and c with their signs from that shape.

First example, slowly: find the CF of (D² − D′² + D − D′)z = 0.

Step 1: group the terms: D² − D′² = (D − D′)(D + D′), and D − D′ is common, so F = (D − D′)(D + D′ + 1).

Step 2: the factor D − D′ has m = 1, c = 0 and contributes φ₁(y + x).

Step 3: write the second factor in the standard shape: D + D′ + 1 = D − (−1)D′ − (−1), so m = −1 and c = −1.

Step 4: it contributes e^−xφ₂(y − x).

Step 5: verify by direct differentiation: for z = e^−xφ(y − x), Dz = −e^−xφ − e^−xφ′ and D′z = e^−xφ′, so (D + D′ + 1)z = 0.

Step 6: CF = φ₁(y + x) + e^−xφ₂(y − x).

Non-example: D² − D′ cannot be written as a product of factors D − mD′ − c with constant m, c. For such an irreducible operator there is no arbitrary function at all; one uses z = Ae^{hx + ky} with F(h, k) = 0, here k = h². The condition that fails is factorisation into linear factors.

Building up: a repeated factor (D − mD′ − c)² contributes e^cx[φ₁(y + mx) + xφ₂(y + mx)], for the same reason as the repeated root of the first lesson.

A factor containing no D, such as βD′ − γ, is the equation βq = γz and contributes e^γy/βφ(x). With these three rules the CF of any operator that splits into linear factors can be written at once.

Where it is used: questions such as (D² − D′² − 3D + 3D′)z = xy + e^{x + 2y} (Example 13 below), and damped or telegraph-type equations, whose lower-order terms produce the exponential factors.

3. Understanding Particular Integrals: Shortcuts and the Failing Case

Why we need it: once the CF is known we still need one function z_p with F(D, D′)z_p = f(x, y), the particular integral (PI). For the common right sides (exponentials, sines, polynomials) there are short rules, and we must know exactly when they break down.

In words: on e^{ax + by}, D acts as multiplication by a and D′ as multiplication by b. So F(D, D′) acts as multiplication by the number F(a, b), and dividing by that number gives the PI e^{ax + by}/F(a, b).

If F(a, b) = 0 the exponential already lies in the CF (the 'failing case'). Then split off the factor that vanishes, D − mD′, and use (D − mD′)⁻¹φ(y + mx) = xφ(y + mx).

First example, slowly: find a PI of (D² − 3DD′ + 2D′²)z = e^{x + 3y} + e^{2x + y}.

Step 1: treat the two terms separately and add the results at the end (the operator is linear).

Step 2: for e^{x + 3y}, a = 1, b = 3 and F(1, 3) = 1 − 9 + 18 = 10 ≠ 0, so PI₁ = e^{x + 3y}/10.

Step 3: for e^{2x + y}, F(2, 1) = 4 − 6 + 2 = 0: the failing case. Indeed e^{2x + y} is a function of y + 2x, which is in the CF.

Step 4: factor F = (D − D′)(D − 2D′) and invert the harmless factor first: (D − D′)⁻¹e^{2x + y} = e^{2x + y}/(2 − 1) = e^{2x + y}.

Step 5: invert the vanishing factor by the failing-case rule: (D − 2D′)⁻¹e^{2x + y} = xe^{2x + y}, so PI₂ = xe^{2x + y}.

Step 6: check PI₂: with E = e^{2x + y}, z_xx = (4 + 4x)E, z_xy = (1 + 2x)E, z_yy = xE, and (4 + 4x) − 3(1 + 2x) + 2x = 1, so F(D, D′)z = E.

Step 7: PI = e^{x + 3y}/10 + xe^{2x + y}.

Non-example: for (D² + D)z = sin(x + y) the operator is not homogeneous. Replacing D² by −1 leaves (D − 1)⁻¹sin(x + y), and D cannot be replaced by a number, because D turns sine into cosine. Rationalise instead: (D + 1)/(D² − 1) gives −(D + 1)sin(x + y)/2 = −[cos(x + y) + sin(x + y)]/2. The condition that fails is homogeneity of degree 2.

Building up: the sine rule works for a homogeneous F of degree 2 because D², DD′, D′² each return a constant times the same sine. The shift rule F(D, D′)(e^{ax + by}V) = e^{ax + by}F(D + a, D′ + b)V handles exponential times anything.

For a polynomial right side the expansion of [F(D, D′)]⁻¹ stops after a few terms; the failing-case rule itself follows from the general method of the next lesson ('Derivation: The General Method and the Failing Case' below).

Where it is used: mixed right sides such as 5 sin(2x + y) + 24(y − x) + e^{3x + 4y} (Example 11 below), where each term needs its own rule and one of them fails.

4. Understanding the General Method for a Particular Integral

Why we need it: right sides such as y cos x, x²sin(x + y) or tan(x + y) fit none of the short rules. One method always works when F splits into linear factors.

In words: (D − mD′)z = f is the Lagrange equation p − mq = f. Its characteristics are the lines y + mx = c, and along such a line z changes with x at the rate f(x, c − mx).

So: replace y by c − mx, integrate with respect to x holding c fixed, and finally put c = y + mx. For several factors, repeat one factor at a time.

First example, slowly: find a PI of (D² − DD′ − 2D′²)z = (y − 1)e^x.

Step 1: factor: m² − m − 2 = (m − 2)(m + 1), so F = (D − 2D′)(D + D′).

Step 2: invert D + D′ (m = −1) first: replace y by c + x, so the integrand is (c + x − 1)e^x.

Step 3: integrate by parts with c fixed: ∫(c + x − 1)e^x dx = (c + x − 1)e^x − e^x = (c + x − 2)e^x.

Step 4: put c = y − x back: (y − 2)e^x.

Step 5: now invert D − 2D′ (m = 2): replace y by c − 2x and integrate: ∫(c − 2x − 2)e^x dx = (c − 2x − 2)e^x + 2e^x = (c − 2x)e^x.

Step 6: put c = y + 2x: PI = ye^x.

Step 7: check: for z = ye^x, z_xx = ye^x, z_xy = e^x, z_yy = 0, so z_xx − z_xy − 2z_yy = (y − 1)e^x.

Non-example: in Step 2, integrating (y − 1)e^x 'with y held constant' gives (y − 1)e^x, but (D + D′)[(y − 1)e^x] = (y − 1)e^x + e^x, which is wrong. Along the characteristic y is not constant (it equals c + x); the step that fails is substituting y = c − mx before integrating.

Building up: the auxiliary equations dx/1 = dy/(−m) = dz/f give y + mx = c and dz = f(x, c − mx) dx, which is the rule ('Derivation: The General Method and the Failing Case' below).

When f = φ(y + mx), the integrand f(x, c − mx) = φ(c) does not depend on x, the integral is xφ(c), and the failing-case rule of the previous lesson appears automatically.

Where it is used: questions with right sides such as y cos x, y sin x or x²sin(x + y) (Examples 4, 14 and 16 below), and as a safe fallback whenever a shortcut is in doubt.

5. Notation and Structure

D = ∂/∂x, D′ = ∂/∂y; the equation is F(D, D′)z = f(x, y) and its general solution is z = CF + PI.

Homogeneous: every term of F has the same total order, e.g. (a₀D² + a₁DD′ + a₂D′²)z = f(x, y).

Idea: 'homogeneous' describes the operator (all terms of the same total order), not the right-hand side. The heat operator D² − D′ is non-homogeneous even when f = 0, and this changes the form of the CF.

The CF of an equation of order n contains n arbitrary functions.

Idea: this mirrors an ODE of order n having n arbitrary constants; z = CF + PI because the operator is linear, so the difference of two solutions solves the equation with f = 0.

6. Complementary Function (homogeneous F)

Put D = m, D′ = 1 to get the auxiliary equation a₀m² + a₁m + a₂ = 0 with roots m₁, m₂.

Distinct roots: CF = φ₁(y + m₁x) + φ₂(y + m₂x).

Repeated root m: CF = φ₁(y + mx) + xφ₂(y + mx).

Reason: (D − mD′)φ(y + mx) = mφ′ − mφ′ = 0.

Why it works: for z = φ(y + mx), D^rD′^sz = m^rφ^(r+s), so (a₀D² + a₁DD′ + a₂D′²)z = (a₀m² + a₁m + a₂)φ″, which vanishes for every φ exactly when m is a root. For a repeated root, (D − mD′)[xφ(y + mx)] = φ, which (D − mD′) then kills, so xφ₂(y + mx) is the second solution.

7. Complementary Function (non-homogeneous F)

If F splits into linear factors, a factor (D − mD′ − c) contributes e^cxφ(y + mx); a repeated factor contributes e^cx[φ₁(y + mx) + xφ₂(y + mx)].

A factor (βD′ − γ) without D contributes e^γy/βφ(x).

If F does not factorise, try z = Ae^{hx + ky} with F(h, k) = 0, giving CF = ∑A e^{hx + ky}.

Idea: an irreducible factor does not produce arbitrary functions, only a family of exponentials whose exponents lie on the curve F(h, k) = 0. For the heat operator D² − D′, k = h² and CF = ∑A e^{hx + h2y}; h = iω gives the decaying modes e^−ω2ycos ωx used in the heat equation topic.

8. Particular Integral

f = e^{ax + by}: PI = e^{ax + by}/F(a, b), provided F(a, b) ≠ 0.

Why it works: D e^{ax + by} = a e^{ax + by} and D′ e^{ax + by} = b e^{ax + by}, so F(D, D′)e^{ax + by} = F(a, b)e^{ax + by}; dividing by the number F(a, b) gives a function that F(D, D′) maps to e^{ax + by}.

Failing case: 1/(D − mD′)^r φ(y + mx) = (x^r/r!) φ(y + mx).

f = sin(ax + by) or cos(ax + by), F homogeneous of degree 2: replace D² by −a², DD′ by −ab, D′² by −b², provided the result is non-zero.

Why it works: each of D², DD′, D′² applied to sin(ax + by) returns −a², −ab, −b² times the same sine, so a degree-2 homogeneous F multiplies it by a constant. A single D or D′ turns sine into cosine, which is why the rule needs F to contain only second-order terms.

f = x^myⁿ: PI = [F(D, D′)]⁻¹x^myⁿ, expanding in ascending powers of D′/D (or D/D′).

f = e^{ax + by}V(x, y): PI = e^{ax + by}[F(D + a, D′ + b)]⁻¹V.

General method: (D − mD′)⁻¹f(x, y) = ∫f(x, c − mx) dx, after which c is replaced by y + mx.

9. Derivation: The General Method and the Failing Case

(D − mD′)z = f(x, y) is the Lagrange equation p − mq = f, with auxiliary equations dx/1 = dy/(−m) = dz/f.

The first pair gives y + mx = c, so along a characteristic y = c − mx and dz = f(x, c − mx) dx.

Hence z = ∫f(x, c − mx) dx + φ(c); the integral, with c replaced by y + mx, is a PI and φ(y + mx) is the CF of this factor. Repeating factor by factor inverts any F that splits into linear factors.

Failing case: if f = φ(y + mx), then f(x, c − mx) = φ(c) is constant in x and the integral is xφ(c) = xφ(y + mx); applying this r times gives (D − mD′)^−rφ(y + mx) = (x^r/r!)φ(y + mx).

Shift rule: D(e^{ax + by}V) = e^{ax + by}(D + a)V and D′(e^{ax + by}V) = e^{ax + by}(D′ + b)V, so F(D, D′)(e^{ax + by}V) = e^{ax + by}F(D + a, D′ + b)V; inverting gives the rule PI = e^{ax + by}[F(D + a, D′ + b)]⁻¹V.

Polynomial right side: for f = x^myⁿ, powers of D′/D beyond n annihilate yⁿ, so the series expansion of [F(D, D′)]⁻¹ stops after finitely many terms and gives an exact polynomial PI.

10. Choosing the Method

If F is homogeneous → CF from the roots of a₀m² + a₁m + a₂ = 0; if F is non-homogeneous but factorises into linear factors → e^cxφ(y + mx) for each factor; if it does not factorise → exponentials with F(h, k) = 0.

If f = e^{ax + by} → e^{ax + by}/F(a, b); if F(a, b) = 0 → the x^r/r! rule for the vanishing factor, or the general method.

If f = sin or cos(ax + by) and F is homogeneous of degree 2 → the substitution rule; if F is not homogeneous → substitute for the second-order terms, then rationalise the remaining first-order operator or try A sin + B cos.

If f is a polynomial → expand [F]⁻¹ in powers of D′/D (or D/D′); if f = e^{ax + by}V → shift rule; if f = φ(ax + by) with F homogeneous → the repeated-integral rule of Section 13.

If nothing else fits (y cos x, x e^y, tan(x + y), ...) → factor F and apply ∫f(x, c − mx) dx one factor at a time.

11. Common Mistakes

Wrong sign in the argument: the factor (D − mD′) gives φ(y + mx); D² − D′² has m = ±1 and CF φ₁(y + x) + φ₂(y − x).

Repeated root without the factor x: (D − D′)²z = 0 has CF φ₁(y + x) + xφ₂(y + x), not φ₁(y + x) + φ₂(y + x), which is only one function.

Dividing by zero: using e^{ax + by}/F(a, b) when F(a, b) = 0; always compute F(a, b) first and show it.

Not putting c back: in ∫f(x, c − mx) dx, c is constant during the integration and must be replaced by y + mx at the end.

Sine rule on a non-homogeneous F: for (D² − D′)z = sin(x + y), D² → −1 leaves −(1 + D′)⁻¹sin(x + y) = −(1 − D′)(1 − D′²)⁻¹sin(x + y) = −[sin(x + y) − cos(x + y)]/2; one cannot simply replace D′ by a number.

Forgetting the CF: 'general solution' means CF + PI; a PI alone earns only part of the marks.

Link: every step copies linear ODEs with constant coefficients (Ordinary Differential Equations): auxiliary equation, CF + PI, the 1/f(D) shortcuts and the failing case; arbitrary functions of y + mx take the place of arbitrary constants.

Link: the discriminant a₁² − 4a₀a₂ of the auxiliary equation is B² − 4AC of the next topic: two real roots means hyperbolic (D² − c²D′² = (D − cD′)(D + cD′) gives D'Alembert's solution), a double root parabolic, complex roots elliptic.

12. Worked Examples

Example 1: (D² − 5DD′ + 6D′²)z = e^{x + y}. Auxiliary m² − 5m + 6 = 0 gives m = 2, 3, and F(1, 1) = 2.

z = φ₁(y + 2x) + φ₂(y + 3x) + (1/2)e^{x + y}.

Example 2: (D² − D′²)z = sin(x + 2y). PI = sin(x + 2y)/(−1 + 4); z = φ₁(y + x) + φ₂(y − x) + (1/3)sin(x + 2y).

Example 3: (D − D′)²z = e^{x + y}. F(1, 1) = 0 and e^{x + y} is a function of y + x, so PI = (x²/2)e^{x + y}; z = φ₁(y + x) + xφ₂(y + x) + (x²/2)e^{x + y}.

Example 4: (D² + DD′ − 6D′²)z = y cos x. Auxiliary m² + m − 6 = 0 gives m = 2, −3, so F = (D − 2D′)(D + 3D′).

(D + 3D′)⁻¹y cos x = ∫(c + 3x)cos x dx = (c + 3x)sin x + 3cos x = y sin x + 3cos x.

(D − 2D′)⁻¹(y sin x + 3cos x) = ∫[(c − 2x)sin x + 3cos x] dx = −(c − 2x)cos x + sin x = sin x − y cos x.

So z = φ₁(y + 2x) + φ₂(y − 3x) + sin x − y cos x.

Example 5: Solve (D² − 2DD′ + D′²)z = 12xy.

Auxiliary (m − 1)² = 0, so CF = φ₁(y + x) + xφ₂(y + x).

