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Partial Differential Equations

From classification of PDEs to the classical equations of mathematical physics — wave, heat, and Laplace — with Fourier methods and Green's functions.

Classification Method of Characteristics Wave Equation Heat Equation Laplace Equation Fourier Analysis

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Classification of Second Order PDEs

Classifying linear second-order PDEs into elliptic, parabolic, and hyperbolic types — the starting point for choosing solution strategies.

📐 The General Second Order PDE

The general linear second-order PDE in two variables is: \[Au_{xx} + 2Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G\] where \(A, B, C\) may depend on \((x,y)\).

Definition — Classification by Discriminant The PDE is classified by the discriminant \(\Delta = B^2 - AC\):

\(\Delta > 0\): Hyperbolic (e.g., wave equation)
\(\Delta = 0\): Parabolic (e.g., heat equation)
\(\Delta < 0\): Elliptic (e.g., Laplace equation)

Theorem — Canonical Forms By an appropriate change of variables, every second-order PDE can be reduced to its canonical form:

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

Example

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

Here \(A=1, B=1, C=-3\). Discriminant: \(\Delta = 1 - (1)(-3) = 4 > 0\). The PDE is hyperbolic. Characteristic equations: \(\frac{dy}{dx} = \frac{B \pm \sqrt{\Delta}}{A} = \frac{1 \pm 2}{1}\), giving characteristics \(y = 3x + c_1\) and \(y = -x + c_2\).

First Order PDEs & Method of Characteristics

Solving first-order PDEs using characteristic curves and the Cauchy problem.

📈 Method of Characteristics

For the quasi-linear first-order PDE \(a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)\), the characteristic equations are:

\[\frac{dx}{a} = \frac{dy}{b} = \frac{du}{c}\]

Theorem — Cauchy Problem for First Order PDE Given initial data \(u = u_0(s)\) on a curve \(\Gamma: (x_0(s), y_0(s))\), the Cauchy problem has a unique local solution provided the transversality condition holds: \[\begin{vmatrix} a & b \\ x_0'(s) & y_0'(s) \end{vmatrix} \neq 0\] on \(\Gamma\). This ensures characteristics are not tangent to the initial curve.

Example

Solve \(u_x + 2u_y = 0\) with \(u(x,0) = \sin x\).

Characteristics: \(\frac{dx}{1} = \frac{dy}{2} = \frac{du}{0}\). From \(du = 0\): \(u\) is constant along characteristics. From \(dx/1 = dy/2\): \(y = 2x + c\), so \(c = y - 2x\) is constant. Thus \(u = f(y-2x)\). Applying initial condition: \(u(x,0) = f(-2x) = \sin x\), so \(f(s) = \sin(-s/2)\). Solution: \(u(x,y) = \sin\!\left(\frac{2x-y}{2}\right)\).

The Wave Equation

D'Alembert's solution, energy methods, and the propagation of waves.

🌊 D'Alembert's Formula & Energy

Theorem — D'Alembert's Formula The solution of the one-dimensional wave equation \(u_{tt} = c^2 u_{xx}\) on \(\mathbb{R}\) with initial conditions \(u(x,0) = f(x)\), \(u_t(x,0) = g(x)\) is: \[u(x,t) = \frac{f(x+ct) + f(x-ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds\]

The solution consists of two traveling waves moving in opposite directions at speed \(c\). The domain of dependence of a point \((x_0,t_0)\) is the interval \([x_0 - ct_0,\; x_0 + ct_0]\).

Theorem — Energy Conservation The total energy of the wave equation is conserved: \[E(t) = \frac{1}{2}\int_{\mathbb{R}}\!\left(u_t^2 + c^2 u_x^2\right)dx = E(0)\] This implies uniqueness of solutions to the Cauchy problem.

Example

Solve \(u_{tt} = 4u_{xx}\), \(u(x,0) = e^{-x^2}\), \(u_t(x,0) = 0\).

Here \(c = 2\), \(f(x) = e^{-x^2}\), \(g(x) = 0\). By D'Alembert: \[u(x,t) = \frac{e^{-(x+2t)^2} + e^{-(x-2t)^2}}{2}\] Two Gaussian pulses propagate in opposite directions at speed 2.

The Heat Equation

Maximum principle, fundamental solution, and uniqueness for the diffusion equation.

🔥 Maximum Principle & Fundamental Solution

Theorem — Maximum Principle (Heat Equation) Let \(u\) satisfy \(u_t = k u_{xx}\) in a bounded domain \(\Omega_T = (0,L)\times(0,T]\). Then the maximum (and minimum) of \(u\) is attained on the parabolic boundary \(\Gamma_T = \{t=0\} \cup \{x=0\} \cup \{x=L\}\).

