Harmonic function in square -PDE

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Hello, I have the solution of a problem but there's something I don't understand

Homework Statement


Find the harmonic function in the square {0<x<1, 0<y<1} with the boundary conditions
u(x,0)=x
u(x,1)=0
ux(0,y)=0
ux(1,y)=y²t


Homework Equations




The Attempt at a Solution


Part1:[/B]
We first solve the problem in the square with BC
u(x,0)=0 ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=y²

The solution is
u(x,y)= ∑ An cosh(nπx) sin(nπy) for n=1,2,....

Part2:
Then solving the problem in the square with BC
u1(x,0)=x ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=0

By separation of variable we solve
X''=-λX with X'(0)=X'(1)=0
Here the solution says that λ=(nπ)² n≥0 with eigenfunctions Xo=constant and Xn=cos(nπx) for n>0

and Y''=λY
For λ=0, we get Y''=0, solving and applying BC Yo(y)=A(y-1)
And for n=1,2,... Yn(y)=A(cosh(nπy)-cothan(nπy)*sinh(nπy))
u2(x,y)=Ao(y-1)+ΣAn (cosh(nπy)-cothan(nπy)*sinh(nπy))*cos(nπx)

What I don't understant, why in part 1, n starts at 1 and not 0 ?
And why in part 2 it starts at n=0, I understand that the sum is not defined for n=0 (coth(0)=oo) but how can I know the values that take n when i solve for the functions X and Y ?

Thanks
 

Answers and Replies

  • #2
LCKurtz
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Hello, I have the solution of a problem but there's something I don't understand

Homework Statement


Find the harmonic function in the square {0<x<1, 0<y<1} with the boundary conditions
u(x,0)=x
u(x,1)=0
ux(0,y)=0
ux(1,y)=y²t


Homework Equations




The Attempt at a Solution


Part1:[/B]
We first solve the problem in the square with BC
u(x,0)=0 ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=y²

The solution is
u(x,y)= ∑ An cosh(nπx) sin(nπy) for n=1,2,....

Part2:
Then solving the problem in the square with BC
u1(x,0)=x ; u(x,1)=0 ; ux(0,y)=0 ; ux(1,y)=0

By separation of variable we solve
X''=-λX with X'(0)=X'(1)=0
Here the solution says that λ=(nπ)² n≥0 with eigenfunctions Xo=constant and Xn=cos(nπx) for n>0

and Y''=λY
For λ=0, we get Y''=0, solving and applying BC Yo(y)=A(y-1)
And for n=1,2,... Yn(y)=A(cosh(nπy)-cothan(nπy)*sinh(nπy))
u2(x,y)=Ao(y-1)+ΣAn (cosh(nπy)-cothan(nπy)*sinh(nπy))*cos(nπx)

What I don't understant, why in part 1, n starts at 1 and not 0 ?
And why in part 2 it starts at n=0, I understand that the sum is not defined for n=0 (coth(0)=oo) but how can I know the values that take n when i solve for the functions X and Y ?

Thanks
In part 2 you have an eigenvalue ##\lambda = 0## with eigenfunctions ##X_0 = 1## and ##Y_0 = y-1##. So your expansion would be$$
\sum_{n=0}^\infty c_nX_n(x)Y_n(y) = c_0(y-1) + \sum_{n=1}^\infty c_n(...)$$I didn't work through part 1 but my guess is that you don't have an eigenvalue ##\lambda = 0##.

As an aside, you might find it instructive working the ##Y## equation to assume a form ##Y = C\sinh(n\pi y) + D\sinh(n\pi(1-y))## instead of the ##\{\sinh ,\cosh\}## pair.
 
  • #3
pasmith
Homework Helper
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What I don't understant, why in part 1, n starts at 1 and not 0 ?
The solution of [itex]Y'' = 0[/itex] which satisfies [itex]Y(0) = 0[/itex] and [itex]Y(1) = 0[/itex] is [itex]Y(y) = 0[/itex].
 
  • #4
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The solution of [itex]Y'' = 0[/itex] which satisfies [itex]Y(0) = 0[/itex] and [itex]Y(1) = 0[/itex] is [itex]Y(y) = 0[/itex].
I get it, thank you !
 

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