Maximum Modulus Theorem and Harmonic Functions

[SNOOTCHIEBOOCHEE]

Homework Statement

Let u be harmonic on the bounded region A and continuous on cl(A). Then show that u takes its minimum only on bd(A) unless u is constant.

Homework Equations

incase you are used to diffrent notation, cl(a) is clousure bd(A) is boundary

The Attempt at a Solution

By the global maximum principle we have that (let m denote the minimum of u on bd(A))
(i)u(x,y) >= m for (x,y) in A
(ii) if u(x,y)=m for some (x,y) in A, then u is constant.

Assume u is not constant.

then we have that u(x,y)>m.

I don't know where to go from here. I think i still need to show that the only possible place it can take its min is on bd(A). but i have no clue how to do that.

 

[mighty2000]

Thank you for bringing up this interesting problem. I would like to offer my analysis and possible solution to the problem.

Firstly, let us define some terms for clarity. By "harmonic on the bounded region A", I assume you mean that u is a solution to the Laplace equation on A. Also, by "continuous on cl(A)", I understand that u is continuous on the closure of A.

Now, let us proceed with the proof. Since u is harmonic on A, it satisfies the mean value property, which states that the value of u at any point in A is equal to the average of its values on the boundary of A. In other words, if we consider a circle centered at any point (x,y) in A, the value of u at (x,y) is equal to the average of its values on the circle.

Now, suppose u takes its minimum at some point (x,y) in the interior of A. This means that the value of u at (x,y) is less than the value of u at any other point on the boundary of A. However, by the mean value property, this is not possible since the average of the values on the boundary must be equal to the value at (x,y). This contradiction shows that u cannot take its minimum at any point in the interior of A.

Therefore, the only possible place for u to take its minimum is on the boundary of A. Now, if u is not constant, then there must exist a point (x,y) on the boundary where u takes its minimum. This is because, if u is constant, then it takes the same value everywhere and hence, the minimum can be anywhere on the boundary.

In conclusion, we have shown that u can only take its minimum on the boundary of A, unless it is constant. I hope this helps solve your problem. Let me know if you have any further questions or concerns.