Partial Differential Equations

Classification, first-order PDEs, wave equation, heat equation, and Laplace equation with full worked solutions for GATE preparation.

Classification First-Order PDE Wave Equation Heat Equation Laplace Equation

0 / 5 units completed0%

Classification of Second-Order PDEs

A second-order linear PDE in two variables is classified as elliptic, parabolic, or hyperbolic based on its discriminant. This classification determines the nature of solutions and appropriate solution techniques.

◆ General Second-Order PDE

Definition — General Form A second-order linear PDE in two independent variables \(x, y\) has the general form: \[A\,u_{xx} + 2B\,u_{xy} + C\,u_{yy} + D\,u_x + E\,u_y + F\,u = G\] where \(A, B, C, D, E, F, G\) may depend on \(x\) and \(y\).

Classification by Discriminant Define the discriminant \(\Delta = B^2 - AC\). Then:

Elliptic if \(\Delta < 0\) (e.g., Laplace equation)
Parabolic if \(\Delta = 0\) (e.g., Heat equation)
Hyperbolic if \(\Delta > 0\) (e.g., Wave equation)

★ Example

Classify the PDE \(u_{xx} + 4u_{xy} + u_{yy} = 0\).

Here \(A = 1\), \(B = 2\), \(C = 1\). The discriminant is: \[\Delta = B^2 - AC = 4 - 1 = 3 > 0\] Since \(\Delta > 0\), the PDE is hyperbolic.

◆ Canonical Forms

Each type of PDE can be reduced to a canonical form via a change of variables using the characteristic curves \(\xi(x,y)\) and \(\eta(x,y)\):

Canonical Forms

Hyperbolic: \(u_{\xi\eta} = \phi(\xi,\eta,u,u_\xi,u_\eta)\)
Parabolic: \(u_{\eta\eta} = \phi(\xi,\eta,u,u_\xi,u_\eta)\)
Elliptic: \(u_{\xi\xi} + u_{\eta\eta} = \phi(\xi,\eta,u,u_\xi,u_\eta)\)

The characteristic curves are found by solving \(A\,dy^2 - 2B\,dx\,dy + C\,dx^2 = 0\).

First-Order PDE & Method of Characteristics

First-order PDEs are solved by the method of characteristics, which reduces the PDE to a system of ODEs along characteristic curves.

◆ Quasi-Linear First-Order PDE

Definition — Quasi-Linear PDE A first-order quasi-linear PDE has the form: \[a(x,y,u)\,u_x + b(x,y,u)\,u_y = c(x,y,u)\]

Lagrange’s Method (Charpit’s Equations) The characteristic equations are: \[\frac{dx}{a} = \frac{dy}{b} = \frac{du}{c}\] The general solution is found by obtaining two independent integrals \(\phi_1(x,y,u) = c_1\) and \(\phi_2(x,y,u) = c_2\), giving \(\Phi(\phi_1, \phi_2) = 0\).

★ Example

Solve \(x\,u_x + y\,u_y = u\) with initial condition \(u(x,1) = x^2\).

The characteristic equations are: \[\frac{dx}{x} = \frac{dy}{y} = \frac{du}{u}\] From \(\frac{dx}{x} = \frac{dy}{y}\): \(\ln x = \ln y + \text{const}\), so \(\frac{x}{y} = c_1\).
From \(\frac{dy}{y} = \frac{du}{u}\): \(\frac{u}{y} = c_2\).
General solution: \(u = y\,f\!\left(\frac{x}{y}\right)\). Applying \(u(x,1) = x^2\): \(f(x) = x^2\).
Solution: \(u = y \cdot \frac{x^2}{y^2} = \frac{x^2}{y}\).

◆ Method of Characteristics for General First-Order PDEs

Definition — General First-Order PDE The general form is \(F(x, y, u, p, q) = 0\) where \(p = u_x\) and \(q = u_y\).

The Cauchy problem prescribes initial data along a curve \(\Gamma\). The solution surface is swept out by the characteristic strips satisfying:

\[\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{du}{pF_p + qF_q} = \frac{-dp}{F_x + pF_u} = \frac{-dq}{F_y + qF_u}\]

Wave Equation & D’Alembert’s Solution

The wave equation models vibrations and wave propagation. D'Alembert's formula gives the explicit solution for the one-dimensional wave equation on the real line.

◆ One-Dimensional Wave Equation

Definition — Wave Equation The one-dimensional wave equation is: \[u_{tt} = c^2\,u_{xx}, \quad x \in \mathbb{R},\; t > 0\] with initial conditions \(u(x,0) = f(x)\) and \(u_t(x,0) = g(x)\).

