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.



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.

  • $\begingroup$ 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

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 $0$.

  • $\begingroup$ $\frac{\partial f}{\partial n} = \nabla{f} \cdot \vec{n}$ 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
  • 1
    $\begingroup$ $\frac{\partial f}{\partial n} := \nabla{f} \cdot \vec{n}$ 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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