6.3: Laplace’s Equation in 2D (2024)

Solution

As with the heat and wave equations, we can solve this problem using the method of separation of variables. Let \(u(x, y)=X(x) Y(y)\). Then, Laplace’s equation becomes \[X^{\prime \prime} Y+X Y^{\prime \prime}=0\nonumber \] and we can separate the \(x\) and \(y\) dependent functions and introduce a separation constant, \(\lambda\), \[\frac{X^{\prime \prime}}{X}=-\frac{Y^{\prime \prime}}{Y}=-\lambda \text {. }\nonumber \] Thus, we are led to two differential equations, \[\begin{align} &X^{\prime \prime}+\lambda X=0\nonumber \\ &Y^{\prime \prime}-\lambda Y=0\label{eq:1} \end{align}\]

From the boundary condition \(u(0, y)=0, u(L, y)=0\), we have \(X(0)=\) \(0, X(L)=0 .\) So, we have the usual eigenvalue problem for \(X(x)\), \[X^{\prime \prime}+\lambda X=0, \quad X(0)=0, X(L)=0 .\nonumber \] The solutions to this problem are given by \[X_{n}(x)=\sin \frac{n \pi x}{L}, \quad \lambda_{n}=\left(\frac{n \pi}{L}\right)^{2}, \quad n=1,2, \ldots\nonumber \]

The general solution of the equation for \(Y(y)\) is given by \[Y(y)=c_{1} e^{\sqrt{\lambda} y}+c_{2} e^{-\sqrt{\lambda} y}\nonumber \] The boundary condition \(u(x, H)=0\) implies \(Y(H)=0 .\) So, we have \[c_{1} e^{\sqrt{\lambda H}}+c_{2} e^{-\sqrt{\lambda H}}=0 .\nonumber \] Thus, \[c_{2}=-c_{1} e^{2 \sqrt{\lambda} H} .\nonumber \] Inserting this result into the expression for \(Y(y)\), we have \[\begin{align} Y(y) &=c_{1} e^{\sqrt{\lambda} y}-c_{1} e^{2 \sqrt{\lambda} H} e^{-\sqrt{\lambda} y}\nonumber \\ &=c_{1} e^{\sqrt{\lambda} H}\left(e^{-\sqrt{\lambda} H} e^{\sqrt{\lambda} y}-e^{\sqrt{\lambda} H} e^{-\sqrt{\lambda} y}\right)\nonumber \\ &=c_{1} e^{\sqrt{\lambda} H}\left(e^{-\sqrt{\lambda}(H-y)}-e^{\sqrt{\lambda}(H-y)}\right)\nonumber \\ &=-2 c_{1} e^{\sqrt{\lambda} H} \sinh \sqrt{\lambda}(H-y)\label{eq:2} \end{align}\]

Since we already know the values of the eigenvalues \(\lambda_{n}\) from the eigenvalue problem for \(X(x)\), we have that the \(y\)-dependence is given by \[Y_{n}(y)=\sinh \frac{n \pi(H-y)}{L} .\nonumber \] So, the product solutions are given by \[u_{n}(x, y)=\sin \frac{n \pi x}{L} \sinh \frac{n \pi(H-y)}{L}, \quad n=1,2, \ldots\nonumber \] These solutions satisfy Laplace’s equation and the three hom*ogeneous boundary conditions and in the problem.

