I am having problems in comprehending the proof of contradiction used by Purcell in his book;

...We can now assert that $W^1$ must be zero at all points in space. For if it is not, it must have a maximum or minimum somewhere-remember that $W$ is zero at infinity as well as on all the conducting boundaries. If $W$ has an extremum at some point $P$, consider a sphere centered on that point. As we saw in chapter2, the average over a sphere of a function that satisfies Laplace's equation is equal to its value at the center. This could not be true if the center is extremum; it must therefore be zero everywhere. $^1 W = \phi(x,y,z) - \psi(x,y,z)$, where the former term is the deduced solution & the later term is the assumed solution in order to proof contradiction.

Extremum means local maximum or local minimum, right? Why can't average be equal to an extrmum value? If it is not equal to the average value, how does it ensure that at all the places $W$ is equal?

  • $\begingroup$ Comment to the question (v1): This mathematical fact is known as the maximum principle for harmonic functions. A (real) harmonic function $f$ cannot have a maximum at an interior point. (Since $-f$ is also a harmonic function, then $f$ cannot have a minimum at an interior point as well.) $\endgroup$ – Qmechanic Jun 26 '15 at 15:01
  • $\begingroup$ Depending on what is troubling you about that statement it might help to recall that the choice of sphere is arbitrary, and that means it is no help assuming a conspiracy to works in a particular geometry: I can pick a smaller sphere... $\endgroup$ – dmckee Jun 26 '15 at 15:02

Well, if you have an extremum (say local maximum) of $W$ at $p$, then you have a small open ball $N$ centered in $p$ such as $W(x)< W(p)$ for some $x\ne p$ in $N$. Therefore $$ \frac{1}{{\rm vol}(N)}\int_N W(x)dx < W(p)~~.\tag{1}$$ assuming we have taken the ball $N$ small enough for $W(x)<W(p)$ for some $x\ne p$ in $N$. But if $\nabla^2W=0$, by the mean value theorem of harmonic functions you know that $$\frac{1}{{\rm vol}(N)}\int_N W(x)dx= W(p)~~,$$ which contradicts the previous result. Similar situation for $-W$, since $-W$ is also harmonic. Therefore there cannot be local extrema (minima or maxima) inside the domain, but they can be at the borders. However, if in the borders $W(x)=0$, then there is no other choice than $W\equiv 0$.

  • $\begingroup$ I'm not sure your proof is 100% rigorous. There are functions $W$ that have local extrema at which the $D^2 W(p)$ matrix is identically zero; for example, the function $W(x,y) = x^4 + y^4$ at $(x,y) = (0,0)$. In such cases, you can't "pick sufficiently small $|x-p|$ so the third term is negligible compared with the second." $\endgroup$ – Michael Seifert Jun 26 '15 at 17:11
  • $\begingroup$ @MichaelSeifert yes, you are right. The argument is not sufficiently general. But I think the definition of local maxima is enough. $\endgroup$ – Enredanrestos Jun 26 '15 at 18:00
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    $\begingroup$ @user36790 The property your book is talking about is the mean value theorem of the harmonic functions. An average is not an extremum value unless all the values are equal. $\endgroup$ – Enredanrestos Jun 26 '15 at 18:02
  • $\begingroup$ @Enredanrestos Right; so long as you can prove that the strict inequality holds, your proof follows. If it's a strict maximum ($W(x) < W(p) \forall x \in N$ excluding $p$), then you're golden. If it's a maximum but not a strict one ($W(x) \leq W(p)$, with equality holding at some points), then you have to be a little more careful. I think in this case you can conclude that $W$ is a constant in $N$, and the rest of the proof follows from there. $\endgroup$ – Michael Seifert Jun 26 '15 at 18:08
  • $\begingroup$ @MichaelSeifert But isn't that the definition of local maximum? with a strict inequality? $\endgroup$ – Enredanrestos Jun 26 '15 at 18:17

Ok, here is an answer in some simple words.

Harmonic functions obey the following property: if you draw a circle around any point $x_0$, and take the average of the function over that circle, that average will be equal to the value of the function at $x_0$.

Now, if $W$ is not constant, it must have a maximum somewhere. Let's call this maximum point $x_0$. Then we can draw any circle centered at $x_0$. The average value of the function on this circle must be equal to the value of the function at $x_0$. We know that at every point on the circle, the $W$ is at MOST equal to $W(x_0)$, since this is the maximum. If at any point on the circle, the function was LESS than $W(x_0)$, the average of the function on the circle couldn't be $W(x_0)$ like we know it has to be*. Thus, the function is equal to $W(x_0)$ at every point on the circle. Since this holds for EVERY CIRCLE centered at $x_0$, we have that the function is in fact constant everywhere.

*A simple way to think of this: Suppose you have a set of numbers that you know has to average to 5, and you know that no number in the set can be bigger than five. If there's a 4 in the set, you know you're doomed, since then there's no way to get the average up to five! Thus, the set just has to be all 5s.


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