PI = (1/D²)(1 − D′/D)⁻²12xy = (1/D²)(1 + 2D′/D + ...)12xy = (1/D²)(12xy + 12x²) = 2x³y + x⁴.

Check: z_xx − 2z_xy + z_yy = (12xy + 12x²) − 12x² + 0 = 12xy; so z = φ₁(y + x) + xφ₂(y + x) + 2x³y + x⁴.

Example 6: Solve (D² − DD′ + D′ − 1)z = cos(x + 2y) + e^y.

F = (D − 1)(D − D′ + 1), so CF = e^xφ₁(y) + e^−xφ₂(y + x).

For cos(x + 2y) put D² = −1, DD′ = −2: F becomes −1 + 2 + D′ − 1 = D′, so PI₁ = (1/D′)cos(x + 2y) = (1/2)sin(x + 2y).

For e^y, F(0, 1) = 0 because the factor D − D′ + 1 vanishes; (D − 1)⁻¹e^y = −e^y, and trying z = Axe^y in (D − D′ + 1)z = −e^y gives A = −1, so PI₂ = −xe^y.

z = e^xφ₁(y) + e^−xφ₂(y + x) + (1/2)sin(x + 2y) − xe^y.

13. More PI Rules, Higher Order and Fitting Data

Rule for φ(ax + by): if F(D, D′) is homogeneous of degree n and F(a, b) ≠ 0, then [F(D, D′)]⁻¹φ(ax + by) = (1/F(a, b)) ∫∫...∫ φ(v) dv ... dv (n times), with v = ax + by.

Failing case: if F has the factor (bD − aD′)^r, then (bD − aD′)^−rφ(ax + by) = (x^r/(b^r r!))φ(ax + by); invert the other factors first (by the rule above) and this factor last.

Order n homogeneous: the auxiliary equation has n roots; a root m of multiplicity k contributes φ₁(y + mx) + xφ₂(y + mx) + ... + x^k−1φ_k(y + mx); factor the cubic by trying m = ±1, ±2.

sin or cos with degree 3: substitute D² → −a², DD′ → −ab, D′² → −b² wherever they occur (D³ = D · D² → −a²D), which leaves a first-degree operator; invert it by a trial A cos(ax + by) or by rationalising.

x^m times sin, cos or e^{ax + by}: factor F and apply (D − mD′)⁻¹f = ∫f(x, c − mx) dx twice, integrating by parts; or use e^{ax + by}[F(D + a, D′ + b)]⁻¹V.

Fitting data: to find the surface through given curves, substitute each curve into the general solution and solve the resulting functional equations for φ₁, φ₂.

Euler–Cauchy type (x²r, xys, y²t, xp, yq with constant coefficients): put x = e^X, y = e^Y; then xD = D_X, x²D² = D_X(D_X − 1), xyDD′ = D_XD_Y, and the equation has constant coefficients.

14. Past UPSC Questions Solved

Example 7 (UPSC 2014, 10 marks): Solve the partial differential equation (2D² − 5DD′ + 2D′²)z = 24(y − x).

Auxiliary 2m² − 5m + 2 = 0 gives m = 2, 1/2, so CF = φ₁(y + 2x) + φ₂(2y + x).

24(y − x) = 24v with v = −x + y, so a = −1, b = 1 and F(a, b) = 2 + 5 + 2 = 9 ≠ 0.

PI = (1/9)∫∫24v dv dv = (1/9)(4v³) = (4/9)(y − x)³; check: z_xx = z_yy = (8/3)v, z_xy = −(8/3)v give (2 + 5 + 2)(8/3)v = 24v.

Answer: z = φ₁(y + 2x) + φ₂(2y + x) + (4/9)(y − x)³.

Example 8 (UPSC 2026, 15 marks): Solve the partial differential equation (2D² − 5DD′ + 2D′²) z = 24(y − x) + sin(y − x), where D ≡ ∂/∂x, D′ ≡ ∂/∂y, z = z(x, y).

The CF and the PI of 24(y − x) are as in the previous example.

For sin(y − x): a = −1, b = 1, so D² → −1, DD′ → −ab = 1, D′² → −1 and F → −2 − 5 − 2 = −9; PI = −(1/9)sin(y − x).

Answer: z = φ₁(y + 2x) + φ₂(2y + x) + (4/9)(y − x)³ − (1/9)sin(y − x).

Example 9 (UPSC 2016, 15 marks): Solve the partial differential equation
∂³z/∂x³ − 2 ∂³z/∂x²∂y − ∂³z/∂x∂y² + 2 ∂³z/∂y³ = e^x+y

Auxiliary m³ − 2m² − m + 2 = (m − 1)(m + 1)(m − 2) = 0, so CF = φ₁(y + x) + φ₂(y − x) + φ₃(y + 2x).

F(1, 1) = 1 − 2 − 1 + 2 = 0, a failing case caused by the factor D − D′.

[(D + D′)(D − 2D′)]⁻¹e^{x + y} = e^{x + y}/(2 · (−1)), and (D − D′)⁻¹e^{x + y} = xe^{x + y} because e^{x + y} is a function of y + x.

Answer: z = φ₁(y + x) + φ₂(y − x) + φ₃(y + 2x) − (x/2)e^{x + y}.

Example 10 (UPSC 2017, 10 marks): Solve (D² − 2DD′ + D′²)z = e^x+2y + x³ + sin 2x,
where D ≡ ∂/∂x, D′ ≡ ∂/∂y, D² ≡ ∂²/∂x², D′² ≡ ∂²/∂y².

Auxiliary (m − 1)² = 0, so CF = φ₁(y + x) + xφ₂(y + x).

e^{x + 2y}: F(1, 2) = (1 − 2)² = 1, PI₁ = e^{x + 2y}; x³: PI₂ = (1/D²)(1 − D′/D)⁻²x³ = (1/D²)x³ = x⁵/20 since D′x³ = 0.

sin 2x: a = 2, b = 0, so D² → −4, DD′ → 0, D′² → 0 and PI₃ = −(1/4)sin 2x.

Answer: z = φ₁(y + x) + xφ₂(y + x) + e^{x + 2y} + x⁵/20 − (1/4)sin 2x.

Example 11 (UPSC 2018, 15 marks): Solve the partial differential equation :
(2D² − 5DD′ + 2D′²)z = 5 sin(2x + y) + 24(y − x) + e^{3x + 4y}
where D ≡ ∂/∂x, D′ ≡ ∂/∂y.

F = (2D − D′)(D − 2D′), CF = φ₁(y + 2x) + φ₂(2y + x).

5 sin(2x + y): D² → −4, DD′ → −2, D′² → −1 gives −8 + 10 − 2 = 0, a failing case, because sin(2x + y) is a function of y + 2x.

(2D − D′)cos(2x + y) = −3 sin(2x + y), so (2D − D′)⁻¹5 sin(2x + y) = −(5/3)cos(2x + y); then (D − 2D′)⁻¹ multiplies by x: PI₁ = −(5x/3)cos(2x + y).

PI₂ = (4/9)(y − x)³ (as for UPSC 2014); e^{3x + 4y}: F(3, 4) = 18 − 60 + 32 = −10, PI₃ = −(1/10)e^{3x + 4y}.

Answer: z = φ₁(y + 2x) + φ₂(2y + x) − (5x/3)cos(2x + y) + (4/9)(y − x)³ − (1/10)e^{3x + 4y}.

Example 12 (UPSC 2020, 10 marks): Solve the partial differential equation : (D³ − 2D²D′ − DD′² + 2D′³)z = e^2x+y + sin(x − 2y); D ≡ ∂/∂x, D′ ≡ ∂/∂y

F = (D − D′)(D + D′)(D − 2D′) as in UPSC 2016, so CF = φ₁(y + x) + φ₂(y − x) + φ₃(y + 2x).

e^{2x + y}: F(2, 1) = 8 − 8 − 2 + 2 = 0 because of D − 2D′; [(D − D′)(D + D′)]⁻¹e^{2x + y} = e^{2x + y}/3, and then (D − 2D′)⁻¹ gives PI₁ = (x/3)e^{2x + y}.

sin(x − 2y): a = 1, b = −2 give D³ → −D, D²D′ → −D′, DD′² → −4D, D′³ → −4D′, so F → −D + 2D′ + 4D − 8D′ = 3(D − 2D′).

(D − 2D′)cos(x − 2y) = −sin(x − 2y) − 4 sin(x − 2y) = −5 sin(x − 2y), so PI₂ = (1/3)(−1/5)cos(x − 2y) = −(1/15)cos(x − 2y).

Answer: z = φ₁(y + x) + φ₂(y − x) + φ₃(y + 2x) + (x/3)e^{2x + y} − (1/15)cos(x − 2y).

Example 13 (UPSC 2021, 15 marks): Find the general solution of the partial differential equation (D² − D′² − 3D + 3D′)z = xy + e^x+2y where D ≡ ∂/∂x and D′ ≡ ∂/∂y.

F = (D − D′)(D + D′) − 3(D − D′) = (D − D′)(D + D′ − 3), so CF = φ₁(y + x) + e^3xφ₂(y − x).

xy: (D + D′ − 3)⁻¹xy = −(1/3)[1 + (D + D′)/3 + (D + D′)²/9]xy = −(1/3)[xy + (x + y)/3 + 2/9].

(D − D′)⁻¹ = (1/D)(1 + D′/D + ...) applied to xy + (x + y)/3 + 2/9 gives x²y/2 + x³/6 + x²/3 + xy/3 + 2x/9, so PI₁ = −(1/54)(9x²y + 3x³ + 6x² + 6xy + 4x).

e^{x + 2y}: F(1, 2) = 0 since D + D′ − 3 → 0; (D + D′ − 3)⁻¹e^{x + 2y} = e^{x + 2y}(D + D′)⁻¹1 = xe^{x + 2y}, then (D − D′)⁻¹(xe^{x + 2y}) = e^{x + 2y}(D − D′ − 1)⁻¹x = −(x + 1)e^{x + 2y}.

The part −e^{x + 2y} = −e^3xe^{2(y − x)} lies in the CF, so PI₂ = −xe^{x + 2y}.

Answer: z = φ₁(y + x) + e^3xφ₂(y − x) − (1/54)(9x²y + 3x³ + 6x² + 6xy + 4x) − xe^{x + 2y}.

Example 14 (UPSC 2022, 15 marks): Find the general solution of the partial differential equation (D² + DD′ − 6D′²) z = x² sin(x + y) where D ≡ ∂/∂x and D′ ≡ ∂/∂y.

F = (D − 2D′)(D + 3D′), so CF = φ₁(y + 2x) + φ₂(y − 3x).

(D + 3D′)⁻¹x²sin(x + y) = ∫x²sin(4x + c) dx with c = y − 3x, which gives g = −(x²/4)cos(x + y) + (x/8)sin(x + y) + (1/32)cos(x + y).

(D − 2D′)⁻¹g = ∫g(x, c − 2x) dx with c = y + 2x, where x + y becomes c − x.

Using ∫x²cos(c − x) dx = −x²sin(c − x) + 2x cos(c − x) + 2 sin(c − x) and ∫x sin(c − x) dx = x cos(c − x) + sin(c − x), the integral is (x²/4)sin(c − x) − (3x/8)cos(c − x) − (13/32)sin(c − x).

Answer: z = φ₁(y + 2x) + φ₂(y − 3x) + (x²/4)sin(x + y) − (3x/8)cos(x + y) − (13/32)sin(x + y) (verified by substitution).

Example 15 (UPSC 2023, 15 marks): Find the surface passing through the two lines z = x = 0 and z − 1 = x − y = 0, and satisfying the partial differential equation ∂²z/∂x² − 4 ∂²z/∂x∂y + 4 ∂²z/∂y² = 0.

The equation is (D − 2D′)²z = 0; auxiliary (m − 2)² = 0, so z = φ₁(y + 2x) + xφ₂(y + 2x).

Line z = x = 0 (the y-axis): 0 = φ₁(y) for every y, so φ₁ ≡ 0.

Line x = y, z = 1: 1 = xφ₂(3x), so φ₂(s) = 3/s.

Answer: z = 3x/(y + 2x), i.e. z(2x + y) = 3x (x = 0 gives z = 0; x = y gives z = 1).

Example 16 (UPSC 2025, 10 marks): Find the solution of the equation (D² + DD′ − 2D²) z = y sin x, where D ≡ ∂/∂x and D′ ≡ ∂/∂y.

[The paper prints (D² + DD′ − 2D²); the intended operator is D² + DD′ − 2D′².]

m² + m − 2 = 0 gives m = 1, −2: CF = φ₁(y + x) + φ₂(y − 2x), F = (D − D′)(D + 2D′).

(D + 2D′)⁻¹y sin x = ∫(c + 2x)sin x dx (c = y − 2x) = −(c + 2x)cos x + 2 sin x = −y cos x + 2 sin x.

(D − D′)⁻¹(−y cos x + 2 sin x) = ∫[−(c − x)cos x + 2 sin x] dx (c = y + x) = −c sin x + x sin x − cos x = −y sin x − cos x.

Answer: z = φ₁(y + x) + φ₂(y − 2x) − y sin x − cos x (check: z_xx + z_xy − 2z_yy = y sin x + cos x − cos x = y sin x).

15. Practice Exercises

Exercise 1 (basic): Write the CF of (D² − 4D′²)z = 0.

Answer: m = ±2, so CF = φ₁(y + 2x) + φ₂(y − 2x).

Exercise 2 (basic): Write the CF of (D² + 2DD′ + D′²)z = 0.

Answer: (m + 1)² = 0, a double root m = −1, so CF = φ₁(y − x) + xφ₂(y − x).

Exercise 3 (basic): Find a PI of (D² − D′²)z = e^{x + 2y}.

Answer: F(1, 2) = 1 − 4 = −3, so PI = −e^{x + 2y}/3.

Exercise 4 (standard): Solve (D² + 3DD′ + 2D′²)z = x + y.

Answer: m = −1, −2; with v = x + y, F(1, 1) = 6 and PI = (1/6)∫∫v dv dv = (x + y)³/36; z = φ₁(y − x) + φ₂(y − 2x) + (x + y)³/36.

Exercise 5 (standard): Solve (D² − DD′)z = cos(x + 2y).

Answer: m = 0, 1; D² → −1, DD′ → −2 give F → 1, so z = φ₁(y) + φ₂(y + x) + cos(x + 2y).

Exercise 6 (standard): Solve (D² − 4DD′ + 4D′²)z = e^{2x + y}.

Answer: F = (D − 2D′)² and e^{2x + y} is a function of y + 2x, so PI = (x²/2)e^{2x + y}; z = φ₁(y + 2x) + xφ₂(y + 2x) + (x²/2)e^{2x + y}.

Exercise 7 (standard): Solve (D² − 2DD′)z = e^2x + x³y.

Answer: CF = φ₁(y) + φ₂(y + 2x); PI = e^2x/4 + (1/D²)(1 + 2D′/D)x³y = e^2x/4 + x⁵y/20 + x⁶/60.

Exercise 8 (UPSC level): Solve (D² + 2DD′ + D′²)z = 2 cos y − x sin y.

Answer: by the general method (factor D + D′ twice) PI = x sin y; z = φ₁(y − x) + xφ₂(y − x) + x sin y.

Exercise 9 (UPSC level): Solve (D² − DD′ − 2D′² + 2D + 2D′)z = sin(2x + y).

Answer: F = (D + D′)(D − 2D′ + 2), CF = φ₁(y − x) + e^−2xφ₂(y + 2x); the second-order part gives −4 + 2 + 2 = 0, so PI = [2(D + D′)]⁻¹sin(2x + y) = −cos(2x + y)/6.

Exercise 10 (UPSC level): Find the surface satisfying ∂²z/∂x² = 6x + 2 that touches z = x³ + y³ along its section by the plane x + y + 1 = 0.

