The Heat Equation — Question 3

Question 3 – Solution

Strategy. Constant steady heat flux requires the gradient to vary inversely with conductivity.

Step 1: Integrate the flux law. Since $(ku_x)_x=0$, the flux $q$ is constant. Integrating $u_x=-q/k(x)$ gives $$T_c-T_h=-\frac{qL}{k_0}\log 2.$$ Consequently $$\boxed{q=\frac{k_0(T_h-T_c)}{L\log 2},\qquad u(x)=T_h-(T_h-T_c)\frac{\log(1+x/L)}{\log 2}.}$$ The flux is positive: heat travels from the hotter left end toward the right.

Step 2: Check the profile and gradient. The logarithm is zero at $0$ and $\log 2$ at $L$, giving the two endpoint values. Also $$u_x=-\frac{T_h-T_c}{L\log 2\,(1+x/L)},\qquad k(x)u_x=-\frac{k_0(T_h-T_c)}{L\log 2}.$$ Thus the product is constant as required. The gradient magnitude is largest at $x=0$ and decreases by a factor of two from left to right.

Step 3: Diagnose the linear candidate. For $\ell=T_h-(T_h-T_c)x/L$, one has $\ell_{xx}=0$, but $$\boxed{(k\ell_x)_x=-\frac{k_0(T_h-T_c)}{L^2}\ne 0.}$$ Its rightward flux grows along the rod, so more heat leaves each small interval than enters. Without a source this profile would cool locally, not remain steady. Using $ku_{xx}=0$ discards the essential term $k'u_x$.

Step 4: Interpret the effective conductivity. The definition gives $k_{\mathrm{eff}}=k_0/\log 2$, while the arithmetic average is $3k_0/2$. In fact $\log 2>2/3$: the function $1/t$ is strictly convex and its average on $[1,2]$ exceeds its midpoint value $2/3$. Hence $$\boxed{k_{\mathrm{eff}}=\frac{k_0}{\log 2}<\frac{3k_0}{2}.}$$ Steady conduction uses the reciprocal average $k_{\mathrm{eff}}=L/\int_0^L k^{-1}\,dx$. The graph uses $\xi=x/L$ and $\theta=(u-T_c)/(T_h-T_c)$.