ODE/PDE- eighenvalues+ eigenfunctions

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Homework Help Overview

The discussion revolves around a Sturm-Liouville eigenvalue problem represented by the ordinary differential equation (ODE) X'' + λX = 0, defined on the interval 0 < x < 1, with specific boundary conditions X(0) = -2X(1) + X'(1) = 0. Participants are exploring the implications of setting λ to zero and the resulting eigenfunctions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the eigenfunctions when λ = 0, questioning the implications of boundary conditions on the constants involved. There is uncertainty about the validity of certain assumptions regarding the eigenfunction X(x) = x and the role of x = 1/2 in the solution.

Discussion Status

There is an ongoing examination of the boundary conditions and their impact on the existence of non-trivial solutions. Some participants have provided insights into the calculations, leading to a recognition of potential errors in reasoning, particularly regarding the application of boundary conditions.

Contextual Notes

Participants are navigating the constraints imposed by the boundary conditions and the specific case of λ = 0, which raises questions about the nature of the solutions and the constants involved.

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Homework Statement


it's already separable, so it's an ODE function.
X''+[tex]\lambda[/tex]*X=0 0<x<1
X(0)=-2X(1)+X'(1)=0


Homework Equations





The Attempt at a Solution



this is a Sturm-Liouville eigenvalue problem.
Now, I know how to solve it and everything, but I'm not sure with one thing.

when I check the case where [tex]\lambda[/tex]=0,
I get C2(-2x+1)=0
so C2 can be anything, correct ?
now what's my eigenfunction ?
X(x)=x is the eigenfunction ?

do I use the x=1/2 somewhere ?

Thanks.
 
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Roni1985 said:

Homework Statement


it's already separable, so it's an ODE function.
X''+[tex]\lambda[/tex]*X=0 0<x<1
X(0)=-2X(1)+X'(1)=0


Homework Equations





The Attempt at a Solution



this is a Sturm-Liouville eigenvalue problem.
Now, I know how to solve it and everything, but I'm not sure with one thing.

when I check the case where [tex]\lambda[/tex]=0,
I get C2(-2x+1)=0
How did you get that?
so C2 can be anything, correct ?
now what's my eigenfunction ?
X(x)=x is the eigenfunction ?

do I use the x=1/2 somewhere ?

Thanks.
 
vela said:
How did you get that?

well, when lamda is zero X''=0
so, X(x)=C1+C2*x
X(0)=C1=0
and by using the second BC, -2*C2*x+C2=0
so to get a nontrivial solution x=1/2, meaning C2 can be any number.

did I do something wrong ?
 
The second BC is at x=1, so you get C2=0. So there's no non-trivial solution that satisfies the boundary conditions when λ=0.
 
vela said:
The second BC is at x=1, so you get C2=0. So there's no non-trivial solution that satisfies the boundary conditions when λ=0.

shoot, you are right :\

forgot to plug in the 1.

Thanks for your help.
 

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