Answer: z = x³ + x² + xf(y) + g(y); equal values and equal x-slopes on x = −1 − y give f(y) = 2(1 + y), g(y) = y³ + (1 + y)², so z = x³ + y³ + (x + y + 1)².

1. Understanding the Type of a Second-Order Equation

Why we need it: second-order equations behave in three very different ways: waves travel (hyperbolic), heat spreads and smooths (parabolic), and steady states settle (elliptic). Each type has its own canonical form and its own kind of boundary data, so the first question about any such equation is 'which type?'.

In words: only the second-order part A u_xx + B u_xy + C u_yy decides the type; first-order terms, u itself and the right side never matter. A, B, C are the full coefficients as written (B is not halved).

The discriminant B² − 4AC is computed at each point: positive means hyperbolic, zero parabolic, negative elliptic (hyperbolic, parabolic, elliptic). When A, B, C depend on x and y the sign, and hence the type, can change from one region to another.

First example, slowly: classify u_xx + 2yu_xy + xu_yy = 0 in the whole plane.

Step 1: read the coefficients: A = 1 (of u_xx), B = 2y (of u_xy), C = x (of u_yy).

Step 2: B² − 4AC = 4y² − 4x = 4(y² − x).

Step 3: find where it vanishes: on the parabola x = y², which divides the plane into two regions.

Step 4: test one point on each side: at (0, 1) the value is 4 > 0, and at (1, 0) it is −4 < 0.

Step 5: conclusion: hyperbolic where x < y², parabolic on x = y², elliptic where x > y² (inside the parabola).

Step 6: notice that adding terms such as 5u_x − u would not change any of this.

Non-example: in uu_xx + u_yy = 0 the coefficient A = u depends on the unknown solution, so B² − 4AC = −4u cannot be evaluated from the equation alone: the type is known only once u is (elliptic where u > 0, hyperbolic where u < 0). The test as stated needs A, B, C to be functions of x and y only.

Building up: the type does not depend on the coordinates: under ξ = ξ(x, y), η = η(x, y) with Jacobian J ≠ 0 the new discriminant is J²(B² − 4AC), and J² > 0 (see 'Why it works' under Classification below).

This is why the names come from conics: AX² + BXY + CY² = 1 is a hyperbola, parabola or ellipse by the same sign, and a change of axes cannot turn one kind of conic into another.

Where it is used: 'classify the equation in different regions' questions, the type of a PDE just formed by elimination (Example 10 of the first topic), and the choice of canonical form in the next two lessons.

2. Understanding Characteristic Curves of a Second-Order Equation

Why we need it: to simplify A u_xx + B u_xy + C u_yy we change to new variables ξ, η. The best choice makes the coefficient of u_ξξ (and if possible of u_ηη) vanish, and the curves ξ = constant that do this are the characteristics.

In words: the new coefficient of u_ξξ is A′ = Aξ_x² + Bξ_xξ_y + Cξ_y². On a level curve ξ = c the slope is dy/dx = −ξ_x/ξ_y, so A′ = 0 becomes A(dy/dx)² − B(dy/dx) + C = 0.

So: solve this quadratic for the slope λ = dy/dx, integrate each ODE dy/dx = λ, and write the result as ξ(x, y) = constant. Two real slopes give two families, one double slope gives one family, and complex slopes give complex characteristics.

First example, slowly: find the characteristics of 3u_xx + 10u_xy + 3u_yy = 0.

Step 1: A = 3, B = 10, C = 3 and B² − 4AC = 100 − 36 = 64 > 0: hyperbolic, so expect two real families.

Step 2: the characteristic equation is 3λ² − 10λ + 3 = 0, that is (3λ − 1)(λ − 3) = 0, so λ = 3 or λ = 1/3.

Step 3: dy/dx = 3 integrates to y − 3x = c, so ξ = y − 3x.

Step 4: dy/dx = 1/3 integrates to 3y − x = c, so η = 3y − x.

Step 5: check A′ for ξ: ξ_x = −3, ξ_y = 1 give 3(9) + 10(−3)(1) + 3(1) = 27 − 30 + 3 = 0.

Step 6: check for η: η_x = −1, η_y = 3 give 3(1) + 10(−1)(3) + 3(9) = 0. Both families are characteristic.

Step 7: the Jacobian ξ_xη_y − ξ_yη_x = (−3)(3) − (1)(−1) = −8 ≠ 0, so (ξ, η) is a genuine change of variables.

Non-example: with the wrong sign, 3λ² + 10λ + 3 = 0 gives λ = −3, −1/3 and ξ = y + 3x. Then A′ = 3(9) + 10(3)(1) + 3(1) = 60 ≠ 0, so u_ξξ survives. The condition that fails is the minus sign in A(dy/dx)² − B(dy/dx) + C = 0.

Building up: the same calculation with η in place of ξ gives C′, so for a hyperbolic equation both A′ and C′ vanish when ξ and η come from the two roots; for a parabolic one only one family exists; for an elliptic one the roots are complex conjugates (next lesson).

For the wave equation u_tt = c²u_xx the characteristics are x ± ct = constant, the paths along which signals travel (see 'Why the minus sign' under Classification below for the derivation of the sign).

Where it is used: every reduction to canonical form, D'Alembert's solution of the wave equation, and the domain of dependence of hyperbolic problems.

3. Understanding the Reduction to Canonical Form

Why we need it: in the right variables a hyperbolic equation becomes u_ξη = (lower-order terms), a parabolic one u_ηη = (...), and an elliptic one u_αα + u_ββ = (...). These canonical forms are easier to recognise and are often integrable, which is why UPSC asks 'reduce to canonical form and hence solve'.

In words: write u(x, y) = U(ξ, η). By the chain rule u_x = ξ_xU_ξ + η_xU_η and u_y = ξ_yU_ξ + η_yU_η; differentiating again gives u_xx, u_xy, u_yy. Substitute, collect the coefficients of U_ξξ, U_ξη, U_ηη, and divide by the surviving one.

With constant coefficients the second derivatives of ξ, η are zero, so no first-order terms appear; with variable coefficients terms such as ξ_xxU_ξ remain and form the right side.

First example, slowly: reduce 3u_xx + 10u_xy + 3u_yy = 0 with ξ = y − 3x, η = 3y − x (previous lesson) and solve it.

Step 1: first derivatives: u_x = −3U_ξ − U_η and u_y = U_ξ + 3U_η.

Step 2: u_xx = 9U_ξξ + 6U_ξη + U_ηη (apply ∂/∂x = −3∂/∂ξ − ∂/∂η to u_x).

Step 3: u_xy = −3U_ξξ − 10U_ξη − 3U_ηη (apply ∂/∂y = ∂/∂ξ + 3∂/∂η to u_x), and u_yy = U_ξξ + 6U_ξη + 9U_ηη.

Step 4: collect: U_ξξ has 27 − 30 + 3 = 0, U_ηη has 3 − 30 + 27 = 0, and U_ξη has 18 − 100 + 18 = −64.

Step 5: so −64U_ξη = 0, and the canonical form is U_ξη = 0. (Check: B′² = 64² equals J²(B² − 4AC) = 64 × 64.)

Step 6: integrate in η: U_ξ = F′(ξ); then in ξ: U = F(ξ) + G(η).

Step 7: return to x, y: u = F(y − 3x) + G(3y − x), with F, G arbitrary.

Non-example: keeping ξ = y − 3x but taking η = 2y − 6x is not allowed: η = 2ξ, the Jacobian is 0, and u would be expressed through one variable only, so u_ξ and u_η cannot be told apart. The condition that fails is J ≠ 0.

Building up: in matrix form the coefficients transform as M′ = J_mMJ_m^T ('Derivation: How the Coefficients Transform' below). For a parabolic equation take ξ from the single family and any η with J ≠ 0 (often η = x); then A′ = B′ = 0 and only U_ηη survives.

For an elliptic equation with complex characteristic ξ = α + iβ, use the real variables α and β; then the coefficients of U_αα and U_ββ become equal and U_αβ disappears, giving the Laplacian.

Where it is used: the 15–20 mark 'reduce to canonical form and hence solve' questions (Examples 7 to 12 below), and D'Alembert's solution, which is exactly U_ξη = 0 for the wave equation.

4. Classification

General form: A u_xx + B u_xy + C u_yy + F(x, y, u, u_x, u_y) = 0, where A, B, C are functions of x, y, not all zero.

At a point: B² − 4AC > 0 hyperbolic, B² − 4AC = 0 parabolic, B² − 4AC < 0 elliptic.

Idea: the names come from conics: replacing u_xx, u_xy, u_yy by X², XY, Y² gives AX² + BXY + CY², a hyperbola, parabola or ellipse by the same discriminant. The type also predicts behaviour: hyperbolic equations carry waves at finite speed along characteristics, parabolic ones diffuse and smooth, and elliptic ones describe equilibria in which every boundary point influences every interior point.

Under ξ = ξ(x, y), η = η(x, y) with Jacobian J = ξ_xη_y − ξ_yη_x ≠ 0, the new coefficients satisfy B′² − 4A′C′ = J²(B² − 4AC), so the type is invariant.

Why it works: write the principal part as the matrix M = [[A, B/2], [B/2, C]], so B² − 4AC = −4 det M. The chain rule (Section 6) shows M′ = J_mMJ_m^T with J_m = [[ξ_x, ξ_y], [η_x, η_y]], hence det M′ = J² det M; since J² > 0 the sign, and so the type, cannot change.

New coefficients: A′ = Aξ_x² + Bξ_xξ_y + Cξ_y², B′ = 2Aξ_xη_x + B(ξ_xη_y + ξ_yη_x) + 2Cξ_yη_y, C′ = Aη_x² + Bη_xη_y + Cη_y².

Characteristic equation: A(dy)² − B dx dy + C(dx)² = 0, that is, A(dy/dx)² − B(dy/dx) + C = 0.

Choosing ξ = constant along a characteristic makes A′ = 0, which is the basis of the reduction.

Why the minus sign: A′ = Aξ_x² + Bξ_xξ_y + Cξ_y² = 0; divide by ξ_y² and use the slope of a level curve ξ = c, dy/dx = −ξ_x/ξ_y. Then ξ_x/ξ_y = −dy/dx and A(dy/dx)² − B(dy/dx) + C = 0, which is the characteristic equation.

5. Reduction to Canonical Form

Hyperbolic: two real families of characteristics ξ(x, y) = c₁, η(x, y) = c₂; canonical form u_ξη = φ(ξ, η, u, u_ξ, u_η), or u_αα − u_ββ = ψ with α = ξ + η, β = ξ − η.

Parabolic: one family ξ(x, y) = c; take η any function with ∂(ξ, η)/∂(x, y) ≠ 0 (often η = x or y); canonical form u_ηη = φ(ξ, η, u, u_ξ, u_η).

Why it works: by invariance B′² − 4A′C′ = 0, and A′ = 0 then forces B′ = 0 as well; only C′u_ηη survives among the second derivatives, and C′ ≠ 0 because the Jacobian is non-zero.

Elliptic: complex conjugate characteristics ξ = α + iβ; take α = Re ξ, β = Im ξ; canonical form u_αα + u_ββ = φ(α, β, u, u_α, u_β).

Why it works: with Q(X, Y) = AX² + BXY + CY², the condition Q(∇ξ) = 0 for ξ = α + iβ splits into real and imaginary parts: Q(∇α) = Q(∇β) and the mixed term vanishes. These are exactly 'coefficient of u_αα = coefficient of u_ββ' and 'no u_αβ term'.

Constant coefficients: if λ₁, λ₂ are roots of Aλ² − Bλ + C = 0, the characteristics are the straight lines y − λ₁x = c₁, y − λ₂x = c₂.

Idea: for the wave equation u_tt = c²u_xx the characteristics are x ± ct = constant; the two through a point (x₀, t₀) cut the initial line at x₀ ± ct₀ and bound its domain of dependence (see the figure above). Data outside that interval cannot affect u(x₀, t₀).

6. Derivation: How the Coefficients Transform

With u_x = ξ_xu_ξ + η_xu_η and u_y = ξ_yu_ξ + η_yu_η, the second-order parts are u_xx ~ ξ_x²u_ξξ + 2ξ_xη_xu_ξη + η_x²u_ηη, u_xy ~ ξ_xξ_yu_ξξ + (ξ_xη_y + ξ_yη_x)u_ξη + η_xη_yu_ηη, u_yy ~ ξ_y²u_ξξ + 2ξ_yη_yu_ξη + η_y²u_ηη (~ means 'apart from first-order terms').

Multiplying by A, B, C and collecting u_ξξ, u_ξη, u_ηη gives exactly the A′, B′, C′ of Section 4; in matrix form M′ = J_mMJ_m^T.

Hyperbolic reduction: take ξ and η from the two real roots of the characteristic equation; then A′ = C′ = 0, and B′ ≠ 0 because B′² = J²(B² − 4AC) > 0. Dividing the transformed equation by B′ leaves u_ξη = φ.

The second form follows from α = ξ + η, β = ξ − η, since then u_ξη = u_αα − u_ββ.

7. Choosing the Method

If A, B, C are constants → solve Aλ² − Bλ + C = 0; the characteristics are y − λx = constant, and for complex λ = p ± iq take α = y − px, β = qx.

If the coefficients vary but the characteristic equation separates (such as y²dy² = x²dx²) → take square roots and integrate each factor separately.

If A = 0 → x = constant is one family; if C = 0 → y = constant is one family.

If the equation is parabolic → ξ from the double root, η = x or y (whichever keeps J ≠ 0).

If the question asks for the type 'in different regions' → factor B² − 4AC, find the curve where it vanishes (that curve is the parabolic set) and test one point on each side.

8. Common Mistakes

Plus sign in the characteristic equation: it is A(dy/dx)² − B(dy/dx) + C = 0, as derived above; the plus sign replaces every slope λ by −λ, so both families come out wrong whenever B ≠ 0.

Mixing conventions: some books write 2B u_xy and test B² − AC; using B² − 4AC with their B misclassifies u_xx + 2u_xy + u_yy = 0 (B = 1 would give −3, 'elliptic', but the equation is parabolic).

Type at one point only: with variable coefficients give the type in every region, including the dividing curve, which is parabolic.

A singular change of variables: ξ = x + y with η = 2x + 2y has J = 0 and is not allowed; always show J ≠ 0.

Dropping first-order terms: with variable ξ, η the terms ξ_xxu_ξ, η_yyu_η survive and give the right side of the canonical form (Example 2).

Link: the test is the classification of the conic AX² + BXY + CY² (Analytic Geometry); in n variables the type is read from the signs of the eigenvalues of the coefficient matrix, exactly as in the reduction of conicoids (Linear Algebra).

Link: the characteristics x ± ct = constant of the wave equation are the paths of D'Alembert's travelling waves f(x − ct), g(x + ct) (Vibrating String topic); a kink in the initial data travels along them and is never smoothed out, unlike in the heat equation.

9. Worked Examples

Example 1 (parabolic): u_xx + 2u_xy + u_yy = 0. B² − 4AC = 4 − 4 = 0; dy/dx = 1, so ξ = y − x; take η = x.

Then u_x = −u_ξ + u_η, u_y = u_ξ, and u_xx + 2u_xy + u_yy = u_ηη = 0.

Hence u = f(ξ) + ηg(ξ) = f(y − x) + xg(y − x).

Example 2 (hyperbolic): u_xx − x²u_yy = 0, x ≠ 0. B² − 4AC = 4x² > 0.

(dy/dx)² = x² gives ξ = y + x²/2, η = y − x²/2.

Then u_x = x(u_ξ − u_η), u_xx = (u_ξ − u_η) + x²(u_ξξ − 2u_ξη + u_ηη), u_yy = u_ξξ + 2u_ξη + u_ηη.

