Differential Equations - Partial: Laplace's Equation - ii

Example 2 Find a solution to the following partial differential equation.
$$\begin{array}{cc}& {\nabla }^2{u}_3=\frac{{\partial }^2{u}_3}{\partial x^2}+\frac{{\partial }^2{u}_3}{\partial y^2}=0\\ & u_3\left(0,y\right)=0u_3\left(L,y\right)=0\\ & u_3\left(x,0\right)=0u_3\left(x,H\right)=f_2\left(x\right)\end{array}$$

Okay, for the first time we’ve hit a problem where we haven’t previous done the separation of variables so let’s go through that. We’ll assume the solution is in the form,
$$u_3\left(x,y\right)=h\left(x\right)\phi \left(y\right)$$We’ll apply this to the homogeneous boundary conditions first since we’ll need those once we get reach the point of choosing the separation constant. We’ll let you verify that the boundary conditions become,
$$h\left(0\right)=0h\left(L\right)=0\phi \left(0\right)=0$$Next, we’ll plug the product solution into the differential equation.
$$\begin{array}{cc}\frac{{\partial }^2}{\partial x^2}\left(h\left(x\right)\phi \left(y\right)\right)+\frac{{\partial }^2}{\partial y^2}\left(h\left(x\right)\phi \left(y\right)\right)& =0\\ \phi \left(y\right)\frac{d^{2}h}{d{x}^2}+h\left(x\right)\frac{d^2\phi }{d{y}^2}& =0\\ \frac{1}{h}\frac{d^{2}h}{d{x}^2}& =-\frac{1}{\phi }\frac{d^2\phi }{d{y}^2}\end{array}$$Now, at this point we need to choose a separation constant. We’ve got two homogeneous boundary conditions on $h$ so let’s choose the constant so that the differential equation for $h$ yields a familiar boundary value problem so we don’t need to redo any of that work. In this case, unlike the $u_4$ case, we’ll need $-\lambda$.
This is a good problem in that is clearly illustrates that sometimes you need $\lambda$ as a separation constant and at other times you need $-\lambda$. Not only that but sometimes all it takes is a small change in the boundary conditions it force the change.
So, after adding in the separation constant we get,
$$\frac{1}{h}\frac{d^{2}h}{d{x}^2}=-\frac{1}{\phi }\frac{d^2\phi }{d{y}^2}=-\lambda$$and two ordinary differential equations that we get from this case (along with their boundary conditions) are,
$$\begin{array}{cccc}& \frac{d^{2}h}{d{x}^2}+\lambda h=0& & \frac{d^2\phi }{d{y}^2}-\lambda \phi =0\\ & h\left(0\right)=0h\left(L\right)=0& & \phi \left(0\right)=0\end{array}$$Now, as we noted above when we were deciding which separation constant to work with we’ve already solved the first boundary value problem. So, the eigenvalues and eigenfunctions for the first boundary value problem are,
$${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2{h}_n\left(x\right)=\sin \left(\frac{n\pi x}{L}\right)n=1,2,3,\dots$$The second differential equation is then,
$$\begin{array}{cc}& \frac{d^2\phi }{d{x}^2}-{\left(\frac{n\pi }{L}\right)}^2\phi =0\\ & \phi \left(0\right)=0\end{array}$$Because the coefficient of the $\phi$ is positive we know that a solution to this is,
$$\phi \left(y\right)=c_1\cosh \left(\frac{n\pi y}{L}\right)+c_2\sinh \left(\frac{n\pi y}{L}\right)$$In this case, unlike the previous example, we won’t need to use a shifted version of the solution because this will work just fine with the boundary condition we’ve got for this. So, applying the boundary condition to this gives,
$$0=\phi \left(0\right)=c_1$$and this solution becomes,
$$\phi \left(y\right)=c_2\sinh \left(\frac{n\pi y}{L}\right)$$The product solution for this case is then,
$$u_n\left(x,y\right)=B_n\sinh \left(\frac{n\pi y}{L}\right)\sin \left(\frac{n\pi x}{L}\right)n=1,2,3,\dots$$The solution to this partial differential equation is then,
$$u_3\left(x,y\right)=\sum _{n=1}^{\infty }{B}_n\sinh \left(\frac{n\pi y}{L}\right)\sin \left(\frac{n\pi x}{L}\right)$$Finally, let’s apply the nonhomogeneous boundary condition to get the coefficients for this solution.
$$u_3\left(x,H\right)=f_2\left(x\right)=\sum _{n=1}^{\infty }{B}_n\sinh \left(\frac{n\pi H}{L}\right)\sin \left(\frac{n\pi x}{L}\right)$$As we’ve come to expect this is again a Fourier sine (although it won’t always be a sine) series and so using previously done work instead of using the orthogonality of the sines to we see that,
$$\begin{array}{cc}{B}_n\sinh \left(\frac{n\pi H}{L}\right)& =\frac{2}{L}{\int }_0^L{f}_2\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dxn=1,2,3,\dots \\ B_n& =\frac{2}{L\sinh \left(\frac{n\pi H}{L}\right)}{\int }_0^L{f}_2\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dxn=1,2,3,\dots \end{array}$$

