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For two length-$L$ slabs with zero exterior temperatures and perfect thermal contact, put ${\alpha }_i=\sqrt{{\rho }_i{c}_i/k_i}$, $u=\omega {\alpha }_{1}L$, and $v=\omega {\alpha }_{2}L$. A separated solution with time factor $e^{-{\omega }^{2}t}$ and $\omega \ne 0$ exists precisely when the displayed interface determinant vanishes. Dividing by $\cos u\cos v$ gives a tangent characteristic equation only away from its poles. Common zeros of both cosines are legitimate normal modes and must be retained. The zero-frequency case gives only the trivial solution of a differential equation with these positive material parameters and exterior boundary conditions.

Source: released examination PDF, bundle pages 230–233.

Solutions to all seven questions, including every optional question and Section C. In Question 6(b), the printed transcendental equation excludes valid separated solutions at common poles of the tangents; the complete characteristic equation and an explicit counterexample are given below.

Substitute the separated solution $y=e^{-\alpha t}F(t)\sin (nx)$ into the damped wave equation. Its time derivatives givewhereas $y_{xx}=-n^2{e}^{-\alpha t}F\sin (nx)$. ThusThe last two conditions follow from $y(x,0)=0$ and $y_t(x,0)=\sin (nx)$. For $0<\alpha <n$, let $\nu =\sqrt{n^2-{\alpha }^2}>0$. The simple harmonic motion equation has solution of a differential equationThis is an underdamped normal mode: its oscillations have the decaying factor $e^{-\alpha t}$.

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.