Laplace and Divergence theorem

draco193
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Homework Statement



Use Divergence theorem to determine an alternate formula for \int\int u \nabla^2 u dx dy dz Then use this to prove laplaces equation \nabla^2 u = 0 is unique. u is given on the boundary.

Homework Equations



u \nabla^2 u = \nabla * (u \nabla u) -(\nabla u)^2

The Attempt at a Solution



Using Divergence theorem, I get that the new equation should be \oint (u \nabla u) *n -\oint (\nabla u)^2 where n is the normal vector.

I wanted to make sure that I had applied this correctly before moving onto the next part.

My plan for the next part would then be to say that u=v-w, where v and w are arbitrary vectors, and show that v =w to show uniqueness of u.
 
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Purely as a TeXnical note: The dot product operator can be written \cdot, and you can get double and triple integral symbols which are properly spaced using \iint and \iiint. Also, differentials in integrals look better if you put a thin space \, before them. So your first equation might be set \iint u \, \nabla^2 u \,dx \,dy \,dz.
 
draco193 said:
Use Divergence theorem to determine an alternate formula for \int\int u \nabla^2 u dx dy dz Then use this to prove laplaces equation \nabla^2 u = 0 is unique. u is given on the boundary.

Please post the original problem statement word for word... If you only specify u on the boundary of a finite region, the solution to \nabla^2 u = 0 is not unique.

Consider a simple counter example: u_1=1 and u_2=\frac{r}{R}...both satisfy Laplace's equation, and both have the same value on the boundary of the sphere r \leq R.
 
@ystael: Thanks. The formatting of the Latex is new stuff to me. :smile:

@gabba: That is the exact problem. I think the idea is to prove that u=v-w=0, and then by the min max principle, u is the only solution.

As an update, the professor said there was a mistake in the book, so we could just treat it as a double integral (not a double with respect to three variables). I also realized that there is no reason to apply divergence to the (\nabla u)^2 term, so my answer for the first part should read

\oint (u\nabla u)\cdot n-\iint (\nabla u)^2
 
draco193 said:
@gabba: That is the exact problem.

You can't prove something that isn't true. Specifying u alone on the boundary will not generate a unique solution (Laplace's equation is a 2nd order DE, and hence requires at least two boundary conditions to be uniquely solved for a given region). If the problem also tells you that \mathbf{\nabla}u is specified on the boundary, then you can use the hint by assuming u_1 and u_2 are both solutions that satisfy the same boundary conditions and then looking at u_3 \equiv u_2 - u_1.

Is this problem from a textbook? If so, what text and what is the problem number?
 
gabbagabbahey said:
You can't prove something that isn't true. Specifying u alone on the boundary will not generate a unique solution (Laplace's equation is a 2nd order DE, and hence requires at least two boundary conditions to be uniquely solved for a given region). If the problem also tells you that \mathbf{\nabla}u is specified on the boundary, then you can use the hint by assuming u_1 and u_2 are both solutions that satisfy the same boundary conditions and then looking at u_3 \equiv u_2 - u_1.

Is this problem from a textbook? If so, what text and what is the problem number?

It is a textbook problem. Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 4th Edition Haberman #2.5.12 a and b.
 
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