Harmonic function in square -PDE

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Homework Help Overview

The discussion revolves around finding a harmonic function within a specified square region, subject to various boundary conditions. The problem involves solving partial differential equations (PDEs) and understanding the implications of boundary conditions on the eigenvalues and eigenfunctions used in the solution process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the separation of variables method applied to solve the PDE. Questions are raised regarding the starting index of the summation for eigenvalues in different parts of the problem, particularly why one part starts at n=1 while another includes n=0. There is also exploration of the implications of boundary conditions on the solutions.

Discussion Status

Some participants have provided insights into the reasoning behind the eigenvalue choices and the nature of the solutions. There is an ongoing exploration of the mathematical concepts involved, with no explicit consensus reached on all points raised.

Contextual Notes

Participants note that the boundary conditions significantly influence the eigenvalues and eigenfunctions, leading to different starting points for the summation in the solutions. The discussion also highlights the importance of understanding the implications of specific boundary conditions on the solutions to the PDEs.

Dassinia
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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
 
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Dassinia said:
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.
 
What I don't understant, why in part 1, n starts at 1 and not 0 ?

The solution of Y&#039;&#039; = 0 which satisfies Y(0) = 0 and Y(1) = 0 is Y(y) = 0.
 
pasmith said:
The solution of Y&#039;&#039; = 0 which satisfies Y(0) = 0 and Y(1) = 0 is Y(y) = 0.
I get it, thank you !
 

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