Definition — Fundamental Solution (Heat Kernel) The fundamental solution of the heat equation \(u_t = ku_{xx}\) on \(\mathbb{R}\) is: \[\Phi(x,t) = \frac{1}{\sqrt{4\pi kt}}\exp\!\left(-\frac{x^2}{4kt}\right), \quad t > 0\] The solution for initial data \(u(x,0) = f(x)\) is \(u(x,t) = \Phi(\cdot,t) * f = \int_{\mathbb{R}}\Phi(x-y,t)f(y)\,dy\).

Key properties of the heat equation:

Infinite speed of propagation: a disturbance at any point is felt everywhere instantly (for \(t>0\)).
Smoothing: solutions become infinitely differentiable for \(t > 0\), regardless of initial data regularity.
Irreversibility: the heat equation is ill-posed backward in time.

Example

Solve \(u_t = u_{xx}\) on \((0,\pi)\) with \(u(0,t)=u(\pi,t)=0\) and \(u(x,0)=\sin x + 3\sin 2x\).

By separation of variables, \(u(x,t) = \sum_{n=1}^{\infty} b_n \sin(nx)e^{-n^2 t}\). Matching initial data: \(b_1=1, b_2=3\), all others zero. \[u(x,t) = \sin(x)e^{-t} + 3\sin(2x)e^{-4t}\] Higher modes decay faster, demonstrating the smoothing property.

Laplace Equation & Harmonic Functions

Properties of harmonic functions, the mean value property, and Green's functions.

🔮 Harmonic Functions & Green's Functions

Definition — Harmonic Function A \(C^2\) function \(u\) is harmonic in a domain \(\Omega\) if \(\Delta u = u_{xx} + u_{yy} = 0\) in \(\Omega\).

Theorem — Mean Value Property If \(u\) is harmonic in \(\Omega\) and \(\overline{B_r(x_0)} \subset \Omega\), then: \[u(x_0) = \frac{1}{|\partial B_r|}\oint_{\partial B_r(x_0)} u\,dS = \frac{1}{|B_r|}\int_{B_r(x_0)} u\,dV\] Conversely, if a continuous function satisfies the mean value property for all balls, it is harmonic.

Theorem — Maximum Principle (Laplace) A non-constant harmonic function in a bounded domain \(\Omega\) attains its maximum and minimum only on the boundary \(\partial\Omega\). This immediately yields uniqueness for the Dirichlet problem.

Definition — Green's Function The Green's function \(G(x,y)\) for the Laplacian on \(\Omega\) satisfies \(-\Delta_y G(x,y) = \delta(x-y)\) in \(\Omega\) with \(G = 0\) on \(\partial\Omega\). The solution to \(-\Delta u = f\) in \(\Omega\), \(u=g\) on \(\partial\Omega\) is: \[u(x) = \int_\Omega G(x,y)f(y)\,dy - \oint_{\partial\Omega} g(y)\frac{\partial G}{\partial \nu_y}\,dS_y\]

Example

Find the Green's function for the Laplacian on the upper half-plane \(\{y > 0\}\) in \(\mathbb{R}^2\).

Use the method of images. For source at \(x_0 = (a,b)\) with \(b>0\), the image point is \(x_0^* = (a,-b)\). \[G(x,x_0) = \frac{1}{2\pi}\ln|x - x_0^*| - \frac{1}{2\pi}\ln|x - x_0| = \frac{1}{4\pi}\ln\frac{(x_1-a)^2+(x_2+b)^2}{(x_1-a)^2+(x_2-b)^2}\] This vanishes on \(\{x_2 = 0\}\) as required.

Fourier Analysis & Transforms in PDE

Fourier series and transforms as fundamental tools for solving PDEs on various domains.

📊 Fourier Series & Fourier Transform

Definition — Fourier Transform The Fourier transform of \(f \in L^1(\mathbb{R})\) is: \[\hat{f}(\xi) = \int_{\mathbb{R}} f(x)e^{-2\pi i\xi x}\,dx\] with inverse \(f(x) = \int_{\mathbb{R}}\hat{f}(\xi)e^{2\pi i\xi x}\,d\xi\).

Theorem — Fourier Transform and PDEs The Fourier transform converts differentiation into multiplication: \(\widehat{f'}(\xi) = 2\pi i\xi\,\hat{f}(\xi)\). For the heat equation \(u_t = ku_{xx}\) on \(\mathbb{R}\), taking the Fourier transform in \(x\) gives \(\hat{u}_t = -4\pi^2k\xi^2\hat{u}\), which has solution \(\hat{u}(\xi,t) = \hat{f}(\xi)e^{-4\pi^2 k\xi^2 t}\).

Theorem — Parseval's Identity For \(f \in L^2(\mathbb{R})\): \(\|f\|_{L^2}^2 = \|\hat{f}\|_{L^2}^2\). The Fourier transform is an isometry on \(L^2\).

