Does the heat equation have a unique solution with these mixed boundary conditions

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Does the heat equation $u_t - u_{xx} = 0$ on the unit square with $\forall 0 \leq x \leq 1: u(x,0)=0$, $\forall 0 \leq t \leq 1: u(0,t)=0$, $\forall 0 \leq t \leq 1: u_x(1,t)=0$ have a unique solution?

Here's my attempt:

Let $u, v$ be two solutions to the above IBVP. Let $w=u-v$. Then $w$ solves the IBVP, $w_t - w_{xx}=0$ and $w(x)=0$ on the boundary of the unit square (*). By the maximum principle this means that $w \leq 0$. Similar $w \geq 0 $ applying the maximum principle to $-w$. Hence $w=0$ and the solution is unique.

(*) This seems wrong to me. $w$ hasn't been specified on the upper edge of the unit square hasn't been specified so $ w $ need not be $0$ on the entire boundary of the unit square.

So I suspect there is no unique solution.

Also what if instead we had the wave/Laplace equation with these same conditions?

Can anyone help me out here? Thanks!

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When you define $\omega$, you know $\omega(x,0)=0$, $\omega(0,t)=0$, and $\omega_x(1,t)=0$; a priori you cannot state $\omega(1,t)=0$, so you cannot conclude $\omega\equiv 0$ from here.

The usual trick for uniqueness with Neumann (or mixed) boundary condition is to use an energy method. We define an auxiliary functional $$ E = \frac{1}{2}\int_{0}^{1}u^2 \mathrm{d}x \geq 0, $$ and observe that if $u$ is a solution to your problem, this energy is dissipated in time: \begin{align} \frac{\mathrm{d}E}{\mathrm{d}t} = \frac{1}{2}\int_{0}^{1}\frac{\partial}{\partial t}u^2 \mathrm{d}x = \int_{0}^{1} uu_t \mathrm{d}x = \int_{0}^{1} uu_{xx} \mathrm{d}x = \left[uu_{x}\right]_{x=0}^{x=1} - \int_{0}^{1} (u_{x})^2 \mathrm{d}x \leq 0. \end{align} Now, you define $\omega=u-v$, and study what happens to its energy. From this you will be able to conclude $\omega\equiv 0$, like you wanted, and show uniqueness.