Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 4

Assume the rod has constant cross-sectional area $A$, no internal source of heat, no bulk motion, and no lateral heat flux. Assume also that each section has a uniform temperature, so that $T$ depends only on $x,t$, and that Fourier's law applies with the stated constant thermal conductivity $k$. The longitudinal heat flux density is then $q=-k{T}_x$.

By conservation of energy, the rate of increase of internal energy in a slice $[x,x+\Delta x]$ equals incoming minus outgoing heat:Divide by $A\Delta x$ and take the limit $\Delta x\to 0$. For a sufficiently differentiable temperature field this gives $\rho c{T}_t=-q_x=k{T}_{xx}$, and henceHere $\rho$ is mass density, $c$ is specific heat capacity, and $\kappa$ is thermal diffusivity.

A steady state has $T_s^{\prime \prime }=0$, so $T_s=a+bx$. The fixed temperature gives $T_s(0)=T_0$, and thermal insulation gives $T_s^{\prime }(L)=0$. ThusSubtracting this steady state produces the homogeneous mixed boundary conditionsThe partial differential equation and boundary conditions apply for $t>0$. The initial condition is attained as $t\downarrow 0$. Its slope at $x=L$ is $-T_0/L$, so it does not satisfy the insulated-end condition at the instant $t=0$. The resulting heat equation solution is smooth for $t>0$ and approaches the initial temperature continuously; its spatial derivative need not be continuous at $(L,0)$.

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.