Differential Equations - Partial: Separation of Variables - iii

Example 4 Use Separation of Variables on the following partial differential equation.$$\begin{array}{cc}& \frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial x^2}\\ & u\left(x,0\right)=f\left(x\right)\frac{\partial u}{\partial t}\left(x,0\right)=g\left(x\right)\\ & u\left(0,t\right)=0u\left(L,t\right)=0\end{array}$$

Now, as with the heat equation the two initial conditions are here only because they need to be here for the problem. We will not actually be doing anything with them here and as mentioned previously the product solution will rarely satisfy them. We will be dealing with those in a later section when we actually go past this first step. Again, the point of this example is only to get down to the two ordinary differential equations that separation of variables gives.
So, let’s get going on that and plug the product solution, $u\left(x,t\right)=\phi \left(x\right)h\left(t\right)$ (we switched the $G$ to an $h$ here to avoid confusion with the $g$ in the second initial condition) into the wave equation to get,
$$\begin{array}{cc}\frac{{\partial }^2}{\partial t^2}\left(\phi \left(x\right)h\left(t\right)\right)& =c^2\frac{{\partial }^2}{\partial x^2}\left(\phi \left(x\right)h\left(t\right)\right)\\ \phi \left(x\right)\frac{d^{2}h}{d{t}^2}& =c^{2}h\left(t\right)\frac{d^2\phi }{d{x}^2}\\ \frac{1}{c^{2}h}\frac{d^{2}h}{d{t}^2}& =\frac{1}{\phi }\frac{d^2\phi }{d{x}^2}\end{array}$$Note that we moved the $c^2$ to the right side for the same reason we moved the $k$ in the heat equation. It will make solving the boundary value problem a little easier.
Now that we’ve gotten the equation separated into a function of only $t$ on the left and a function of only $x$ on the right we can introduce a separation constant and again we’ll use $-\lambda$ so we can arrive at a boundary value problem that we are familiar with. So, after introducing the separation constant we get,
$$\frac{1}{c^{2}h}\frac{d^{2}h}{d{t}^2}=\frac{1}{\phi }\frac{d^2\phi }{d{x}^2}=-\lambda$$
The two ordinary differential equations we get are then,
$$\frac{d^{2}h}{d{t}^2}=-\lambda c^{2}h\frac{d^2\phi }{d{x}^2}=-\lambda \phi$$The boundary conditions in this example are identical to those from the first example and so plugging the product solution into the boundary conditions gives,
$$\phi \left(0\right)=0\phi \left(L\right)=0$$Applying separation of variables to this problem gives,
$$\begin{array}{cc}\frac{d^{2}h}{d{t}^2}=-\lambda c^{2}h& \frac{d^2\phi }{d{x}^2}=-\lambda \phi \\ & \phi \left(0\right)=0\phi \left(L\right)=0\end{array}$$

Next, let’s take a look at the 2-D Laplace’s Equation.

Example 5 Use Separation of Variables on the following partial differential equation.$$\begin{array}{cc}& \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=00\le x\le L0\le y\le H\\ & u\left(0,y\right)=g\left(y\right)u\left(L,y\right)=0\\ & u\left(x,0\right)=0u\left(x,H\right)=0\end{array}$$

This problem is a little (well actually quite a bit in some ways) different from the heat and wave equations. First, we no longer really have a time variable in the equation but instead we usually consider both variables to be spatial variables and we’ll be assuming that the two variables are in the ranges shown above in the problems statement. Note that this also means that we no longer have initial conditions, but instead we now have two sets of boundary conditions, one for $x$and one for $y$.
Also, we should point out that we have three of the boundary conditions homogeneous and one nonhomogeneous for a reason. When we get around to actually solving this Laplace’s Equation we’ll see that this is in fact required in order for us to find a solution.
For this problem we’ll use the product solution,
$$u\left(x,y\right)=h\left(x\right)\phi \left(y\right)$$It will often be convenient to have the boundary conditions in hand that this product solution gives before we take care of the differential equation. In this case we have three homogeneous boundary conditions and so we’ll need to convert all of them. Because we’ve already converted these kind of boundary conditions we’ll leave it to you to verify that these will become,
$$h\left(L\right)=0\phi \left(0\right)=0\phi \left(H\right)=0$$Plugging this into the differential equation and separating gives,
$$\begin{array}{cc}\frac{{\partial }^2}{\partial x^2}\left(h\left(x\right)\phi \left(y\right)\right)+\frac{{\partial }^2}{\partial y^2}\left(h\left(x\right)\phi \left(y\right)\right)& =0\\ \phi \left(y\right)\frac{d^{2}h}{d{x}^2}+h\left(x\right)\frac{d^2\phi }{d{y}^2}& =0\\ \frac{1}{h}\frac{d^{2}h}{d{x}^2}& =-\frac{1}{\phi }\frac{d^2\phi }{d{y}^2}\end{array}$$Okay, now we need to decide upon a separation constant. Note that every time we’ve chosen the separation constant we did so to make sure that the differential equation
$$\frac{d^2\phi }{d{y}^2}+\lambda \phi =0$$would show up. Of course, the letters might need to be different depending on how we defined our product solution (as they’ll need to be here). We know how to solve this eigenvalue/eigenfunction problem as we pointed out in the discussion after the first example. However, in order to solve it we need two boundary conditions.
So, for our problem here we can see that we’ve got two boundary conditions for $\phi \left(y\right)$ but only one for $h\left(x\right)$ and so we can see that the boundary value problem that we’ll have to solve will involve $\phi \left(y\right)$ and so we need to pick a separation constant that will give use the boundary value problem we’ve already solved. In this case that means that we need to choose $\lambda$ for the separation constant. If you’re not sure you believe that yet hold on for a second and you’ll soon see that it was in fact the correct choice here.
Putting the separation constant gives,
$$\frac{1}{h}\frac{d^{2}h}{d{x}^2}=-\frac{1}{\phi }\frac{d^2\phi }{d{y}^2}=\lambda$$The two ordinary differential equations we get from Laplace’s Equation are then,
$$\frac{d^{2}h}{d{x}^2}=\lambda h-\frac{d^2\phi }{d{y}^2}=\lambda \phi$$and notice that if we rewrite these a little we get,
$$\frac{d^{2}h}{d{x}^2}-\lambda h=0\frac{d^2\phi }{d{y}^2}+\lambda \phi =0$$We can now see that the second one does now look like one we’ve already solved (with a small change in letters of course, but that really doesn’t change things).
So, let’s summarize up here.
$$\begin{array}{cccc}\frac{d^{2}h}{d{x}^2}-\lambda h& =0& \frac{d^2\phi }{d{y}^2}+\lambda \phi & =0\\ h\left(L\right)& =0& \phi \left(0\right)& =0\phi \left(H\right)=0\end{array}$$