The remaining boundary condition, \(u(x, 0)=f(x)\), still needs to be satisfied. Inserting \(y=0\) in the product solutions does not satisfy the boundary condition unless \(f(x)\) is proportional to one of the eigenfunctions \(X_{n}(x)\). So, we first write down the general solution as a linear combination of the product solutions, \[u(x, y)=\sum_{n=1}^{\infty} a_{n} \sin \frac{n \pi x}{L} \sinh \frac{n \pi(H-y)}{L} .\label{eq:3}\]

Now we apply the boundary condition, \(u(x, 0)=f(x)\), to find that \[f(x)=\sum_{n=1}^{\infty} a_{n} \sinh \frac{n \pi H}{L} \sin \frac{n \pi x}{L} .\label{eq:4}\] Defining \(b_{n}=a_{n} \sinh \frac{n \pi H}{L}\), this becomes \[f(x)=\sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi x}{L} .\label{eq:5}\] We see that the determination of the unknown coefficients, \(b_{n}\), is simply done by recognizing that this is a Fourier sine series. The Fourier coefficients are easily found as \[b_{n}=\frac{2}{L} \int_{0}^{L} f(x) \sin \frac{n \pi x}{L} d x .\label{eq:6}\]

Since \(a_{n}=b_{n} / \sinh \frac{n \pi H}{L}\), we can finish solving the problem. The solution is \[u(x,y)=\sum\limits_{n=1}^\infty a_n\sin\frac{n\pi x}{L}\sinh\frac{n\pi (H-y)}{L},\label{eq:7}\] where \[a_{n}=\frac{2}{L \sinh \frac{n \pi H}{L}} \int_{0}^{L} f(x) \sin \frac{n \pi x}{L} d x\label{eq:8}\]

Solution

The problem with this example is that none of the boundary conditions are hom*ogeneous. This means that the corresponding eigenvalue problems will not have the hom*ogeneous boundary conditions which Sturm-Liouville theory in Section 4 needs. However, we can express this problem in terms of four different problems with nonhom*ogeneous boundary conditions on only one side of the rectangle.

In Figure \(\PageIndex{3}\) we show how the problem can be broken up into four separate problems for functions \(u_{i}(x, y), i=1, \ldots, 4\). Since the boundary conditions and Laplace’s equation are linear, the solution to the general problem is simply the sum of the solutions to these four problems, \[u(x, y)=u_{1}(x, y)+u_{2}(x, y)+u_{3}(x, y)+u_{4}(x, y) .\nonumber \] Then, this solution satisfies Laplace’s equation, \[\nabla^{2} u(x, y)=\nabla^{2} u_{1}(x, y)+\nabla^{2} u_{2}(x, y)+\nabla^{2} u_{3}(x, y)+\nabla^{2} u_{4}(x, y)=0,\nonumber \] and the boundary conditions. For example, using the boundary conditions defined in Figure \(\PageIndex{3}\), we have for \(y=0\), \[u(x, 0)=u_{1}(x, 0)+u_{2}(x, 0)+u_{3}(x, 0)+u_{4}(x, 0)=f_{1}(x) .\nonumber \] The other boundary conditions can also be shown to hold.

We can solve each of the problems in Figure \(\PageIndex{3}\) quickly based on the solution we obtained in the last example. The solution for \(u_{1}(x, y)\), which satisfies the boundary conditions \[\begin{gathered} u_{1}(0, y)=0, \quad u_{1}(L, y)=0, \quad 0<y<H, \\ u_{1}(x, 0)=f_{1}(x), \quad u_{1}(x, H)=0, \quad 0<x<L, \end{gathered}\nonumber \] is the easiest to write down. It is given by \[u_{1}(x, y)=\sum_{n=1}^{\infty} a_{n} \sin \frac{n \pi x}{L} \sinh \frac{n \pi(H-y)}{L} .\label{eq:9}\] where \[a_{n}=\frac{2}{L \sinh \frac{n \pi H}{L}} \int_{0}^{L} f_{1}(x) \sin \frac{n \pi x}{L} d x\label{eq:10}\]

For the boundary conditions \[\begin{gathered} u_{2}(0, y)=0, \quad u_{2}(L, y)=0, \quad 0<y<H, \\ u_{2}(x, 0)=0, \quad u_{2}(x, H)=f_{2}(x), \quad 0<x<L . \end{gathered}\nonumber \] the boundary conditions for \(X(x)\) are \(X(0)=0\) and \(X(L)=0\). So, we get the same form for the eigenvalues and eigenfunctions as before: \[X_{n}(x)=\sin \frac{n \pi x}{L}, \quad \lambda_{n}=\left(\frac{n \pi}{L}\right)^{2}, n=1,2, \ldots .\nonumber \]

