Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 2

For $v=(t+1{)}^{2}u$, the derivatives areSubstitution cancels the two terms $2(t+1)u$, giving the diffusion equationThis is a scalar integrating factor for a reaction-diffusion equation, since $\exp [\int 2dt/(t+1)]=(t+1{)}^2$ up to a constant. We use $t\ge 0$, so the factor never vanishes. A classical solution at the initial corners additionally requires compatible data, $h(0)=0$ and $h(\pi )=f(0)$.

The transformed endpoint values are $u(0,t)=0$ and $u(\pi ,t)=3$. Apply boundary lifting with $u=3x/\pi +w$. The lifted profile is linear, so $w_t=w_{xx}$ with homogeneous Dirichlet boundary conditions andFor separation of variables, put $w=X(x)T(t)$. Then $X^{\prime \prime }/X=T^{\prime }/T=-\lambda$. The endpoint conditions give nonzero spatial modes only for $\lambda =n^2$, $n=1,2,\dots$, with $X_n=\sin nx$ and $T_n=e^{-n^{2}t}$. Henceby orthogonality of the sine modes. Only the $n=2$ mode is present. Undoing both substitutions givesAt $t=0$ this has the prescribed initial profile, and at $x=\pi$ it is $3(t+1{)}^2$. Differentiating also verifies the original evolution equation directly.

The nonstationary part of $u$ decays exponentially, whereas the integrating factor grows quadratically. For fixed $x>0$,ConsequentlyAt $x=0$, both expressions are identically zero; the ratio formulation is understood for $x>0$. The diffusion modes have died away, leaving the lifted linear profile multiplied by the reaction-induced amplitude factor.