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How can i show uniqueness of

$$ \Delta u= \int _\Omega u(y)dy \phantom{2} ,\phantom{2} x \in \Omega $$ $$ u=0 \phantom{2}, \phantom{2} x \in \partial\Omega $$

I suppose that there are two solutions $u_1$ and $u_2$ such that $w=u_1 - u_2$. Applying Green first identity , i cannot get $w=0$

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Multiply the equation by $u$ and integrate over $\Omega$. Integrating by parts on the left hand side we have $$ -\int_\Omega |\nabla u|^2 = \left(\int_\Omega u\right)^2. $$ Note that the left hand side is non-positive and the right hand side is non-negative. Thus it must be the case that $u \equiv 0$.

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  • $\begingroup$ How does the right hand side become $ \int_\Omega u \int_\Omega u = \left(\int_\Omega u\right)^2. $ $\endgroup$ – express78 Mar 31 '17 at 19:38
  • $\begingroup$ Think of $\int_\Omega u(y) dy$ as a constant, so when we integrate (say, integrate in $z$) we have $\int_\Omega (\int_\Omega u(y) dy) u(z) dz = \int_\Omega u(y)dy \cdot \int_\Omega u(z)dz.$ $\endgroup$ – Matt Mar 31 '17 at 19:42
  • $\begingroup$ You're right. I got the point. Thanks :) $\endgroup$ – express78 Mar 31 '17 at 19:46
  • $\begingroup$ You're welcome! Good luck! $\endgroup$ – Matt Mar 31 '17 at 19:47

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