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If $\int\limits_V f \; \mathrm dV = 0$ can we say that $f=0$ everywhere? Or what conditions are there on concluding this.

In particular I want to solve the PDE $\nabla^2 f=f^3$ on the region $$D=\{(x,y)\in\mathbb{R}|x^2+y^2<1\}$$ with $f=0$ on the boundary.

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    Seb note that $f$ could change sign ! But if in addition you have that $f$ is positive then you right2011-10-31
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    If this holds for only one given domain $V$, no chance unless you add the condition that $f\geqslant0$ everywhere. If this holds for every subdomain $V$, you win. In both cases the conclusion is that $f=0$ except on a set of measure zero--unless $f$ has some regularity such as being continuous, and then this set of measure zero is in fact empty.2011-10-31
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    Multiply both sides by $f$ and integrate by parts to get $\int_V \|\nabla f\|^2 dx = -\int_V f^4 dx$. Since $\|\nabla f\|$ and $f^4$ are both non-negative, this implies that $f=0$ is the only solution.2011-11-04

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