Mean Value Property in partial differential equations

[ajax2000]

Homework Statement

S is a ball of radius 1 in R^2;
Δu=0 in S
u=g in ∂S, g(x1,x2)>1 for any (x1,x2) in ∂S. Show that for any r satisfying 0<r<1 there is a point (x1,x2) in S such that u(x1, x2) >=1.

Homework Equations

using mean value formula: ∫u(y)dy=1/Vr^n(∫u(y)dy)

The Attempt at a Solution

g(x1,x2)=(1/2π)∫g dS, g(x1,x2)>1 on the boundary

(1/2π)∫0 to 2π g dS=1/2π(2π*g^2/2), 2π cancels out, leaving g(x1,x2)=g^2/2>1
u=g on the boundary so g(x1,x2)=u(x1,x2)>1.

I don't think I did this right, I might have to use u(0,0), any hints will be appreciated

 

[mighty2000]

!

Your attempt at a solution is a good start, but there are a few things that need to be clarified and corrected. First, the mean value formula you used is not applicable in this case, as it is only valid for functions that are harmonic in the entire domain. In this problem, the function u is only harmonic inside the ball S, not in the entire space R^2.

To solve this problem, we can use the maximum principle for harmonic functions. This principle states that the maximum value of a harmonic function inside a domain is always attained on the boundary of the domain. Since u=g on the boundary of the ball S, we know that the maximum value of u inside S is equal to the maximum value of g on the boundary.

Now, we are given that g(x1,x2)>1 for any (x1,x2) in ∂S. This means that the maximum value of g on the boundary is greater than 1. Therefore, the maximum value of u inside S must also be greater than 1. This proves that for any r satisfying 0<r<1, there must exist a point (x1,x2) in S such that u(x1,x2)>=1.

I hope this helps clarify the solution to this problem. Good luck with your studies!