Laplace's equation in a rectangle

Branko Ćurgus

Boundary value problem

Here we consider the following boundary value problem: Let $K$ and $L$ be positive real numbers. Let $f_1$ and $f_2$ be real functions defined on $[0,K]$ and let $g_1$ and $g_2$ be real functions defined on $[0,L]$. Find the real function $u$ defined on a rectangle $\bigl\{(x,y) : 0 \leq x \leq K, 0 \leq y \leq L \bigr\}$ which satisfies the Laplace PDE $$\label{eqBVPR} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$ and the boundary conditions \begin{alignat}{2} \label{eqBVPR1} u(x,0) & = f_1(x), & \qquad u(x,L) & = f_2(x), \qquad 0 \leq x \leq K, \\ \label{eqBVPR2} u(0,y) & = g_1(y), & \qquad u(K,y) & = g_2(y), \qquad 0 \leq y \leq L. \\ \end{alignat}
Split the given problem in two problems

• Problem 1. Solve the PDE \begin{equation*} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \end{equation*} subject to the boundary conditions \begin{alignat}{2} u(x,0) & = 0, & \qquad u(x,L) & = 0, \qquad 0 \leq x \leq K, \\ u(0,y) & = g_1(y), & \qquad u(K,y) & = g_2(y), \qquad 0 \leq y \leq L. \\ \end{alignat}
• Problem 2. Solve the PDE \begin{equation*} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0. \end{equation*} subject to the boundary conditions \begin{alignat}{2} u(x,0) & = f_1(x), & \qquad u(x,L) & = f_2(x), \qquad 0 \leq x \leq K, \\ u(0,y) & = 0, & \qquad u(K,y) & = 0, \qquad 0 \leq y \leq L. \\ \end{alignat}

Solving Problem 1

• First we use separation of variables to find a "few" solutions of the homogeneous problem: $$\label{eqHBVP} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \quad u(x,0) = 0, \quad u(x,L) = 0, \quad 0 \leq x \leq K.$$
• Substituting $u(x,y) = A(x)B(y)$ in PDE yields $A^{\prime\prime}(x) B(y) + A(x) B^{\prime\prime}(y) = 0.$ Separating variables further yields $\frac{A^{\prime\prime}(x)}{A(x)} = - \frac{B^{\prime\prime}(y)}{B(y)} = \lambda$ where $\lambda$ is a constant to be determined.
• The last expression leads to two ordinary differential equations $A^{\prime\prime}(x) = \lambda A(x), \qquad - B^{\prime\prime}(y) = \lambda B(y).$ Further, the boundary conditions \begin{equation*} A(x)B(0) = 0, \qquad A(x)B(L) = 0, \qquad 0 \leq x \leq K, \end{equation*} yield the boundary conditions for $B$: $B(0) = 0, \quad B(L) = 0$.
• Thus we have to solve two ordinary differential equations problems. The first one is an eigenvalue problem for $B$ $$\label{eqEBVB} - B^{\prime\prime}(y) = \lambda B(y), \qquad B(0) = 0, \quad B(L) = 0.$$ The second one is just an ODE involving $A$: $$\label{eqODEA} A^{\prime\prime}(x) = \lambda_n A(x).$$
• We have solved the eigenvalue problem \eqref{eqEBVB} earlier. The solutions are $$\label{eqEBVBs} \lambda_n = \left( \frac{n \pi}{L} \right)^2, \qquad \sin\left( \frac{n \pi}{L} y \right), \qquad n \in {\mathbb N}.$$
• With $\lambda$-s from \eqref{eqEBVBs} we solve the ODE \eqref{eqODEA}. A popular fundamental set of solutions of \eqref{eqODEA} is $\cosh\left( \frac{n \pi}{L} x \right), \qquad \sinh\left( \frac{n \pi}{L} x \right), \qquad n \in {\mathbb N}.$ However, since the important values for $x$ are $0$ and $K$ we will choose the following fundamental set of solutions of \eqref{eqODEA}: $$\label{eqODEAs} \dfrac{\sinh\left( \dfrac{n \pi}{L} (K- x) \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)}, \qquad \frac{\sinh\left( \dfrac{n \pi}{L} x \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)}, \qquad n \in {\mathbb N}.$$ Why do we choose these solutions? We choose these solutions since they take nice values at $0$ and at $K$. The value of the first solution at $0$ is $1$ and at $K$ is $0$. The value of the second solution at $0$ is $0$ and at $K$ is $1$. There are no better solutions for our task.
• Finally we have a "few" solutions of \eqref{eqHBVP}: $$\label{eqHBVPs} \sin\left( \dfrac{n \pi}{L} y \right)\dfrac{\sinh\left( \dfrac{n \pi}{L} (K- x) \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)}, \qquad \sin\left( \frac{n \pi}{L} y \right)\, \dfrac{\sinh\left( \dfrac{n \pi}{L} x \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)}, \qquad n \in {\mathbb N}.$$
• By the superposition principle, for arbitrary $a_n$ and $b_n$ the function $$\label{eqBVPs} u_1(x,y) = \sum_{n=1}^{\infty} a_n \sin\left( \dfrac{n \pi}{L} y \right)\dfrac{\sinh\left( \dfrac{n \pi}{L} (K- x) \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)} + \sum_{n=1}^{\infty} b_n \sin\left( \frac{n \pi}{L} y \right)\, \dfrac{\sinh\left( \dfrac{n \pi}{L} x \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)}$$ is also a solution of \eqref{eqHBVP}.
• Next we choose $a_n$ and $b_n$ such that $u_1(0,y) = g_1(y)$ and $u_1(K,y) = g_2(y)$. Substituting $x=0$ and $x=K$ in \eqref{eqBVPs} we get $\sum_{n=1}^{\infty} a_n \sin\left( \dfrac{n \pi}{L} y \right) = g_1(y)$ and $\sum_{n=1}^{\infty} b_n \sin\left( \frac{n \pi}{L} y \right) = g_2(y)$
• Using the orthogonality of the sine functions in \eqref{eqEBVBs} we get $a_n = \frac{2}{L} \int_0^L g_1(y) \sin\left( \dfrac{n \pi}{L} y \right) dy, \qquad n \in {\mathbb N},$ and $b_n = \frac{2}{L} \int_0^L g_2(y) \sin\left( \dfrac{n \pi}{L} y \right) dy, \qquad n \in {\mathbb N}.$

