Heat Equation with Non-Zero Temperature Boundaries — Question 7

Question 7 – Solution

Strategy. Switching off a boundary step is a delayed negative step; its instantaneous boundary effect does not erase the stored interior heat.

Step 1: Superpose the two boundary events. Linearity gives $$\boxed{u=A+B S(x,t)\quad(0<t<\tau),\qquad u=A+B[S(x,t)-S(x,t-\tau)]\quad(t>\tau).}$$ Each term solves the heat equation in its time interval. Before the switch, the left trace is $A+B$; afterward the two unit left traces cancel and leave $A$. Both right traces are always $A$. The initial interior trace is $A$.

Step 2: Identify the post-switch initial data. The stationary lines cancel after the switch, leaving $$u-A=\frac{2B}{\pi}\sum_{n\ge 1} \frac{1-e^{-n^2\pi^2\tau}}n e^{-n^2\pi^2(t-\tau)}\sin(n\pi x).$$ These are the sine coefficients of $B S(x,\tau)$ on the open interval. For every interior $x$, $S(x,t-\tau)\to 0$ as $t\downarrow\tau$, so the interior trace agrees with the pre-switch value. The left boundary itself jumps from $A+B$ to $A$; no single continuous trace at that corner can match both. For times strictly after the switch, all differentiated series converge uniformly by Gaussian decay. The switching trace is also recovered in $L^2$.

Step 3: Track the retained heat. For $t>\tau$, integration yields $$H_{\mathrm{ex}}(t)=\frac{4B}{\pi^2}\sum_{n\text{ odd}} \frac{1-e^{-n^2\pi^2\tau}}{n^2}e^{-n^2\pi^2(t-\tau)}.$$ Its derivative is $-4B\sum_{n\text{ odd}}(1-e^{-n^2\pi^2\tau}) e^{-n^2\pi^2(t-\tau)}<0$. This equals $[u_x]_0^1$ after the switch. At any later fixed starting time the exponentially damped coefficients are summable, which proves uniform convergence to the equilibrium $u=A$ as $t\to\infty$.

Step 4: Interpret the finite pulse. The profiles before the switch have left value two; those after it have left value one. The right value is one throughout. At the switch the excess heat is $B\int_0^1 S(x,\tau)\,dx>0$, as follows from the positive derivative of step-response heat and its zero initial value. It must subsequently diffuse out. A changed boundary value is not an instantaneous reset of the interior.