Heat on $[-1,1]$.

mark

Consider the heat problem

$$u_t = k u_{xx}, \: u(0,t) = 0, \: u(2,0) = 0, \: u(x,0) = \begin{cases} M & |x-1|<\varepsilon \\ 0 & \text{else}, \end{cases}$$

where $M>0$ is probably a large parameter, $\varepsilon>0$ is probably a small parameter, and $k>0$ is rather plain.

1. Use a Fourier series to find an analytic expression for the solution involving the parameters $M$, $\varepsilon$, and $k$.
2. Pick a kinda large number for $M$ and a kinda small number of $\varepsilon$. Then sketch reasonable plots for $u(x,t)$ for $t=0$, $t=0.1$, $t=1$, and $t=10$.
3. Using your choices of $M$ and $\varepsilon$ together with a numerical integrator, compute an estimate to $u(0,0.1)$ that's probably valid to 3 digits past the decimal point.

AbS

(1) As we have shown in numerous previous problems, separation of variables leads to the solution $u(x,t) = X(x)T(t),$ where $$T(t)=e^{-k\lambda t}$$ and $$X(x)=a\cos\left(\sqrt{\lambda}x\right) + b\sin\left(\sqrt{\lambda}x\right).$$ From our boundary conditions, we know $X(0) = a = 0,$ so $$X(x)=b\sin\left(\sqrt{\lambda}x\right).$$

We also know $X(2)=b\sin\left(\sqrt{\lambda}x\right) = 0,$ so $$\sqrt{\lambda}) = \frac{n\pi}{2}.$$

Thus, our full solution is $$u(x,t)=\sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2 }{4}kt}$$\sin\left(\frac{n\pi x}{2}\right).$$

In order to satisfy our initial conditions, we must find values of $b_n$ such that $$b_n = \int_0^2f(x)\sin\left(\frac{n\pi x}{2}\right)dx,$$ where $u(x,0)=f(x).$ For $$f(x) = \begin{cases} M & |x-1|<\varepsilon \\ 0 & \text{else}, \end{cases}$$

assuming $1-\varepsilon>0$ and $1+\varepsilon<2,$

$$b_n = M\int_{1-\varepsilon}^{1+\varepsilon}\sin\left(\frac{n\pi x}{2}\right)dx = \frac{2M}{n\pi}\left[-\cos\left(\frac{n\pi x}{2}\right)\right]_{1-\varepsilon}^{1+\varepsilon} = \frac{2M}{n\pi}\left(\cos\left(\frac{(1-\varepsilon)n\pi }{2}\right) - \cos\left(\frac{(1+\varepsilon)n\pi }{2}\right)\right).$$

Hence,

$$u(x,t) = \frac{2M}{\pi}\sum_{n=1}^{\infty}\frac{\left(\cos\left(\frac{(1-\varepsilon)n\pi }{2}\right) - \cos\left(\frac{(1+\varepsilon)n\pi }{2}\right)\right)e^{-\frac{n^2\pi^2 }{4}kt}sin\left(\frac{n\pi x}{2}\right)}{n}$$

(2)

(3) We know from our boundary conditions that $u(0,t)=0,$ so $$u(0,0.1)=0.$$ Using the method of lines yields the expected $u(0,0.1) = 0.$