So (u_ξ − u_η) − 4x²u_ξη = 0; with x² = ξ − η the canonical form is u_ξη = (u_ξ − u_η)/(4(ξ − η)).

Example 3 (elliptic): u_xx + x²u_yy = 0, x ≠ 0. (dy/dx)² = −x² gives y ± ix²/2 = c, so α = y, β = x²/2.

Then u_x = xu_β, u_xx = u_β + x²u_ββ, u_yy = u_αα, so x²(u_αα + u_ββ) + u_β = 0 and u_αα + u_ββ = −u_β/(2β).

Example 4 (hyperbolic, variable coefficients): Reduce y²u_xx − x²u_yy = 0 (x, y > 0) to canonical form.

B² − 4AC = 4x²y² > 0; y²(dy/dx)² = x² gives y dy = ±x dx, so ξ = y² − x², η = y² + x².

u_x = 2x(u_η − u_ξ), u_y = 2y(u_ξ + u_η); substituting, y²u_xx − x²u_yy = −16x²y²u_ξη + 2ξu_η − 2ηu_ξ.

Since 4x²y² = η² − ξ², the canonical form is u_ξη = (ξu_η − ηu_ξ)/(2(η² − ξ²)).

Example 5 (elliptic, constant coefficients): Reduce u_xx + 4u_xy + 5u_yy = 0 to canonical form.

B² − 4AC = 16 − 20 = −4 < 0; λ² − 4λ + 5 = 0 gives λ = 2 ± i, so the characteristics are y − (2 ± i)x = c; take α = y − 2x, β = x.

u_x = −2u_α + u_β, u_y = u_α; u_xx = 4u_αα − 4u_αβ + u_ββ, u_xy = −2u_αα + u_αβ, u_yy = u_αα.

Sum: (4 − 8 + 5)u_αα + (−4 + 4)u_αβ + u_ββ = 0, so the canonical form is u_αα + u_ββ = 0.

Mixed type: the Tricomi equation yu_xx + u_yy = 0 is elliptic for y > 0, parabolic on y = 0 and hyperbolic for y < 0.

Standard types: the wave equation u_tt = c²u_xx is hyperbolic, the heat equation u_t = ku_xx is parabolic, and the Laplace equation u_xx + u_yy = 0 is elliptic.

Example 6 (find all regions): Classify (1 − x²)u_xx − 2xyu_xy + (1 − y²)u_yy = 0 at every point of the plane.

B² − 4AC = 4x²y² − 4(1 − x²)(1 − y²) = 4(x² + y² − 1).

So the equation is elliptic inside the unit circle x² + y² < 1, parabolic on x² + y² = 1 and hyperbolic outside it.

10. Choosing η and Integrating the Canonical Form

Parabolic: after ξ from the single characteristic, take η = x or η = y, whichever makes J = ξ_xη_y − ξ_yη_x ≠ 0; express the leftover coefficients in ξ, η.

Coefficient A = 0: the characteristic equation A dy² − B dx dy + C dx² = 0 then has the factor dx, so x = constant is one family of characteristics.

Chain rule in full: u_xx = ξ_x²u_ξξ + 2ξ_xη_xu_ξη + η_x²u_ηη + ξ_xxu_ξ + η_xxu_η (similarly u_xy, u_yy); the terms ξ_xxu_ξ, η_xxu_η produce the first-order part of the canonical form.

Solving the canonical form: u_ηη = g(η) integrates twice in η with 'constants' f(ξ), g(ξ); ηu_ηη + u_η = 0 is ∂(ηu_η)/∂η = 0; u_ξη + u_η/ξ = 0 is ∂(ξu_η)/∂ξ = 0; an equation of the form (∂_y + α)(∂_x + β)u = 0 is solved by putting w = u_x + βu.

11. Past UPSC Questions Solved

Example 7 (UPSC 2015, 15 marks): Reduce the second-order partial differential equation x² ∂²u/∂x² − 2xy ∂²u/∂x∂y + y² ∂²u/∂y² + x ∂u/∂x + y ∂u/∂y = 0 into canonical form. Hence, find its general solution.

A = x², B = −2xy, C = y²: B² − 4AC = 0, parabolic; the characteristic equation is (x dy/dx + y)² = 0, so ξ = xy; take η = x (J = −x ≠ 0).

u_x = yu_ξ + u_η, u_y = xu_ξ, u_xx = y²u_ξξ + 2yu_ξη + u_ηη, u_xy = u_ξ + xyu_ξξ + xu_ξη, u_yy = x²u_ξξ.

Second-order part = x²u_ηη − 2xyu_ξ; first-order part xu_x + yu_y = 2xyu_ξ + xu_η; the sum is x²u_ηη + xu_η = 0.

Canonical form ηu_ηη + u_η = 0, i.e. ∂(ηu_η)/∂η = 0, so ηu_η = f(ξ) and u = f(ξ)log η + g(ξ).

Answer: canonical form u_ηη = −(1/η)u_η with ξ = xy, η = x; general solution u = f(xy)log x + g(xy).

Example 8 (UPSC 2017, 15 marks): Reduce the equation
y² ∂²z/∂x² − 2xy ∂²z/∂x∂y + x² ∂²z/∂y² = (y²/x) ∂z/∂x + (x²/y) ∂z/∂y
to canonical form and hence solve it.

A = y², B = −2xy, C = x²: B² − 4AC = 0, parabolic; (y dy/dx + x)² = 0 gives ξ = x² + y²; take η = x.

z_x = 2xz_ξ + z_η, z_y = 2yz_ξ, z_xx = 2z_ξ + 4x²z_ξξ + 4xz_ξη + z_ηη, z_xy = 4xyz_ξξ + 2yz_ξη, z_yy = 2z_ξ + 4y²z_ξξ.

LHS = 2(x² + y²)z_ξ + y²z_ηη (the z_ξξ, z_ξη terms cancel); RHS = (y²/x)(2xz_ξ + z_η) + (x²/y)(2yz_ξ) = 2(x² + y²)z_ξ + (y²/x)z_η.

So z_ηη = z_η/η (with η = x² instead, this becomes z_ηη = 0); then ∂(z_η/η)/∂η = 0, z_η = ηf(ξ) and z = (η²/2)f(ξ) + g(ξ).

Answer: canonical form z_ηη = (1/η)z_η (ξ = x² + y², η = x); solution z = x²f(x² + y²) + g(x² + y²).

Example 9 (UPSC 2019, 20 marks): Reduce the following second order partial differential equation to canonical form and find the general solution :
∂²u/∂x² − 2x ∂²u/∂x∂y + x² ∂²u/∂y² = ∂u/∂y + 12x.

A = 1, B = −2x, C = x²: B² − 4AC = 0, parabolic; (dy/dx + x)² = 0 gives ξ = y + x²/2; take η = x.

u_x = xu_ξ + u_η, u_y = u_ξ, u_xx = u_ξ + x²u_ξξ + 2xu_ξη + u_ηη, u_xy = xu_ξξ + u_ξη, u_yy = u_ξξ.

LHS = u_ξ + u_ηη, so u_ξ + u_ηη = u_ξ + 12x: canonical form u_ηη = 12η.

Integrating twice in η: u = 2η³ + ηf(ξ) + g(ξ).

Answer: u = 2x³ + xf(y + x²/2) + g(y + x²/2).

Example 10 (UPSC 2022, 15 marks): Reduce the following partial differential equation to a canonical form and hence solve it: yu_xx + (x + y)u_xy + xu_yy = 0

A = y, B = x + y, C = x: B² − 4AC = (x − y)² > 0 for x ≠ y, hyperbolic; y(dy/dx)² − (x + y)dy/dx + x = (y dy/dx − x)(dy/dx − 1) = 0 gives ξ = y − x, η = y² − x².

u_x = −u_ξ − 2xu_η, u_y = u_ξ + 2yu_η; u_xx = u_ξξ + 4xu_ξη + 4x²u_ηη − 2u_η, u_xy = −u_ξξ − 2(x + y)u_ξη − 4xyu_ηη, u_yy = u_ξξ + 4yu_ξη + 4y²u_ηη + 2u_η.

Substituting, the u_ξξ and u_ηη coefficients vanish, u_ξη gets 8xy − 2(x + y)² = −2(x − y)² and u_η gets 2(x − y).

With x − y = −ξ: canonical form u_ξη + (1/ξ)u_η = 0, i.e. ∂(ξu_η)/∂ξ = 0, so ξu_η = F′(η) and u = F(η)/ξ + G(ξ).

Answer: u_ξη = −u_η/ξ (ξ = y − x, η = y² − x²); u = F(y² − x²)/(y − x) + G(y − x). UPSC 2026 (Q6a) set the same equation.

Example 11 (UPSC 2023, 15 marks): Reduce the partial differential equation ∂²z/∂y² − ∂²z/∂x∂y + ∂z/∂x − (∂z/∂y)(1 + 1/x) + z/x = 0 to canonical form.

Coefficients: A = 0 (z_xx), B = −1 (z_xy), C = 1 (z_yy), so B² − 4AC = 1 > 0, hyperbolic everywhere.

Characteristics: A dy² − B dx dy + C dx² = dx(dy + dx) = 0, so ξ = x, η = x + y.

z_x = z_ξ + z_η, z_y = z_η, z_xy = z_ξη + z_ηη, z_yy = z_ηη; substituting, z_ηη − z_ξη − z_ηη + z_ξ + z_η − z_η(1 + 1/ξ) + z/ξ = 0.

So z_ξη = z_ξ − (1/ξ)(z_η − z); it factorises as (∂/∂ξ + 1/ξ)(z_η − z) = 0, which gives z = e^{x + y}G(x) + H(x + y)/x.

Answer: canonical form ∂²z/∂ξ∂η = ∂z/∂ξ − (1/ξ)(∂z/∂η − z) with ξ = x, η = x + y; general solution z = e^{x + y}G(x) + H(x + y)/x.

Example 12 (UPSC 2024, 15 marks): Solve the partial differential equation
∂/∂y (∂φ/∂x + φ) + 2x²y (∂φ/∂x + φ) = 0
by transforming it to the canonical form.

Expanded: φ_xy + φ_y + 2x²yφ_x + 2x²yφ = 0; A = C = 0, B = 1, so B² − 4AC = 1 > 0 (hyperbolic) and the characteristics are x = const, y = const.

With ξ = x, η = y the equation is already canonical: φ_ξη = −φ_η − 2ξ²η(φ_ξ + φ).

Put w = φ_x + φ: w_y + 2x²yw = 0, so w = f(x)e^−x2y2.

Then ∂(e^xφ)/∂x = e^xf(x)e^−x2y2; integrate in x from a fixed x₀.

Answer: φ(x, y) = e^−x[g(y) + ∫_x0^x f(s)e^{s − s2y2} ds], f and g arbitrary.

12. Practice Exercises

Exercise 1 (basic): Classify u_xx + 4u_xy + 4u_yy = 0.

Answer: B² − 4AC = 16 − 16 = 0: parabolic everywhere.

Exercise 2 (basic): Classify u_xx + u_xy + u_yy = 0.

Answer: B² − 4AC = 1 − 4 = −3 < 0: elliptic everywhere.

Exercise 3 (basic): Classify u_xx − yu_yy = 0.

Answer: B² − 4AC = 4y: hyperbolic for y > 0, parabolic on y = 0, elliptic for y < 0.

Exercise 4 (standard): Reduce x²u_xx − y²u_yy = 0 (x, y > 0) to canonical form.

Answer: x²λ² = y² gives ξ = xy, η = y/x; the canonical form is u_ξη = u_η/(2ξ).

Exercise 5 (standard): Reduce u_xx + 2u_xy + 5u_yy = 0 to canonical form.

Answer: λ² − 2λ + 5 = 0 gives λ = 1 ± 2i; with α = y − x, β = 2x the equation becomes 4(u_αα + u_ββ) = 0, i.e. u_αα + u_ββ = 0.

Exercise 6 (standard): Reduce u_xx − 2u_xy + u_yy = 0 to canonical form and solve it.

Answer: parabolic, λ = −1 (double): ξ = x + y, η = x give u_ηη = 0, so u = f(x + y) + xg(x + y).

Exercise 7 (standard): Reduce u_xx + 5u_xy + 6u_yy = 0 to canonical form and solve it.

Answer: λ = 2, 3: ξ = y − 2x, η = y − 3x give u_ξη = 0, so u = f(y − 2x) + g(y − 3x).

Exercise 8 (UPSC level): Reduce y²u_xx + x²u_yy = 0 (x, y > 0) to canonical form.

Answer: elliptic; y dy = ±ix dx gives α = y², β = x², and the canonical form is u_αα + u_ββ = −u_α/(2α) − u_β/(2β).

Exercise 9 (UPSC level): Reduce u_xx − 2 sin x u_xy − cos²x u_yy − cos x u_y = 0 to canonical form and find its general solution.

Answer: B² − 4AC = 4 > 0; λ = −sin x ± 1 gives ξ = y + x − cos x, η = y − x − cos x; the first-order terms cancel and u_ξη = 0, so u = f(y + x − cos x) + g(y − x − cos x).

Exercise 10 (UPSC level): Reduce 4u_xx + 12u_xy + 9u_yy = 9 to canonical form and find its general solution.

Answer: parabolic, λ = 3/2 (double): ξ = 2y − 3x, η = x give 4u_ηη = 9, so u = 9x²/8 + xf(2y − 3x) + g(2y − 3x).

1. Understanding Separation of Variables and the Eigenvalue Problem

Why we need it: the heat, wave and Laplace equations on a bounded interval come with boundary conditions such as 'the ends are kept at 0'. General solutions with arbitrary functions are of little help there; we need a method that builds the boundary conditions in from the start.

In words: try a product u = X(x)T(t), one factor for each variable. For u_t = ku_xx this gives XT′ = kX″T, and dividing by kXT gives T′/(kT) = X″/X.

The left side depends only on t and the right side only on x; they can be equal for all x and t only if both equal one constant, written −λ (the separation constant).

The boundary conditions u(0, t) = u(L, t) = 0 become X(0) = X(L) = 0. The problem X″ + λX = 0 with these end values has non-zero solutions only for special λ, the eigenvalues, and these solutions are the eigenfunctions.

First example, slowly: solve u_t = 2u_xx, 0 < x < 1, u(0, t) = u(1, t) = 0, u(x, 0) = 5 sin 3πx.

Step 1: with u = X(x)T(t): T′/(2T) = X″/X = −λ, so X″ + λX = 0 and T′ = −2λT.

Step 2: the ends give X(0) = X(1) = 0 (otherwise T ≡ 0 and u ≡ 0).

Step 3: λ ≤ 0 gives only X = 0 (the case analysis is in 'Why it works' under Separation of Variables below). For λ = k² > 0, X = A cos kx + B sin kx; X(0) = 0 gives A = 0, and X(1) = 0 gives sin k = 0, so k = nπ.

Step 4: so the eigenvalues are λ_n = n²π² and the eigenfunctions are sin nπx, n = 1, 2, 3, ...

Step 5: the time factor is T = e^−2n2π2t, so each mode is e^−2n2π2tsin nπx.

Step 6: the initial data is already one mode, n = 3 with coefficient 5, so u = 5e^−18π2tsin 3πx.

Step 7: check: u_t = −18π²u and 2u_xx = 2(−9π²)u = −18π²u; both ends give 0, and t = 0 gives 5 sin 3πx.

Non-example: if one end is held at u(0, t) = 1, a product would need X(0)T(t) = 1 for all t, which forces T to be constant and destroys the time dependence. Separation needs homogeneous boundary conditions; with non-zero end values, first subtract the steady state (see 'Ends at fixed non-zero temperatures' below).