The remaining hom*ogeneous boundary condition is now \(Y(0)=0\). Recalling that the equation satisfied by \(Y(y)\) is \[Y^{\prime \prime}-\lambda Y=0,\nonumber \] we can write the general solution as \[Y(y)=c_{1} \cosh \sqrt{\lambda} y+c_{2} \sinh \sqrt{\lambda} y .\nonumber \]

Requiring \(Y(0)=0\), we have \(c_{1}=0\), or \[Y(y)=c_{2} \sinh \sqrt{\lambda} y .\nonumber \] Then, the general solution is \[u_{2}(x, y)=\sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi x}{L} \sinh \frac{n \pi y}{L} .\label{eq:11}\]

We now force the nonhom*ogeneous boundary condition, \(u_{2}(x, H)=f_{2}(x)\), \[f_{2}(x)=\sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi x}{L} \sinh \frac{n \pi H}{L} .\label{eq:12}\] Once again we have a Fourier sine series. The Fourier coefficients are given by \[b_{n}=\frac{2}{L \sinh \frac{n \pi H}{L}} \int_{0}^{L} f_{2}(x) \sin \frac{n \pi x}{L} d x .\label{eq:13}\]

Next we turn to the problem with the boundary conditions \[\begin{gathered} u_{3}(0, y)=g_{1}(y), \quad u_{3}(L, y)=0, \quad 0<y<H, \\ u_{3}(x, 0)=0, \quad u_{3}(x, H)=0, \quad 0<x<L . \end{gathered}\nonumber \] In this case the pair of hom*ogeneous boundary conditions \(u_{3}(x, 0)=0, \quad u_{3}(x, H)=\) 0 lead to solutions \[Y_{n}(y)=\sin \frac{n \pi y}{H}, \quad \lambda_{n}=-\left(\frac{n \pi}{H}\right)^{2}, \quad n=1,2 \ldots .\nonumber \] The condition \(u_{3}(L, 0)=0\) gives \(X(x)=\sinh \frac{n \pi(L-x)}{H}\).

The general solution satisfying the hom*ogeneous conditions is \[u_{3}(x, y)=\sum_{n=1}^{\infty} c_{n} \sin \frac{n \pi y}{H} \sinh \frac{n \pi(L-x)}{H} .\label{eq:14}\] Applying the nonhom*ogeneous boundary condition, \(u_{3}(0, y)=g_{1}(y)\), we obtain the Fourier sine series \[g_{1}(y)=\sum_{n=1}^{\infty} c_{n} \sin \frac{n \pi y}{H} \sinh \frac{n \pi L}{H} .\label{eq:15}\] The Fourier coefficients are found as \[c_{n}=\frac{2}{H \sinh \frac{n \pi L}{H}} \int_{0}^{H} g_{1}(y) \sin \frac{n \pi y}{H} d y .\label{eq:16}\]

Finally, we can find the solution \[\begin{gathered} u_{4}(0, y)=0, \quad u_{4}(L, y)=g_{2}(y), \quad 0<y<H, \\ u_{4}(x, 0)=0, \quad u_{4}(x, H)=0, \quad 0<x<L . \end{gathered}\nonumber \] Following the above analysis, we find the general solution \[u_{4}(x, y)=\sum_{n=1}^{\infty} d_{n} \sin \frac{n \pi y}{H} \sinh \frac{n \pi x}{H} .\label{eq:17}\] The nonhom*ogeneous boundary condition, \(u(L, y)=g_{2}(y)\), is satisfied if \[g_{2}(y)=\sum_{n=1}^{\infty} d_{n} \sin \frac{n \pi y}{H} \sinh \frac{n \pi L}{H} .\label{eq:18}\] The Fourier coefficients, \(d_{n}\), are given by \[d_{n}=\frac{2}{H \sinh \frac{n \pi L}{H}} \int_{0}^{H} g_{1}(y) \sin \frac{n \pi y}{H} d y .\label{eq:19}\]

