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Let $D\subset \mathbb{R}^3$. Let $D$ be a connected subset of $\mathbb{R}^3$. Show that if there is a solution of the system of equations \begin{equation} \Delta u=f \text{ in } D, \frac{du}{dn}=g\text{ on boundary of } D, \end{equation} then $\int_D\ f \ dV=\int_{\text{boundary of} D} g \ ds$.

My partial answer:

Assume that $\int_D\ f \ dV\neq\int_{\text{boundary of} D} g \ dS$ and $u$ is the solution of the system of equation, then

\begin{equation} \int_D \ \Delta u \ dV= \int_D\ f \ dV\neq\int_{\text{boundary of} D} g \ dS= \int_{\text{boundary of} D} \frac{du}{dn} \ dS. \end{equation} This contradicts Green's first identity.

Please let me know that idea of my answer is correct. Is it possible to prove this question without using a contradiction.

dmtri
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The solution is correct. It can be phrased in a more direct way, if your statement is made less "negative". Instead of saying "there does not exist a solution unless equality holds", we could say "if there exists a solution, then equality holds". Then the proof is direct: let $u$ be a solution then use Green's identity and finally conclude that the equality holds.

In any case, yours is not really a proof by contradiction (reductio ad absurdum), but rather a proof of contrapositive statement.

dmtri
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    @dmtri: Your edit (to the Question above) largely obviates the Accepted Answer here. Normally I would reject it on that basis. But (as the Answer points out) rephrasing the Question really does improve its clarity, and neither the OP of the Question or the Answer is currently active, so I think it is worthwhile in this restricted case. – hardmath Mar 10 '19 at 20:04