Building up: the same separation for u_tt = c²u_xx gives T″ + c²λT = 0, so the modes oscillate as cos(nπct/L) and sin(nπct/L); for Laplace's equation the second factor becomes sinh or cosh. General initial data are handled by adding modes, the subject of the next lesson.

Where it is used: every vibrating-string, heat-conduction and steady-plate question, and, through the eigenvalue idea, the Fourier series of Real Analysis and the eigenvectors of Linear Algebra.

2. Understanding Fitting the Initial Data by a Fourier Series

Why we need it: real initial data, such as a constant temperature or a plucked shape, is not a single sine. We must write it as a sum of the eigenfunctions so that each piece can be evolved separately.

In words: because the PDE and the boundary conditions are linear and homogeneous, any sum of modes u = ∑b_ne^{−kn2π2t/L2}sin(nπx/L) is again a solution (superposition).

At t = 0 this must equal f(x): f(x) = ∑b_nsin(nπx/L), a Fourier sine series. The coefficients are picked out by orthogonality: ∫₀^L sin(mπx/L)sin(nπx/L) dx is 0 for m ≠ n and L/2 for m = n.

So multiplying both sides by sin(mπx/L) and integrating kills every term except one, and b_m = (2/L)∫₀^L f(x)sin(mπx/L) dx.

First example, slowly: solve u_t = u_xx, 0 < x < π, u(0, t) = u(π, t) = 0, u(x, 0) = π − x.

Step 1: here L = π and k = 1, so the modes are e^−n2tsin nx and u = ∑b_ne^−n2tsin nx.

Step 2: at t = 0 we need π − x = ∑b_nsin nx on 0 < x < π, so b_n = (2/π)∫₀^π(π − x)sin nx dx.

Step 3: integrate by parts, differentiating π − x and integrating sin nx: the integral is [−(π − x)cos nx/n]₀^π − ∫₀^π cos nx/n dx.

Step 4: the bracket is 0 − (−π/n) = π/n, and the last integral is sin nx/n² from 0 to π, which is 0. So the integral is π/n.

Step 5: hence b_n = (2/π)(π/n) = 2/n for every n.

Step 6: u(x, t) = 2∑_n=1^∞ (1/n)e^−n2tsin nx.

Step 7: notice what the formula says: the higher modes die fastest, so for large t, u ≈ 2e^−tsin x, a single arch shrinking to 0. (At x = 0 the series gives 0, not π: the data jump there, and the series takes the mean of the end values of the odd extension.)

Non-example: expanding π − x in a cosine series instead would give terms cos nx, which are not zero at x = 0 and x = π, so the boundary conditions u(0, t) = u(π, t) = 0 would fail. The eigenfunctions must be those of the boundary conditions: sines for fixed ends, cosines (with the constant term) for insulated ends.

Building up: for the wave equation there are two data, u(x, 0) = f and u_t(x, 0) = g; f fixes the cosine-in-time coefficients A_n, and g fixes B_n with the extra factor L/(nπc) (see 'Why B_n has the factor 1/(nπc)' below).

The series converges to f wherever f is smooth, and to the average of the one-sided limits at a jump; the 'Useful integrals' below save time in the exam.

Where it is used: the 15–20 mark rod and string questions with data such as x, x(L − x), a constant or a triangle (Examples 5, 8, 9, 12 and 13 below).

3. Understanding D'Alembert's Solution of the Wave Equation

Why we need it: on a very long string (the whole line) there are no ends and no eigenvalues. D'Alembert's formula gives the solution in closed form and shows what the wave equation really does: it moves shapes.

In words: the characteristics of u_tt = c²u_xx are x + ct = constant and x − ct = constant, and the general solution is F(x + ct) + G(x − ct): one shape moving left and one moving right, both with speed c and without change of form.

Fitting the initial displacement f and velocity g gives u = [f(x + ct) + f(x − ct)]/2 + (1/(2c))∫_{x − ct}^{x + ct} g(s) ds. The value at (x, t) depends only on the data in [x − ct, x + ct], the domain of dependence.

First example, slowly: solve u_tt = u_xx on −∞ < x < ∞ with u(x, 0) = 1/(1 + x²) and u_t(x, 0) = 0.

Step 1: identify c = 1, f(x) = 1/(1 + x²) (a single bump of height 1 at x = 0) and g = 0 (released from rest).

Step 2: since g = 0 the integral term vanishes, so u = [f(x + t) + f(x − t)]/2.

Step 3: substitute: u = (1/2)[1/(1 + (x + t)²) + 1/(1 + (x − t)²)].

Step 4: check t = 0: u = (1/2)(2f(x)) = f(x). Check u_t at t = 0: it is [f′(x) − f′(x)]/2 = 0.

Step 5: interpret: the bump splits into two bumps of height 1/2; at t = 3 they are centred at x = 3 and x = −3.

Step 6: a sample value: u(0, 1) = (1/2)(1/2 + 1/2) = 1/2.

Step 7: with a non-zero velocity, for example g(x) = cos x and c = 1, the extra term is (1/2)∫_{x − t}^{x + t} cos s ds = (1/2)[sin(x + t) − sin(x − t)] = cos x sin t.

Non-example: on a finite string 0 < x < L with fixed ends the formula cannot be used as it stands, because x ± ct soon leaves (0, L) where f and g are given. The condition that fails is that the domain is the whole line; the remedy is to extend f and g as odd, 2L-periodic functions, which reproduces the Fourier series solution.

Building up: in characteristic variables ξ = x + ct, η = x − ct the equation becomes u_ξη = 0, so u = F(ξ) + G(η); the two data then fix F and G up to a constant that cancels ('Derivation: The Three Equations and D'Alembert's Formula' below).

Where it is used: infinite-string questions, checking separated solutions (sin x sin ct = [cos(x − ct) − cos(x + ct)]/2 is a pair of travelling waves), and the finite speed of propagation that distinguishes waves from heat.

4. Understanding Laplace's Equation on a Rectangle and a Disc

Why we need it: after a long time a heated plate reaches a steady state, u_t = 0, and its temperature satisfies u_xx + u_yy = 0 with given values on the whole boundary. There is no time variable, so the role of time is taken by the second space variable.

In words: separate u = X(x)Y(y) to get X″/X = −Y″/Y = constant. Choose the sign so that the factor oscillates (sin) in the direction in which both boundary values are zero; the other factor is then sinh or cosh, chosen to vanish on the remaining zero side.

For a disc use polar coordinates: the angular factor must be 2π-periodic (cos nθ, sin nθ), and the radial factor rⁿ is kept while r⁻ⁿ and log r are dropped because u must stay finite at the centre.

First example, slowly: solve u_xx + u_yy = 0 in the square 0 < x < π, 0 < y < π with u = 0 on x = 0, x = π and y = π, and u(x, 0) = sin 2x.

Step 1: the two zero sides x = 0 and x = π face each other, so X must oscillate: X″ + n²X = 0 with X(0) = X(π) = 0 gives X = sin nx.

Step 2: then Y″ = n²Y, and Y must vanish on the third zero side y = π; the combination that does this is Y = sinh(n(π − y)).

Step 3: so u = ∑c_nsin nx sinh(n(π − y)); at y = 0 this is ∑c_nsinh(nπ)sin nx, which must equal sin 2x.

Step 4: only n = 2 survives, with c₂sinh 2π = 1.

Step 5: u = sin 2x sinh(2(π − y))/sinh 2π.

Step 6: check: u_xx = −4u and u_yy = 4u add to 0; x = 0 and x = π give 0, y = π gives sinh 0 = 0, and y = 0 gives sin 2x.

Step 7: notice that the temperature decays away from the heated side y = 0, roughly like e^−2y.

Non-example: with the other sign, X″ = k²X gives X = A cosh kx + B sinh kx, and X(0) = 0, X(π) = 0 force A = 0 and B sinh kπ = 0, so X ≡ 0. The condition that fails is that the oscillating factor must be the one with two zero boundary values.

Building up: if several sides carry non-zero data, solve one problem per side and add (linearity). In a disc the same separation gives (r/a)ⁿ(a_ncos nθ + b_nsin nθ) with a_n, b_n the Fourier coefficients of the boundary data.

The centre value is then a₀/2, the mean of the boundary values, and the maximum principle shows that the solution is unique ('Maximum principle' below).

Where it is used: steady temperatures in plates and discs, the annulus question of UPSC 2018 (Example 11 below), and potential problems in Fluid Dynamics and Complex Analysis, where harmonic functions are real parts of analytic functions.

5. Separation of Variables

Assume u = X(x)T(t), substitute, divide by XT and set both sides equal to a separation constant −λ.

Idea: after dividing, one side depends only on x and the other only on t; a function of x can equal a function of t for all x and t only if both are the same constant. The method needs linear homogeneous boundary conditions on the x-boundaries, so that each product term satisfies them separately.

With X(0) = X(L) = 0, the problem X″ + λX = 0 has eigenvalues λ_n = n²π²/L² and eigenfunctions sin(nπx/L), n = 1, 2, 3, ...; λ ≤ 0 gives only X = 0.

Why it works: for λ = 0, X = A + Bx and X(0) = X(L) = 0 give A = B = 0; for λ = −k² < 0, X = A cosh kx + B sinh kx, X(0) = 0 gives A = 0 and B sinh kL = 0 gives B = 0. For λ = k² > 0, X = A cos kx + B sin kx, so A = 0 and sin kL = 0, i.e. k = nπ/L.

Superpose the product solutions and find the coefficients from the initial data using Fourier sine (or cosine) series.

Idea: superposition is allowed because the PDE and the boundary conditions are linear and homogeneous. The coefficients are picked out by orthogonality: ∫₀^L sin(mπx/L) sin(nπx/L) dx = 0 for m ≠ n and L/2 for m = n, so multiplying the series by sin(mπx/L) and integrating isolates one coefficient.

6. Vibrating String (Wave Equation)

u_tt = c²u_xx, 0 < x < L, with c² = T/ρ (tension over mass per unit length); u(0, t) = u(L, t) = 0, u(x, 0) = f(x), u_t(x, 0) = g(x).

Solution: u = ∑_n=1^∞ [A_ncos(nπct/L) + B_nsin(nπct/L)] sin(nπx/L).

A_n = (2/L)∫₀^L f(x) sin(nπx/L) dx, B_n = (2/(nπc))∫₀^L g(x) sin(nπx/L) dx.

The n-th normal mode has frequency nc/(2L); the fundamental frequency is c/(2L).

Idea: the n-th mode sin(nπx/L) has n − 1 interior nodes and vibrates at n times the fundamental frequency (see the figure above); this is why a string sounds a note with harmonics. Plucking at a node of mode n leaves that mode out (Section 9).

D'Alembert's solution (infinite string): u = [f(x + ct) + f(x − ct)]/2 + (1/(2c))∫_{x − ct}^{x + ct} g(s) ds.

Example 1: u_tt = u_xx, 0 < x < π, fixed ends, u(x, 0) = 0, u_t(x, 0) = sin 2x. Only B₂ ≠ 0 and 2B₂ = 1, so u = (1/2)sin 2t sin 2x.

Example 2: string released from rest with f(x) = kx(L − x): A_n = 8kL²/(n³π³) for odd n and 0 for even n, B_n = 0.

Example 3: Solve u_tt = c²u_xx, −∞ < x < ∞, with u(x, 0) = x², u_t(x, 0) = sin x.

By D'Alembert: u = [(x + ct)² + (x − ct)²]/2 + (1/(2c))∫_{x − ct}^{x + ct} sin s ds = x² + c²t² + (1/(2c))[cos(x − ct) − cos(x + ct)].

So u = x² + c²t² + (1/c)sin x sin ct (check: u_tt = 2c² − c sin x sin ct = c²u_xx).

7. Heat Equation

u_t = ku_xx (k the diffusivity), 0 < x < L, u(0, t) = u(L, t) = 0, u(x, 0) = f(x).

Idea: mode n decays like e^{−kn2π2t/L2}, so the higher modes, which carry the sharp corners of f, die out first; the temperature profile smooths quickly and soon looks like the single arch sin(πx/L) shrinking to zero (see the figure above). Unlike the wave equation, time cannot be run backwards.

Solution: u = ∑ b_n e^{−kn2π2t/L2} sin(nπx/L), with b_n = (2/L)∫₀^L f(x) sin(nπx/L) dx.

Insulated ends (u_x = 0 at x = 0, L): u = a₀/2 + ∑ a_n e^{−kn2π2t/L2} cos(nπx/L), a_n = (2/L)∫₀^L f(x) cos(nπx/L) dx; u tends to the mean initial temperature a₀/2.

Ends at fixed non-zero temperatures u(0, t) = T₁, u(L, t) = T₂: write u = T₁ + (T₂ − T₁)x/L + v, where v solves the problem with zero end values.

Why it works: w(x) = T₁ + (T₂ − T₁)x/L satisfies w″ = 0 and the end conditions, so it is the steady state; v = u − w then has zero end values and initial data f(x) − w(x), which must be expanded in the sine series.

Example 4: u_t = u_xx, 0 < x < π, u(0, t) = u(π, t) = 0, u(x, 0) = 3 sin x + 2 sin 3x. Then u = 3e^−tsin x + 2e^−9tsin 3x.

Example 5: A rod 0 ≤ x ≤ L has initial temperature 100 throughout, and both ends are kept at 0 for t > 0; find u(x, t) for u_t = ku_xx.

b_n = (2/L)∫₀^L 100 sin(nπx/L) dx = (200/(nπ))(1 − cos nπ), which is 400/(nπ) for odd n and 0 for even n.

u = (400/π)∑_{n odd} (1/n)e^{−kn2π2t/L2} sin(nπx/L), which tends to 0 as t → ∞.

8. Laplace Equation

u_xx + u_yy = 0 describes steady temperature and potential; its solutions are harmonic functions.

Idea: a harmonic function equals the average of its values on any circle centred at a point (mean value property), so it can have no interior peak or pit; it is the shape of a soap film or a stretched membrane fixed along a wire frame.

Rectangle 0 < x < a, 0 < y < b, with u = 0 on x = 0, x = a, y = 0 and u(x, b) = f(x): u = ∑ c_n sin(nπx/a) sinh(nπy/a), where c_n sinh(nπb/a) = (2/a)∫₀^a f(x) sin(nπx/a) dx.

Idea: choose the sign of the separation constant so that the factor oscillates (sin) in the direction with two zero boundary conditions; the other factor is then sinh or cosh, and sinh(nπy/a) is the one vanishing at y = 0. If the non-zero data are on y = 0 instead, use sinh(nπ(b − y)/a).

Example 6: f(x) = sin(πx/a) gives u = sin(πx/a) sinh(πy/a)/sinh(πb/a).

Disc r < R in polar form u_rr + u_r/r + u_θθ/r² = 0 with u(R, θ) = f(θ): u = a₀/2 + ∑ (r/R)ⁿ(a_ncos nθ + b_nsin nθ), where a_n, b_n are the Fourier coefficients of f.

Terms r⁻ⁿ and log r are discarded inside the disc because u must stay bounded at r = 0.

Example 7: Solve u_rr + u_r/r + u_θθ/r² = 0 in r < 1 with u(1, θ) = cos²θ.

cos²θ = 1/2 + (1/2)cos 2θ, so a₀/2 = 1/2, a₂ = 1/2 and all other coefficients vanish.

u = 1/2 + (r²/2)cos 2θ = (1 + x² − y²)/2; the value 1/2 at the centre is the mean of the boundary values.

Maximum principle: a function harmonic in a bounded domain and continuous on its closure attains its maximum and minimum on the boundary, and if it is non-constant, only there; hence the Dirichlet problem has at most one solution.

Why it works: if u₁, u₂ solve the same Dirichlet problem, w = u₁ − u₂ is harmonic and zero on the boundary; its maximum and minimum are both 0, so w ≡ 0.

