# ODE/PDE- eighenvalues+ eigenfunctions

#### Roni1985

1. The problem statement, all variables and given/known data
it's already separable, so it's an ODE function.
X''+$$\lambda$$*X=0 0<x<1
X(0)=-2X(1)+X'(1)=0

2. Relevant equations

3. 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 $$\lambda$$=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.

#### vela

Staff Emeritus
Science Advisor
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1. The problem statement, all variables and given/known data
it's already separable, so it's an ODE function.
X''+$$\lambda$$*X=0 0<x<1
X(0)=-2X(1)+X'(1)=0

2. Relevant equations

3. 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 $$\lambda$$=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.

#### Roni1985

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 ?

#### vela

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Science Advisor
Homework Helper
Education Advisor
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.

#### Roni1985

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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