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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    $\begingroup$ Isn't it easier to just use maximum principle? $\endgroup$ Commented Aug 3, 2020 at 15:07
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    $\begingroup$ Anyway, for the method you choose: $u(x,R) \frac{du}{dy}(x,R) = \frac12 \frac{d}{dy} |u(x,y)|^2 \Big]_{y = R}$. If the lim of $u(x,y)$ tends to zero, there exists some $R$ for which the integral of this derivative is negative. $\endgroup$ Commented Aug 3, 2020 at 15:10
  • $\begingroup$ @Willie Wong the maximum principale is available if $\Omega$ a bounded domain theoreme 2.17 math.ucdavis.edu/~hunter/pdes/ch2.pdf $\endgroup$
    – Pascal
    Commented Aug 3, 2020 at 17:25
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    $\begingroup$ Let $\Omega_t = \Omega \cap \{ y < t\}$. Apply the maximum principle for $u$ on $\Omega_t$. Take $t\to \infty$. $\endgroup$ Commented Aug 3, 2020 at 18:02
  • $\begingroup$ @Willie Wong I tried to apply the maximum principle with your indication with $ \Omega_t $ but without success. Can you give your answer because I do not see why it is easy and especially that A.Ermenko says that there is a counter example. $\endgroup$
    – Pascal
    Commented Aug 4, 2020 at 4:19

1 Answer 1


If you want the positive answer you should state your last condition more carefully. For example, add that $u$ is bounded, or that $u(x+iy)$ tends to $0$ as $y\to\infty$ UNIFORMLY with respect to $x$.

As you presently stated, the answer is negative. I sketch the construction of a counterexample.

  1. There exists a non-zero entire function, real on the real line and such that $f(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta$. (See, for example, my answer to this question, which explains how to construct $f$.)

  2. $v(z)=\Im f(z)$ is a non-zero harmonic function in the upper half-plane, equal to $0$ on the real line and $u(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta\in(0,\pi)$.

  3. Take without loss of generality $a=0,\; b=\pi$, then $e^{-i(z-\pi)}$ maps your strip $0<z<\pi$ into the upper half-plane, with a removed half-disk. So the function $w(z)=v(e^{-i(z-\pi)})$ is harmonic, zero on infinite sides of your half-strip and satisfies the property at $\infty$: $w(x+iy)\to 0$ as $y\to+\infty$ for every $x\in(0,\pi)$. Notice that this function is unbounded. To satisfy the last requirement, that $u(x)=0$ for $0<x<\pi$, set $u(z)=w(z)-w_1(z)$, where $w_1(z)$ is the solution of Dirichlet problem matching the boundary values of $w$ on the finite part of the boundary and bounded in your strip.

Remark. Since every simply connected domain other than the plane is conformally equivalent to the unit disk, your question is equivalent to the following. Suppose that $u$ is harmonic in the unit disk and $\lim_{z\to\zeta} u(z)=0$ for all $\zeta\in\{ \zeta:|\zeta|=1\}\backslash\{1\}$, and moreover $u(z)\to 0$ along any non-tangential segment ending at $1$, that is $u(1-re^{i\theta})\to 0,\; r\to 0$ for every $\theta\in(-\pi/2,\pi/2)$. Does it follow that $u=0$? The answer is NO.

  • $\begingroup$ @ A.Eremenko I think you are right, assumptions are missing, I edited my post. I have not yet understood the construction of a counter example $\endgroup$
    – Pascal
    Commented Aug 4, 2020 at 4:55
  • $\begingroup$ @cerise: I do not see how you edited your post, but if you add one of the conditions that I suggested, the counterexample will loose its meaning. $\endgroup$ Commented Aug 4, 2020 at 5:26
  • $\begingroup$ @ A.Eremenko I edited to continue my reasoning and didn't add anything as additional assumptions. I noticed that the answer is positive to the question if $\displaystyle ||u(.,R)||_{L^2]0,\pi[}\to^{R\to\infty} 0$ $\endgroup$
    – Pascal
    Commented Aug 4, 2020 at 5:32
  • $\begingroup$ conflicting opinions of experts, does not help me. $\endgroup$
    – Pascal
    Commented Aug 4, 2020 at 11:00
  • $\begingroup$ @cerise: your conclusion that $\| u(,.R)\|_2\to 0,\; R\to\infty$ is incorrect: it does not hold in my example. From your assumptions AS STATED it does NOT follow that $u=0$. Look in the literature about "passing to the limit under the integral sign". $\endgroup$ Commented Aug 4, 2020 at 14:58

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