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I am trying to give a solution to this question but I'm getting stuck.

Let $f$ be an harmonic function $V \to \mathbb R$ where $V$ is a solid in $\mathbb R^3$ bounded by a solid surface $S$ with normal $\vec n$. Show that

$$ \iint_S \frac{\partial f}{\partial n} \ dS = 0$$

For the first part I said that a harmonic function is a function that satisfies $$\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=0$$ I think this is right.

Then for the next part I tried to use the divergence theorem to say

$$\iint_S \frac{\partial f}{\partial n}dS=\iiint_R \text {div}(\frac{\partial f}{\partial n})dV$$ then I get stuck working out $$\text {div}(\frac{\partial f}{\partial n})$$

I know it is the dot product between $\nabla$ and $f$ but I don't know how to split $f$ into a vector to take the dot product considering I don't even know what $f$ is.

If anyone here could help guide me that would be fantastic.

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Hints:

This question is mostly about getting the definitions straight. Once that's done you should find the one 'computational' step relatively straight forward.

The integral $$ \iint_S \frac{\partial f}{\partial\vec n} \ dS$$ is, by definition of the directional derivative $\partial/\partial \vec n$, equal to

$$\iint_S \nabla f \cdot \vec n \ dS$$

Now apply the divergence theorem to turn this into an integral over the volume $V$.

Note also that as $V$ is a region in $\mathbb R^3$, the Laplacian of $f$ is $$\nabla^2 f = f_{xx} + f_{yy} + f_{zz},$$ not just the first two terms as in your question.

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  • $\begingroup$ Okay so I get why $$\iint_S \nabla f \cdot \vec{n} dS=0$$ using the divergence theorem I think. (Because $\nabla f \cdot \vec{n}=\nabla f$ as they are parallel) and then we just use the fact that since it is harmonic when we calculate the divergence we get it's $0$ via the Laplacian but I don't understand how you switched the integral in the first place, what defintion are you using I don't get the physical interpretation and the equality of the two integrals sorry. $\endgroup$ – James Jul 22 '15 at 21:06
  • $\begingroup$ A few things (a) We don't know if $\nabla f$ and $\vec n$ are parallel. There is no reason they should be. (b) The equation $\nabla f \cdot \vec n = \nabla f$ doesn't make sense as the LHS is a scalar and the RHS is a vector. (c) Just to clarify, using the divergence theorem we have that $$\iint_S \nabla f \cdot \vec n \ dS = \iiint_V \operatorname{div}(\nabla f) \ dV = \iiint_V \nabla^2 f \ dV $$ As $f$ is harmonic, $\nabla^2 f = 0$ and thus the RHS integral is zero. $\endgroup$ – Simon S Jul 22 '15 at 21:10
  • $\begingroup$ As to why $\displaystyle \frac{\partial f}{\partial \vec n} = \nabla f \cdot \vec n$: this is the definition of the directional derivative. Take the gradient $\nabla f$ then project that onto the direction $\vec n$ (which is a unit vector, being the unit normal to $S$.) $\endgroup$ – Simon S Jul 22 '15 at 21:10
  • $\begingroup$ Okay I think I understand better now thanks. $\endgroup$ – James Jul 22 '15 at 21:17
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Divergence theorem tells us that $$ \iiint_V div(F) \, dV = \iint_S F \cdot \vec{n} \, dS $$ Note that $\frac{\partial f}{\partial n} = \nabla{f} \cdot \vec{n}$ so $$ \iint_S \frac{\partial f}{\partial n} dS = \iint_S \nabla{f} \cdot \vec{n} \, dS = \iiint_V div(\nabla{f}) \, dV = \iiint_V \nabla^2{f} \, dV$$ The condition you've written for a harmonic function is a statement for a function of 2 variables of the general condition that $\nabla^2{f} = 0$ so the integrand is $0$ and hence the whole integral is $0$.

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  • $\begingroup$ $\frac{\partial f}{\partial n} = \nabla{f} \cdot \vec{n}$ I don't follow this (and only this) step. Could you explain why this is the case maybe? Thanks. What does $\frac{\partial f}{\partial n}$ even mean in this context? $\endgroup$ – James Jul 22 '15 at 21:08
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    $\begingroup$ $\frac{\partial f}{\partial n} := \nabla{f} \cdot \vec{n}$ is one possible definition of $\frac{\partial f}{\partial n}$ and works perfectly well. If you aren't happy with this you can take the alternative definition that $$\frac{\partial f}{\partial n}(\vec{x}) := \lim_{t \to 0} \frac{f(\vec{x} + t \vec{n}) - f(\vec{x})}{t}$$ and prove by applying the chain rule that these two are equivalent. $\endgroup$ – Rhys Steele Jul 22 '15 at 21:19
  • $\begingroup$ Good idea on the directional derivative. @James, you should convince yourself the limit RJS has written down is the same as $\nabla f \cdot\vec n$. $\endgroup$ – Simon S Jul 22 '15 at 21:51

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