9. Plucked Strings, Decay Terms and Useful Integrals

String plucked at x = a to height h and released from rest: A_n = 2hL²sin(nπa/L)/(n²π²a(L − a)), B_n = 0; modes with sin(nπa/L) = 0 are absent.

Decay term: u_t = ku_xx − γu becomes the heat equation for v under u = e^−γtv, with the same boundary and initial data.

Useful integrals on 0 to L: ∫x sin(nπx/L) dx = (−1)ⁿ⁺¹L²/(nπ); ∫sin(nπx/L) dx = L(1 − (−1)ⁿ)/(nπ); ∫x(L − x)sin(nπx/L) dx = 2L³(1 − (−1)ⁿ)/(n³π³); and ∫₀¹ x²sin nπx dx = (−1)ⁿ⁺¹/(nπ) + 2((−1)ⁿ − 1)/(nπ)³.

Data not vanishing at the ends (such as u(x, 0) = x): the formulas are unchanged; the sine series converges to the data in 0 < x < L but not at x = L, and a remark on this earns credit.

10. Laplace Equation in Polar Coordinates: Disc and Annulus

Separation: V = R(r)Θ(θ) gives Θ″ + n²Θ = 0 and r²R″ + rR′ − n²R = 0 (an Euler equation); 2π-periodicity in θ makes n = 0, 1, 2, ...

R = A_nrⁿ + B_nr⁻ⁿ for n ≥ 1 and A₀ + B₀log r for n = 0; Θ = cos nθ, sin nθ (or e^±inθ).

Disc r ≤ a: drop r⁻ⁿ and log r (boundedness at r = 0); exterior r ≥ a: drop rⁿ and log r (boundedness at ∞).

Annulus a ≤ r ≤ b: keep both powers and log r, expand both boundary functions in Fourier series, and solve a 2 × 2 system for each n; with u(a, θ) = 0 the radial factor is (r/a)ⁿ − (a/r)ⁿ.

11. Infinite Domains and Uniqueness

Fourier transform: F(k) = ∫_−∞^∞ f(x)e^−ikx dx, f(x) = (1/(2π))∫_−∞^∞ F(k)e^ikx dk; f″ transforms to −k²F when f, f′ → 0 at ±∞, and the transform of a convolution ∫f(ξ)g(x − ξ) dξ is F(k)G(k).

Green's first identity: ∬_S (w∇²w + |∇w|²) dx dy = ∮_Γ w ∂w/∂n ds; it proves uniqueness for the Dirichlet problem (w = 0 on Γ) and, up to a constant, for the Neumann problem (∂w/∂n = 0 on Γ).

12. Derivation: The Three Equations and D'Alembert's Formula

Wave equation: take the piece of string between x and x + δx under tension T, with small slope u_x. The net vertical force is T[u_x(x + δx, t) − u_x(x, t)] ≈ Tu_xxδx, and Newton's law gives ρδx u_tt = Tu_xxδx, so u_tt = c²u_xx with c² = T/ρ.

Heat equation: Fourier's law makes the heat flux −κu_x; the heat gained by a slab of width δx per unit time is κ[u_x(x + δx) − u_x(x)] ≈ κu_xxδx, and this equals cρδx u_t (c the specific heat, ρ the density), so u_t = ku_xx with k = κ/(cρ).

In the steady state u_t = 0, and in two dimensions the same balance gives u_xx + u_yy = 0: Laplace's equation is the time-independent heat equation.

D'Alembert's formula: with ξ = x + ct, η = x − ct the wave equation becomes u_ξη = 0 (previous topic), so u = F(x + ct) + G(x − ct).

The data give F(x) + G(x) = f(x) and c[F′(x) − G′(x)] = g(x), so F(x) − G(x) = (1/c)∫₀^x g(s) ds + K.

Hence F = f/2 + (1/(2c))∫₀^x g + K/2 and G = f/2 − (1/(2c))∫₀^x g − K/2; adding F(x + ct) and G(x − ct), K cancels and the two integrals combine to (1/(2c))∫_{x − ct}^{x + ct} g(s) ds, which is the formula of Section 6.

Why B_n has the factor 1/(nπc): u_t(x, 0) = ∑B_n(nπc/L) sin(nπx/L) = g(x), so B_n(nπc/L) = (2/L)∫g sin(nπx/L) dx, which gives B_n = (2/(nπc))∫₀^L g(x) sin(nπx/L) dx.

13. Choosing the Method

If the domain is the whole line → D'Alembert for the wave equation, the Fourier transform for the heat equation; if it is a finite interval → separation of variables.

If the ends are held at zero → sine series; if the ends are insulated → cosine series with the constant term; if the ends are at non-zero constants → subtract the steady state first.

If a rectangle has non-zero data on more than one side → split into one problem per side and add the solutions.

If the region is a disc, an annulus or the outside of a circle → polar coordinates with rⁿ, r⁻ⁿ and log r, keeping only the bounded terms.

If the equation has an extra term −γu → put u = e^−γtv first; if the data are already a finite sum of sines → read off the coefficients without integrating.

14. Common Mistakes

Non-zero end temperatures: expanding f itself in a sine series; it is f − w (initial data minus steady state) that has to be expanded.

Insulated ends: forgetting a₀/2, which is the final temperature of the rod.

Wrong sinh factor: with data on y = 0 and u = 0 on y = b, the factor must vanish at y = b, so it is sinh(nπ(b − y)/a), not sinh(nπy/a).

Parity slips: 1 − cos nπ = 1 − (−1)ⁿ is 2 for odd n and 0 for even n; writing it as 2 for all n doubles the series.

Wrong limits in D'Alembert: the integral of g runs from x − ct to x + ct, not from 0 to x + ct.

Keeping unbounded terms: r⁻ⁿ and log r inside a disc, or rⁿ outside it.

Link: the coefficients are Fourier series coefficients (Real Analysis), and the problem X″ + λX = 0, X(0) = X(L) = 0 is an eigenvalue problem like Ax = λx in Linear Algebra, with orthogonal eigenfunctions in place of orthogonal eigenvectors.

Link: harmonic functions are the real parts of analytic functions (Complex Analysis): the answer of Example 7, (1 + x² − y²)/2, is Re((1 + z²)/2); the velocity potential of an irrotational incompressible flow also satisfies ∇²φ = 0 (Fluid Dynamics).

15. Past UPSC Questions Solved

Example 8 (UPSC 2014, 15 marks): Solve ∂²u/∂t² = ∂²u/∂x², 0 < x < 1, t > 0, given that
(i) u(x, 0) = 0, 0 ≤ x ≤ 1
(ii) ∂u/∂t (x, 0) = x², 0 ≤ x ≤ 1
(iii) u(0, t) = u(1, t) = 0, for all t

Here c = 1, L = 1, u(x, 0) = 0, so A_n = 0 and u = ∑B_nsin nπt sin nπx with nπB_n = 2∫₀¹ x²sin nπx dx.

∫₀¹ x²sin nπx dx = (−1)ⁿ⁺¹/(nπ) + 2((−1)ⁿ − 1)/(nπ)³ (integrate by parts twice).

So B_n = (2/(n²π²))[(−1)ⁿ⁺¹ + 2((−1)ⁿ − 1)/(n²π²)]: 2/(n²π²) − 8/(n⁴π⁴) for odd n and −2/(n²π²) for even n.

Answer: u(x, t) = ∑_n=1^∞ (2/(n²π²))[(−1)ⁿ⁺¹ + 2((−1)ⁿ − 1)/(n²π²)] sin nπx sin nπt.

Example 9 (UPSC 2015, 15 marks): Find the solution of the initial-boundary value problem
u_t − u_xx + u = 0, 0 < x < l, t > 0
u(0, t) = u(l, t) = 0, t ≥ 0
u(x, 0) = x(l − x), 0 < x < l

Put u = e^−tv: u_t − u_xx + u = e^−t(v_t − v_xx), so v_t = v_xx with v(0, t) = v(l, t) = 0, v(x, 0) = x(l − x).

v = ∑b_ne^−n2π2t/l2sin(nπx/l) with b_n = (2/l)∫₀^l x(l − x)sin(nπx/l) dx = 4l²(1 − (−1)ⁿ)/(n³π³).

So b_n = 8l²/(n³π³) for odd n and 0 for even n.

Answer: u(x, t) = (8l²/π³)∑_{n odd} (1/n³)e^{−(1 + n2π2/l2)t}sin(nπx/l).

Example 10 (UPSC 2017, 10 marks): Let Γ be a closed curve in xy-plane and let S denote the region bounded by the curve Γ. Let
∂²w/∂x² + ∂²w/∂y² = f(x, y) ∀ (x, y) ∈ S.
If f is prescribed at each point (x, y) of S and w is prescribed on the boundary Γ of S, then prove that any solution w = w(x, y), satisfying these conditions, is unique.

Let w₁, w₂ be two solutions and w = w₁ − w₂; then ∇²w = f − f = 0 in S and w = 0 on Γ.

Green's first identity (the divergence theorem for w∇w): ∬_S (w∇²w + w_x² + w_y²) dx dy = ∮_Γ w ∂w/∂n ds.

Both ∇²w = 0 in S and w = 0 on Γ, so ∬_S (w_x² + w_y²) dx dy = 0.

The integrand is continuous and non-negative, so w_x = w_y = 0 in S; w is constant, and continuity up to Γ, where w = 0, gives w ≡ 0 (alternatively, apply the maximum principle to the harmonic w).

Answer: w₁ = w₂ throughout S, so the solution is unique. Hence proved.

Example 11 (UPSC 2018, 20 marks): A thin annulus occupies the region 0 < a ≤ r ≤ b, 0 ≤ θ ≤ 2π. The faces are insulated. Along the inner edge the temperature is maintained at 0°, while along the outer edge the temperature is held at T = K cos(θ/2), where K is a constant. Determine the temperature distribution in the annulus.

Steady state: u_rr + u_r/r + u_θθ/r² = 0 in a ≤ r ≤ b, with u(a, θ) = 0 and u(b, θ) = K cos(θ/2), 0 ≤ θ ≤ 2π.

Fourier coefficients of cos(θ/2) on (0, 2π): (1/π)∫₀^2π cos(θ/2)cos nθ dθ = 0 for all n ≥ 0, and (1/π)∫₀^2π cos(θ/2)sin nθ dθ = 8n/(π(4n² − 1)).

So K cos(θ/2) = (8K/π)∑_n=1^∞ n sin nθ/(4n² − 1) for 0 < θ < 2π, and only the sine terms are needed (no constant or log r term).

Separated solutions (A_nrⁿ + B_nr⁻ⁿ)sin nθ with A_naⁿ + B_na⁻ⁿ = 0 give the radial factor (r/a)ⁿ − (a/r)ⁿ; dividing by its value at r = b matches the outer edge.

Answer: u(r, θ) = (8K/π)∑_n=1^∞ [n/(4n² − 1)] · [(r/a)ⁿ − (a/r)ⁿ]/[(b/a)ⁿ − (a/b)ⁿ] · sin nθ (across θ = 0 the boundary value jumps between −K and K, and the series gives the mean 0 there).

Example 12 (UPSC 2020, 20 marks): One end of a tightly stretched flexible thin string of length l is fixed at the origin and the other at x = l. It is plucked at x = l/3 so that it assumes initially the shape of a triangle of height h in the x-y plane. Find the displacement y at any distance x and at any time t after the string is released from rest. Take, (horizontal tension)/(mass per unit length) = c².

Initial shape y(x, 0) = 3hx/l for 0 ≤ x ≤ l/3 and 3h(l − x)/(2l) for l/3 ≤ x ≤ l; the string starts from rest, so B_n = 0 and y = ∑A_nsin(nπx/l)cos(nπct/l).

Integrating by parts on each piece, the terms at x = l/3 cancel and only the jump in slope (3h/l + 3h/(2l) = 9h/(2l)) survives: A_n = (2/l)(l/(nπ))²(9h/(2l))sin(nπ/3).

So A_n = (9h/(n²π²))sin(nπ/3), the plucked-string formula with a = l/3.

Answer: y(x, t) = (9h/π²)∑_n=1^∞ (1/n²)sin(nπ/3)sin(nπx/l)cos(nπct/l); the modes n = 3, 6, 9, ... are absent.

Example 13 (UPSC 2023, 20 marks): Solve the partial differential equation a² ∂²u/∂x² = ∂²u/∂t², 0 < x < L, t > 0 subject to the conditions
u(0, t) = 0, u(L, t) = 0, t > 0
u(x, 0) = x, (∂u/∂t)_t=0 = 1, 0 < x < L

u = ∑[A_ncos(nπat/L) + B_nsin(nπat/L)]sin(nπx/L), with A_n = (2/L)∫₀^L x sin(nπx/L) dx = 2L(−1)ⁿ⁺¹/(nπ).

(nπa/L)B_n = (2/L)∫₀^L sin(nπx/L) dx = 2(1 − (−1)ⁿ)/(nπ), so B_n = 4L/(n²π²a) for odd n and 0 for even n.

[The data u(x, 0) = x and u_t(x, 0) = 1 do not vanish at x = L, so the series represents them only in 0 < x < L.]

Answer: u(x, t) = (2L/π)∑_n=1^∞ ((−1)ⁿ⁺¹/n)cos(nπat/L)sin(nπx/L) + (4L/(π²a))∑_{n odd} (1/n²)sin(nπat/L)sin(nπx/L).

Example 14 (UPSC 2024, 20 marks): Show that the solution of the two-dimensional Laplace's equation
∂²φ(x, y)/∂x² + ∂²φ(x, y)/∂y² = 0, x ∈ (−∞, ∞), y ≥ 0
subject to the boundary condition
φ(x, 0) = f(x), x ∈ (−∞, ∞),
along with φ(x, y) → 0 for |x| → ∞ and y → ∞ can be written in the form
φ(x, y) = (y/π) ∫_−∞^∞ f(ξ) dξ/[y² + (x − ξ)²].

Take the Fourier transform in x: Φ(k, y) = ∫_−∞^∞ φ(x, y)e^−ikx dx; since φ, φ_x → 0 as |x| → ∞, φ_xx transforms to −k²Φ and the PDE becomes Φ_yy − k²Φ = 0.

Φ = A(k)e^−|k|y + B(k)e^|k|y; φ → 0 as y → ∞ forces B = 0, and y = 0 gives A = F(k), the transform of f.

The inverse of e^−|k|y is g(x, y) = (1/(2π))∫e^−|k|ye^ikx dk = (1/(2π))[1/(y − ix) + 1/(y + ix)] = y/(π(x² + y²)).

Since Φ = F(k) · e^−|k|y, the convolution theorem gives φ(x, y) = ∫f(ξ)g(x − ξ, y) dξ.

Answer: φ(x, y) = (y/π)∫_−∞^∞ f(ξ) dξ/[y² + (x − ξ)²], Poisson's integral for the half-plane. Hence proved.

Example 15 (UPSC 2026, 15 marks): Two-dimensional harmonic equation in plane polar coordinates (r, θ) takes the form:
∂²V/∂r² + (1/r) ∂V/∂r + (1/r²) ∂²V/∂θ² = 0
Deduce that it has solutions of the form (Arⁿ + Br⁻ⁿ) e^±inθ, where A, B, n are constants. Determine V if it satisfies the two-dimensional harmonic equation in the region 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π and satisfies the conditions:
(i) V remains finite as r → 0,
(ii) V = ∑_n c_n cos(nθ) on r = a.

Put V = R(r)Θ(θ) and multiply by r²/(RΘ): (r²R″ + rR′)/R = −Θ″/Θ = n² (a separation constant).

Θ″ + n²Θ = 0 gives Θ = e^±inθ; r²R″ + rR′ − n²R = 0 is an Euler equation, and R = r^m gives m² = n², so R = Arⁿ + Br⁻ⁿ.

