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

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.