Introduction to Partial Differential Equations

The general second-order linear PDE in two variables is

$A u_{xx} + B u_{xy} + C u_{yy} + D u_x + E u_y + F u = G$

• Elliptic ($B^2 - 4AC \lt 0$): e.g., Laplace”s equation $u_{xx} + u_{yy} = 0$.
• Parabolic ($B^2 - 4AC = 0$): e.g., the heat equation $u_t = \alpha^2 u_{xx}$.
• Hyperbolic ($B^2 - 4AC > 0$): e.g., the wave equation $u_{tt} = c^2 u_{xx}$.

$u_t = \alpha^2 u_{xx}, \quad 0 \lt x \lt L, \quad t > 0$

With boundary conditions $u(0, t) = u(L, t) = 0$ and initial condition $u(x, 0) = f(x)$.

Consider a thin rod of length $L$ with uniform cross-section and density $\rho$. Let $u(x, t)$ be the Temperature at position $x$ and time $t$. By Fourier’s law of heat conduction, the heat flux Through a cross-section is proportional to the negative temperature gradient:

$q = -\kappa u_x$

Where $\kappa$ is the thermal conductivity. Conservation of energy on $[x, x + \Delta x]$:

$\rho c \frac{\partial u}{\partial t} \Delta x = q(x) - q(x + \Delta x) = -\kappa u_x(x) + \kappa u_x(x + \Delta x)$

Dividing by $\Delta x$ and taking $\Delta x \to 0$:

$\rho c \, u_t = \kappa u_{xx} \implies u_t = \frac{\kappa}{\rho c} u_{xx} = \alpha^2 u_{xx}$

Where $\alpha^2 = \kappa/(\rho c)$ is the thermal diffusivity.

Assume $u(x, t) = X(x)T(t)$. Substituting:

$X T' = \alpha^2 X'' T \implies \frac{T'}{\alpha^2 T} = \frac{X''}{X} = -\lambda$

This gives two ODEs:

$X'' + \lambda X = 0, \quad X(0) = X(L) = 0$ $T' + \alpha^2 \lambda T = 0$

The boundary value problem for $X$ has solutions only for $\lambda_n = (n\pi/L)^2$ $n = 1, 2, 3, \ldots$With $X_n(x) = \sin(n\pi x/L)$.

The corresponding $T_n(t) = e^{-\alpha^2 (n\pi/L)^2 t}$.

By superposition:

$u(x, t) = \sum_{n=1}^{\infty} b_n \sin\frac{n\pi x}{L} e^{-\alpha^2 (n\pi/L)^2 t}$

Where $b_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\, dx$ (the sine series coefficients of $f$).

Problem. Solve $u_t = u_{xx}$ for $0 \lt x \lt \pi$, $t > 0$With $u(0, t) = u(\pi, t) = 0$ And $u(x, 0) = \sin(2x) + 3\sin(5x)$.

Solution. Here $\alpha = 1$ and $L = \pi$. The initial condition is already a sine series.

$\lambda_n = n^2$, $X_n = \sin(nx)$, $T_n = e^{-n^2 t}$.

$u(x, t) = e^{-4t}\sin(2x) + 3e^{-25t}\sin(5x)$. $\blacksquare$

$u_{tt} = c^2 u_{xx}, \quad 0 \lt x \lt L, \quad t > 0$

With boundary conditions $u(0, t) = u(L, t) = 0$ And initial conditions $u(x, 0) = f(x)$ $u_t(x, 0) = g(x)$.

