Solving the Heat Equation — Question 6

Question 6 – Solution

Strategy. Use half-integer modes for the mixed boundary problem; integer sine modes would solve a different problem.

Step 1: Evaluate the coefficient integral. For $f=x(2L-x)$, $f(0)=0$, $f'(L)=0$ and $f''=-2$. Two integrations by parts, using $\cos(k_nL)=0$, give $$\int_0^L f\sin(k_nx)\,dx=\frac{2}{k_n^3},\qquad \boxed{u(x,t)=\frac 4L\sum_{n=0}^\infty \frac{e^{-\kappa k_n^2t}\sin(k_nx)}{k_n^3}.}$$ The vanishing boundary terms use the derivative condition at $L$, not a zero value of $f(L)$, which is actually $L^2$.

Step 2: Verify the complete solution. The coefficients are summable, so the initial series is uniformly convergent; the supplied Fourier theorem identifies it as $f$. Summable domination gives uniform convergence to $f$ as $t\downarrow 0$. For $t\ge\tau>0$, differentiated series converge uniformly by exponential decay and verify the PDE. At zero, all sine factors vanish; at $L$, all derivative cosine factors vanish. The initial function also satisfies these value/first-derivative data, although higher corner compatibility is a separate issue.

Step 3: Compute heat loss through the open thermal boundary. Since $\int_0^L\sin(k_nx)\,dx=1/k_n$, $$H(t)=\frac 4L\sum_{n\ge 0}\frac{e^{-\kappa k_n^2t}}{k_n^4},\qquad H(0)=\frac{2L^3}{3}.$$ For $t>0$, $H'=-(4\kappa/L)\sum e^{-\kappa k_n^2t}/k_n^2=-\kappa u_x(0,t)$. This is $\kappa[u_x]_0^L$ because the right derivative is zero. Every term in the loss sum is positive; insulation blocks flux at the right endpoint only.

Step 4: Extract the slowest mode rigorously. Multiplying the series by $e^{\kappa k_0^2t}$ leaves the first term and a remainder bounded by $e^{-\kappa(k_1^2-k_0^2)t}\sum_{n\ge 1}4/(Lk_n^3)$, which tends to zero. Hence uniformly, $$\boxed{e^{\kappa k_0^2t}u\longrightarrow \frac{32L^2}{\pi^3}\sin(\pi x/(2L)).}$$ The slowest rate is $\kappa\pi^2/(4L^2)$, one quarter of the Dirichlet–Dirichlet rate $\kappa\pi^2/L^2$. Equal material properties and length do not imply equal decay rates when the boundary conditions differ.