Prove that |u|<=1: Laplace Eq. on [0,1]^2, Boundary Cond.

[Mechmathian]

1. We look at a Laplace equation ( \Delta u(x,y) =o) on a square [0, 1]* [0, 1]
If we know that u_{x=o}= siny , u_{x=1}= cosy
u_y|_{y=0}= 0 , u_y|_{y=1}= 0 we differentiate here by y. proove that |u|<=1.

The Attempt at a Solution

We now know that the maximum of u has to be on the boundary. If it is greater then one, then it has to be on either y=0 or y =1.

 

[Mindscrape]

Yes, that is true and I would say good enough for an applied math course, with just a little bit more about the maximum value theorem (or whatever it is called) to justify.

 

[Mechmathian]

OK)) Could you tell me what that "little bit " is)?

 

[Dick]

The solution to any old PDE doesn't satisfy the maximum principle. The solutions to the Laplace equation do, because they are a special kind of function with a special name. I think mindscrape just wants to you point to the theorem that says that the maximum is on the boundary.

 

[Mechmathian]

Guys, I have solved it!
No more help needed on it!