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I need some help with the following question. Any help is greatly appreciated. Thank you very much.

Consider the Neumann problem $$ \begin{eqnarray*} \Delta u =& f(x,y,z) &\text{in } D \\ \frac{\partial u}{\partial \mathbf n} =& 0 &\text{on } \operatorname{bdy} D \end{eqnarray*} $$

  1. What can we surely add to any solution to get another solution? So we do not have uniqueness.

  2. Use the divergence theorem and the PDE to show that $$ \iiint\limits_{D} f(x,y,z) \ \mathrm{d}x \ \mathrm{d}y \ \mathrm{d}z = 0 $$ is a necessary condition for the Neumann problem to have a solution.

  3. Give a physical interpretation for part (1.) and/or part (2.) for either heat flow or diffusion.

I just don't really know what to do exactly.

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Well, first let's look at (a), as it's quite simple. If $u$ is a solution, what can you add to $u$ so that it is still a solution? I'll give you a big hint: what do first-year calculus students forget to put at the end of an indefinite integral? As for part (b), if $f=\Delta u=\nabla\cdot\nabla u$, how might we go about writing $\iiint_D f dV$ as one side of the divergence theorem? – anon Sep 21 '11 at 3:04
up vote 3 down vote accepted

For the first part, note that any of the two equations use differentiation of $u$. If you add a constant to $u$ then when you differentiate, that constant vanishes, and therefore the equations still hold.

For the second part, note the following well known formulas from calculus; maybe they were presented in your pde course:

Integration by parts: $$\int_\Omega \frac{\partial u}{\partial x_i}v d \Omega=\int_\Gamma uv\nu_i d \Gamma-\int_\Omega u \frac{\partial v}{\partial x_i} d \Omega,$$ where $d\Omega$ is standard integration and $d \Gamma$ is surface integration. Summing this for $v=v_i$ each $i=1..n$ this leads to

$$\int_\Omega \nabla u \cdot v d \Omega=\int_\Gamma (uv)\cdot \nu d \Gamma-\int_\Omega u ( \nabla \cdot v) d \Omega,$$

and taking $v=\nabla v $ in the last formula you get Green's Formula. $$ \int \nabla u \cdot \nabla v d\Omega =\int_\Gamma u \nabla v \cdot \mu d \Gamma-\int_\Omega u \Delta v d \Omega$$

Taking $u\equiv 1$ in the second formula you get the divergence theorem which plainly states that $$0 = \int_\Gamma v \cdot \nu d \Gamma-\int_\Omega \text{div v}d \Omega $$

These formulas are basic, so you better keep them in mind. To solve point two, just note that putting $v=\nabla u$ in the last formula, or putting $u=1$ in Green's formula you get that

$$0=\int_\Gamma \frac{\partial u}{\partial n}d \nu- \int_\Omega \Delta u d \Omega$$

Now, since $\Delta u=f$ and $\frac{\partial u}{\partial n}=0$, replacing these two in the equality gets you to the result.

A nice video about the divergence theorem is presented here

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