Question.Let $Ω$ be a bounded Connected on $R^3$ with smooth boundary $\partial{Ω}$.Let $u$ be a harmonic function on $Ω$ with continuous derivatives on $Ω\cup \partial{Ω}$ prove that. $$\iint_V \ {\partial{u}}/\partial{n} \, dA=0$$ $V= \partial{Ω}$ Doesnt state what $n$ is .
My answer.(this was an exams question on pdes today)
: Let $F=u+i0$ Since $u$ and $0$ are harmonic $F$ is analytic since $u$ and $0$ satisfy Riemann-cauchy eq .Since $F$ is analytic its derivatives are so. So $\partial{F}/\partial{n}$ is analytic so the integral in turn is zero along the curve of the boundary $\partial{Ω}$ (Cauchy's theorem for analytic functions on closed curves) of the boundary $\partial{Ω}$ Now since the contour integral is zero Using (green-stokes) theorem which transpose's the contour integral on an open connected set to the region of the boundary of the set into a double integral i get the result requested to prove.
Now i Know i made mistakes i answered this at the very last moment and it was supposed to be a pde's exams.Not a complex analysis one .
But Is my answer correct?Or at least intuitively correct?And how would you say it if it is theoretical correct.?
Is there any problem because my set is on $R^3$ not $R^2$ as im used to?
How would grade this answer ?/2 (2/2 equals 100% correct)
Your answer doesn't make much sense - complex analysis does not apply in any sense when the dimension is odd. I'm guessing $\partial/\partial n$ denotes the normal derivative along the boundary, in which case the divergence theorem provides the answer: since $u$ is harmonic, $\nabla u$ is divergence-free, and thus $$\iint_{\partial \Omega} \frac{\partial u}{\partial n} = \iint_{\partial \Omega}\nabla u \cdot n=\int_\Omega {\rm div}\ \nabla u = \int_\Omega \Delta u = 0.$$