# Laplace equation

maria clara

## Homework Statement

I need to solve Laplace equation in the domain D= 0 < x,y < pi

Neumann boundary conditions are given:

du/dx(0,y)=du/dx(pi,y)=0
du/dy(x,pi)=x^2-pi^2/3+1
du/dy(x,0)=1

2. The attempt at a solution

first, we check that the integral of directional derivative of u on the edge of D is zero. This should be the necessary condition for the existence of a solution to the problem. the condition is satisfied (the integral is zero indeed), but why is this condition sufficient and not only necessary?

Anyway, assuming a solution does exist, I propose a solution of the form u=X(x)Y(t), and solving the appropriate SL system I find that the solution sould be of the form:
u(x,y)= A+sigma{Cos(nx)*[Cosh(ny)+Sinh(ny)]}

but the solution should have the form

u(x,y)= A+By+sigma{Cos(nx)*[Coshny+Cosh[n(y-pi)]}

where does the By come from? and does Cosh[n(y-pi)] equal Sinhny? It doesn't make sense because for example if we have y=pi , Cosh[n(y-pi)] = 1, while Sinh(ny) gives an infinite number of answers, depending on n.

Homework Helper

## Homework Statement

I need to solve Laplace equation in the domain D= 0 < x,y < pi

Neumann boundary conditions are given:

du/dx(0,y)=du/dx(pi,y)=0
du/dy(x,pi)=x^2-pi^2/3+1
du/dy(x,0)=1

2. The attempt at a solution

first, we check that the integral of directional derivative of u on the edge of D is zero. This should be the necessary condition for the existence of a solution to the problem. the condition is satisfied (the integral is zero indeed), but why is this condition sufficient and not only necessary?
I don't understand this. What makes you think this condition is sufficient?

Anyway, assuming a solution does exist, I propose a solution of the form u=X(x)Y(t), and solving the appropriate SL system I find that the solution sould be of the form:
u(x,y)= A+sigma{Cos(nx)*[Cosh(ny)+Sinh(ny)]}

but the solution should have the form

u(x,y)= A+By+sigma{Cos(nx)*[Coshny+Cosh[n(y-pi)]}

where does the By come from? and does Cosh[n(y-pi)] equal Sinhny? It doesn't make sense because for example if we have y=pi , Cosh[n(y-pi)] = 1, while Sinh(ny) gives an infinite number of answers, depending on n.

The By is because of that "+1" on the du/dy at both y= 0 and y=$\pi$.

maria clara
Well, this is what they write in the answer, that here the condition is "necessary and sufficient"....
I guess it is not merely a mistake, because this is not the first time I encounter this argument.

any suggestions regarding the problem in the equation:
cosh[n(y-pi)]=sinh(ny)
?

Belgium 12
It's to satisfy the boundary conditions at 0 and pi.
solutions are u(x,y)=A exp(ny) + Bexp(-ny)
or =a sinh(ny)+bcosh(ny)
or =cosh(ny)+cosh n(y-pi) this satisfy thr BC at 0 and pi
The laplace equation is invariant under translation
belgium 12