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+$\lambda$m=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 +/- $\lambda$i

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

y'=-B$\lambda$sin($\lambda$x)+C$\lambda$cos($\lambda$x)

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

y'(0)=0 gives
0=$\lambda$C, so C=0

y'(L)=0 gives
0=-$\lambda$Bsin($\lambda$L)

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
$\lambda$L=n$\pi$ giving $\lambda$=n$\pi$/L

which then gives the eigen function
y=A+Bcos((n$\pi$x)/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+$\lambda$m=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 +/- $\lambda$i

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

y'=-B$\lambda$sin($\lambda$x)+C$\lambda$cos($\lambda$x)

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

y'(0)=0 gives
0=$\lambda$C, so C=0

y'(L)=0 gives
0=-$\lambda$Bsin($\lambda$L)

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
$\lambda$L=n$\pi$ giving $\lambda$=n$\pi$/L

which then gives the eigen function
y=A+Bcos((n$\pi$x)/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.