A scalar field satisfying the Laplace's Equation is zero everywhere Q: A scalar field $A(y_1,y_2,y_3)$ is equal to zero on a closed surface $S$ in $\mathcal{R}$. If $A$ satisfies the Laplace equation $A_{,ii}=0$ in $\mathcal{R}$, show that $A=0$ everywhere in $\mathcal{R}$.
I have to show the above statement is true using tensor notations. To me it looks like solving a differential equation that is
\begin{align*}
A&=0\qquad\text{on } S\subset\mathcal{R}=\mathbb{R}^3\\
\nabla^2A&=0\qquad\text{in }\mathcal{R}=\mathbb{R}^3\\
\end{align*}
Is my idea correct? If so, how do I go about solving this? Don't you need at least two conditions to solve a second order PDE?

I'm going to assume that $S$ is smooth closed surface that is embedded into $\mathbb R^3$. The smoothness assumption can be relaxed to $C^1$. Also, I think the assumption that $S$ is embedded is natural in this context. If you do actually want to consider the case $S$ is not embedded then let me know.
Proof:
Since non-orientable surfaces cannot be embedded into $\mathbb R^3$, $S$ must be orientable. Then, by the Jordan-Brouwer separation theorem, $\mathbb R^3 \setminus S$ has two connected components: one bounded which we will call $\Omega$ and one unbounded. For a proof of the Jordan-Brouwer separation theorem in the case $S$ is smooth see The Jordan-Brouwer Separation Theorem for Smooth Hypersurfaces by Lima and references therein for the $C^1$ case.
Since $A$ is harmonic in $\Omega$ and $A=0$ on $\partial \Omega=S$, by the maximum principle $A \equiv 0$ in $\Omega$. But then, since $A$ is harmonic in all of $\mathbb R^3$, it is analytic in $\mathbb R^3$. Since $A$ vanishes in $\Omega$ and $A$ is analytic in all of $\mathbb R^3$, it follows from unique continuation that $A \equiv 0$ in $\mathbb R^3$.   $\square$
