The Heat Equation — Question 8

Question 8 – Solution

Strategy. A small time-dependent perturbation makes the maximum argument strict, while a growing test field exposes a missing source.

Step 1: Prove the upper bound. For $\varepsilon>0$, $w$ is nonpositive initially and at the spatial endpoints. If it has a positive maximum on the space-time rectangle, that maximum occurs at an interior spatial point and a time $t_*>0$. There $w_{xx}\leq 0$ and $w_t\geq 0$: at the latest time, use the derivative from earlier times; at an interior time it is zero. Yet $$w_t-\alpha w_{xx}=-\varepsilon<0,$$ a contradiction. One can first apply the argument on any shorter time rectangle and pass to $T$ by continuity if needed. Thus $u\leq M+\varepsilon t$, and letting $\varepsilon\downarrow 0$ gives $u\leq M$.

Step 2: Prove the lower bound. Apply the same argument to $-u$ with upper bound zero to obtain $-u\leq 0$. Hence $\boxed{0\leq u\leq M}$ throughout the rectangle. The theorem does not say $u_t\leq 0$ at every point. At a local temperature minimum with positive curvature, the equation gives $u_t=\alpha u_{xx}>0$. Redistribution can warm a cooler location without exceeding the allowed global range.

Step 3: Diagnose the growing sine. Write $\mu=(\pi/L)^2$. Then $v_t=\alpha\mu v$ and $v_{xx}=-\mu v$, so $$\boxed{v_t-\alpha v_{xx}=2\alpha\mu v\ne 0}$$ in the spatial interior. It satisfies zero endpoint values and the allowed initial profile, but for $t>0$ its midpoint exceeds $M$. This violates the conclusion because it fails the source-free PDE hypothesis.

Step 4: Restore the missing physics or correct the time factor. The required source is $s=C(v_t-\alpha v_{xx})=2k\mu v>0$ in the interior. With that source, the earlier source-free maximum argument no longer applies. Alternatively, $$\boxed{\widetilde v(x,t)=M e^{-\alpha\mu t}\sin(\pi x/L)}$$ has $\widetilde v_t=-\alpha\mu\widetilde v=\alpha\widetilde v_{xx}$, the same initial profile and zero endpoints. It remains between zero and $M$.