Hence V = (Arⁿ + Br⁻ⁿ)e^±inθ; single-valuedness makes n an integer (n = 0 gives A + B log r).

In 0 ≤ r ≤ a, finiteness at r = 0 forces B = 0 and removes log r, so V = ∑ rⁿ(A_ncos nθ + B_n′sin nθ); at r = a, A_naⁿ = c_n and B_n′ = 0.

Answer: V(r, θ) = ∑_n c_n(r/a)ⁿcos nθ.

16. Practice Exercises

Exercise 1 (basic): Find the eigenvalues and eigenfunctions of X″ + λX = 0, X(0) = 0, X′(L) = 0.

Answer: λ_n = ((2n − 1)π/(2L))² and X_n = sin((2n − 1)πx/(2L)), n = 1, 2, 3, ... (λ ≤ 0 gives only X = 0).

Exercise 2 (basic): Solve u_t = u_xx, 0 < x < π, u(0, t) = u(π, t) = 0, u(x, 0) = 4 sin 2x − sin 5x.

Answer: u = 4e^−4tsin 2x − e^−25tsin 5x.

Exercise 3 (basic): Use D'Alembert's formula to solve u_tt = 4u_xx on the whole line with u(x, 0) = 0, u_t(x, 0) = 2x.

Answer: c = 2: u = (1/4)∫_{x − 2t}^{x + 2t} 2s ds = (1/4)[(x + 2t)² − (x − 2t)²] = 2xt.

Exercise 4 (standard): A rod 0 ≤ x ≤ 2 with u_t = 3u_xx has its ends kept at 0 and initial temperature 5. Find u(x, t).

Answer: b_n = ∫₀² 5 sin(nπx/2) dx = 10(1 − (−1)ⁿ)/(nπ), so u = (20/π)∑_{n odd} (1/n)e^−3n2π2t/4sin(nπx/2).

Exercise 5 (standard): Solve u_t = u_xx, 0 < x < π, with insulated ends u_x(0, t) = u_x(π, t) = 0 and u(x, 0) = cos²x.

Answer: cos²x = 1/2 + (1/2)cos 2x, so u = 1/2 + (1/2)e^−4tcos 2x, which tends to the mean 1/2.

Exercise 6 (standard): A string 0 < x < π with c = 2 and fixed ends starts from u(x, 0) = 0 with velocity u_t(x, 0) = sin x + 3 sin 2x. Find u.

Answer: u = ∑B_nsin 2nt sin nx with 2nB_n equal to the coefficients of the data: u = (1/2)sin 2t sin x + (3/4)sin 4t sin 2x.

Exercise 7 (standard): Solve u_t = u_xx, 0 < x < 1, u(0, t) = 0, u(1, t) = 10, u(x, 0) = 10x + 3 sin 2πx.

Answer: the steady state is 10x and v = u − 10x starts from 3 sin 2πx, so u = 10x + 3e^−4π2tsin 2πx.

Exercise 8 (UPSC level): Solve u_xx + u_yy = 0 in the rectangle 0 < x < a, 0 < y < b with u = 0 on x = 0, y = 0 and y = b, and u(a, y) = y(b − y).

Answer: u = ∑c_nsinh(nπx/b)sin(nπy/b) with c_nsinh(nπa/b) = 4b²(1 − (−1)ⁿ)/(n³π³); so u = (8b²/π³)∑_{n odd} sinh(nπx/b)sin(nπy/b)/(n³sinh(nπa/b)).

Exercise 9 (UPSC level): Find the steady temperature in the annulus 1 ≤ r ≤ 2 when u(1, θ) = 0 and u(2, θ) = sin θ.

Answer: try u = (Ar + B/r)sin θ: A + B = 0 and 2A + B/2 = 1 give A = 2/3, B = −2/3, so u = (2/3)(r − 1/r)sin θ.

Exercise 10 (UPSC level): A string of length L with fixed ends is released from rest in the shape u(x, 0) = μ sin³(πx/L). Find the displacement.

Answer: sin³θ = (3 sin θ − sin 3θ)/4, so u = (μ/4)[3 sin(πx/L)cos(πct/L) − sin(3πx/L)cos(3πct/L)].

1. Formation of PDEs and Lagrange's Method

Problem 1 (10 marks): Form the partial differential equation by eliminating the arbitrary function φ from φ(x² + y² + z², z² − 2xy) = 0.

Let u = x² + y² + z², v = z² − 2xy; the eliminant is Pp + Qq = R with P = ∂(u, v)/∂(y, z), Q = ∂(u, v)/∂(z, x), R = ∂(u, v)/∂(x, y).

P = u_yv_z − u_zv_y = (2y)(2z) − (2z)(−2x) = 4z(x + y).

Q = u_zv_x − u_xv_z = (2z)(−2y) − (2x)(2z) = −4z(x + y).

R = u_xv_y − u_yv_x = (2x)(−2x) − (2y)(−2y) = 4(y² − x²).

Dividing 4z(x + y)p − 4z(x + y)q = 4(y − x)(y + x) by 4(x + y) gives the equation.

Answer: z(p − q) = y − x.

Problem 2 (15 marks): Find the integral surface of (2xy − 1)p + (z − 2x²)q = 2(x − yz) which passes through the line x = 1, z = 0.

Multipliers x, y, 1/2: x(2xy − 1) + y(z − 2x²) + (x − yz) = 0, so x dx + y dy + dz/2 = 0 and x² + y² + z = c₁.

Multipliers z, 1, x: z(2xy − 1) + (z − 2x²) + 2x(x − yz) = 0, so z dx + dy + x dz = 0 and xz + y = c₂.

On the line x = 1, z = 0, y = t: c₁ = 1 + t² and c₂ = t, so c₁ = 1 + c₂².

Replacing c₁, c₂: x² + y² + z = 1 + (xz + y)², which simplifies to x² + z − 1 = x²z² + 2xyz.

Check: at x = 1, z = 0 both sides are 0 for every y, so the whole line lies on the surface.

Answer: x² + y² + z = 1 + (xz + y)², that is x²(1 − z²) − 2xyz + z − 1 = 0.

2. Nonlinear First-Order Equations and the Cauchy Problem

Problem 3 (15 marks): Use Charpit's method to find a complete integral of px + qy = pq.

F = px + qy − pq: F_x = p, F_y = q, F_z = 0, F_p = x − q, F_q = y − p.

Charpit's equations include −dp/(F_x + pF_z) = −dq/(F_y + qF_z), that is dp/p = dq/q, so p = aq.

Substituting: aqx + qy = aq², so q = (ax + y)/a and p = ax + y.

dz = (ax + y)dx + ((ax + y)/a)dy = (1/a)(ax + y)d(ax + y), so z = (ax + y)²/(2a) + b.

Check: px + qy = (ax + y)(x + y/a) = (ax + y)²/a = pq.

Answer: z = (ax + y)²/(2a) + b.

Problem 4 (10 marks): Find the complete and singular integrals of z = px + qy + √(1 + p² + q²).

This is Clairaut's form, so the complete integral is z = ax + by + √(1 + a² + b²).

Let S = √(1 + a² + b²); differentiating with respect to a and b: x + a/S = 0 and y + b/S = 0.

So a = −xS, b = −yS, and x² + y² = (a² + b²)/S² = 1 − 1/S².

Then z = −S(x² + y²) + S = S(1 − x² − y²) = 1/S = √(1 − x² − y²), which satisfies the PDE (p = −x/z, q = −y/z).

Answer: complete integral z = ax + by + √(1 + a² + b²); singular integral x² + y² + z² = 1 (with z > 0).

Problem 5 (15 marks): Solve the Cauchy problem uu_x + u_y = 1 with u = s/2 on the line x = y = s, and state where the solution fails.

Characteristic equations: dx/dt = u, dy/dt = 1, du/dt = 1, with x = y = s and u = s/2 at t = 0.

J = x₀′b − y₀′a = 1 · 1 − 1 · (s/2) = 1 − s/2, which is non-zero for s ≠ 2.

Solving: u = s/2 + t, y = s + t, x = s + st/2 + t²/2.

With t = y − s: x = s − sy/2 + y²/2, so s = (2x − y²)/(2 − y).

u = s/2 + t = y − s/2 = (4y − y² − 2x)/(2(2 − y)); on x = y = s this gives (2s − s²)/(2(2 − s)) = s/2, as required.

Answer: u = (4y − y² − 2x)/(2(2 − y)), valid for y ≠ 2; at the point s = 2 of the data line J = 0 (the line touches a characteristic projection), and the formula breaks down on y = 2.

3. Linear Equations with Constant Coefficients

Problem 6 (10 marks): Solve (D² + DD′ − 6D′²)z = cos(2x + y).

Auxiliary m² + m − 6 = 0 gives m = 2, −3: CF = φ₁(y + 2x) + φ₂(y − 3x).

The usual rule fails: D² → −4, DD′ → −2, D′² → −1 gives −4 − 2 + 6 = 0, because y + 2x is an argument of the CF.

PI = (D − 2D′)⁻¹(D + 3D′)⁻¹cos(2x + y); since (D + 3D′)sin(2x + y) = 5cos(2x + y), (D + 3D′)⁻¹cos(2x + y) = (1/5)sin(2x + y).

By the failing-case rule, (D − 2D′)⁻¹(1/5)sin(y + 2x) = (x/5)sin(y + 2x).

Check with w = 2x + y: z_xx + z_xy − 6z_yy = [(4cos w − 4x sin w) + (cos w − 2x sin w) + 6x sin w]/5 = cos w.

Answer: z = φ₁(y + 2x) + φ₂(y − 3x) + (x/5)sin(2x + y).

Problem 7 (15 marks): Solve (D − D′ − 1)(D − D′ − 2)z = e^{2x − y} + x.

Each factor has the form D − mD′ − c with m = 1 (c = 1 and c = 2), so CF = e^xφ₁(y + x) + e^2xφ₂(y + x).

For e^{2x − y}: D → 2, D′ → −1 gives (2 + 1 − 1)(2 + 1 − 2) = 2, so PI₁ = (1/2)e^{2x − y}.

For x, write E = D − D′: 1/((E − 1)(E − 2)) = (1/2)(1 − E)⁻¹(1 − E/2)⁻¹ = (1/2)(1 + (3/2)E + ...).

Since Ex = 1 and E²x = 0, PI₂ = (1/2)(x + 3/2) = x/2 + 3/4.

Check: (E² − 3E + 2)(x/2 + 3/4) = 0 − 3/2 + x + 3/2 = x.

Answer: z = e^xφ₁(y + x) + e^2xφ₂(y + x) + (1/2)e^{2x − y} + x/2 + 3/4.

4. Classification and Canonical Forms

Problem 8 (15 marks): Reduce u_xx − 2u_xy − 3u_yy = 0 to canonical form, and solve it with u(x, 0) = 3x², u_y(x, 0) = 0.

B² − 4AC = 4 + 12 = 16 > 0, so the equation is hyperbolic; λ² + 2λ − 3 = 0 gives λ = 1, −3, so ξ = y − x, η = y + 3x.

u_x = −u_ξ + 3u_η, u_y = u_ξ + u_η; then u_xx − 2u_xy − 3u_yy = −16u_ξη, so u_ξη = 0 and u = f(y − x) + g(y + 3x).

Data: f(−x) + g(3x) = 3x² and f′(−x) + g′(3x) = 0; differentiating the first, −f′(−x) + 3g′(3x) = 6x.

Adding: 4g′(3x) = 6x, so g′(s) = s/2 and g(s) = s²/4 + C; then f′(r) = 3r/2 and f(r) = 3r²/4 − C.

u = (3/4)(y − x)² + (1/4)(y + 3x)² = 3x² + y²; check: 6 − 0 − 3(2) = 0, u(x, 0) = 3x², u_y(x, 0) = 0.

Answer: canonical form u_ξη = 0 with ξ = y − x, η = y + 3x; u = 3x² + y².

Problem 9 (10 marks): Show that x²u_xx + 2xyu_xy + y²u_yy = 0 is parabolic, reduce it to canonical form for x ≠ 0, and find its general solution.

B² − 4AC = 4x²y² − 4x²y² = 0 at every point, so the equation is parabolic everywhere.

The characteristic equation x²(dy/dx)² − 2xy(dy/dx) + y² = 0, that is (x dy/dx − y)² = 0, gives ξ = y/x; take η = x.

u_x = −(y/x²)u_ξ + u_η, u_y = u_ξ/x, so the Euler operator E = x∂/∂x + y∂/∂y becomes η∂/∂η.

The given operator equals E² − E = η²∂²/∂η² + η∂/∂η − η∂/∂η = η²∂²/∂η², so the equation becomes η²u_ηη = 0.

Integrating twice in η: u = f(ξ) + ηg(ξ).

Answer: canonical form u_ηη = 0 with ξ = y/x, η = x; general solution u = f(y/x) + xg(y/x).

5. Wave, Heat and Laplace Equations

Problem 10 (15 marks): A string of length L with fixed ends is pulled aside at its midpoint to a height h and released from rest. Find the displacement u(x, t).

Initial shape f(x) = 2hx/L for 0 ≤ x ≤ L/2 and f(x) = 2h(L − x)/L for L/2 ≤ x ≤ L; g(x) = 0, so B_n = 0.

A_n = (2/L)[∫₀^L/2 (2hx/L)sin(nπx/L) dx + ∫_L/2^L (2h(L − x)/L)sin(nπx/L) dx].

Integrating by parts, the terms in cos(nπ/2) cancel and the bracket equals (4hL/(n²π²))sin(nπ/2), so A_n = (8h/(n²π²))sin(nπ/2).

u = (8h/π²)∑_n=1^∞ (sin(nπ/2)/n²)sin(nπx/L)cos(nπct/L); only odd n appear, with alternating signs.

Answer: u(x, t) = (8h/π²)[sin(πx/L)cos(πct/L) − (1/9)sin(3πx/L)cos(3πct/L) + (1/25)sin(5πx/L)cos(5πct/L) − ...].

Problem 11 (15 marks): Solve u_t = u_xx, 0 < x < π, t > 0, with insulated ends u_x(0, t) = u_x(π, t) = 0 and u(x, 0) = x.

Separation with X′(0) = X′(π) = 0 gives X = cos nx, n = 0, 1, 2, ..., and T = e^−n2t.

a₀ = (2/π)∫₀^π x dx = π.

a_n = (2/π)∫₀^π x cos nx dx = (2/π)(cos nπ − 1)/n², which is −4/(πn²) for odd n and 0 for even n ≥ 2.

So u = π/2 − (4/π)∑_{n odd} e^−n2tcos nx/n².

Answer: u(x, t) = π/2 − (4/π)[e^−tcos x + e^−9tcos 3x/9 + e^−25tcos 5x/25 + ...], which tends to the mean temperature π/2 as t → ∞.

Problem 12 (15 marks): Solve u_xx + u_yy = 0 in the square 0 < x < π, 0 < y < π, with u = 0 on x = 0, x = π and y = 0, and u(x, π) = x(π − x).

Separation with X(0) = X(π) = 0 gives X = sin nx and Y″ = n²Y; Y(0) = 0 gives Y = sinh ny, so u = ∑ c_n sin nx sinh ny.

At y = π: c_n sinh nπ = (2/π)∫₀^π x(π − x)sin nx dx = 4(1 − cos nπ)/(πn³).

So c_n = 8/(πn³ sinh nπ) for odd n and c_n = 0 for even n.

Answer: u(x, y) = (8/π)∑_{n odd} sin nx sinh ny/(n³ sinh nπ).

Problem 13 (10 marks): Find the steady temperature u(r, θ) in the disc r < a when u(a, θ) = sin³θ, and the temperature at the centre.

sin³θ = (3 sin θ − sin 3θ)/4, so b₁ = 3/4, b₃ = −1/4 and all other Fourier coefficients are 0.

