MHB The eigenvalues are real and that the eigenfunctions are orthogonal

[mathmari]

Hey!

We have the Sturm-Liouville problem $\displaystyle{Lu=\lambda u}$.

I am looking at the following proof that the eigenvalues are real and that the eigenfunctions are orthogonal and I have some questions...

$\displaystyle{Lu_i=\lambda_iu_i}$

$\displaystyle{Lu_j=\lambda_ju_j \Rightarrow Lu_j^*=\lambda_j^*u_j^*}$

$\displaystyle{\int_a^b(u_j^*Lu_i-u_iLu_j^*)dx=\int_a^b((\lambda_i-\lambda_j^*)u_iu_j^*)dx=(\lambda_i-\lambda_j^*)(u_j,u_i)}$

Since the operator $L$ is self-adjoint, the left side of the relation above is equal to $0$.

$\displaystyle{(\lambda_i-\lambda_j^*)(u_j, u_i)=0} $

• For $i=j:$ $(u_i,u_i)=\int u_i^*u_i dx=\int |u_i|^2 dx \geq 0$. So that is is equal to $0$, it should be $\lambda_i=\lambda_i^* \Rightarrow \lambda_i \in \mathbb{R}$
  
• For $i \neq j \Rightarrow \lambda_i \neq \lambda_j$, so that the relation $(\lambda_i-\lambda_j^*)(u_j, u_i)=0$ stands, it should be $(u_j, u_i)=0$. In this case th eigenfunctions are orthogonal.

Could you explain why at the beginning we have taken the integral:
$\displaystyle{\int_a^b(u_j^*Lu_i-u_iLu_j^*)dx}$?? (Wondering)

And also why does it stand that "since the operator $L$ is self-adjoint, the left side of the relation above is equal to $0$."?? (Wondering)$ \left ( \text{ We have defined the dot product as : } \displaystyle{(v,u)=\int_a^b v^* u dx} \right ) $

 

[I like Serena]

Hi! (Happy)

$\displaystyle{\int_a^b(u_j^*Lu_i-u_iLu_j^*)dx=\int_a^b((\lambda_i-\lambda_j^*)u_iu_j^*)dx=(\lambda_i-\lambda_j^*)(u_j,u_i)}$

Since the operator $L$ is self-adjoint, the left side of the relation above is equal to $0$.Could you explain why at the beginning we have taken the integral:
$\displaystyle{\int_a^b(u_j^*Lu_i-u_iLu_j^*)dx}$?? (Wondering)

And also why does it stand that "since the operator $L$ is self-adjoint, the left side of the relation above is equal to $0$."?? (Wondering)$ \left ( \text{ We have defined the dot product as : } \displaystyle{(v,u)=\int_a^b v^* u dx} \right ) $

The left hand side is equal to $(u_j, Lu_i) - (Lu_j, u_i)$.
Apparently this was chosen because the definition of a self-adjoint operator $A$ is that $(Ax, y)=(x,Ay)$. In other words, that means that it is 0.

The fact that $L$ is self-adjoint is not given in your problem statement, so I'll have to assume that this is given in some context of your problem. (Wondering)

 

[mathmari]

The left hand side is equal to $(u_j, Lu_i) - (Lu_j, u_i)$.
Apparently this was chosen because the definition of a self-adjoint operator $A$ is that $(Ax, y)=(x,Ay)$. In other words, that means that it is 0.

The fact that $L$ is self-adjoint is not given in your problem statement, so I'll have to assume that this is given in some context of your problem. (Wondering)

Ahaa! I got it! Thanks a lot for your answer! (Mmm)