In this lecture, we'll prove uniqueness for two variants of the heat equation. The different variants you should consider as as follows (all posed on $x\in[0, L]$): $$\begin{aligned} u_t &= \kappa u_{xx}, \\ u(x, 0) &= 0, \end{aligned}$$ with either Dirichlet conditions [D] $$\begin{aligned} u &= T_0 \quad (x = 0), \\ u(L, t) &= T_1\quad (x = L), \end{aligned}$$ or Neumann conditions [N] $$\begin{aligned} u_x &= G_0 \quad (x = 0), \\ u_x &= G_1\quad (x = L), \end{aligned}$$ or mixed conditions [M] $$\begin{gathered} A_0 u + B_0 u_x = H_0 \quad (x = 0), \\ A_1 u + B_1 u_x = H_1 \quad (x = L). \end{gathered}$$

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 $w(0, t) = 0 = w(L, t)$ [*]. 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. We have the three statements:

1. $E'(t) \leq 0$ for all time,
2. $E(0) = 0$,
3. $E(t) \geq 0$ for all time,

and you would conclude that it has to remain at its initial value, and therefore $$E(t) \equiv 0$$ for all time. Looking at the form of the integrand, you would conclude that the only way this occurs is if the integrand is itself zero, or $$w^2(x, t) = 0$$ for all $x\in[0, L]$ and for all $t \geq 0$. So $u(x, t) \equiv v(x, t)$ and the solutions must be the same.

The proof of uniqueness for the Neumann condition [N] is done identically to the above, except in the step marked with [*] we use the fact that $w_x(0, t) = 0 = w_x(L, t)$. But you would arrive at the same conclusions as above.