If $u$ is harmonic and vanishes on the boundary, then $u\equiv 0$.

Question

Let's assume $u$ is harmonic on $D=\{x^2+y^2\leq1\}$. If $u=0$ on $\partial D$ then $u\equiv 0$ on $D$.

$\textbf{Proof}$: Let's use Green's theorem and set $P=-u\frac{\partial u}{\partial y}$ and $Q=u\frac{\partial u}{\partial x}$.

Then

$$1. \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} =\left(\frac{\partial u}{\partial x}\right)^2+ u\left(\frac{\partial u}{\partial x}\right)^2+ \left(\frac{\partial u}{\partial y}\right)^2+ u\left(\frac{\partial u}{\partial y}\right)^2 =\left(\frac{\partial u}{\partial x}\right)^2+ \left(\frac{\partial u}{\partial y}\right)^2. $$

By Green's theorem,
$$ 0=\int_{\partial D} -u\frac{\partial u}{\partial y}dx+u\frac{\partial u}{\partial x}dy=\int \int_{D} \left( \left(\frac{\partial u}{\partial x}\right)^2+ \left(\frac{\partial u}{\partial y}\right)^2\right)dxdy. $$

Because $u\in C^2$, we get

$$ \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2=0. $$

So $u\equiv c$ on $D$ but on $\partial D$ we have $u=0$. So $u\equiv 0$.

What have we done on the first move for $1$.?

Answer 1

In the second line of your proof, it should be:

\begin{align*} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} &=\left(\frac{\partial u}{\partial x}\right)^2+ u\frac{\partial^2 u}{\partial x^2}+ \left(\frac{\partial u}{\partial y}\right)^2+ u \frac{\partial^2 u}{\partial y^2} \\ &= \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 + u \underbrace{\left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right)}_{0} \\ &=\left(\frac{\partial u}{\partial x}\right)^2+ \left(\frac{\partial u}{\partial y}\right)^2 \end{align*} since $u$ is harmonic on $D$, i.e., $u$ must satisfy Laplace's equation.

Answer 2

We have used the product rule; e.g.

$$ \partial_x Q = \partial (u \partial_x u) = (\partial_x u)^2 + u \partial_x^2 u. $$

(Your notation seems a bit off at that point.)