# Uniqueness of $\Delta u= \int _\Omega u(y)dy$ , $x \in \Omega$

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 $w=0$

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 $u \equiv 0$.
• How does the right hand side become $\int_\Omega u \int_\Omega u = \left(\int_\Omega u\right)^2.$ – express78 Mar 31 '17 at 19:38
• Think of $\int_\Omega u(y) dy$ as a constant, so when we integrate (say, integrate in $z$) we have $\int_\Omega (\int_\Omega u(y) dy) u(z) dz = \int_\Omega u(y)dy \cdot \int_\Omega u(z)dz.$ – Matt Mar 31 '17 at 19:42