Is the Sturm Liouville Operator Symmetric?

[Dassinia]

Hello, I'm solving the previous exams and I have a problem with an exercise:

Homework Statement

q(x) a real function defined in [0,1] and continuous
L a sturm Liouville operator :
Lf(x)=f''(x)+q(x)*f(x)

f ∈ C²([0,1]) with f(0)=0 and f'(1)=0.

Is L a symetric operator relative to the scalar product defined as
(f,g)=∫f(x)*g(x) dx from 0 to 1 ?

I just want to be sure I have to show that (Lf(x),Lg(x))=(Lg(x),Lf(x)) ( or not equal) ?

Thanks

 

[Dick]

Hello, I'm solving the previous exams and I have a problem with an exercise:

Homework Statement

q(x) a real function defined in [0,1] and continuous
L a sturm Liouville operator :
Lf(x)=f''(x)+q(x)*f(x)

f ∈ C²([0,1]) with f(0)=0 and f'(1)=0.

Is L a symetric operator relative to the scalar product defined as
(f,g)=∫f(x)*g(x) dx from 0 to 1 ?

I just want to be sure I have to show that (Lf(x),Lg(x))=(Lg(x),Lf(x)) ( or not equal) ?

Thanks

No, you want to show (Lf(x),g(x))=(f(x),Lg(x)).

 

[Dassinia]

Hello,
So if I want to prove that (Lf(x),g(x))=(f(x),Lg(x))
The operator is applied on functions which properties are f ∈ C²([0,1]) with f(0)=0 and f'(1)=0.
So when I calculate
(Lf(x),g(x))
I can use the fact that g'(1)=g'(0)=0 ?
Thanks

 

[Dick]

Hello,
So if I want to prove that (Lf(x),g(x))=(f(x),Lg(x))
The operator is applied on functions which properties are f ∈ C²([0,1]) with f(0)=0 and f'(1)=0.
So when I calculate
(Lf(x),g(x))
I can use the fact that g'(1)=g'(0)=0 ?
Thanks

I think you have g'(1)=g(0)=0. That's a little different.

 

[Dassinia]

Hello,
I'm sorry it's f'(0)=f'(1)=0
So if I want to show that it is symmetric I have to calculate this part
∫f''(x)g(x) from 0 to 1
By integration
=[ f(x)*f'(x) ] - ∫g'(x)f'(x) dx
= - ∫g'(x)f'(x) dx
=-[g'(x)*f(x)] + ∫f(x)*g''(x)dx
=∫f(x)*g''(x)dx

So if ∫f''(x)g(x)=∫f(x)*g''(x)dx
We can rewrite
(Lf(x),g(x))= ∫f''(x)g(x) + ∫g(x)f(x)g(x) dx
=∫f(x)*g''(x)dx + ∫g(x)f(x)g(x) dx
=(f(x),Lg(x)) ?

 

[Dick]

Hello,
I'm sorry it's f'(0)=f'(1)=0
So if I want to show that it is symmetric I have to calculate this part
∫f''(x)g(x) from 0 to 1
By integration
=[ f(x)*f'(x) ] - ∫g'(x)f'(x) dx
= - ∫g'(x)f'(x) dx
=-[g'(x)*f(x)] + ∫f(x)*g''(x)dx
=∫f(x)*g''(x)dx

So if ∫f''(x)g(x)=∫f(x)*g''(x)dx
We can rewrite
(Lf(x),g(x))= ∫f''(x)g(x) + ∫g(x)f(x)g(x) dx
=∫f(x)*g''(x)dx + ∫g(x)f(x)g(x) dx
=(f(x),Lg(x)) ?

Well, yes it's integration by parts. I don't think that's a very clear presentation though.