Heat on $\mathbb R$
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.
- Use an integral involving the heat kernel to write down an analytic expression for the solution involving the parameters $M$ and $\varepsilon$.
- 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 $t=10$.
- 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.
Comments
(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
$$u(0,0.1) \approx 1.7263957699071146.$$