Let $u$ an harmonique function on $\Omega=(a,b)\times (0,+\infty)$ and boundary conditions :

$\displaystyle u(a,y)=u(b,y)=0,\quad\forall y\geq 0$

$\displaystyle u(x,0)=0,\,\lim_{y\to +\infty} u(x,y)=0 \quad \forall x\in (a,b)$

Can we conclude that $\quad u=0$ on $\Omega$ ?

My adempt

Let $$\Omega_{R}=(a,b)\times (0,R),\forall R>0$$By IBP, i show that $$\int_{\Omega_R}u\Delta u=\int_a^{b}u(x,R)\frac{du}{dy}(x,R)dx-\int_{\Omega_R}|\nabla u|^2 $$ Thus $$\forall R>0,\quad \int_{\Omega_R}|\nabla u|^2=\int_a^bu(x,R)\frac{du}{dy}(x,R)dx$$

I need help to cointinuous ( For example to show $\int_{\Omega}|\nabla u|^2=0$)

**edit Continuing the initial reasoning,** with $a=0$ and $b=\pi$ as suggested by A Ermenko

$\Big(\int_{\Omega_R}|\nabla u|^2\Big)^2=\Big(\int_0^{\pi}u(x,R)\frac{du}{dy}(x,R)dx\Big)^2\\ \leq \int_0^{\pi}u(x,R)^2dx\int_0^{\pi}\Big(\frac{du}{dy}(x,R)\Big)^2dx ,\mbox{by Cauchy–Schwarz inequality }\\\\$

$\leq \int_0^{\pi}u(x,R)^2dx \int_0^{\pi}|\nabla u|^2(x,R)dx \\\\$

$= \int_0^{\pi}u(x,R)^2dx\int_{]0,\pi[\times\{R\}}|\nabla u|^2 \\\\$

$\leq \int_0^{\pi}u(x,R)^2dx \int_{\Omega_R}|\nabla u|^2,\mbox{because} ]0,\pi[\times\{R\}\subset\Omega_R$

Then $$\int_{\Omega_R}|\nabla u|^2\leq \int_0^{\pi}u(x,R)^2.$$ I can only conclude if $\displaystyle ||u(.,R)||_{L^2]0,\pi[}\to^{R\to\infty} 0$

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