Differential Equations - Partial: Laplace's Equation - i

The next partial differential equation that we’re going to solve is the 2-D Laplace’s equation,

$${\nabla }^{2}u=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$$

A natural question to ask before we start learning how to solve this is does this equation come up naturally anywhere? The answer is a very resounding yes! If we consider the 2‑D heat equation,

$$\frac{\partial u}{\partial t}=k{\nabla }^{2}u+\frac{Q}{cp}$$

We can see that Laplace’s equation would correspond to finding the equilibrium solution (i.e. time independent solution) if there were not sources. So, this is an equation that can arise from physical situations.

How we solve Laplace’s equation will depend upon the geometry of the 2-D object we’re solving it on. Let’s start out by solving it on the rectangle given by $0\le x\le L$,$0\le y\le H$. For this geometry Laplace’s equation along with the four boundary conditions will be,

$$\begin{array}{}\begin{array}{cc}& {\nabla }^{2}u=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0\\ & u\left(0,y\right)=g_1\left(y\right)u\left(L,y\right)=g_2\left(y\right)\\ & u\left(x,0\right)=f_1\left(x\right)u\left(x,H\right)=f_2\left(x\right)\end{array}& \text{(1)}\end{array}$$

One of the important things to note here is that unlike the heat equation we will not have any initial conditions here. Both variables are spatial variables and each variable occurs in a 2^nd order derivative and so we’ll need two boundary conditions for each variable.

Next, let’s notice that while the partial differential equation is both linear and homogeneous the boundary conditions are only linear and are not homogeneous. This creates a problem because separation of variables requires homogeneous boundary conditions.

To completely solve Laplace’s equation we’re in fact going to have to solve it four times. Each time we solve it only one of the four boundary conditions can be nonhomogeneous while the remaining three will be homogeneous.

The four problems are probably best shown with a quick sketch so let’s consider the following sketch.

Now, once we solve all four of these problems the solution to our original system, $\text{(1)}$, will be,

$$u\left(x,y\right)=u_1\left(x,y\right)+u_2\left(x,y\right)+u_3\left(x,y\right)+u_4\left(x,y\right)$$

Because we know that Laplace’s equation is linear and homogeneous and each of the pieces is a solution to Laplace’s equation then the sum will also be a solution. Also, this will satisfy each of the four original boundary conditions. We’ll verify the first one and leave the rest to you to verify.

$$u\left(x,0\right)=u_1\left(x,0\right)+u_2\left(x,0\right)+u_3\left(x,0\right)+u_4\left(x,0\right)=f_1\left(x\right)+0+0+0=f_1\left(x\right)$$

In each of these cases the lone nonhomogeneous boundary condition will take the place of the initial condition in the heat equation problems that we solved a couple of sections ago. We will apply separation of variables to each problem and find a product solution that will satisfy the differential equation and the three homogeneous boundary conditions. Using the Principle of Superposition we’ll find a solution to the problem and then apply the final boundary condition to determine the value of the constant(s) that are left in the problem. The process is nearly identical in many ways to what we did when we were solving the heat equation.

We’re going to do two of the cases here and we’ll leave the remaining two for you to do.

Example 1 Find a solution to the following partial differential equation.$$\begin{array}{cc}& {\nabla }^2{u}_4=\frac{{\partial }^2{u}_4}{\partial x^2}+\frac{{\partial }^2{u}_4}{\partial y^2}=0\\ & u_4\left(0,y\right)=g_1\left(y\right)u_4\left(L,y\right)=0\\ & u_4\left(x,0\right)=0u_4\left(x,H\right)=0\end{array}$$

