= Solution
{c}
For uniqueness, subtract two solutions and call their difference $w$. It has zero boundary and initial data, with $w_t=w_{xx}+aw_x+bw$. Its squared <L2 norm> obeys the <energy estimate>
$$
\frac12\frac d{dt}\int_{-1}^1w^2\,dx
=-\int_{-1}^1w_x^2\,dx+\frac a2[w^2]_{-1}^1+\int_{-1}^1bw^2\,dx\leq0.
$$
The boundary term vanishes and $b\leq0$. 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 \b[false for an arbitrary coefficient $b(x,t)\leq0$]. 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 $x\ne0$.
With the additional assumption $b(-x,t)=b(x,t)$, the intended argument is valid. For $a=0$, <reflection> $\widetilde T(x,t)=T(-x,t)$ solves the same equation with the same initial and boundary data. Uniqueness then proves $\boxed{T(-x,t)=T(x,t)}$. If $T_0\ne0$, 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>.
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