The Heat Equation — Question 7

Question 7 – Solution

Strategy. A manufactured temperature field determines its source and boundary data; those data cannot be prescribed independently afterward.

Step 1: Compute the interior residual. The derivatives are $u_t=a$, $u_x=b(L-2x)$ and $u_{xx}=-2b$. Substitution requires $$\boxed{s=Ca+2kb.}$$ This is a positive spatially and temporally constant source, in $\mathrm{W/m^3}$. It supplies both the rising stored energy and heat lost at the ends.

Step 2: State the compatible data. At $t=0$, $u(x,0)=T_a+bx(L-x)$. Both endpoint temperatures equal $T_a+at$. The endpoint derivatives are $bL$ and $-bL$, neither zero, so this field does not satisfy insulated-end conditions when $b>0$. Prescribing insulation in addition to the realized end temperatures would contradict the candidate rather than add useful information.

Step 3: Verify total energy accounting. The outward fluxes are $ku_x(0)=kbL$ and $-ku_x(L)=kbL$. The stored heat relative to $T_a$ is $$E=CA\int_0^L(u-T_a)\,dx =CA\left(atL+\frac{bL^3}{6}\right),\qquad E'=CAaL.$$ Total source power minus total outward power is $$\boxed{AL(Ca+2kb)-2AkbL=CAaL=E'.}$$ This independently checks the source sign and both boundary signs.

Step 4: Remove the source consistently. For $s=0$, the formula can satisfy the PDE only if $$\boxed{a=-\frac{2kb}{C}.}$$ Its initially warm curved profile then decreases at a constant rate while both endpoint temperatures follow the same decreasing law. If $a$ retains a different value, the residual $Ca+2kb$ is nonzero at every interior point. Boundary values alone cannot repair that local equation while keeping the same candidate field. Changing boundary data generally requires changing the field itself.