Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A friend of mine presented me the following problem a couple days ago:

Let $S$ in $\mathbb{R}^2$ be a square and $u$ a continuous harmonic function on the closure of $S$. Show that the average of $u$ over the perimeter of $S$ is equal to the average of $u$ over the union of two diagonals.

I recall that the 'standard' mean value property of harmonic functions is proven over a sphere using greens identities. I've given this one some thought but I haven't come up with any ideas of how to proceed. It's driving me crazy! Maybe it has something to do with the triangles resulting from a diagonal? Any ideas?

share|improve this question
    
Please, could you define average of $u$? –  Tomás Apr 25 '13 at 18:27
    
I would assume its defined in the same sense as it is here: en.wikipedia.org/wiki/Harmonic_function#The_mean_value_property –  Jax Apr 25 '13 at 18:40
1  
The average of a function $u$ over a set $E$ is assumed to be $\frac{1}{|E|} \int_E u$. In this case, the assumption would be that we mean 1-dimensional Hausdorff measure in the usual way on these line segments. –  Ray Yang Apr 25 '13 at 18:50

2 Answers 2

This is a re-interpretation of Ray Yang's answer, which also shows how the result can be generalized to other polygons. Introduce the function $$v(x,y)=(1-\max(|x|,|y|))^+ ,\qquad (x,y)\in\mathbb R^2$$ This is a compactly supported Lipschitz function with support $Q=[-1,1]\times [-1,1]$. Its graph is a pyramid with $Q$ as the base.

If $u$ is harmonic in a neighborhood of $Q$, then integration by parts yields $$0=\int_{\mathbb R^2} v\,\Delta u = \int_{\mathbb R^2} u\,\Delta v \tag{1}$$ By considering $u(\alpha x,\alpha y)$ with $\alpha\to 1^-$, we extend (1) to functions continuous in $Q$ and harmonic in its interior.

It remains to observe that $\Delta v$ is the distribution composed of

  • the linear measure on $\partial Q$
  • $-\sqrt{2}$ times the linear measure on the diagonals of $Q$

This follows from considering the discontinuities of the normal derivative of $v$ across the aforementioned lines; elsewhere $v$ is harmonic. One can also save the trouble of calculating the factor of $-\sqrt{2}$ by using the fact that $\int_{\mathbb R^2}\Delta v=0$.


It should be clear that there is nothing special about the square and its diagonals: any piecewise linear compactly supported function gives rise to a similar identity. For example, one can build a pyramid on top of any regular polygon $P$ and conclude that the average of a harmonic function along $\partial P$ is equal to its average over the union of segments connecting the vertices of $P$ to its center. No computations of slopes are necessary: it's clear that $(\Delta v)^+$ and $(\Delta v)^-$ are constant multiples of linear measure, and the identity $\int_{\mathbb R^2}\Delta v=0$ gives all the information we need.

share|improve this answer

Consider the isosceles right triangles formed from two sides of the square and a diagonal.

Let's consider the first such triangle. Call the sides $L_1, L_2$ and $H_1$ for legs and hypotenuse; for sake of convenience, our square is the unit square, so we give the triangle $T_1$, legs $L_1$ from $(0,0)$ to $(1,0)$ and $L_2$ from $(1,0)$ to $(1,1)$; the hypotenuse $H_1$ clearly runs from $(0,0)$ to $(1,1)$.

Now consider the function $\phi(x) = |x-y|$, and we're going to use integration by parts.

$$ \int_{\partial T_1} u \phi_{\nu} = \int_{\partial T_1} \phi u_{\nu} $$ since $u$ and $\phi$ are both harmonic in the interior of the triangle $T$. Now, $$ \int_{\partial T_1} u \phi_{\nu} = - \sqrt{2} \int_{H_1} u + \int_{L_1} u + \int_{L_2} u = \int_{L_1} x u_\nu + \int_{L_2} (1-y) u_\nu = \int_{\partial T_1} \phi u_\nu $$

Perform the same construction on the other triangle $T_2$ with hypotenuse $H_1$, but with legs $L_3$ from (0,0) to (0,1) and $L_4$ from (0,1) to (1,1). We get $$ \int_{\partial T_2} u \phi_{\nu} = - \sqrt{2} \int_{H_1} u + \int_{L_3} u + \int_{L_4} u = \int_{L_3} y u_\nu + \int_{L_4} (1-x) u_\nu = \int_{\partial T_2} \phi u_\nu $$

Now consider the function $\psi(x) = |x+y-1|$ on the triangle $T_3$ formed by $L_1$ as above, $L_3$ from (0,0) to (0,1), and $H_2$ from (1,0) to (0,1). $$ \int_{\partial T_3} u \psi_{\nu} = - \sqrt{2} \int_{H_2} u + \int_{L_3} u + \int_{L_1} u = \int_{L_3} (1-y) u_\nu + \int_{L_1} (1-x) u_\nu = \int_{\partial T_3} \psi u_\nu $$

Finally, on the triangle $T_4$ formed by $L_4$, $L_2$, and $H_2$ we have $$ \int_{\partial T_4} u \psi_{\nu} = - \sqrt{2} \int_{H_2} u + \int_{L_2} u + \int_{L_4} u = \int_{L_2} y u_\nu + \int_{L_4} x u_\nu = \int_{\partial T_4} \psi u_\nu $$

Summing all these terms together, we get $$ -2 \sqrt{2} \int_{H_1 \cup H_2} u + 2 \int_{\partial S} u = \int_{\partial S} u_\nu$$ Since $u$ is harmonic, this must equal 0, which tells us that the average over the diagonals is the average over the perimeter.

share|improve this answer
    
I originally posted something incorrect which took some time to correct. I hope this is okay, but have not had time to proofread it carefully. –  Ray Yang Apr 25 '13 at 18:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.