Unit 6: Elliptic Boundary Value Problems: Laplace & Poisson Equations and Harmonic Functions

1. The Universal Role of Elliptic Potential Equations

Elliptic partial differential equations govern time-independent, equilibrium, and steady-state phenomena throughout physics and engineering. The premier models are:

• Laplace's Equation (Homogeneous Potential): $\nabla^2 u = 0$
• Poisson's Equation (Inhomogeneous Potential with Sources): $\nabla^2 u = f(\mathbf{x})$

Solutions $u$ of Laplace's equation in an open set $\Omega \subset \mathbb{R}^n$ are termed harmonic functions.

2. Physical Manifestations

1. Electrostatics (Maxwell's Equations)

In electrostatic equilibrium, Faraday's law $\nabla \times \mathbf{E} = \mathbf{0}$ implies the electric field is conservative, represented as the negative gradient of an electrostatic scalar potential $\Phi$:

Gauss's law in differential form relates electric flux to charge density $\rho(\mathbf{x})$:

In charge-free vacuum ($\rho = 0$), the electrostatic potential satisfies Laplace's equation $\nabla^2 \Phi = 0$.

2. Newtonian Gravitation

The gravitational field $\mathbf{g} = -\nabla V_g$ generated by a mass density distribution $\mu(\mathbf{x})$ obeys Gauss's law for gravity $\nabla \cdot \mathbf{g} = -4\pi G \mu$, leading directly to Poisson's equation:

3. Steady-State Heat Conduction

When a conducting medium reaches thermal equilibrium ($u_t = 0$), the heat equation $\rho c_p u_t = k \nabla^2 u + Q$ reduces to:

If no internal heat generation occurs ($Q = 0$), the steady temperature field satisfies $\nabla^2 u = 0$.

4. Irrotational Incompressible Fluid Flow (Potential Flow)

For an ideal fluid with zero vorticity ($\nabla \times \mathbf{v} = \mathbf{0}$), there exists a velocity potential $\phi$ such that $\mathbf{v} = \nabla \phi$. Incompressibility requires $\nabla \cdot \mathbf{v} = 0$, giving:

1. The Mean Value Property (MVP)

Theorem 6.1 (Gauss's Mean Value Property for Harmonic Functions): Let $u \in C^2(\Omega)$ be harmonic in an open domain $\Omega \subseteq \mathbb{R}^n$. For any ball $B_R(\mathbf{x}_0) = \{ \mathbf{x} : \|\mathbf{x} - \mathbf{x}_0\| < R \} \subset \subset \Omega$:

1. Spherical Mean Value Property: The value at the center is the average over the boundary sphere:

In $\mathbb{R}^2$ ($n = 2$): $u(x_0, y_0) = \frac{1}{2\pi} \int_0^{2\pi} u(x_0 + R\cos\theta, y_0 + R\sin\theta)\,d\theta$.

1. Ball Mean Value Property: The value at the center is the volume average over the solid ball:

Proof: Fix $\mathbf{x}_0 \in \Omega$ and define for $0 < r < R$:

Differentiating with respect to $r$ under the integral sign:

By Green's first identity (Divergence Theorem):

Since $u$ is harmonic, $\nabla^2 u = 0$, so $\phi'(r) = 0$ for all $r \in (0, R)$. Thus $\phi(r)$ is constant. Taking the limit as $r \to 0^+$:

Hence $\phi(R) = u(\mathbf{x}_0)$. Integrating $r^{n-1}\phi(r)$ from $0$ to $R$ immediately yields the solid ball average. $\blacksquare$

2. The Strong Maximum Principle

Theorem 6.2 (Strong Maximum Principle for Harmonic Functions): Let $\Omega \subset \mathbb{R}^n$ be a connected open domain (a region). If $u \in C^2(\Omega)$ is harmonic and attains its global maximum (or minimum) at an interior point $\mathbf{x}_0 \in \Omega$, then $u$ is identically constant throughout $\Omega$.

Proof: Let $M = \sup_\Omega u$. Suppose there exists $\mathbf{x}_0 \in \Omega$ such that $u(\mathbf{x}_0) = M$. Consider the set:

1. Since $u$ is continuous, $E = u^{-1}(\{M\})$ is closed in $\Omega$.
2. We show $E$ is open in $\Omega$:

Let $\mathbf{y} \in E$. Since $\Omega$ is open, choose $r > 0$ such that $B_r(\mathbf{y}) \subset \Omega$. By the Ball Mean Value Property:

Since $M$ is the global maximum, $M - u(\mathbf{z}) \ge 0$ everywhere. The integral of a continuous non-negative function vanishes if and only if the integrand is identically zero:

Thus $B_r(\mathbf{y}) \subseteq E$, so $E$ is open! Since $\Omega$ is connected and $E$ is non-empty, open, and closed in $\Omega$, topological connectedness forces $E = \Omega$. Therefore, $u(\mathbf{x}) \equiv M$ everywhere in $\Omega$. $\blacksquare$

3. Liouville's Theorem for Harmonic Functions

Theorem 6.3 (Liouville's Theorem): If $u: \mathbb{R}^n \to \mathbb{R}$ is harmonic on the entire space $\mathbb{R}^n$ and bounded ($|u(\mathbf{x})| \le M$ for all $\mathbf{x} \in \mathbb{R}^n$), then $u$ is constant.