The solution to the general problem is given by the sum of these four solutions. \[\begin{align} u(x, y)=& \sum_{n=1}^{\infty}\left[\left(a_{n} \sinh \frac{n \pi(H-y)}{L}+b_{n} \sinh \frac{n \pi y}{L}\right) \sin \frac{n \pi x}{L}\right.\nonumber \\ &\left.+\left(c_{n} \sinh \frac{n \pi(L-x)}{H}+d_{n} \sinh \frac{n \pi x}{H}\right) \sin \frac{n \pi y}{H}\right],\label{eq:20} \end{align}\] where the coefficients are given by the above Fourier integrals.

Solution

Separation of variable proceeds as usual. Let \(u(r, \theta)=R(r) \Theta(\theta)\). Then \[\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial(R \Theta)}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2}(R \Theta)}{\partial \theta^{2}}=0,\label{eq:23}\] or \[\Theta \frac{1}{r}\left(r R^{\prime}\right)^{\prime}+\frac{1}{r^{2}} R \Theta^{\prime \prime}=0 .\label{eq:24}\] Diving by \(u(r, \theta)=R(r) \Theta(\theta)\), multiplying by \(r^{2}\), and rearranging, we have \[\frac{r}{R}\left(r R^{\prime}\right)^{\prime}=-\frac{\Theta^{\prime \prime}}{\Theta}=\lambda .\label{eq:25}\]

Since this equation gives a function of \(r\) equal to a function of \(\theta\), we set the equation equal to a constant. Thus, we have obtained two differential equations, which can be written as \[\begin{align} r\left(r R^{\prime}\right)^{\prime}-\lambda R &=0\label{eq:26} \\ \Theta^{\prime \prime}+\lambda \Theta &=0\label{eq:27} \end{align}\]

We can solve the second equation subject to the periodic boundary conditions in the \(\theta\) variable. The reader should be able to confirm that \[\Theta(\theta)=a_{n} \cos n \theta+b_{n} \sin n \theta, \quad \lambda=n^{2}, n=0,1,2, \ldots\nonumber \] is the solution. Note that the \(n=0\) case just leads to a constant solution.

Inserting \(\lambda=n^{2}\) into the radial equation, we find \[r^{2} R^{\prime \prime}+r R^{\prime}-n^{2} R=0 .\nonumber \] This is a Cauchy-Euler type of ordinary differential equation. Recall that we solve such equations by guessing a solution of the form \(R(r)=r^{m}\). This leads to the characteristic equation \(m^{2}-n^{2}=0\). Therefore, \(m=\pm n\). So, \[R(r)=c_{1} r^{n}+c_{2} r^{-n} .\nonumber \] Since we expect finite solutions at the origin, \(r=0\), we can set (. Thus, the general solution is \[u(r, \theta)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \theta+b_{n} \sin n \theta\right) r^{n} .\label{eq:28}\] Note that we have taken the constant term out of the sum and put it into a familiar form.

Now we can impose the remaining boundary condition, \(u(a, \theta)=f(\theta)\), or \[f(\theta)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \theta+b_{n} \sin n \theta\right) a^{n} .\label{eq:29}\] This is a Fourier trigonometric series. The Fourier coefficients can be determined using the results from Chapter 4 : \[\begin{align} &a_{n}=\frac{1}{\pi a^{n}} \int_{-\pi}^{\pi} f(\theta) \cos n \theta d \theta, \quad n=0,1, \ldots,\label{eq:30} \\ &b_{n}=\frac{1}{\pi a^{n}} \int_{-\pi}^{\pi} f(\theta) \sin n \theta d \theta \quad n=1,2 \ldots .\label{eq:31} \end{align}\]