Consider a string of length $L$ under tension $T$. Let $u(x, t)$ be the vertical displacement. For A small segment $[x, x + \Delta x]$Newton’s second law in the vertical direction gives:

$\rho \Delta x \, u_{tt} = T\sin\theta(x + \Delta x) - T\sin\theta(x)$

For small displacements, $\sin\theta \approx \tan\theta = u_x$ So:

$\rho \, u_{tt} = T \frac{u_x(x + \Delta x) - u_x(x)}{\Delta x} \xrightarrow{\Delta x \to 0} T u_{xx}$

$u_{tt} = \frac{T}{\rho} u_{xx} = c^2 u_{xx}, \quad c = \sqrt{T/\rho}$

Separation of variables $u(x, t) = X(x)T(t)$ gives:

$X'' + \lambda X = 0, \quad T'' + c^2 \lambda T = 0$

With $\lambda_n = (n\pi/L)^2$:

$X_n(x) = \sin\frac{n\pi x}{L}, \quad T_n(t) = a_n \cos\frac{cn\pi t}{L} + b_n \sin\frac{cn\pi t}{L}$

$u(x, t) = \sum_{n=1}^{\infty} \sin\frac{n\pi x}{L}\left(a_n \cos\frac{cn\pi t}{L} + b_n \sin\frac{cn\pi t}{L}\right)$

Where $a_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\, dx$ and $b_n = \frac{2}{cn\pi}\int_0^L g(x)\sin\frac{n\pi x}{L}\, dx$.

For the wave equation on $-\infty \lt x \lt \infty$:

$u(x, t) = \frac{f(x + ct) + f(x - ct)}{2} + \frac{1}{2c}\int_{x - ct}^{x + ct} g(s)\, ds$

This represents the solution as a superposition of right-moving and left-moving waves.

$u_{xx} + u_{yy} = 0$

On a domain $\Omega \subseteq \mathbb{R}^2$With boundary conditions on $\partial\Omega$.

Theorem 8.1 (Maximum Principle). A harmonic function $u$ (satisfying Laplace’s equation) on a Bounded domain attains its maximum and minimum on the boundary.

Theorem 8.2 (Uniqueness). The Dirichlet problem for Laplace’s equation has at most one solution.

Proof. If $u_1$ and $u_2$ are two solutions with the same boundary data, then $v = u_1 - u_2$ is Harmonic with $v = 0$ on $\partial\Omega$. By the maximum principle, $v \equiv 0$. $\blacksquare$

Problem. A string of length $\pi$ with fixed ends is plucked: $u(x, 0) = x(\pi - x)$ $u_t(x, 0) = 0$. Find $u(x, t)$.

Solution. With $c = 1$ and $L = \pi$: $a_n = \frac{2}{\pi}\int_0^{\pi} x(\pi - x)\sin(nx)\, dx$ $b_n = 0$ (since $g = 0$).

Integrating by parts twice:

$\int_0^{\pi} x(\pi - x)\sin(nx)\, dx = \left[-\frac{x(\pi - x)\cos(nx)}{n}\right]_0^{\pi} + \frac{1}{n}\int_0^{\pi}(\pi - 2x)\cos(nx)\, dx$

$= 0 + \frac{1}{n}\left[\frac{(\pi - 2x)\sin(nx)}{n}\right]_0^{\pi} + \frac{2}{n^2}\int_0^{\pi}\sin(nx)\, dx$

$= 0 + \frac{2}{n^2}\left[-\frac{\cos(nx)}{n}\right]_0^{\pi} = \frac{2}{n^3}(1 - (-1)^n)$

For even $n$: $a_n = 0$. For odd $n = 2k + 1$: $a_n = \frac{2}{\pi} \cdot \frac{4}{n^3} = \frac{8}{\pi n^3}$.

$u(x, t) = \frac{8}{\pi}\sum_{k=0}^{\infty} \frac{\sin((2k+1)x)}{(2k+1)^3}\cos((2k+1)t)$. $\blacksquare$

Problem. Solve $u_{xx} + u_{yy} = 0$ on $0 \lt x \lt \pi$, $0 \lt y \lt 1$ with $u(0, y) = u(\pi, y) = u(x, 1) = 0$ and $u(x, 0) = f(x) = x(\pi - x)$.

A Sturm-Liouville problem consists of the ODE

$(p(x)y')' + [\lambda w(x) - q(x)]y = 0$

On $[a, b]$ with homogeneous boundary conditions, where $p, w > 0$ and $p, p', q, w$ are continuous.

