Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V / 6

Conservation of thermal energy and Fourier's law give the heat equationsThe density times the specific heat capacity, ${\rho }_i{c}_i$, is the volumetric heat capacity, and $k_i$ is the thermal conductivity.

The condition $T_1(-L,t)=0$ holds the left exterior face at a fixed zero reference temperature; $T_2(L,t)=0$ does the same for the right exterior face. The condition $T_1(0_{-},t)=T_2(0_{+},t)$ expresses perfect thermal contact, with no temperature jump at the interface. Finally,expresses continuity of the signed heat flux in the positive $x$ direction: the interface neither stores nor produces heat.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

Choose the positive weightsThese are the volumetric heat capacities. The heat equations and integration by parts giveThe exterior boundary terms vanish because the temperatures there are zero. At the thermal interface, the remaining term isby continuity of temperature and heat flux. Hence the weighted energy estimate isBoth weights are positive, so $I\ge 0$. Zero initial temperatures give $I(0)=0$, forcing $I(t)=0$ for all $t>0$. Each temperature therefore vanishes throughout its slab. Applying this argument to a difference also proves uniqueness for arbitrary common initial data in this thermal interface problem.