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My question is nearly the same as this one. I would like to know how to prove the uniqueness of $$\begin{cases} u_{xx}+u_{yy}=0,\;(x,y)\in\mathbb{R}^2 \\ u(x,0)=0,\;x\in\mathbb{R} \\ u_y(x,0)=0,\; x\in\mathbb{R} \end{cases}$$ without using d'Alembert formula and complex variable. Any hint?

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user39756
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It suffices to show that $u$ vanishes identically in $\mathbb R^2.$ To this end, let us show that $u$ is odd and also even in $y.$ We need the following lemma, probably it is a good exercise (in which we note that $u$ is analytic).

Lemma. Let $\Omega\subset\mathbb R^2$ be a bounded domain (an open and connected subset). Suppose $u\in C^2(\Omega)\cap C^0(\overline\Omega)$ is a harmonic function. If $u$ vanishes in some open ball $B\subset\Omega,$ then $u$ must be zero in the whole $\Omega.$

For any $R>0,$ we set $B_R^+=\{(x,y)\in B_R(0);y>0\}.$ Suppose that $u$ solves the Laplace equation above, we define a new harmonic function $u^\ast$ via reflection principle, that is, $$u^\ast(x,y)=\cases{u(x,y),&if $(x,y)\in\overline{B_R^+}$,\cr -u(x,-y),&if $(x,y)\in B_R(0)\setminus B_R^+,$ }$$then $u^\ast:B_R(0)\to\mathbb R$ is a harmonic function, and $u^\ast|_{B_R^+}=u.$ Our lemma above shows that $u=u^\ast$ in $B_R(0).$ Sending $R\to\infty,$ this yields that $u$ is odd in $y.$ Similarly, $u_y$ is odd in $y,$ which implies that $u$ is even in $y.$ Hence we complete the proof.

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