Key properties:

1. The eigenvalues are real and form an infinite increasing sequence $\lambda_1 \lt \lambda_2 \lt \cdots \to \infty$.
2. Eigenfunctions corresponding to distinct eigenvalues are orthogonal with respect to the weight $w(x)$: $\int_a^b y_m(x) y_n(x) w(x)\, dx = 0$ for $m \neq n$.
3. The eigenfunctions form a complete set in the weighted $L^2$ space.

Remark. The boundary value problems encountered in the heat and wave equations ($X'' + \lambda X = 0$ with $X(0) = X(L) = 0$) are special cases of Sturm-Liouville problems With $p = 1$, $q = 0$, $w = 1$.

When the boundary specifies the derivative (heat flux) rather than the value, we have Neumann Conditions. For the heat equation:

$u_x(0, t) = 0, \quad u_x(L, t) = 0$

(insulated ends). The separation of variables gives $X'(0) = X'(L) = 0$Yielding eigenvalues $\lambda_0 = 0$ with $X_0 = 1$ And $\lambda_n = (n\pi/L)^2$ for $n \geq 1$ with $X_n = \cos(n\pi x/L)$.

The solution is

$u(x, t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\frac{n\pi x}{L} e^{-\alpha^2 (n\pi/L)^2 t}$

Where $a_n = \frac{2}{L}\int_0^L f(x)\cos\frac{n\pi x}{L}\, dx$.

Remark. As $t \to \infty$All exponential terms decay, and $u(x, t) \to a_0/2$The average Value of the initial temperature. Physically, an insulated rod reaches a uniform steady-state Temperature.

Problem. Solve $u_t = u_{xx}$ for $0 \lt x \lt \pi$, $t > 0$With $u(0, t) = u(\pi, t) = 0$ And $u(x, 0) = x(\pi - x)$.

Problem. Solve $u_{tt} = 4u_{xx}$ for $-\infty \lt x \lt \infty$ with $u(x, 0) = e^{-x^2}$ and $u_t(x, 0) = 0$.

Mistake 1: Confusing the classification of PDEs. The discriminant $B^2 - 4AC$ determines whether a PDE is elliptic, parabolic, or hyperbolic. Each type has different properties and requires different solution methods. Do not assume that all PDEs can be solved by the same method.

Mistake 2: Forgetting boundary conditions in separation of variables. Separation of variables requires both initial and boundary conditions to determine the solution uniquely. Forgetting boundary conditions leads to an incomplete solution. Always specify both types of conditions.

Mistake 3: Assuming that all PDEs have unique solutions. The existence and uniqueness of solutions depend on the type of PDE, the boundary conditions, and the regularity of the data. Do not assume that a solution exists or is unique without checking the appropriate conditions.

Mistake 4: Confusing the heat equation with the wave equation. The heat equation $u_t = \alpha^2 u_{xx}$ describes diffusion and has solutions that smooth out over time. The wave equation $u_{tt} = c^2 u_{xx}$ describes oscillations and has solutions that propagate without damping. Do not confuse the two; they model different physical phenomena.

Mistake 5: Assuming that superposition always applies. Superposition applies to linear PDEs but not to nonlinear PDEs. Do not assume that you can add solutions of a nonlinear PDE to get another solution; superposition only works for linear equations.

Partial differential equations extend ODEs to functions of multiple variables, describing phenomena that spread through space and time. While an ODE tracks a single particle’s trajectory, a PDE tracks an entire field, like temperature分布 across a metal plate. The three canonical types have distinct personalities: elliptic equations like Laplace’s describe steady states, parabolic equations like the heat equation describe diffusion toward equilibrium, and hyperbolic equations like the wave equation describe propagation without dissipation. The method of separation of variables works by decomposing complex behavior into simpler modes, like breaking a musical chord into individual notes.

• Fourier Series: Fourier series provide the eigenfunction expansions needed to solve PDEs by separation of variables.

• Second-Order Linear ODEs: The spatial ODEs arising from separation of variables are second-order linear equations with boundary conditions.

• Laplace Transforms: The Laplace transform can solve the heat equation in the time variable, converting the PDE to an ODE.

• Vector Calculus: The gradient, divergence, and curl operators appearing in PDE formulations are central concepts in vector calculus.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism