Heat Equation with Non-Zero Temperature Boundaries — Question 5

Question 5 – Solution

Strategy. A boundary lifting with nonzero time derivative creates a source in the transformed equation; a persistent lag balances that source.

Step 1: Keep the source created by the lifting. For $w=u-\ell-rt$, $$w_t=\kappa w_{xx}-r,\qquad w(0,t)=w(L,t)=0,\qquad w(x,0)=0.$$ The line $\ell+rt$ has time derivative $r$ and zero second spatial derivative, so it fails the original source-free PDE when $r>0$.

Step 2: Find the lag profile and the initial correction. A fixed $P$ must satisfy $\kappa P''=r$, $P(0)=P(L)=0$, giving $P=-r x(L-x)/(2\kappa)$. The remaining transient has initial data $-P$. Its odd sine coefficients are $4rL^2/(\kappa\pi^3n^3)$. Thus $$\boxed{u=\ell+rt+P+\frac{4rL^2}{\kappa\pi^3} \sum_{n\text{ odd}}\frac{e^{-\kappa(n\pi/L)^2t}\sin(n\pi x/L)}{n^3}.}$$

Step 3: Verify all data and the moving-frame limit. The field $\ell+rt+P$ satisfies $u_t=\kappa u_{xx}=r$; its added transient satisfies the homogeneous heat equation. All correction terms vanish at the endpoints. At $t=0$, the summable sine expansion equals $-P$ uniformly and recovers $\ell$. For $t\ge\tau>0$, the exponential transient permits all termwise derivatives. Initial corner smoothness is not implied: the boundary time derivative is $r$, whereas $\kappa\ell''=0$. As $t\to\infty$, the transient tends uniformly to zero, so $u-\ell-rt\to P$ uniformly.

Step 4: Quantify the persistent lag and heat input. The midpoint offset is $P(L/2)=-rL^2/(8\kappa)$, and the mean offset is $L^{-1}\int P=-rL^2/(12\kappa)$. Their negatives are the positive lags behind the boundary line. Since $P'(L)-P'(0)=rL/\kappa$, the eventual flux balance is $H'=\kappa[u_x]_0^L\to rL$, also obtained by differentiating the integrated solution. Boundary values grow indefinitely; it is the offset, not the absolute temperature, that approaches a fixed state.