Laplace and Poisson Equations

Definition Let $\Omega \subseteq \mathbb{R}^2$ be an open set and let $u(x, y)$ have continuous second partial derivatives in $\Omega$, with $u_{xx} = \partial^2 u/\partial x^2$ and $u_{yy} = \partial^2 u/\partial y^2$. The Laplacian of $u$ is

and Laplace’s equation is $\Delta u = 0$ in $\Omega$. A function with continuous second partial derivatives that satisfies Laplace’s equation in $\Omega$ is called harmonic in $\Omega$. In three variables the Laplacian is $u_{xx} + u_{yy} + u_{zz}$, and the maximum principle, uniqueness and compatibility results below carry over, with volume and surface integrals in place of area and boundary integrals. The equation is linear and homogeneous, so sums and constant multiples of harmonic functions are harmonic. Its principal part $u_{xx} + u_{yy}$ makes it elliptic, and since it contains no time variable it comes with boundary conditions only, never with initial conditions.

Definition Let $\Omega \subseteq \mathbb{R}^2$ be open and let $f$ be a given function on $\Omega$, called the source term. Poisson’s equation is

and Laplace’s equation is the case $f = 0$. This page uses the plus sign convention written here. Many texts write $-\Delta u = f$ instead, and then the sign of every source is reversed. A boundary-value problem adds conditions on the boundary $\partial\Omega$. The Dirichlet problem prescribes the values $u = g$ on $\partial\Omega$. The Neumann problem prescribes the outward normal derivative

where $\nu$ is the outward unit normal. In a heat model, $\partial u/\partial \nu$ is proportional to the rate at which heat flows in across the boundary, and the steady state of $u_t = \kappa\Delta u + q$ with an internal heating rate $q$ satisfies $\Delta u = -q/\kappa$. So in this convention a positive heat supply gives a negative $f$.

Definition On the rectangle $0 < x < \pi$, $0 < y < 1$, consider the Dirichlet problem

Substituting $u = X(x)Y(y)$, with $X' = dX/dx$ and $Y' = dY/dy$, and dividing by $XY$ gives $X''/X = -Y''/Y$. The left side depends only on $x$ and the right side only on $y$, so both equal a constant $-\lambda$. The zero side conditions give the Dirichlet eigenvalue problem $X'' + \lambda X = 0$ with $X(0) = X(\pi) = 0$, so $\lambda = n^2$ and $X = \sin nx$ for $n = 1, 2, 3, \ldots$ The vertical factor then solves $Y'' - n^2 Y = 0$ with $Y(0) = 0$, so it is a multiple of $\sinh ny$. Superposition gives

where the coefficients are the Fourier sine coefficients of the top data,

chosen so that the series equals $g$ at $y = 1$. If $g$ is continuous and piecewise smooth with $g(0) = g(\pi) = 0$, so that the top data meet the zero sides continuously at the corners, the series converges uniformly on the closed rectangle, it can be differentiated term by term inside, and its sum is the solution. When several sides carry nonzero data, solve one such problem for each side and add the results.

Definition In polar coordinates $x = r\cos\theta$, $y = r\sin\theta$, the Laplacian at points with $r > 0$ is

Consider the Dirichlet problem on the unit disk, $\Delta u = 0$ for $r < 1$ with $u(1, \theta) = g(\theta)$. Substituting a separated solution $u = R(r)\Theta(\theta)$, with $R' = dR/dr$ and $\Theta' = d\Theta/d\theta$, and multiplying by $r^2/(R\Theta)$ splits the equation as $(r^2R'' + rR')/R = -\Theta''/\Theta = \lambda$ for a constant $\lambda$. The angular factor satisfies $\Theta'' + \lambda\Theta = 0$ and must be $2\pi$-periodic, which gives the periodic eigenvalues $\lambda = n^2$, with $\Theta = 1$ for $n = 0$ and $\Theta = \cos n\theta$ or $\sin n\theta$ for $n = 1, 2, 3, \ldots$ The radial factor then solves the Cauchy-Euler equation

The trial $R = r^m$ gives $m^2 = n^2$. For $n \ge 1$ the roots $m = \pm n$ give the solutions $r^n$ and $r^{-n}$ on $r > 0$. For $n = 0$ the root $m = 0$ is repeated, and the solutions are $1$ and $\ln r$. The center is not part of the boundary, but a solution must be smooth there, and this rules out $r^{-n}$ and $\ln r$, which are unbounded near $r = 0$. What remains is

where $a_n$ and $b_n$ are the Fourier coefficients of $g$ on $[-\pi, \pi]$,

Each term $r^n\cos n\theta$ or $r^n\sin n\theta$ is a polynomial in $x$ and $y$, so nothing goes wrong at the center. If $g$ is continuous, piecewise smooth and $2\pi$-periodic, the series converges uniformly on the closed disk and its sum is the solution.

Definition Let $\Omega \subseteq \mathbb{R}^2$ be a bounded open set with boundary $\partial\Omega$ and closure $\overline{\Omega} = \Omega \cup \partial\Omega$, and let $u$ be continuous on $\overline{\Omega}$ and harmonic in $\Omega$. The maximum principle states that

In its strong form, if $\Omega$ is also connected and $u$ takes its maximum or its minimum at a point of $\Omega$, then $u$ is constant. Behind it is the mean value property. If the closed disk of radius $\rho$ about a point lies in $\Omega$, the value of $u$ at that point is the average of $u$ over the circle of radius $\rho$ around it. Uniqueness for the Dirichlet problem follows. For given $f$ and $g$, the problem $\Delta u = f$ in $\Omega$ with $u = g$ on $\partial\Omega$ has at most one solution that is twice continuously differentiable in $\Omega$ and continuous on $\overline{\Omega}$. If $u_1$ and $u_2$ are two such solutions, then $w = u_1 - u_2$ is harmonic with $w = 0$ on $\partial\Omega$, so the principle gives $0 \le w \le 0$. The same argument gives continuous dependence on the data. Two solutions with the same $f$ and boundary values $g_1$ and $g_2$ satisfy

throughout $\Omega$. Uniqueness says nothing about existence, which needs further conditions on $\Omega$ and on the data.

Definition Let $\Omega \subseteq \mathbb{R}^2$ be a bounded, connected open set with a smooth boundary and outward unit normal $\nu$. The pure Neumann problem prescribes the normal derivative on all of $\partial\Omega$,

If $u$ is a solution with continuous second derivatives up to the boundary, the divergence theorem applied to $\nabla u$ gives the Neumann compatibility condition

Without it there is no solution. For smooth data that satisfy it there is a solution, and it is unique up to a constant. Every constant $C$ has $\Delta C = 0$ and $\partial C/\partial \nu = 0$, so $u + C$ solves the same problem. Conversely, the difference $w$ of two solutions has $\Delta w = 0$ and $\partial w/\partial \nu = 0$, and Green’s identity gives

so $w$ is constant on the connected set $\Omega$. On a set made of several separate pieces, a different constant could be added on each piece. An extra normalization, such as zero mean $\int_\Omega u\,dA = 0$ or a prescribed value at one point, selects a single solution. Dirichlet data leave no such freedom, because they fix the values on the boundary.