Boundary Value Problem; Eigenvalues and Eigenfunctions

[Pinedas42]

Homework Statement

Find the eigenvalues and eigenfunction for the BVP:
y'''+\lambda^2y'=0

y(0)=0, y'(0)=0, y'(L)=0

Homework Equations

m^3+\lambdam=0, auxiliary equation

The Attempt at a Solution

3 cases \lambda=0, \lambda<0, \lambda>0
this first 2 give y=0 always, as the only solution.

\lambda>0 solution attempt

m^3+\lambda^2m=0
m(m^2+\lambda^2)=0
roots:
m=0, and +/- \lambdai

general solution:
y=A+Bcos(\lambdax)+Csin(\lambdax)
Where A, B, and C are constants

y'=-B\lambdasin(\lambdax)+C\lambdacos(\lambdax)

y(0)=0 gives
0=A+B, or A=-B

y'(0)=0 gives
0=\lambdaC, so C=0

y'(L)=0 gives
0=-\lambdaBsin(\lambdaL)

The only solution I find from these data is y=0, which seems kind of off since no eigenfunction/values are found. From what I've read/studied so far when \lambda>0 there is always an eigen function/value.

The alternative I've considered is to consider B≠0 and having the eigenvalue be
\lambdaL=n\pi giving \lambda=n\pi/L

which then gives the eigen function
y=A+Bcos((n\pix)/L)

 

[LCKurtz]

Homework Statement

Find the eigenvalues and eigenfunction for the BVP:
y'''+\lambda^2y'=0

y(0)=0, y'(0)=0, y'(L)=0

Homework Equations

m^3+\lambdam=0, auxiliary equation

The Attempt at a Solution

3 cases \lambda=0, \lambda<0, \lambda>0
this first 2 give y=0 always, as the only solution.

\lambda>0 solution attempt

m^3+\lambda^2m=0
m(m^2+\lambda^2)=0
roots:
m=0, and +/- \lambdai

general solution:
y=A+Bcos(\lambdax)+Csin(\lambdax)
Where A, B, and C are constants

y'=-B\lambdasin(\lambdax)+C\lambdacos(\lambdax)

y(0)=0 gives
0=A+B, or A=-B

y'(0)=0 gives
0=\lambdaC, so C=0

y'(L)=0 gives
0=-\lambdaBsin(\lambdaL)

The only solution I find from these data is y=0, which seems kind of off since no eigenfunction/values are found.

That's wrong, your next paragraph is the correct procedure

The alternative I've considered is to consider B≠0 and having the eigenvalue be
\lambdaL=n\pi giving \lambda=n\pi/L

which then gives the eigen function
y=A+Bcos((n\pix)/L)

And since A = -B, you have $B(-1+\cos(\frac {n\pi x} L))$. You could leave off the B and write $y_n =-1+\cos(\frac {n\pi x} L)$.

 

[Pinedas42]

OK, thank you!
I had thought that there always had to be a function for \lambda>0, but I wasn't sure and I couldn't find any literature specifically mentioning it.