Solving Problem 2

• Notice that Problem 1 and Problem 2 are symmetric in the sense that the roles of $x$ and $y$ are reversed.
• Using this symmetry we can immediately read that a few solutions of the homogeneous boundary value problem $$\label{eqHBVP2} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \quad u(0,y) = 0, \quad u(K,y) = 0, \quad 0 \leq y \leq L.$$ are $$\label{eqHBVP2s} \sin\left( \dfrac{n \pi}{K} x \right)\dfrac{\sinh\left( \dfrac{n \pi}{K} (L - y) \right)}{\sinh\left( \dfrac{n \pi}{K} L \right)}, \qquad \sin\left( \frac{n \pi}{K} x \right)\, \dfrac{\sinh\left( \dfrac{n \pi}{K} y \right)}{\sinh\left( \dfrac{n \pi}{K} L \right)}, \qquad n \in {\mathbb N}.$$
• By the superposition principle, for arbitrary $c_n$ and $d_n$ the function $$\label{eqBVP2s} u_2(x,y) = \sum_{n=1}^{\infty} c_n \sin\left( \dfrac{n \pi}{K} x \right)\dfrac{\sinh\left( \dfrac{n \pi}{K} (L - y) \right)}{\sinh\left( \dfrac{n \pi}{K} L \right)} + \sum_{n=1}^{\infty} d_n \sin\left( \frac{n \pi}{K} x \right)\, \dfrac{\sinh\left( \dfrac{n \pi}{K} y \right)}{\sinh\left( \dfrac{n \pi}{K} L \right)}$$ is also a solution of \eqref{eqHBVP2}.
• Next we choose $c_n$ and $d_n$ such that $u_2(x,0) = f_1(x)$ and $u_2(x,L) = f_2(x)$. Substituting $y=0$ and $y=L$ in \eqref{eqBVP2s} we get $\sum_{n=1}^{\infty} c_n \sin\left( \dfrac{n \pi}{K} x \right) = f_1(x)$ and $\sum_{n=1}^{\infty} d_n \sin\left( \frac{n \pi}{K} x \right) = f_2(x)$
• Using the orthogonality of the sine functions we get $c_n = \frac{2}{K} \int_0^K f_1(x) \sin\left( \dfrac{n \pi}{K} x \right) dx, \qquad n \in {\mathbb N},$ and $d_n = \frac{2}{K} \int_0^K f_2(x) \sin\left( \dfrac{n \pi}{K} x \right) dx, \qquad n \in {\mathbb N}.$

The solution of the boundary value problem

The solution of the given boundary value problem \eqref{eqBVPR}, \eqref{eqBVPR1}, \eqref{eqBVPR2} is the sum $u_1(x,y)+u_2(x,y)$, that is \begin{multline*} u(x,y) = \sum_{n=1}^{\infty} a_n \sin\left( \dfrac{n \pi}{L} y \right)\dfrac{\sinh\left( \dfrac{n \pi}{L} (K- x) \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)} + \sum_{n=1}^{\infty} b_n \sin\left( \frac{n \pi}{L} y \right)\, \dfrac{\sinh\left( \dfrac{n \pi}{L} x \right)}{\sinh\left( \dfrac{n \pi}{L} K \right)} \\ + \sum_{n=1}^{\infty} c_n \sin\left( \dfrac{n \pi}{K} x \right)\dfrac{\sinh\left( \dfrac{n \pi}{K} (L - y) \right)}{\sinh\left( \dfrac{n \pi}{K} L \right)} + \sum_{n=1}^{\infty} d_n \sin\left( \frac{n \pi}{K} x \right)\, \dfrac{\sinh\left( \dfrac{n \pi}{K} y \right)}{\sinh\left( \dfrac{n \pi}{K} L \right)} \end{multline*} with the coefficients $a_n, b_n, c_n, d_n$ calculated by \begin{align*} a_n & = \frac{2}{L} \int_0^L g_1(y) \sin\left( \dfrac{n \pi}{L} y \right) dy, \qquad n \in {\mathbb N}, \\ b_n & = \frac{2}{L} \int_0^L g_2(y) \sin\left( \dfrac{n \pi}{L} y \right) dy, \qquad n \in {\mathbb N}, \\ c_n & = \frac{2}{K} \int_0^K f_1(x) \sin\left( \dfrac{n \pi}{K} x \right) dx, \qquad n \in {\mathbb N}, \\ d_n & = \frac{2}{K} \int_0^K f_2(x) \sin\left( \dfrac{n \pi}{K} x \right) dx, \qquad n \in {\mathbb N}. \end{align*}
Mathematica implementation of the solution

The above solution is implemented in this Mathematica notebook.