Differential Equations - Partial: Solving The Heat Equation - iii

Example 4 Find a solution to the following partial differential equation.$$\begin{array}{cc}& \frac{\partial u}{\partial t}=k\frac{{\partial }^{2}u}{\partial x^2}\\ & u\left(x,0\right)=f\left(x\right)\frac{\partial u}{\partial x}\left(0,t\right)=0\frac{\partial u}{\partial x}\left(L,t\right)=0\end{array}$$

We applied separation of variables to this problem in Example 2 of the previous section. So, after assuming that our solution is in the form,
$$u\left(x,t\right)=\phi \left(x\right)G\left(t\right)$$and applying separation of variables we get the following two ordinary differential equations that we need to solve.
$$\begin{array}{cc}\frac{dG}{dt}=-k\lambda G& \frac{d^2\phi }{d{x}^2}+\lambda \phi =0\\ & \frac{d\phi }{dx}\left(0\right)=0\frac{d\phi }{dx}\left(L\right)=0\end{array}$$We solved the boundary value problem in Example 2 of the Eigenvalues and Eigen functions section of the previous chapter for $L=2\pi$ so as with the first example in this section we’re not going to put a lot of explanation into the work here. If you need a reminder on how this works go back to the previous chapter and review the example we worked there. Let’s get going on the three cases we’ve got to work for this problem.
$\underset{–}{\lambda >0}$
The solution to the differential equation is,
$$\phi \left(x\right)=c_1\cos \left(\sqrt{\lambda }x\right)+c_2\sin \left(\sqrt{\lambda }x\right)$$Applying the first boundary condition gives,
$$0=\frac{d\phi }{dx}\left(0\right)=\sqrt{\lambda }{c}_2\Rightarrow c_2=0$$The second boundary condition gives,
$$0=\frac{d\phi }{dx}\left(L\right)=-\sqrt{\lambda }{c}_1\sin \left(L\sqrt{\lambda }\right)$$Recall that $\lambda >0$ and so we will only get non-trivial solutions if we require that,
$$\sin \left(L\sqrt{\lambda }\right)=0\Rightarrow L\sqrt{\lambda }=n\pi n=1,2,3,\dots$$The positive eigenvalues and their corresponding eigenfunctions of this boundary value problem are then,
$${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2{\phi }_n\left(x\right)=\cos \left(\frac{n\pi x}{L}\right)n=1,2,3,\dots$$$\underset{–}{\lambda =0}$
The general solution is,
$$\phi \left(x\right)=c_1+c_{2}x$$Applying the first boundary condition gives,
$$0=\frac{d\phi }{dx}\left(0\right)=c_2$$Using this the general solution is then,
$$\phi \left(x\right)=c_1$$and note that this will trivially satisfy the second boundary condition. Therefore $\lambda =0$ is an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is,
$$\phi \left(x\right)=1$$$\underset{–}{\lambda <0}$
The general solution here is,
$$\phi \left(x\right)=c_1\cosh \left(\sqrt{-\lambda }x\right)+c_2\sinh \left(\sqrt{-\lambda }x\right)$$Applying the first boundary condition gives,
$$0=\frac{d\phi }{dx}\left(0\right)=\sqrt{-\lambda }{c}_2\Rightarrow c_2=0$$The second boundary condition gives,
$$0=\frac{d\phi }{dx}\left(L\right)=\sqrt{-\lambda }{c}_1\sinh \left(L\sqrt{-\lambda }\right)$$We know that $L\sqrt{-\lambda }\ne 0$ and so $\sinh \left(L\sqrt{-\lambda }\right)\ne 0$. Therefore, we must have $c_1=0$ and so, this boundary value problem will have no negative eigenvalues.
So, the complete list of eigenvalues and eigenfunctions for this problem is then,
$$\begin{array}{cccc}{\lambda }_n& ={\left(\frac{n\pi }{L}\right)}^2& {\phi }_n\left(x\right)& =\cos \left(\frac{n\pi x}{L}\right)n=1,2,3,\dots \\ {\lambda }_0& =0& {\phi }_0\left(x\right)& =1\end{array}$$and notice that we get the ${\lambda }_0=0$ eigenvalue and its eigen function if we allow $n=0$ in the first set and so we’ll use the following as our set of eigenvalues and eigenfunctions.
$${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2{\phi }_n\left(x\right)=\cos \left(\frac{n\pi x}{L}\right)n=0,1,2,3,\dots$$The time problem here is identical to the first problem we looked at so,
$$G\left(t\right)=c{e}^{-k{\left(\frac{n\pi }{L}\right)}^{2}t}$$Our product solutions will then be,
$$u_n\left(x,t\right)=A_n\cos \left(\frac{n\pi x}{L}\right)e^{-k{\left(\frac{n\pi }{L}\right)}^{2}t}n=0,1,2,3,\dots$$and the solution to this partial differential equation is,
$$u\left(x,t\right)=\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)e^{-k{\left(\frac{n\pi }{L}\right)}^{2}t}$$If we apply the initial condition to this we get,
$$u\left(x,0\right)=f\left(x\right)=\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)$$and we can see that this is nothing more than the Fourier cosine series for $f\left(x\right)$on $0\le x\le L$ and so again we could use the orthogonality of the cosines to derive the coefficients or we could recall that we’ve already done that in the previous chapter and know that the coefficients are given by,
$$A_n=\{\begin{array}{cc}\frac{1}{L}{\int }_0^{L}f\left(x\right)dx& n=0\\ \frac{2}{L}{\int }_0^{L}f\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx& n\ne 0\end{array}$$

The last example that we’re going to work in this section is a little different from the first two. We are going to consider the temperature distribution in a thin circular ring. We will consider the lateral surfaces to be perfectly insulated and we are also going to assume that the ring is thin enough so that the temperature does not vary with distance from the center of the ring.

So, what does that leave us with? Let’s set $x=0$ as shown below and then let $x$ be the arc length of the ring as measured from this point.

We will measure $x$ as positive if we move to the right and negative if we move to the left of $x=0$. This means that at the top of the ring we’ll meet where $x=L$ (if we move to the right) and $x=-L$ (if we move to the left). By doing this we can consider this ring to be a bar of length 2$L$ and the heat equation that we developed earlier in this chapter will still hold.

At the point of the ring we consider the two “ends” to be in perfect thermal contact. This means that at the two ends both the temperature and the heat flux must be equal. In other words we must have,

$$u\left(-L,t\right)=u\left(L,t\right)\frac{\partial u}{\partial x}\left(-L,t\right)=\frac{\partial u}{\partial x}\left(L,t\right)$$

If you recall from the section in which we derived the heat equation we called these periodic boundary conditions. So, the problem we need to solve to get the temperature distribution in this case is,