Let's show uniqueness of solutions to the heat equation with Dirichlet boundary conditions.

Let $u$, $v$ be distinct solutions and set $w = u - v$. Then $w$ satisfies $$ \begin{aligned} w_t &= \kappa w_{xx}, \\ w(0, t) &= T_0 - T_0 = 0, \\ w(L, t) &= 0, \\ w(x, 0) &= 0, \end{aligned} $$

Essentially this is a heat problem where the system begins at zero heat, has BCs with zero heat everywhere; it is sensible that the only solution is trivial. To prove it, we can manipulate the equation in the following way.

$$ \begin{aligned} w_t &= \kappa w_{xx} \\ \Longrightarrow w w_t &= \kappa w w_{xx} \\ \Longrightarrow \frac{\partial}{\partial t} \frac{1}{2}w^2 &= \kappa w w_{xx} \\ \Longrightarrow \frac{\partial}{\partial t} \int_0^L \frac{1}{2}w^2 \, \mathrm{d}x &= \kappa \int_0^L w w_{xx} \, \mathrm{d}x \\ \end{aligned} $$

Let's call the quantity $$ E(t) = \int_0^L \frac{1}{2}w^2 \, \mathrm{d}x, $$ the energy.

Now use the fact that $\partial_x(w w_x) = w_x^2 + w w_{xx}$ to re-write the RHS. We then get after integrating once, $$ E'(t) = \kappa w w_x \biggr\rvert_0^L - \kappa \int_0^L w_x^2 \, \mathrm{d}x = - \kappa \int_0^L w_x^2 \, \mathrm{d}x $$ where the second equality occurs because $w = 0$ at $x = 0$ and $x = L$. So in the end, we have the fact that $$ E'(t) = - \kappa \int_0^L w_x^2 \, \mathrm{d}x \leq 0, $$ so the energy is always decreasing. But note that $E'(0) = 0$ since $w(x, 0) = 0$. Finally, note that $E(t)$ is always $\geq 0$ by its form (the integral of a squared quantity). So the energy is always decreasing, begins from zero, and can never be negative. Therefore $$ E(t) \equiv 0 $$ for all time.