Example

Use Fourier series to solve \(u_{tt} = c^2u_{xx}\) on \((0,L)\) with \(u(0,t)=u(L,t)=0\), \(u(x,0)=f(x)\), \(u_t(x,0)=0\).

Separation gives \(u(x,t)=\sum_{n=1}^{\infty}b_n\sin\!\left(\frac{n\pi x}{L}\right)\cos\!\left(\frac{n\pi ct}{L}\right)\) where \(b_n = \frac{2}{L}\int_0^L f(x)\sin\!\left(\frac{n\pi x}{L}\right)dx\). Each term represents a standing wave (normal mode) with frequency \(\omega_n = n\pi c/L\).

Key Takeaways

Classification (elliptic/parabolic/hyperbolic) determines the nature of solutions and appropriate boundary conditions.
D'Alembert's formula gives an explicit solution for the wave equation; energy conservation ensures uniqueness.
The heat equation has a maximum principle, infinite propagation speed, and smoothing effect.
Harmonic functions satisfy the mean value property and maximum principle — cornerstones of elliptic theory.
The Fourier transform converts PDEs into algebraic or ODE problems in the frequency domain.
Green's functions encode the geometry of the domain and reduce PDE solving to integration.

Practice Problems

Problem 1

Classify the PDE \(x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = 0\). Does the type change with \((x,y)\)?

Show Solution ▼

\(A=x^2, B=xy, C=y^2\). Discriminant: \(\Delta = (xy)^2 - x^2 y^2 = 0\). The PDE is parabolic everywhere. The type does not change.

Problem 2

Solve \(xu_x + yu_y = u\) with \(u(x,1) = x^2\) using the method of characteristics.

Show Solution ▼

Characteristics: \(\frac{dx}{x} = \frac{dy}{y} = \frac{du}{u}\). From the first two: \(y/x = c_1\). From first and third: \(u/x = c_2\). So \(u = xF(y/x)\). From \(u(x,1) = x^2\): \(xF(1/x) = x^2\), so \(F(s) = 1/s\). Thus \(u = x \cdot \frac{x}{y} = \frac{x^2}{y}\). Alternatively, \(u(x,y)=x^2/y\).

Problem 3

Using D'Alembert's formula, find \(u(2,1)\) for \(u_{tt} = u_{xx}\), \(u(x,0) = \cos x\), \(u_t(x,0) = \sin x\).

Show Solution ▼

\(c=1\). \(u(2,1) = \frac{\cos 3 + \cos 1}{2} + \frac{1}{2}\int_1^3 \sin s\,ds = \frac{\cos 3 + \cos 1}{2} + \frac{-\cos 3 + \cos 1}{2} = \cos 1 \approx 0.5403\).

Problem 4

Show that if \(u\) is harmonic and bounded in all of \(\mathbb{R}^2\), then \(u\) is constant. (Liouville's theorem for harmonic functions.)

Show Solution ▼

By the mean value property, for any two points \(x_1, x_2\) and large radius \(R\): \(|u(x_1)-u(x_2)| \le \frac{C}{R^2}\cdot |B_R|\cdot\frac{|x_1-x_2|}{R}\|u\|_\infty \to 0\) as \(R\to\infty\). Alternatively, apply the maximum principle on balls of increasing radius: \(\sup u = \inf u\) in the limit, so \(u\) is constant.

Problem 5

Find the Fourier transform of the heat kernel \(\Phi(x,t) = \frac{1}{\sqrt{4\pi kt}}e^{-x^2/(4kt)}\).

Show Solution ▼

\(\hat{\Phi}(\xi,t) = e^{-4\pi^2 k\xi^2 t}\). This is a Gaussian in frequency space — the wider the Gaussian in space (larger \(t\)), the narrower in frequency, reflecting the smoothing property.

Self-Assessment Quiz

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

A Elliptic

B Parabolic

C Hyperbolic

D None of the above

2. In D'Alembert's formula, the domain of dependence of \((x_0, t_0)\) is:

A All of \(\mathbb{R}\)

B \([x_0 - ct_0,\; x_0 + ct_0]\)

C The single point \(x_0\)

D \((-\infty, x_0-ct_0] \cup [x_0+ct_0, \infty)\)

3. The heat equation has which distinctive property?

A Finite propagation speed

B Infinite propagation speed and smoothing

C Time reversibility

D Energy conservation

4. The mean value property is a characterization of:

A Harmonic functions

B Solutions of the heat equation

C Analytic functions only

D Subharmonic functions

5. The Fourier transform converts \(\frac{\partial^2 u}{\partial x^2}\) into:

A \(2\pi i\xi\,\hat{u}\)

B \(-4\pi^2\xi^2\,\hat{u}\)

C \(\frac{d^2\hat{u}}{d\xi^2}\)

D \(\xi^2\hat{u}\)