 Solution

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 $b=0$.