Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 2

For separation of variables, take $n=T(t)X(x)Y(y)Z(z)$ and divide the reaction-diffusion equation by this product wherever it is nonzero:Each spatial term depends on a different variable and must be constant. Write $X^{\prime \prime }=-k_x^{2}X$ and similarly for the other two factors. The Dirichlet boundary conditions give $X(0)=X(L)=0$. A zero separation constant permits only a linear function and hence the zero solution; a positive value of $X^{\prime \prime }/X$ likewise permits no nonzero function vanishing at both endpoints. The nontrivial choices are thereforeThe temporal ordinary differential equation isso, absorbing its constant multiplier into $a_{p,q,r}$,The superposition principle and completeness of the Fourier sine series give the stated threefold expansion with these factors. For an initial density $n_0$, orthogonality also determines the constants:

The fastest temporal growth occurs for the lowest spatial eigenvalue, with $p=q=r=1$. Its growth rate isFor a generic initial condition its coefficient is nonzero, so this term determines the long-time behavior. In fact, every nonzero, nonnegative initial density has $a_{111}>0$, since the product of the three first sine eigenfunctions is strictly positive in the interior. ConsequentlyFor $L<L_{\mathrm{crit}}$ every term decays. At equality the first term is stationary and all higher terms decay: equality does not produce exponential growth. This is a reaction-diffusion equation threshold arising from competition between production and loss through the Dirichlet boundary condition.

The separation of variables solution with Dirichlet boundary conditions on the cuboid isThe constants are fixed by the initial density through Fourier sine series. The fastest term again has $p=q=r=1$. When $L=2L_{\mathrm{crit}}$, its growth rate becomesThus the precise condition for no exponentially growing term isFor example, $\lambda =1/4$ gives ${\lambda }^{-2}+2\lambda =33/2>12$, while $\lambda =6$ gives $12+1/36>12$: either a sufficiently flat or a sufficiently elongated cuboid suppresses growth despite its unchanged volume. The function ${\lambda }^{-2}+2\lambda$ decreases to its minimum $3$ at $\lambda =1$, then increases without bound. Hence there are two positive threshold values, with growth only between them; equality leaves one stationary term and all other terms decaying.