Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 2

Write a nonzero separated solution as $T=X(x)Y(y)S(t)$. Division of the heat equation by this product givesThe two spatial factors solve independent constant-coefficient eigenvalue problems. Zero endpoint values in $x$ requireThe insulating conditions requireThe zero cosine mode is the nonzero constant function and must be retained; there is no nonzero zero-index sine mode. The temporal equation is $S^{\prime }=-\lambda {\pi }^2(n^2/a^2+m^2/b^2)S$. Superposing the complete sine-cosine basis yieldsThe constants are fixed by expansion of the initial profile. Orthogonality gives sine norm $a/2$; the cosine norm is $b$ at $m=0$ and $b/2$ otherwise. This is the heat equation on a rectangle with two insulated sides.

First decompose the initial $y$ dependence using ${\sin }^2(2\pi y/b)=[1-\cos (4\pi y/b)]/2$. Only the cosine indices $m=0$ and $m=4$ occur. Expand the remaining factor in a Fourier sine series:For $k_n=n\pi /a$, two integrations by parts giveThus even $n$ vanish, while $b_n=8a^2/(n^3{\pi }^3)$ for odd $n$. The nonzero double-series coefficients are $A_{n0}=b_n/2$ and $A_{n4}=-b_n/2$. Their separate heat-mode decay factors giveAt $t=0$ the bracket is $2{\sin }^2(2\pi y/b)$, which recovers the initial profile with the displayed sine coefficients. Each term satisfies the mixed Dirichlet boundary conditions and Neumann boundary conditions.

The slowest surviving mode has $n=1,m=0$. HenceIt is independent of $y$: insulation allows the constant transverse mode, while every nonconstant transverse mode decays faster. The next possible relative decay rates are $8\lambda {\pi }^2/a^2$ from $n=3,m=0$ and $16\lambda {\pi }^2/b^2$ from $n=1,m=4$.