Solving the Heat Equation — Question 1

Question 1 – Solution

Strategy. Different spatial modes decay at different rates; local temperature and total heat need not have the same monotonicity.

Step 1: Evolve each initial mode. Since $(\sin nx)''=-n^2\sin nx$, the solution is $$\boxed{u=e^{-t}\sin x+\tfrac 12e^{-4t}\sin 2x.}$$ Termwise differentiation of this finite sum gives $u_t=u_{xx}$. Both sines vanish at the endpoints, and setting $t=0$ gives the prescribed field. The formula is smooth on the closed space interval for all $t\ge 0$.

Step 2: Prove nonnegativity without a plot. Factor the solution as $u=e^{-t}\sin x[1+e^{-3t}\cos x]$. For $0\le x\le\pi$ and $t\ge 0$, $\sin x\ge 0$ and the bracket is nonnegative. At positive time the bracket is strictly positive; hence the interior temperature is positive. This sign statement concerns values, not their time derivatives. Diffusion can transport heat into a cooler interior region.

Step 3: Find the local warming interval. At $x=5\pi/6$, $$u=\tfrac 12e^{-t}-\tfrac{\sqrt 3}{4}e^{-4t},\qquad u_t=-\tfrac 12e^{-t}+\sqrt 3e^{-4t}.$$ Thus $u_t(x,0)=\sqrt 3-1/2>0$. Factoring out $e^{-4t}$ shows that the derivative changes sign exactly when $e^{3t}=2\sqrt 3$. The unique maximum occurs at $$\boxed{t_*=\tfrac 13\log(2\sqrt 3).}$$ Before $t_*$ the derivative is positive; afterward it is negative. This gives a concrete counterexample to pointwise monotone cooling.

Step 4: Compare with the global heat balance. Integration gives $H(t)=2e^{-t}$, since the second sine has zero integral. Also $u_x=e^{-t}\cos x+e^{-4t}\cos 2x$, so $H'(t)=[u_x]_0^\pi=-2e^{-t}<0$. The right endpoint derivative minus the left is the signed flux balance obtained by integrating the PDE. Total heat falls even while a neighborhood of $5\pi/6$ initially warms.