1. Dirichlet Problem on a Rectangle $[0, a] \times [0, b]$

Consider steady-state temperature on a rectangular plate with three edges grounded at zero and top edge at potential $f(x)$:

Separating variables $u(x, y) = X(x) Y(y)$:

• Spatial ODE in $x$: $X'' + \lambda X = 0, X(0) = X(a) = 0 \implies \lambda_n = (n\pi/a)^2, X_n(x) = \sin\left(\frac{n\pi x}{a}\right)$.
• ODE in $y$: $Y_n''(y) - \left(\frac{n\pi}{a}\right)^2 Y_n(y) = 0, Y_n(0) = 0 \implies Y_n(y) = \sinh\left(\frac{n\pi y}{a}\right)$.

General solution:

At $y = b$:

Therefore:

2. Dirichlet Problem in Polar Coordinates for the Disk

In polar coordinates $(r, \theta)$, Laplace's equation is:

Inside a disk of radius $R$ with boundary condition $u(R, \theta) = h(\theta)$: Separating variables $u(r, \theta) = R(r) \Theta(\theta)$:

• Angular equation: $\Theta_n(\theta) = A_n \cos(n\theta) + B_n \sin(n\theta)$ (by $2\pi$-periodicity).
• Radial equation (Euler-Cauchy): $r^2 R'' + r R' - n^2 R = 0$.

Solutions are $R(r) = C r^n + D r^{-n}$. Since the solution must be bounded at the origin $r = 0$, we must set $D = 0$ for all $n \ge 0$. Thus, the general series solution inside the disk is:

where $a_n, b_n$ are the standard Fourier coefficients of the boundary function $h(\theta)$.

1. Summation of the Fourier Series into an Integral Kernel

Substitute the Fourier coefficient formulas into the disk series solution:

Exchanging summation and integration:

Using the cosine subtraction identity:

Summing the geometric series using Euler's identity: Let $\rho = r/R < 1$ and $\psi = \theta - \phi$.

Using the sum of geometric series $\sum_{n=0}^\infty z^n = \frac{1}{1 - z}$:

Restoring $\rho = r/R$:

Theorem 6.4 (Poisson Integral Formula for the Disk): For any continuous boundary profile $h(\phi)$ on $\partial B_R$, the unique harmonic function in $B_R$ is given by:

The kernel $P(r, \theta; R, \phi) = \frac{R^2 - r^2}{R^2 - 2Rr\cos(\theta - \phi) + r^2}$ is the Poisson Kernel for the disk.

2. Poisson Integral Formula for the Upper Half-Plane

For the upper half-plane $y > 0$ with boundary condition $u(x, 0) = f(x)$ on the real axis $\mathbb{R}$:

where $P_y(x - s) = \frac{y}{\pi[(x - s)^2 + y^2]}$ is the Cauchy distribution / Poisson kernel for the half-plane.

1. Dirichlet Energy Functional and Dirichlet's Principle

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain with smooth boundary $\partial \Omega$. Consider the space of functions $V = \{ w \in C^1(\bar{\Omega}) : w|_{\partial\Omega} = g \}$. Define the Dirichlet energy functional:

Theorem 6.5 (Dirichlet's Principle): A function $u \in V$ minimizes the energy functional $E[w]$ over all candidate functions in $V$ if and only if $u$ is harmonic in $\Omega$:

Proof: Let $u \in V$ be harmonic. For any other function $w \in V$, let $v = w - u$. Then $v \in C^1(\bar{\Omega})$ and $v|_{\partial\Omega} = g - g = 0$. Now evaluate $E[w] = E[u + v]$:

Applying Green's First Identity to the cross term:

1. On the boundary $\partial\Omega$, $v = 0$, so the surface integral vanishes.
2. In the interior, $\nabla^2 u = 0$ since $u$ is harmonic.

Thus the cross term vanishes identically!

Equality holds if and only if $\|\nabla v\| = 0 \implies v = \text{constant}$. Since $v = 0$ on $\partial\Omega$, $v \equiv 0 \implies w = u$. Therefore, the harmonic function $u$ is the unique global minimizer of Dirichlet energy. $\blacksquare$

2. Uniqueness of the Neumann Boundary Value Problem

For the Neumann problem:

1. Compatibility Condition: By the Divergence Theorem:

A solution exists if and only if the total boundary flux integrates to zero.

2. Uniqueness up to an Additive Constant:

If $u_1$ and $u_2$ are solutions, $w = u_1 - u_2$ satisfies $\nabla^2 w = 0$ and $\frac{\partial w}{\partial\mathbf{n}} = 0$. By Green's identity:

Thus $\nabla w = \mathbf{0} \implies w(\mathbf{x}) = C$. Solutions are unique up to an arbitrary additive constant.