On the orthogonality of Sturm-Liouville eigenvectors

[mjordan2nd]

From what I understand, solutions to the Sturm-Liouville differential equation (SLDE) are considered to be orthogonal because of the following statement:

$$\left( \lambda_m-\lambda_n \right) \int_a^b w(x) y_m(x)y_n(x) dx = 0$$

My first question involves the assumptions that go into this equation. One of the assumptions that go into this equation is that the solutions to the SLDE satisfy the Dirichlet, Neumann, or mixed homogeneous boundary conditions, correct? If the boundary conditions were inhomogeneous then the above equation would not necessarily be true, correct? Is it then correct to say that solutions to the SLDE are only orthogonal if they satisfy homogeneous boundary conditions?

My second question involves the case when $\lambda_m=\lambda_n$. Since the SLDE is a second order ordinary differential equation there should be two linearly independent solutions for each eigenvalue. So even if $\lambda_m=\lambda_n$, that doesn't necessarily mean $y_m=y_n$. In this case, it is not clear that these two solutions are orthogonal. I can buy that every eigensubspace of the SLDE is orthogonal to the others, however what about two vectors belonging to the same subspace?

Thanks

 

[Orodruin]

If the boundary conditions were inhomogeneous then the above equation would not necessarily be true, correct? Is it then correct to say that solutions to the SLDE are only orthogonal if they satisfy homogeneous boundary conditions?

Having homogeneous boundary conditions are a prerequisite of Sturm-Liouville's theorem. You cannot make superpositions of solutions with inhomogeneous boundary conditions and get new solutions fulfilling the same boundary conditions.

Since the SLDE is a second order ordinary differential equation there should be two linearly independent solutions for each eigenvalue.

No, this is not true. For each possible value of the eigenvalue, there will be two independent solutions. However, the boundary conditions will take care of this and determine whether or not a non-trivial solution is possible for that eigenvalue.

 

[mjordan2nd]

Thank you for your response!

However, the boundary conditions will take care of this and determine whether or not a non-trivial solution is possible for that eigenvalue.

I am unclear about what you mean. Do you mean to say that the boundary conditions will reduce the number of nontrivial solutions per eigenvalue to one? Because this is the only way I can see that the equation I posted above proves that all eigenvectors are mutually orthogonal.

 

[Orodruin]

Yes. They will also tell you what eigenvalues are allowed. My suggestion is to work it out in the absolutely easiest case of the SL operator $-d^2/dx^2$ on the interval $[0,1]$.