Theorem — D’Alembert’s Formula The solution of the Cauchy problem for the wave equation is: \[u(x,t) = \frac{1}{2}\bigl[f(x+ct) + f(x-ct)\bigr] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds\]

★ Example

Solve \(u_{tt} = 4u_{xx}\), \(u(x,0) = \sin x\), \(u_t(x,0) = 0\).

Here \(c = 2\), \(f(x) = \sin x\), \(g(x) = 0\). By D'Alembert's formula: \[u(x,t) = \frac{1}{2}[\sin(x+2t) + \sin(x-2t)]\] Using the sum-to-product identity: \[u(x,t) = \sin x \cos 2t\]

◆ Wave Equation on a Bounded Domain

For the wave equation on \([0,L]\) with homogeneous Dirichlet boundary conditions \(u(0,t) = u(L,t) = 0\), we use separation of variables:

Fourier Series Solution \[u(x,t) = \sum_{n=1}^{\infty}\left(A_n \cos\frac{n\pi c t}{L} + B_n \sin\frac{n\pi c t}{L}\right)\sin\frac{n\pi x}{L}\] where \(A_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx\) and \(B_n = \frac{2}{n\pi c}\int_0^L g(x)\sin\frac{n\pi x}{L}\,dx\).

Heat Equation — Separation of Variables & Fourier Series

The heat equation models diffusion processes. Separation of variables combined with Fourier series is the standard approach for bounded domains.

◆ Heat Equation on a Finite Rod

Definition — Heat Equation \[u_t = k\,u_{xx}, \quad 0 < x < L,\; t > 0\] with boundary conditions \(u(0,t) = u(L,t) = 0\) and initial condition \(u(x,0) = f(x)\).

Theorem — Fourier Series Solution Assuming \(u(x,t) = X(x)\,T(t)\), separation yields: \[X'' + \lambda X = 0,\quad T' + k\lambda T = 0\] The solution is: \[u(x,t) = \sum_{n=1}^{\infty} B_n\,e^{-k(n\pi/L)^2 t}\sin\frac{n\pi x}{L}\] where \(B_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx\).

★ Example

Solve \(u_t = u_{xx}\), \(0 < x < \pi\), \(u(0,t)=u(\pi,t)=0\), \(u(x,0) = \sin 2x\).

Here \(k=1\), \(L=\pi\). The initial condition is already a single Fourier mode with \(n=2\), so \(B_2 = 1\) and all other \(B_n = 0\).
Solution: \(u(x,t) = e^{-4t}\sin 2x\).

◆ Heat Kernel on the Real Line

Fundamental Solution For the heat equation on \(\mathbb{R}\), the solution is given by convolution with the heat kernel: \[u(x,t) = \frac{1}{\sqrt{4\pi k t}}\int_{-\infty}^{\infty} f(s)\,\exp\!\left(-\frac{(x-s)^2}{4kt}\right)ds\]

The heat kernel \(\Phi(x,t) = \frac{1}{\sqrt{4\pi k t}}\exp\!\left(-\frac{x^2}{4kt}\right)\) is the fundamental solution, satisfying \(\Phi_t = k\Phi_{xx}\) and \(\Phi(x,0) = \delta(x)\).

Laplace Equation — Boundary Value Problems & Maximum Principle

The Laplace equation describes steady-state phenomena. The maximum principle and boundary value problems are central to its study.

◆ Laplace Equation & Harmonic Functions

Definition — Laplace Equation The Laplace equation in two dimensions is: \[\nabla^2 u = u_{xx} + u_{yy} = 0\] A \(C^2\) function satisfying Laplace's equation is called harmonic.

Maximum Principle If \(u\) is harmonic on a bounded domain \(\Omega\) and continuous on \(\overline{\Omega}\), then: \[\max_{\overline{\Omega}} u = \max_{\partial\Omega} u \quad\text{and}\quad \min_{\overline{\Omega}} u = \min_{\partial\Omega} u\] Moreover, if \(u\) attains its maximum (or minimum) at an interior point, then \(u\) is constant.

◆ Laplace Equation on a Rectangle & Disk

★ Example — Dirichlet Problem on a Rectangle

Solve \(\nabla^2 u = 0\) on \(0 < x < a\), \(0 < y < b\) with \(u(x,0)=f(x)\), \(u(x,b)=0\), \(u(0,y)=0\), \(u(a,y)=0\).

Using separation of variables: \[u(x,y) = \sum_{n=1}^{\infty} B_n \frac{\sinh\frac{n\pi(b-y)}{a}}{\sinh\frac{n\pi b}{a}}\sin\frac{n\pi x}{a}\] where \(B_n = \frac{2}{a}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx\).

