Mean value of a harmonic function on a square

[bndnchrs]

Homework Statement

The idea is to prove that the average of a harmonic function over a square is the same as the average over its diagonals.

Homework Equations

Really, none, other than the mean value theorem, that is the value of the function at a point is the same as the average of the function over the boundary of a ball at any radius from it.

The Attempt at a Solution

I'm not really sure how to get here. I've thought about using the compactness of the square to my advantage but that doesn't guarantee anything about the collection of balls and wouldn't do my much good anyways, at least I can't see how. I know of a proof using complex analytic functions but this isn't applicable here, I'm looking strictly at real-defined functions. An important thing in that proof is that the integral over a closed contour is zero, but I can't assume that for a real function... right?

Steps in the right direction are appreciated, I wan't to work this one out but I need a nudge.

 

[mighty2000]

Thank you for your question. The average of a harmonic function over a square is indeed the same as the average over its diagonals. This is a well-known property of harmonic functions and can be proven using the mean value theorem. However, let me provide some steps that may help you in your proof:

1. Recall that a harmonic function is one that satisfies Laplace's equation, which is ∇²f = 0. This means that the function has the same value at any point as the average of its values over any closed surface surrounding that point.

2. Consider a square with side length a and let f(x,y) be a harmonic function over this square. Let the coordinates of the center of the square be (x0,y0).

3. Using the mean value theorem, we can write the average of f over the square as:
f(x0,y0) = 1/a² ∫∫f(x,y)dxdy, where the integral is taken over the square.

4. Now, let us divide the square into four smaller squares by drawing diagonals. Each of these smaller squares has side length a/√2.

5. Using the same argument as in step 3, we can write the average of f over each of these smaller squares as:
f(x0,y0) = 1/(a/√2)² ∫∫f(x,y)dxdy, where the integral is taken over each smaller square.

6. Since the integrals in steps 3 and 5 are taken over different areas, we can equate them and get:
∫∫f(x,y)dxdy = (√2/a)² ∫∫f(x,y)dxdy.

7. Simplifying, we get:
∫∫f(x,y)dxdy = 2/a² ∫∫f(x,y)dxdy

8. This means that the average of f over the square is equal to twice the average over each of the smaller squares.

9. Repeating this process for each of the four smaller squares, we can see that the average over the entire square is equal to four times the average over each of the smaller squares.

10. But since each of the smaller squares has the same average, we can conclude that the average over the entire square is equal to the average over the diagonals.

I hope this helps in your proof