The Heat Equation — Question 1

Question 1 – Solution

Strategy. Balance total heat power before dividing by the local cross-sectional area.

Step 1: Derive the conservation law. The stored thermal energy relative to a fixed reference has time derivative $\int_a^b CAu_t\,dx$. Inflow minus outflow plus generation gives $$\int_a^b CAu_t\,dx=Q(a,t)-Q(b,t)+\int_a^b As\,dx.$$ Since this holds on every interval, with Fourier’s law it yields $$\boxed{CAu_t=-Q_x+As=(kAu_x)_x+As.}$$ Each local term has units $\mathrm{W/m}$: $C$ is $\mathrm{J/(m^3K)}$, $A$ is $\mathrm{m^2}$ and $u_t$ is $\mathrm{K/s}$. Each integrated term, including $Q(a)-Q(b)$, is in watts.

Step 2: Identify the geometric contribution. Dividing by $CA$ gives $$\boxed{u_t=\frac{k}{C}u_{xx}+\frac{(kA)'}{CA}u_x+\frac{s}{C}.}$$ The diffusivity $k/C$ has units $\mathrm{m^2/s}$. For a homogeneous constant-area rod without a source, this is $u_t=\alpha u_{xx}$ with constant $\alpha=k/C$. More generally, the first-derivative term vanishes when $kA$ is constant; constant diffusivity additionally requires $k/C$ constant.

Step 3: Evaluate the nondimensional example. Here $u_x=1$ and $u_{xx}=0$, but $$\boxed{Q(x)=-e^x,\qquad Au_t=-Q_x=e^x,\qquad u_t=1.}$$ The signed power is negative because heat moves toward decreasing temperature, which here is the negative $x$ direction. Its magnitude increases toward the larger cross section.

Step 4: Interpret the nonzero storage. Flux per unit area is $q=-ku_x=-1$, constant in this example. Total power is $Q=Aq=-e^x$, which is not constant. More heat enters a small interval through its larger right face than leaves through its smaller left face, so energy accumulates. Zero curvature alone is a steady-state test only under the coefficient and geometry assumptions that remove the extra term.