Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Seb note that $f$ could change sign ! But if in addition you have that $f$ is positive then you right –  Leandro Oct 31 '11 at 2:56
6  
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. –  Did Oct 31 '11 at 2:58
    
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. –  Jeff Nov 4 '11 at 1:06
add comment

2 Answers

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.

share|improve this answer
    
Or, both of those could be negative. –  Dhaivat Pandya Dec 4 '11 at 2:47
    
$\lVert \nabla f\rVert$ tends to be non-negative rather than negative, Dhaivat. –  Mariano Suárez-Alvarez Dec 4 '11 at 6:59
add comment

Let me extend the boundary beyond its original limit, filling up the newly added space by f=0.

$\int\int\int \nabla^2 f dV=\int\int\int \nabla.(\nabla f) dV$

$=\int\int (\nabla f).n dS=0$

Therefore, $\int\int\int \nabla^2 f dV=0$

But $\int\int\int \nabla^2 f dV$ for the additional space is zero. Therefore the integral is zero for the original space also.

We may now consider a sub-volume[V'] inside the original 3D region. Let the surface/boundary of this sub-volume be S'. We may try to perform the same trick as before but the difficulty is that f in general is not zero on the boundary.A continuous extension with f=0 will not be possible. This time we may carry out an infinitesimally small extension so the gradf becomes zero on the extended boundary. Again the volume integral is zero.

The integral seems to vanish for an arbitrary volume

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.