Solving Laplace Equation on Square: Proving |u|<=1

[Mechmathian]

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

 

[DavidWhitbeck]

A harmonic function that is not constant has to achieve it's max on the boundary, else it's constant. This is because you can redefine it as an analytic function on the complex plane (since it's harmonic) and apply the maximum modulus principle.

Your solution is then bounded by the bounds it attains on the boundary, which are sin and cos, so they are bounded by 1 which means the modulus of the function itself must be bounded by 1 everywhere (if not then you can find a point in the interior where the function achieves a global max and then forces the function to be constant).

I could be completely off though, it's been along time.

 

[Mechmathian]

Thank you for the answer. It might be right, but I have some questions
1) What are the other 2 boundary conditions given for?
2) I didn't quite understand what you mean by redefining the function as an analytical one on a complex plane.. There does exist a theorem that if u is harmonic, then there exists a harmonic function v, such that u+iv is analytical..

 

[Mechmathian]

Actually what you have said is right.. The second question is taken out. The first one is still unclear to me

 

[Mechmathian]

what I meant to say is that if the maximum is greater then one then it has to be on the other two sides of the square, we just eliminated the first two sides..

 

[Mechmathian]

Solved.