Heat on $\mathbb R$

mark

Consider the heat problem

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

where $M>0$ is probably a large parameter and $\varepsilon>0$ is probably a small parameter.

1. Use an integral involving the heat kernel to write down an analytic expression for the solution involving the parameters $M$ and $\varepsilon$.
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) The solution to the unbounded heat problem $u_t = ku_xx$, where $u(x,0) = f(x)$ is

$$u(x,t) = \int_{-\infty}^{\infty}f(y)G(y-x,t)dy = \frac{1}{\sqrt{4\pi k}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{t}}f(y)e^{-\frac{(y-x)^2}{4kt}}dy .$$

For

$$f(x) = \begin{cases} M & |x|<\varepsilon \\ 0 & \text{else}, \end{cases},$$

$$u(x,t) = \frac{M}{\sqrt{4\pi k}}\int_{-\varepsilon}^{\varepsilon}\frac{1}{\sqrt{t}}e^{-\frac{(y-x)^2}{4kt}}dy .$$

(2)

(3) Setting $M=5$, $\varepsilon = 0.2$, and $k=1$, we find

$$u(0,0.1) \approx 1.7263957699071146.$$