Basic sturm liouville boundary problems

[Pi Face]

Homework Statement

Compute the eigenvalues/functions of the given regular S-L problem

f''(x)+λf(x)=0
0<x<π
f(0)=0
f'(π)=0

2. The attempt at a solution

First off, why is π not included in the given boundary if it tells you f'(x) at π?

Now for my attempt:

assuming λ=0
f''(x)+0*f(x)=0
f''(x)=0
f'(x)=a, and f'(π)=0 so a=0
f(x)=ax+b, but a=0 so f(x)=b, and f(0)=0 so b also =0
therefore
f(x)=ax+b=0+0=0
so this doesn't really help

assuming λ>0
f''(x)+λf(x)=0
char eqn: r^2+λ=0
r=±√(λ)i
f(x)=c1*cos(√(λ)x)+c2*sin(√(λ)x)
f(0)=0 so plug in 0 ->f(0)=c1*cos(0)+c2*sin(0)c1=c1=0, therefore c1=0
now we are left with
f(x)=c2*sin(√(λ)x)
f'(x)=c2*√(λ)*sin(√(λ)x)
f'(π)=c2*√(λ)*sin(√(λ)π)=0
we can conclude than c2=0 or λ=0 but that wouldn't get us anywhere
instead, we assume
sin(√(λ)π)=0
√(λ)π=n*π (where n=1,2,...)
√(λ)=n
λ=n^2

so for λ>0 we have an eigenvalue of λ=n^2 where n=1,2,... and an eigenfunction of f(x)=c2*sin(nx) (plug n^2 into λ)

the last case is λ<0
f''(x)+λf(x)=0
char eqn: r^2+λ=0
r=±√(λ)
f(x)=c1*e^(√(λ)x)+c2*x*e^(√(λ)x)
f(0)=c1*e^0+c2*0*e^0=c1=0, therefore c1=0
we are left with
f(x)=c2*x*e^(√(λ)x)
f'(x)=c2*x*√(λ)*e^(√(λ)x)+c2*e^(√(λ)x)
f'(π)=c2*π*√(λ)*e^(√(λ)π)+c2*e^(√(λ)π)=0
we can factor it to make it a little more manageable
f'(π)=c2*e^(√(λ)π)*(√(λ)π+1)
once again, we can see that c2 and λ=0, but are trivial
assume that the last term =0
√(λ)π+1=0
√(λ)π=-1
√(λ)=-1/π
λ=1/π^2
(here, can I specify that only the negative root is used and not the positive one?)

so this gives us an eigenvalue of λ=1/π^2 and an eigenfunction of f(x)=c2*x*e^(±x/π)


The answer in the back of the book is
λ=(2n-1)^2 /4
f(x)=sin((2n-1)x/2) n=1,2...

My answer looks nothing like that so what am I misunderstanding?

 

[LCKurtz]

Homework Statement

Compute the eigenvalues/functions of the given regular S-L problem

f''(x)+λf(x)=0
0<x<π
f(0)=0
f'(π)=0

2. The attempt at a solution

First off, why is π not included in the given boundary if it tells you f'(x) at π?

Now for my attempt:

assuming λ=0
f''(x)+0*f(x)=0
f''(x)=0
f'(x)=a, and f'(π)=0 so a=0
f(x)=ax+b, but a=0 so f(x)=b, and f(0)=0 so b also =0
therefore
f(x)=ax+b=0+0=0
so this doesn't really help

assuming λ>0
f''(x)+λf(x)=0
char eqn: r^2+λ=0
r=±√(λ)i
f(x)=c1*cos(√(λ)x)+c2*sin(√(λ)x)
f(0)=0 so plug in 0 ->f(0)=c1*cos(0)+c2*sin(0)c1=c1=0, therefore c1=0
now we are left with
f(x)=c2*sin(√(λ)x)
f'(x)=c2*√(λ)*sin(√(λ)x)

That should be a cosine.

f''(x)+λf(x)=0
char eqn: r^2+λ=0
r=±√(λ)
f(x)=c1*e^(√(λ)x)+c2*x*e^(√(λ)x)

You don't have a repeated root here. It would be much clearer to state this case as $\lambda = -\mu^2 < 0$ so the equation is $f''(x) -\mu^2f(x) = 0$ so your characteristic equation is $r^2-\mu^2=0$. This will give you the solution pair $\{e^{\mu x}, e^{-\mu x}\}$ but it is easier to work with $\{ \cosh μx, \sinh μx\}$. Try it and you will find no nontrivial solutions.

 

[Pi Face]

Wow such careless errors haha, but thanks for finding where I went wrong. I got the answer and everything seems good!