Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V / 6

For uniqueness, subtract two solutions and call their difference $w$. It has zero boundary and initial data, with $w_t=w_{xx}+a{w}_x+bw$. Its squared L2 norm obeys the energy estimateThe boundary term vanishes and $b\le 0$. A nonnegative quantity starting at zero must stay zero, proving uniqueness in the usual classical or corresponding energy class.

The printed evenness claim for $a=0$ is false for an arbitrary coefficient $b(x,t)\le 0$. Take $T_0>0$ and $b(x,t)=-2-x$. The strong maximum principle makes the solution positive in the interior for $t>0$. If it were even, $T_t$ and $T_{xx}$ would both be even. Subtracting the equation at $x$ and $-x$ would give $0=[b(x,t)-b(-x,t)]T(x,t)=-2xT(x,t)$, impossible for $x\ne 0$.

With the additional assumption $b(-x,t)=b(x,t)$, the intended argument is valid. For $a=0$, reflection $\tilde{T}(x,t)=T(-x,t)$ solves the same equation with the same initial and boundary data. Uniqueness then proves $\boxed{T(-x,t)=T(x,t)}$. If $T_0\ne 0$, initial and boundary values disagree at the two time-zero corners; the initial condition is understood in the interior or in the L2 norm limit.

With $x=2z-1$, the chain rule gives $T_{xx}=V_{zz}/4$. ThusSeparation of variables gives the eigenfunctions $\sin (n\pi z)$ with time factors $e^{-n^2{\pi }^{2}t/4}$. The Fourier sine coefficients of the initial constant areOnly odd modes survive, soReturning to $x$, $\sin ((2k+1)\pi (x+1)/2)=(-1{)}^k\cos ((2k+1)\pi x/2)$, henceFor every $t>0$ the exponential coefficients give uniform convergence of the differentiated series on later-time compact sets, verifying the PDE and boundary conditions. As $t\downarrow 0$, the solution tends to $T_0$ at every interior point and in L2 norm, while remaining zero at the endpoints. The expression is visibly even in $x$, consistent with the corrected symmetry argument for $b=0$.