Heat Equation with Non-Zero Temperature Boundaries — Question 3

Question 3 – Solution

Strategy. Subtract a stationary solution of the forced equation, not just the line joining its boundary values.

Step 1: Solve the stationary forced problem. The equation $\kappa S''+Q=0$ and the endpoint data give $$\boxed{S=\ell+\frac{Q}{2\kappa}x(L-x).}$$ It is strictly concave, with critical point $x_*=L/2+\kappa(B-A)/(QL)$. This is strictly interior exactly when $|B-A|<QL^2/(2\kappa)$. If equality holds, the maximum is at an endpoint; otherwise the maximum is at the endpoint in the direction of the larger value.

Step 2: Construct the transient from the actual initial difference. For $v=u-S$, the PDE and boundaries are homogeneous and $v(x,0)=-Qx(L-x)/(2\kappa)$. Two integrations by parts give the nonzero coefficients $-4QL^2/(\kappa\pi^3n^3)$ for odd $n$. Therefore $$\boxed{u=S-\frac{4QL^2}{\kappa\pi^3} \sum_{n\text{ odd}}\frac{e^{-\kappa(n\pi/L)^2t}\sin(n\pi x/L)}{n^3}.}$$

Step 3: Check the PDE, trace and maintained fluxes. The transient solves the homogeneous heat equation; $\kappa S''+Q=0$ then verifies the forced equation for $u$. The transient vanishes at the ends. Summable coefficients and Fourier convergence recover $u(\cdot,0)=\ell$ uniformly. Gaussian decay permits differentiation for $t>0$. The initial linear profile has $\kappa\ell''+Q=Q$, inconsistent with a zero endpoint time derivative at an initial corner, so full corner smoothness is not asserted. For $j_S=-\kappa S'$, one has $j_S(L)-j_S(0)=QL$. Hence the steady balance $0=j_S(0)-j_S(L)+QL$ accounts for all generated heat.

Step 4: Interpret the internally heated profile. In the example, $S=1+x+2x(1-x)$, $x_*=3/4$ and $S(x_*)=17/8$. The final maximum exceeds both boundary values because heat is generated inside the rod. The source-free maximum principle cannot be applied unchanged to this forced equation. The transient decays uniformly toward the plotted stationary field.