Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 5

Take a bar of constant cross-sectional area $A$, thermally insulated along its sides, with no internal heat production. Fourier's law says that the heat flux in the positive $x$ direction isThe thermal conductivity $k(x)$ is positive. On a small slice $[x,x+\Delta x]$, conservation of energy givesHere $\rho c_p$ is the volumetric heat capacity, the product of mass density and specific heat capacity. Divide by $A\Delta x$, take $\Delta x\to 0$, and use $\rho c_p=1$. The resulting heat equation isFor differentiable $k$, this is $T_t=k{T}_{xx}+k^{\prime }{T}_x$; retaining the divergence form also expresses conservation of energy correctly across an interface where $k$ jumps.

Inward heat flux at the left end points in the positive $x$ direction, so the boundary conditions areFor uniform thermal conductivity, the steady heat equation gives $T_s^{\prime \prime }=0$. The heat flux is constant and equal to $Q$, so $T_s^{\prime }=-Q/k$. Imposing the right-end temperature gives

Write $T=T_s+T_u$. The transient satisfies the heat equation and homogeneous mixed boundary conditionsThe left condition is a Neumann boundary condition and the right one a Dirichlet boundary condition. Separation of variables, $T_u=X(x)G(t)$, givesThe condition $X^{\prime }(0)=0$ removes the sine term. The condition $X(2\ell )=0$ then selectsThere is no nonzero mode with $\mu =0$, because a linear $X$ satisfying both conditions is zero. There is no growing mode either: integration by parts gives ${\int }_0^{2\ell }(X^{\prime }{)}^{2}dx={\mu }^2{\int }_0^{2\ell }{X}^{2}dx$, so every nonzero mode has ${\mu }^2>0$.

These Sturm-Liouville eigenfunctions have orthogonality and squared normExpand the initial transient in this half-integer Fourier cosine series. Its coefficients areThe integration formula follows by integration by parts; the cosine at the endpoint is zero. Thus the half-integer cosine expansion of a linear profile givesAt $t=0$ the Fourier series reproduces $-T_s$ everywhere, including the endpoints; its $n^{-2}$ coefficient decay gives uniform convergence. For every $t>0$ the exponential factors allow termwise differentiation and verify the heat equation and both boundary conditions. The sudden switch-on of nonzero heat flux is incompatible with the spatial derivative of the constant initial temperature at the corner $(0,0)$. Accordingly, the positive-time boundary derivative must not be inferred by differentiating the initial Fourier series term by term at that corner.

For the stated inward heating, take $Q\ge 0$ and put $\tau ={\pi }^{2}kt/(16{\ell }^2)$. Since $|\cos ({\mu }_{n}x)|\le 1$, the transient from ii satisfiesEquality holds at $x=0$, where every cosine is one and all coefficients have the same sign. Thus the maximum is attained at the heated end, and its exact value and late-time form areThis is dominance of the slowest-decaying eigenfunction when $kt\gg {\ell }^2$. More explicitly, the omitted relative correction is nonnegative and bounded by $({\pi }^2/8-1)e^{-8\tau }$, using ${\sum }_{n\ge 0}(2n+1{)}^{-2}={\pi }^2/8$. For a signed imposed heat flux, the maximum-deviation formula uses $|Q|$.

In a steady state the heat equation says $(k{T}_s^{\prime }{)}^{\prime }=0$, so the positive-$x$ heat flux $q=-k{T}_s^{\prime }$ is the same constant $Q$ in both materials. Assume perfect thermal contact: temperature is continuous at $x=\ell$, and conservation of energy requires continuous heat flux there. Integrating from the right endpoint givesThe two formulas agree at the interface. Their slopes differ, but $-k_1{T}_s^{\prime }({\ell }^{-})=-k_2{T}_s^{\prime }({\ell }^{+})=Q$. The total temperature drop is $Q\ell (1/k_1+1/k_2)$, the sum of the two layers' contributions in steady heat conduction through layers.