Each boundary term sin nθ extends to the bounded harmonic function (r/a)ⁿsin nθ inside the disc.

At r = 0 the solution is 0, which equals the mean of sin³θ over a period (mean value property).

Answer: u(r, θ) = (3r/(4a))sin θ − (r³/(4a³))sin 3θ; u = 0 at the centre.

Problem 14 (10 marks): For u_tt = c²u_xx, 0 < x < L, with u(0, t) = u(L, t) = 0, show that E(t) = (1/2)∫₀^L (u_t² + c²u_x²) dx is constant, and deduce that the initial-boundary value problem has at most one smooth solution.

dE/dt = ∫₀^L (u_tu_tt + c²u_xu_xt) dx = c²∫₀^L (u_tu_xx + u_xu_xt) dx = c²∫₀^L ∂(u_tu_x)/∂x dx.

This equals c²[u_tu_x]₀^L = 0, because u = 0 at both ends for all t gives u_t(0, t) = u_t(L, t) = 0.

If u₁, u₂ are two solutions with the same data, w = u₁ − u₂ solves the problem with zero data, so E_w(t) = E_w(0) = 0.

The integrand is continuous and non-negative, so w_t = w_x = 0; w is constant and w(0, t) = 0 gives w ≡ 0.

Answer: dE/dt = 0, so energy is conserved, and the solution is unique.

1. Families of Surfaces and Formation of PDEs

2024 Paper II, Q5(a), 10 marks: Show that if f and g are arbitrary functions of their respective arguments, then u = f(x − kt + iαy) + g(x − kt − iαy), is a solution of
∂²u/∂x² + ∂²u/∂y² = (1/C²) ∂²u/∂t², where α² = 1 − k²/C².

2023 Paper II, Q5(a), 10 marks: By eliminating the arbitrary functions f and g from z = f(x² − y) + g(x² + y), form partial differential equation.

2022 Paper II, Q5(a), 10 marks: It is given that the equation of any cone with vertex at (a, b, c) is f((x − a)/(z − c), (y − b)/(z − c)) = 0. Find the differential equation of the cone.

2021 Paper II, Q5(a), 10 marks: Obtain the partial differential equation by eliminating arbitrary function f from the equation f(x + y + z, x² + y² + z²) = 0.

2020 Paper II, Q5(a), 10 marks: Form a partial differential equation by eliminating the arbitrary functions f(x) and g(y) from z = y f(x) + x g(y) and specify its nature (elliptic, hyperbolic or parabolic) in the region x > 0, y > 0.

2019 Paper II, Q5(a), 10 marks: Form a partial differential equation of the family of surfaces given by the following expression :
ψ(x² + y² + 2z², y² − 2zx) = 0.

2018 Paper II, Q5(a), 10 marks: Find the partial differential equation of the family of all tangent planes to the ellipsoid : x² + 4y² + 4z² = 4, which are not perpendicular to the xy plane.

2. Lagrange's Method for Quasilinear First-Order PDEs

2026 Paper II, Q5(a), 10 marks: Find the integral surface of the equations dx/(xz − y) = dy/(yz − x) = dz/(1 − z²).

2024 Paper II, Q7(a), 15 marks: Find the integral surface of the following quasi-linear equation
(y − φ) ∂φ/∂x + (φ − x) ∂φ/∂y = x − y,
which passes through the curve φ = 0, xy = 1 and through the circle x + y + φ = 0, x² + y² + φ² = a².

2020 Paper II, Q6(a), 15 marks: Find the integral surface of the partial differential equation : (x − y)y² ∂z/∂x + (y − x)x² ∂z/∂y = (x² + y²)z that contains the curve : xz = a³, y = 0 on it.

2018 Paper II, Q6(a), 15 marks: Find the general solution of the partial differential equation :
(y³x − 2x⁴)p + (2y⁴ − x³y)q = 9z(x³ − y³),
where p = ∂z/∂x, q = ∂z/∂y, and find its integral surface that passes through the curve : x = t, y = t², z = 1.

2016 Paper II, Q5(a), 10 marks: Find the general equation of surfaces orthogonal to the family of spheres given by x² + y² + z² = cz.

2016 Paper II, Q5(e), 10 marks: Find the general integral of the partial differential equation
(y + zx)p − (x + yz)q = x² − y².

2015 Paper II, Q5(a), 10 marks: Solve the partial differential equation (y² + z² − x²)p − 2xyq + 2xz = 0 where p = ∂z/∂x and q = ∂z/∂y.

2015 Paper II, Q6(a), 15 marks: Solve for the general solution p cos(x + y) + q sin(x + y) = z, where p = ∂z/∂x and q = ∂z/∂y.

3. Cauchy's Method of Characteristics and Nonlinear First-Order PDEs

2025 Paper II, Q7(a), 15 marks: Find the complete integral of z(p² − q²) = x − y; p ≡ ∂z/∂x, q ≡ ∂z/∂y.

2025 Paper II, Q8(a), 15 marks: Find the characteristics of the partial differential equation
p² + q² = 2; p ≡ ∂z/∂x, q ≡ ∂z/∂y
and determine the integral surface which passes through x = 0, z = y.

2021 Paper II, Q8(a), 15 marks: Find a complete integral of the partial differential equation p = (z + qy)² by using Charpit’s method.

2020 Paper II, Q7(a), 15 marks: Find the solution of the partial differential equation : z = (1/2)(p² + q²) + (p − x)(q − y); p ≡ ∂z/∂x, q ≡ ∂z/∂y which passes through the x-axis.

2019 Paper II, Q6(a), 15 marks: Solve the first order quasilinear partial differential equation by the method of characteristics :
x ∂u/∂x + (u − x − y) ∂u/∂y = x + 2y in x > 0, −∞ < y < ∞ with u = 1 + y on x = 1.

2017 Paper II, Q6(a), 15 marks: Find a complete integral of the partial differential equation
2(pq + yp + qx) + x² + y² = 0.

2016 Paper II, Q6(a), 15 marks: Determine the characteristics of the equation z = p² − q², and find the integral surface which passes through the parabola 4z + x² = 0, y = 0.

4. Linear Second-Order PDEs with Constant Coefficients

2026 Paper II, Q7(a), 15 marks: Solve the partial differential equation (2D² − 5DD′ + 2D′²) z = 24(y − x) + sin(y − x), where D ≡ ∂/∂x, D′ ≡ ∂/∂y, z = z(x, y).

2025 Paper II, Q5(a), 10 marks: Find the solution of the equation (D² + DD′ − 2D²) z = y sin x, where D ≡ ∂/∂x and D′ ≡ ∂/∂y.

2023 Paper II, Q6(a), 15 marks: Find the surface passing through the two lines z = x = 0 and z − 1 = x − y = 0, and satisfying the partial differential equation ∂²z/∂x² − 4 ∂²z/∂x∂y + 4 ∂²z/∂y² = 0.

2022 Paper II, Q7(a), 15 marks: Find the general solution of the partial differential equation (D² + DD′ − 6D′²) z = x² sin(x + y) where D ≡ ∂/∂x and D′ ≡ ∂/∂y.

2021 Paper II, Q7(a), 15 marks: Find the general solution of the partial differential equation (D² − D′² − 3D + 3D′)z = xy + e^x+2y where D ≡ ∂/∂x and D′ ≡ ∂/∂y.

2020 Paper II, Q5(d), 10 marks: Solve the partial differential equation : (D³ − 2D²D′ − DD′² + 2D′³)z = e^2x+y + sin(x − 2y); D ≡ ∂/∂x, D′ ≡ ∂/∂y

2018 Paper II, Q7(a), 15 marks: Solve the partial differential equation :
(2D² − 5DD′ + 2D′²)z = 5 sin(2x + y) + 24(y − x) + e^{3x + 4y}
where D ≡ ∂/∂x, D′ ≡ ∂/∂y.

2017 Paper II, Q5(a), 10 marks: Solve (D² − 2DD′ + D′²)z = e^x+2y + x³ + sin 2x,
where D ≡ ∂/∂x, D′ ≡ ∂/∂y, D² ≡ ∂²/∂x², D′² ≡ ∂²/∂y².

2016 Paper II, Q7(a), 15 marks: Solve the partial differential equation
∂³z/∂x³ − 2 ∂³z/∂x²∂y − ∂³z/∂x∂y² + 2 ∂³z/∂y³ = e^x+y

2015 Paper II, Q5(b), 10 marks: Solve (D² + DD′ − 2D′²)u = e^x+y, where D = ∂/∂x and D′ = ∂/∂y.

2014 Paper II, Q5(a), 10 marks: Solve the partial differential equation (2D² − 5DD′ + 2D′²)z = 24(y − x).

5. Classification and Canonical Forms of Second-Order PDEs

2026 Paper II, Q6(a), 20 marks: Reduce the partial differential equation y ∂²z/∂x² + (x + y) ∂²z/∂x∂y + x ∂²z/∂y² = 0 to canonical form and hence solve it.

2024 Paper II, Q8(a), 15 marks: Solve the partial differential equation
∂/∂y (∂φ/∂x + φ) + 2x²y (∂φ/∂x + φ) = 0
by transforming it to the canonical form.

2023 Paper II, Q8(a), 15 marks: Reduce the partial differential equation ∂²z/∂y² − ∂²z/∂x∂y + ∂z/∂x − (∂z/∂y)(1 + 1/x) + z/x = 0 to canonical form.

2022 Paper II, Q8(a), 15 marks: Reduce the following partial differential equation to a canonical form and hence solve it: yu_xx + (x + y)u_xy + xu_yy = 0

2019 Paper II, Q7(c), 20 marks: Reduce the following second order partial differential equation to canonical form and find the general solution :
∂²u/∂x² − 2x ∂²u/∂x∂y + x² ∂²u/∂y² = ∂u/∂y + 12x.

2017 Paper II, Q7(a), 15 marks: Reduce the equation
y² ∂²z/∂x² − 2xy ∂²z/∂x∂y + x² ∂²z/∂y² = (y²/x) ∂z/∂x + (x²/y) ∂z/∂y
to canonical form and hence solve it.

2015 Paper II, Q8(a), 15 marks: Reduce the second-order partial differential equation x² ∂²u/∂x² − 2xy ∂²u/∂x∂y + y² ∂²u/∂y² + x ∂u/∂x + y ∂u/∂y = 0 into canonical form. Hence, find its general solution.

2014 Paper II, Q6(a), 15 marks: Reduce the equation ∂²z/∂x² = x² ∂²z/∂y² to canonical form.

6. Vibrating String, Heat Equation and Laplace Equation

2026 Paper II, Q8(a), 15 marks: Two-dimensional harmonic equation in plane polar coordinates (r, θ) takes the form:
∂²V/∂r² + (1/r) ∂V/∂r + (1/r²) ∂²V/∂θ² = 0
Deduce that it has solutions of the form (Arⁿ + Br⁻ⁿ) e^±inθ, where A, B, n are constants. Determine V if it satisfies the two-dimensional harmonic equation in the region 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π and satisfies the conditions:
(i) V remains finite as r → 0,
(ii) V = ∑_n c_n cos(nθ) on r = a.

2025 Paper II, Q6(a), 20 marks: Solve ∂²u/∂x² + ∂²u/∂y² = 0 for a rectangular plate subject to the boundary conditions
u(0, y) = 0, u(a, y) = 0
u(x, 0) = 0, u(x, b) = f(x)

2024 Paper II, Q6(a), 20 marks: Show that the solution of the two-dimensional Laplace's equation
∂²φ(x, y)/∂x² + ∂²φ(x, y)/∂y² = 0, x ∈ (−∞, ∞), y ≥ 0
subject to the boundary condition
φ(x, 0) = f(x), x ∈ (−∞, ∞),
along with φ(x, y) → 0 for |x| → ∞ and y → ∞ can be written in the form
φ(x, y) = (y/π) ∫_−∞^∞ f(ξ) dξ/[y² + (x − ξ)²].

2023 Paper II, Q7(c), 20 marks: Solve the partial differential equation a² ∂²u/∂x² = ∂²u/∂t², 0 < x < L, t > 0 subject to the conditions
u(0, t) = 0, u(L, t) = 0, t > 0
u(x, 0) = x, (∂u/∂t)_t=0 = 1, 0 < x < L

2022 Paper II, Q6(a), 20 marks: Solve the heat equation ∂u/∂t = ∂²u/∂x², 0 < x < l, t > 0 subject to the conditions
u(0, t) = u(l, t) = 0
u(x, 0) = x(l − x), 0 ≤ x ≤ l

2021 Paper II, Q6(a), 20 marks: Solve the wave equation a² ∂²u/∂x² = ∂²u/∂t², 0 < x < L, t > 0 subject to the conditions u(0, t) = 0, u(L, t) = 0, u(x, 0) = (1/4)x(L − x), ∂u/∂t|_t=0 = 0

2020 Paper II, Q8(a), 20 marks: One end of a tightly stretched flexible thin string of length l is fixed at the origin and the other at x = l. It is plucked at x = l/3 so that it assumes initially the shape of a triangle of height h in the x-y plane. Find the displacement y at any distance x and at any time t after the string is released from rest. Take, (horizontal tension)/(mass per unit length) = c².

2018 Paper II, Q8(c), 20 marks: A thin annulus occupies the region 0 < a ≤ r ≤ b, 0 ≤ θ ≤ 2π. The faces are insulated. Along the inner edge the temperature is maintained at 0°, while along the outer edge the temperature is held at T = K cos(θ/2), where K is a constant. Determine the temperature distribution in the annulus.

2017 Paper II, Q5(d), 10 marks: Let Γ be a closed curve in xy-plane and let S denote the region bounded by the curve Γ. Let
∂²w/∂x² + ∂²w/∂y² = f(x, y) ∀ (x, y) ∈ S.
If f is prescribed at each point (x, y) of S and w is prescribed on the boundary Γ of S, then prove that any solution w = w(x, y), satisfying these conditions, is unique.

2017 Paper II, Q8(a), 20 marks: Given the one-dimensional wave equation
∂²y/∂t² = c² ∂²y/∂x²; t > 0,
where c² = T/m, T is the constant tension in the string and m is the mass per unit length of the string.
(i) Find the appropriate solution of the above wave equation.
(ii) Find also the solution under the conditions
y(0, t) = 0, y(l, t) = 0 for all t
and [∂y/∂t]_t=0 = 0, y(x, 0) = a sin(πx/l), 0 < x < l, a > 0.

2016 Paper II, Q8(a), 20 marks: Find the temperature u(x, t) in a bar of silver of length 10 cm and constant cross-section of area 1 cm². Let density ρ = 10·6 g/cm³, thermal conductivity K = 1·04 cal/(cm sec °C) and specific heat σ = 0·056 cal/g °C. The bar is perfectly isolated laterally, with ends kept at 0°C and initial temperature f(x) = sin(0·1πx) °C. Note that u(x, t) follows the heat equation u_t = c²u_xx, where c² = K/(ρσ).

2015 Paper II, Q7(a), 15 marks: Find the solution of the initial-boundary value problem
u_t − u_xx + u = 0, 0 < x < l, t > 0
u(0, t) = u(l, t) = 0, t ≥ 0
u(x, 0) = x(l − x), 0 < x < l

2014 Paper II, Q7(a), 15 marks: Find the deflection of a vibrating string (length = π, ends fixed, ∂²u/∂t² = ∂²u/∂x²) corresponding to zero initial velocity and initial deflection f(x) = k(sin x − sin 2x).

2014 Paper II, Q8(a), 15 marks: Solve ∂²u/∂t² = ∂²u/∂x², 0 < x < 1, t > 0, given that
(i) u(x, 0) = 0, 0 ≤ x ≤ 1
(ii) ∂u/∂t (x, 0) = x², 0 ≤ x ≤ 1
(iii) u(0, t) = u(1, t) = 0, for all t