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Partial differential equations basics

Mathematics (B.Sc) · Differential equations · free study notes for exams, placements and interviews

In short

A partial differential equation (PDE) is an equation relating an unknown function of two or more independent variables to its partial derivatives — for example the heat equation ∂u/∂t = c²∂²u/∂x². Unlike ordinary differential equations, a PDE's general solution contains arbitrary functions rather than arbitrary constants, so boundary and initial conditions decide the answer. The B.Sc / CBCS syllabus covers formation, first-order equations (Lagrange, Charpit, Clairaut), second-order linear equations with constant coefficients, classification into elliptic, parabolic and hyperbolic types, and solution of the wave, heat and Laplace equations by separation of variables. It carries heavy weight in GATE MA, CSIR-UGC NET and JAM, and underlies every modelling or simulation role — CFD and FEA at L&T, Tata Consultancy Research, ISRO and Ansys users, option pricing in BFSI analytics, and image/signal processing in Indian product firms.

Explained

Definitions, order, degree and classification by linearity

Formation of a PDE

Two routes, both standard two-mark starters:

  1. By eliminating arbitrary constants. Given z = f(x, y, a, b), differentiate once with respect to x and y to get p and q, then eliminate a and b among the three equations → a first-order PDE.
  2. By eliminating an arbitrary function. Given φ(u, v) = 0 where u, v are functions of x, y, z, differentiating and eliminating ∂φ/∂u, ∂φ/∂v gives the Lagrange form Pp + Qq = R with P, Q, R the Jacobians ∂(u,v)/∂(y,z), ∂(u,v)/∂(z,x), ∂(u,v)/∂(x,y).
  3. Rule of thumb: eliminating n arbitrary constants gives order n at most; eliminating one arbitrary function of two arguments gives a first-order PDE; eliminating two arbitrary functions typically gives a second-order PDE.

First-order PDEs: linear and nonlinear methods

dp/(f_x + p f_z) = dq/(f_y + q f_z) = dz/(−p f_p − q f_q) = dx/(−f_p) = dy/(−f_q). Any one integral giving a relation between p, q and the variables is combined with the original equation to get p and q, then z follows from dz = p dx + q dy.

Second-order linear PDEs with constant coefficients

Classification, canonical forms and characteristics

The three classical equations and separation of variables

Key points

Formulas and examples

p = ∂z/∂x, q = ∂z/∂y, r = ∂²z/∂x², s = ∂²z/∂x∂y, t = ∂²z/∂y² (Monge notation; z = z(x, y))
Lagrange: Pp + Qq = R → dx/P = dy/Q = dz/R ; integrals u = c₁, v = c₂ → φ(u, v) = 0
Charpit: dp/(f_x + p f_z) = dq/(f_y + q f_z) = dz/(−p f_p − q f_q) = dx/(−f_p) = dy/(−f_q)
Clairaut: z = px + qy + f(p, q) → z = ax + by + f(a, b)
Homogeneous f(D, D′)z = 0 with roots m₁ ≠ m₂: z = φ₁(y + m₁x) + φ₂(y + m₂x); repeated m: φ₁(y + mx) + x φ₂(y + mx)
Classification of A u_xx + B u_xy + C u_yy + … = 0 by Δ = B² − 4AC: Δ > 0 hyperbolic, Δ = 0 parabolic, Δ < 0 elliptic
Wave equation u_tt = c² u_xx, c = √(T/ρ) in m/s (T in N, ρ in kg/m)
Heat equation u_t = α u_xx, α = k/(ρ c_p) in m²/s (k in W/m·K, ρ in kg/m³, c_p in J/kg·K)
Laplace u_xx + u_yy = 0 ; Poisson u_xx + u_yy = f(x, y)
D'Alembert: u(x, t) = ½[f(x + ct) + f(x − ct)] + (1/2c) ∫ from (x − ct) to (x + ct) of g(s) ds, with u(x,0) = f, u_t(x,0) = g
String, both ends fixed at x = 0, L: u(x, t) = Σ_{n≥1} [bₙ sin(nπx/L) cos(nπct/L)], bₙ = (2/L) ∫₀L f(x) sin(nπx/L) dx
Bar with ends at 0 °C: u(x, t) = Σ_{n≥1} bₙ sin(nπx/L) e−α n²π²t/L²

Example 1 — Formation by eliminating constants. Form the PDE from z = ax + by + ab.

Example 2 — Lagrange's equation. Solve (y − z)p + (z − x)q = x − y.

Example 3 — Second-order linear PDE, CF and PI. Solve (D² − 3DD′ + 2D′²) z = e2x+3y.

Example 4 — Classification. Classify u_xx + 4u_xy + 4u_yy + u_x = 0 and x²u_xx − y²u_yy = 0 (x, y ≠ 0).

Example 5 — Vibrating string, numbers. A steel wire of length L = 0.60 m under tension T = 120 N has linear density ρ = 3.0 × 10⁻³ kg/m; it is plucked so that u(x, 0) = sin(πx/L) m (idealised unit-amplitude first mode) with zero initial velocity, ends fixed.

Example 6 — Heat equation, decay of modes. A copper bar L = 1.0 m, α = 1.11 × 10⁻⁴ m²/s, both ends held at 0 °C, initial u(x, 0) = 100 sin(πx/L) °C.

FeatureHyperbolic (wave)Parabolic (heat)Elliptic (Laplace)
Model equationu_tt = c²u_xxu_t = αu_xxu_xx + u_yy = 0
Δ = B² − 4AC> 0= 0< 0
Real characteristicsTwo familiesOne familyNone
Data neededInitial u, u_t + boundaryInitial u + boundaryBoundary only
Signal speedFinite (c)InfiniteNo time variable
SmoothnessKinks propagateSolution smoothedAnalytic inside
Typical numericsExplicit, CFL-limitedExplicit (r ≤ ½) or Crank–NicolsonIterative solvers (SOR, multigrid)
AspectODEPDE
Independent variablesOneTwo or more
General solutionArbitrary constantsArbitrary functions
Extra dataInitial/boundary valuesConditions on curves or surfaces
Typical methodIntegrating factor, operator DCharacteristics, separation, transforms
# Explicit finite-difference check of the heat equation (stability r <= 0.5)
import numpy as np
L, T_end, alpha = 1.0, 600.0, 1.11e-4
nx = 21; dx = L/(nx-1)
dt = 0.4*dx*dx/alpha                 # r = 0.4 < 0.5, stable
nt = int(T_end/dt)
u = 100*np.sin(np.pi*np.linspace(0, L, nx))
u[0] = u[-1] = 0.0
for _ in range(nt):
    u[1:-1] += 0.4*(u[2:] - 2*u[1:-1] + u[:-2])
print(u[nx//2])                      # ~51.8 degC, matches the series solution

Where it is used

Interview questions

Common mistakes

Related topics

Ordinary differential equations · Fourier series and transforms · Laplace transforms · Numerical solution of PDEs (finite differences) · Vector calculus and Green's/Stokes' theorems · Sturm–Liouville and boundary-value problems

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