Okay, we’ve worked two of the four cases that would need to be solved in order to completely solve $\text{(1)}$. As we’ve seen each case was very similar and yet also had some differences. We saw the use of both separation constants and that sometimes we need to use a “shifted” solution in order to deal with one of the boundary conditions.

Before moving on let’s note that we used prescribed temperature boundary conditions here, but we could just have easily used prescribed flux boundary conditions or a mix of the two. No matter what kind of boundary conditions we have they will work the same.

As a final example in this section let’s take a look at solving Laplace’s equation on a disk of radius $a$ and a prescribed temperature on the boundary. Because we are now on a disk it makes sense that we should probably do this problem in polar coordinates and so the first thing we need to so do is write down Laplace’s equation in terms of polar coordinates.

Laplace’s equation in terms of polar coordinates is,

$${\nabla }^{2}u=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2}$$

Okay, this is a lot more complicated than the Cartesian form of Laplace’s equation and it will add in a few complexities to the solution process, but it isn’t as bad as it looks. The main problem that we’ve got here really is that fact that we’ve got a single boundary condition. Namely,

$$u\left(a,\theta \right)=f\left(\theta \right)$$

This specifies the temperature on the boundary of the disk. We are clearly going to need three more conditions however since we’ve got a 2^nd derivative in both $r$ and $\theta$.

When we solved Laplace’s equation on a rectangle we used conditions at the end points of the range of each variable and so it makes some sense here that we should probably need the same kind of conditions here as well. The range on our variables here are,

$$0\le r\le a-\pi \le \theta \le \pi$$

Note that the limits on $\theta$ are somewhat arbitrary here and are chosen for convenience here. Any set of limits that covers the complete disk will work, however as we’ll see with these limits we will get another familiar boundary value problem arising. The best choice here is often not known until the separation of variables is done. At that point you can go back and make your choices.

Okay, we now need conditions for $r=0$ and $\theta =\pm \pi$. First, note that Laplace’s equation in terms of polar coordinates is singular at $r=0$ (i.e. we get division by zero). However, we know from physical considerations that the temperature must remain finite everywhere in the disk and so let’s impose the condition that,

$$\left|u\left(0,\theta \right)\right|<\infty$$

This may seem like an odd condition and it definitely doesn’t conform to the other boundary conditions that we’ve seen to this point, but it will work out for us as we’ll see.

Now, for boundary conditions for $\theta$ we’ll do something similar to what we did for the 1‑D head equation on a thin ring. The two limits on $\theta$ are really just different sides of a line in the disk and so let’s use the periodic conditions there. In other words,

$$u\left(r,-\pi \right)=u\left(r,\pi \right)\frac{\partial u}{\partial r}\left(r,-\pi \right)=\frac{\partial u}{\partial r}\left(r,\pi \right)$$

With all of this out of the way let’s solve Laplace’s equation on a disk of radius $a$.