Math 309B/C Winter 2012, Homework 6 Examples


There are three basic forms of the heat transfer problem that we will deal with solving. Below, I have examples of how to recognize and solve each one.
  1. Solve the following heat conduction problem: \[ \left\{\begin{array}{cc} 4u_{xx}=u_{t} & t>0, \;\; x\in [0,1] \\ u(0,t)=0, u(1,t)=1 \\ u(x,0)=\sin \pi x+x\end{array}\right.\] Note that one of the boundary conditions on $u$ is nonzero, i.e. $u(0,t)\neq 0$, so this is a nonhomogeneous problem. Hence, what we do is first subtract the line $L(x)$ that equals $0$ at $0$ and $1$ at $1$ (that is, it has the same boundary values as $u(x,t)$ in the $x$ variable) and solve the homogeneous problem \[ \left\{\begin{array}{cc} 4w_{xx}=w_{t} & t>0, \;\; x\in [0,1] \\ w(0,t)=w(1,t)=0 \\ w(x,0)=(\sin \pi x+x)-L(x)\end{array}\right.\] If we solve for what $L(x)$ must be, it is $L(x)=x$ (this is the only line such that $L(0)=0$ and $L(1)=1$, so the initial condition on $w$ is actually $w(x,0)=\sin x$. This homogeneous problem has solution \[w(x,t)=\sum_{n=1}^{\infty}b_{n}e^{-\frac{4 \pi^{2}n^{2}}{1^{2}}t}\sin n\pi x\] where \[b_{n}=\frac{2}{1}\int_{0}^{1}\sin \pi x \sin \frac{n\pi x}{1}=2\int_{0}^{1}\sin \pi x \sin n\pi x.\] Notice that $\sin\pi x\sin n\pi x$ is an even function on the interval $[-1,1]$ (since it is the product of two odd functions). Hence, this integral is actually \[b_{n}=2\int_{0}^{1}\sin \pi x \sin n\pi x=\int_{-1}^{1}\sin \pi x\sin n\pi x.\] Now, if you remember from when we first talked about Fourier series, we computed that \[\int_{-L}^{L}\sin \frac{m\pi x}{L}\sin\frac{n\pi x}{L}dx=\left\{\begin{array}{cc} L & m=n \\ 0 & m\neq n.\end{array}\right.\] Thus, $b_{n}=0$ if $n\neq 1$ and $b_{1}=1$. Thus, the solution to the homogeneous problem is \[w(x,t)=e^{-\frac{4 \pi^{2}n^{2}}{1^{2}}t}\sin n\pi x .\] Now, back to the nonhomogeneous problem. All we need to do is add the line $L(x)$ back on, so our final solution is \[u(x,t)=L(x)+w(x,t)=x+e^{-\frac{4 \pi^{2}n^{2}}{1^{2}}t}\sin n\pi x.\]
  2. Solve the following heat conduction problem: \[ \left\{\begin{array}{cc} 16u_{xx}=u_{t} & t>0, \;\; x\in [0,3] \\ u_{x}(0,t)=u_{x}(3,t)=0 \\ u(x,0)=3-x \end{array}\right.\] Because the boundary conditions are in terms of the $x$ derivative (i.e. the presence of the $u_{x}$), this is an insulation problem, which has solution \[u(x,t)=\frac{c_{0}}{2}+\sum_{n=1}^{\infty} c_{n} e^{-\frac{16\pi^{2}n^{2}}{3^{2}}}\cos \frac{n\pi x}{3}\] where \[c_{0}=\frac{2}{3}\int_{0}^{3}(3-x)dx=3\] and \begin{align*} c_{n} & =\frac{2}{3}\int_{0}^{3}(3-x)\cos \frac{n\pi x}{3} \\ & = 2\int_{0}^{3}\cos \frac{n\pi x}{3} - \frac{2}{3}\int_{0}^{3}x\cos\frac{n\pi x}{3} \\ & = \frac{6}{n\pi}\sin \frac{n\pi x}{3}|_{0}^{3} - \frac{2}{3}\left(\frac{3x}{n\pi} \sin\frac{n\pi x}{3}|_{0}^{3}-\frac{3}{n\pi}\int_{0}^{3}\sin\frac{n\pi x}{3}\right)\\ = 0-\frac{2}{3}\left(0+\left(\frac{3}{n\pi}\right)^{2}\cos\frac{n\pi x}{3}|_{0}^{3}\right) =-\frac{6}{n^{2}\pi^{2}}(\cos n\pi -1) \end{align*} and so the final solution is \[ u(x,t)= \frac{3}{2}-\sum_{n=1}^{\infty}\frac{6}{n^{2}\pi^{2}}(\cos n\pi -1)e^{-\frac{16n^{2}\pi^{2}}{3^{2}}t}\]