We’ll start by assuming that our solution will be in the form,
$$u_4\left(x,y\right)=h\left(x\right)\phi \left(y\right)$$and then recall that we performed separation of variables on this problem (with a small change in notation) back in Example 5 of the Separation of Variables section. So from that problem we know that separation of variables yields the following two ordinary differential equations that we’ll need to solve.
$$\begin{array}{cccc}& \frac{d^{2}h}{d{x}^2}-\lambda h=0& & \frac{d^2\phi }{d{y}^2}+\lambda \phi =0\\ & h\left(L\right)=0& & \phi \left(0\right)=0\phi \left(H\right)=0\end{array}$$Note that in this case, unlike the heat equation we must solve the boundary value problem first. Without knowing what $\lambda$is there is no way that we can solve the first differential equation here with only one boundary condition since the sign of $\lambda$will affect the solution.
Let’s also notice that we solved the boundary value problem in Example 1 of Solving the Heat Equation and so there is no reason to resolve it here. Taking a change of letters into account the eigenvalues and eigenfunctions for the boundary value problem here are,
$${\lambda }_n={\left(\frac{n\pi }{H}\right)}^2{\phi }_n\left(y\right)=\sin \left(\frac{n\pi y}{H}\right)n=1,2,3,\dots$$Now that we know what the eigenvalues are let’s write down the first differential equation with $\lambda$ plugged in.
$$\begin{array}{cc}\frac{d^{2}h}{d{x}^2}-{\left(\frac{n\pi }{H}\right)}^{2}h& =0\\ h\left(L\right)& =0\end{array}$$Because the coefficient of $h\left(x\right)$ in the differential equation above is positive we know that a solution to this is,
$$h\left(x\right)=c_1\cosh \left(\frac{n\pi x}{H}\right)+c_2\sinh \left(\frac{n\pi x}{H}\right)$$However, this is not really suited for dealing with the $h\left(L\right)=0$ boundary condition. So, let’s also notice that the following is also a solution.
$$h\left(x\right)=c_1\cosh \left(\frac{n\pi \left(x-L\right)}{H}\right)+c_2\sinh \left(\frac{n\pi \left(x-L\right)}{H}\right)$$You should verify this by plugging this into the differential equation and checking that it is in fact a solution. Applying the lone boundary condition to this “shifted” solution gives,
$$0=h\left(L\right)=c_1$$The solution to the first differential equation is now,
$$h\left(x\right)=c_2\sinh \left(\frac{n\pi \left(x-L\right)}{H}\right)$$and this is all the farther we can go with this because we only had a single boundary condition. That is not really a problem however because we now have enough information to form the product solution for this partial differential equation.
A product solution for this partial differential equation is,
$$u_n\left(x,y\right)=B_n\sinh \left(\frac{n\pi \left(x-L\right)}{H}\right)\sin \left(\frac{n\pi y}{H}\right)n=1,2,3,\dots$$The Principle of Superposition then tells us that a solution to the partial differential equation is,
$$u_4\left(x,y\right)=\sum _{n=1}^{\infty }{B}_n\sinh \left(\frac{n\pi \left(x-L\right)}{H}\right)\sin \left(\frac{n\pi y}{H}\right)$$and this solution will satisfy the three homogeneous boundary conditions.
To determine the constants all we need to do is apply the final boundary condition.
$$u_4\left(0,y\right)=g_1\left(y\right)=\sum _{n=1}^{\infty }{B}_n\sinh \left(\frac{n\pi \left(-L\right)}{H}\right)\sin \left(\frac{n\pi y}{H}\right)$$Now, in the previous problems we’ve done this has clearly been a Fourier series of some kind and in fact it still is. The difference here is that the coefficients of the Fourier sine series are now,
$$B_n\sinh \left(\frac{n\pi \left(-L\right)}{H}\right)$$instead of just $B_n$. We might be a little more tempted to use the orthogonality of the sines to derive formulas for the $B_n$, however we can still reuse the work that we’ve done previously to get formulas for the coefficients here.
Remember that a Fourier sine series is just a series of coefficients (depending on $n$ times a sine. We still have that here, except the “coefficients” are a little messier this time that what we saw when we first dealt with Fourier series. So, the coefficients can be found using exactly the same formula from the Fourier sine series section of a function on $0\le y\le H$ we just need to be careful with the coefficients.
$$\begin{array}{cc}{B}_n\sinh \left(\frac{n\pi \left(-L\right)}{H}\right)& =\frac{2}{H}{\int }_0^H{g}_1\left(y\right)\sin \left(\frac{n\pi y}{H}\right)dyn=1,2,3,\dots \\ B_n& =\frac{2}{H\sinh \left(\frac{n\pi \left(-L\right)}{H}\right)}{\int }_0^H{g}_1\left(y\right)\sin \left(\frac{n\pi y}{H}\right)dyn=1,2,3,\dots \end{array}$$The formulas for the $B_n$ are a little messy this time in comparison to the other problems we’ve done but they aren’t really all that messy.

Okay, let’s do one of the other problems here so we can make a couple of points.