Poisson’s Formula for the Disk For the Dirichlet problem on a disk of radius \(R\) with boundary data \(u(R,\theta) = h(\theta)\): \[u(r,\theta) = \frac{1}{2\pi}\int_0^{2\pi} \frac{R^2 - r^2}{R^2 - 2Rr\cos(\theta-\phi) + r^2}\,h(\phi)\,d\phi\]

★ Key Takeaways

Classify PDEs using the discriminant \(\Delta = B^2 - AC\): elliptic, parabolic, or hyperbolic.
First-order PDEs are solved via Lagrange's characteristic method: \(\frac{dx}{a} = \frac{dy}{b} = \frac{du}{c}\).
D'Alembert's formula gives the explicit solution for the wave equation on the real line.
Separation of variables with Fourier series is the standard tool for heat and wave equations on bounded domains.
The maximum principle guarantees harmonic functions achieve extrema only on the boundary.

📝 Practice Problems

Problem 1

Classify the PDE \(3u_{xx} + 4u_{xy} + u_{yy} - 2u_x = 0\).

Show Solution ▼

Here \(A=3\), \(B=2\), \(C=1\). Discriminant \(\Delta = 4 - 3 = 1 > 0\). The PDE is hyperbolic.

Problem 2

Solve the first-order PDE \(u_x + 2u_y = u\) with \(u(x,0) = e^{3x}\).

Show Solution ▼

Characteristics: \(\frac{dx}{1} = \frac{dy}{2} = \frac{du}{u}\). From the first pair, \(y - 2x = c_1\). From the first and third, \(u\,e^{-x} = c_2\). So \(u = e^x f(y-2x)\). Applying IC: \(f(-2x) = e^{3x} \cdot e^{-x}\cdot e^{x}\), wait let's redo. \(u = e^x f(y - 2x)\). At \(y=0\): \(e^{3x} = e^x f(-2x)\), so \(f(-2x) = e^{2x}\), meaning \(f(s) = e^{-s}\). Thus \(u = e^x \cdot e^{-(y-2x)} = e^{3x - y}\).

Problem 3

Use D'Alembert's formula to solve \(u_{tt} = 9u_{xx}\), \(u(x,0) = e^{-x^2}\), \(u_t(x,0) = 0\).

Show Solution ▼

With \(c = 3\) and \(g = 0\): \(u(x,t) = \frac{1}{2}\left[e^{-(x+3t)^2} + e^{-(x-3t)^2}\right]\).

Problem 4

Solve \(u_t = u_{xx}\) on \(0 < x < 1\) with \(u(0,t) = u(1,t) = 0\) and \(u(x,0) = x(1-x)\).

Show Solution ▼

\(u(x,t) = \sum_{n=1}^{\infty} B_n e^{-n^2\pi^2 t}\sin(n\pi x)\) where \(B_n = 2\int_0^1 x(1-x)\sin(n\pi x)\,dx = \frac{4}{n^3\pi^3}[1-(-1)^n]\). Only odd \(n\) contribute: \(B_n = \frac{8}{n^3\pi^3}\) for \(n\) odd.

Problem 5

Let \(u\) be harmonic on the unit disk with \(u = 3\) on the boundary. What is \(u\) everywhere inside?

Show Solution ▼

By the maximum principle, \(u\) achieves its max and min on the boundary, where it is constantly 3. Thus \(u \equiv 3\) on the entire closed disk.

🎯 Interactive Quiz

1. The PDE \(u_{xx} - 6u_{xy} + 9u_{yy} = 0\) is:

A Hyperbolic

B Parabolic

C Elliptic

D Cannot be classified

2. In D'Alembert's solution for the wave equation \(u_{tt} = c^2 u_{xx}\) with \(g(x) = 0\), the solution represents:

A Two waves traveling in opposite directions

B A standing wave

C An exponentially decaying wave

D A stationary profile

3. The characteristic equations for \(2u_x + 3u_y = u^2\) are:

A \(\frac{dx}{3} = \frac{dy}{2} = \frac{du}{u^2}\)

B \(\frac{dx}{2} = \frac{dy}{3} = \frac{du}{u^2}\)

C \(\frac{dx}{2} = \frac{dy}{3} = \frac{du}{u}\)

D \(\frac{dx}{4} = \frac{dy}{9} = \frac{du}{u^2}\)

4. The solution of the heat equation \(u_t = k\,u_{xx}\) on \([0,L]\) with homogeneous Dirichlet BC decays at a rate proportional to:

A \(1/t\)

B \(e^{-k(\pi/L)^2 t}\)

C \(1/t^2\)

D \(\cos(kt)\)

5. By the maximum principle, if \(u\) is harmonic on a connected bounded domain \(\Omega\) and achieves its maximum at an interior point, then:

A \(u\) is bounded

B \(u\) is constant

C \(u = 0\)

D Nothing can be concluded