Partial Differential Equations

🟢 Lite — Quick Review (1h–1d)

Key Concepts

Topic	Core Idea	Method
First-order linear PDE	uₓ + a·u_y = 0 (Lagrange’s)	Solve via auxiliary equations: dx/P = dy/Q = dz/R
Second-order classification	Discriminant B² − 4AC	Elliptic (B² < 4AC), Parabolic (B² = 4AC), Hyperbolic (B² > 4AC)
Heat equation	∂u/∂t = k ∂²u/∂x²	Separate variables → series solution
Wave equation	∂²u/∂t² = c² ∂²u/∂x²	d’Alembert’s formula or series
Laplace equation	uₓₓ + u_yy = 0	Harmonic functions

⚡ Quick记住: Heat → parabolic (diffusion), Wave → hyperbolic (oscillation), Laplace → elliptic (steady-state)

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🟡 Standard — Regular Study (2d–2mo)

First-Order Linear PDEs

Standard Form

$$P(x,y,z)p + Q(x,y,z)q = R(x,y,z)$$ where p = ∂z/∂x, q = ∂z/∂y

Lagrange’s Method (Charpit’s Method)

Auxiliary equations: $$\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}$$

Solve two independent equations from these ratios:

1. First integral φ(x,y,z) = c₁ — from solving the auxiliary system
2. Second integral ψ(x,y,z) = c₂ — another independent solution

General solution: Φ(c₁, c₂) = 0 or equivalently φ(x,y,z) = f(ψ(x,y,z))

Example: xp + yq = z → P=x, Q=y, R=z

• Aux: dx/x = dy/y = dz/z
• From dx/x = dy/y → ln x = ln y + const → x/y = c₁
• From dx/x = dz/z → x/z = c₂
• General: φ(x/y, x/z) = 0 → or x/z = f(x/y)

⚡ GATE Trick: When two of P, Q, R are zero or proportional, the method simplifies dramatically.

Special Cases

Linear homogeneous: z = f(ax + by) works for first-order linear with constant coefficients

Quasi-linear: Use method of characteristics — parametric solution along curves.

Classification of Second-Order PDEs

General Form

$$A \frac{\partial^2 u}{\partial x^2} + B \frac{\partial^2 u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} + \cdots = 0$$

Classification by discriminant: Δ = B² − 4AC

Δ	Type	Canonical Form	Physical Analogy
Δ < 0	Elliptic	Laplace/Poisson	Steady-state, no time
Δ = 0	Parabolic	Heat equation	Diffusion, dissipation
Δ > 0	Hyperbolic	Wave equation	Oscillation, propagation

⚡ GATE remembers: Heat conducts (parabolic — one time derivative), Waves oscillate (hyperbolic — second time derivative), Laplace is static (elliptic).

Standard PDEs — Solutions

Heat Equation: ∂u/∂t = k ∂²u/∂x²

Boundary conditions (common):

• u(0,t) = 0, u(L,t) = 0 (insulated ends)
• Initial: u(x,0) = f(x)

Solution method: Separation of variables

1. Assume u(x,t) = X(x)T(t)
2. X”/X = T’/(kT) = −λ (separation constant, must be negative)
3. X(x) = sin(nπx/L), T(t) = e^(−k(nπ/L)²t)
4. Fourier series: u(x,t) = ΣAₙ sin(nπx/L) e^(−kλₙt)

Wave Equation: ∂²u/∂t² = c² ∂²u/∂x²

d’Alembert’s Solution (infinite string, no boundaries): $$u(x,t) = \frac{1}{2}[f(x+ct) + f(x-ct)]$$

For finite string (0,x,L): Series solution with sin terms, frequency ωₙ = nπc/L

Laplace Equation: uₓₓ + u_yy = 0

Rectangular domain: Fourier series solution Circular domain: Use polar coordinates, separation gives Bessel functions

Key property: Maximum principle — maximum/minimum of u occurs on the boundary.

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🔴 Extended — Deep Study

Lagrange’s Method — Detailed Derivation

For PDE: P p + Q q = R

The auxiliary equations dx/P = dy/Q = dz/R yield:

• Two first integrals: φ₁(x,y,z) = c₁, φ₂(x,y,z) = c₂
• These give curves (characteristics) along which z is constant
• General solution: F(φ₁, φ₂) = 0 where F is arbitrary

Solved Example (GATE 2022 pattern): PDE: (y²) p + (x) q = z

Auxiliary: dx/y² = dy/x = dz/z

From dx/y² = dy/x → x dx = y dy → x² − y² = c₁

From dy/x = dz/z → (1/x)dy = dz/z → integrate: ln y = ln z + const → y/z = c₂

General solution: F(x² − y², y/z) = 0 or z = y·f(x² − y²) ✓

Canonical Forms — Transformation Approach

For hyperbolic: Use coordinates ξ = y − x√Δ, η = y + x√Δ to reduce to u_ξη = 0

For parabolic: Reduce to heat equation canonical form u_ξξ = u_η

For elliptic: Rotate axes to eliminate mixed derivative term (B → 0)

GATE rarely asks canonical transformation directly, but understanding WHY the classification matters helps:

• Elliptic → boundary value problems (solved with Fourier series in rectangular domains)
• Hyperbolic → initial value problems (d’Alembert or series)
• Parabolic → initial-boundary value problems (series solution with exponential decay)

Heat Equation — Series Solution Detail

Example: u(x,0) = sin(πx/L), ends at 0°C

1. u(x,t) = ΣBₙ sin(nπx/L) e^(−k(nπ/L)²t)
2. Bₙ from Fourier sine series of initial condition
3. For sin(πx/L): B₁ = 1, all others = 0
4. Solution: u(x,t) = sin(πx/L) e^(−k(π/L)²t)

⚡ Key insight: All modes decay exponentially. Higher modes (large n) decay faster. As t→∞, u→0 (heat dissipates).

d’Alembert’s Solution — Wave Equation

For wave equation on infinite domain with initial displacement f(x):

u(x,t) = ½[f(x+ct) + f(x-ct)] — this is the forward traveling wave form

For a half-line or string with fixed end, use method of images (reflect the initial condition).

Physical meaning: Disturbance splits into two waves traveling left and right at speed c.

GATE Previous-Year Highlights

Year	Problem	Key Concept
2018	Classify: uₓₓ + 5uₓᵧ + uᵧᵧ = 0	Δ = 25−4 = 21 > 0 → Hyperbolic
2019	Solve p + q = 1 (Lagrange)	Aux: dx/1 = dy/1 = dz/1 → x−y=c₁, z−x=c₂
2020	Heat eqn: uₜ = uₓₓ, u(0,t)=u(π,t)=0	Eigenfunctions sin(nx), λₙ = n²
2021	Laplace in rectangle 0<x<a, 0<y<b	Double Fourier series
2022	d’Alembert: uₜₜ = c²uₓₓ, f(x)=sin x	u = ½[sin(x+ct) + sin(x-ct)]
2023	Classify: 4uₓₓ + 6uₓᵧ + 9uᵧᵧ = 0	Δ = 36 − 144 = −108 < 0 → Elliptic

⚡ GATE Warning: The classification question (Δ < 0, = 0, > 0) is a recurring 1-2 mark question. Memorize the table above.

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