Sturm-Liouville Eigenvalue problem

[dochalladay32]

Homework Statement

We are to show that for 0<β<1, eigenvalues are strictly positive and for β>1, we have to determine how many negative eigenvalues there are.

u''+λ²u=0, u(0)=0, βu(π)-u'(π) = 0

Homework Equations

I've already shown that the eigenvalues are determined by tan(λπ)=λ/β (was told this is correct).

Some how just from this, we are supposed to determine what I listed in part 1. I tried using the Rayliegh quotient, but that is no good as you can't definitely say what β should and shouldn't be to make it positive in this case.

The Attempt at a Solution

We know that L is self-adjoint so our eigenvalues, λ², are real. Therefore λ is either strictly real or strictly imaginary. I was told that if you plug in λ=i*ω, leading to negative eigenvalues, this would force β>1. And then if you assume λ is real, this would force β<1. I do not see why this is true at all. If you assume λ is imaginary, you wind up with

tanh(ωπ)=ω/β,

and this has solutions for β<1, which indicates that you still have negative eigenvalues for β<1, which I was told should not be possible.

Can anyone explain what I was supposed to be seeing when I put a complex or real number in, or have an alternate way of explaining it?

EDIT: Professor finally admitted there was an error. We should actually be using the interval [0,1], not [0,π]. tan(λ)=λ/β, which is easy to show now what beta you need.

 

[mighty2000]

I would approach this problem by first analyzing the equation and determining what information is given and what is being asked. From the given equations, it is clear that the problem involves finding the eigenvalues of a self-adjoint operator L, which can be done by solving the differential equation u''+λ2u=0 with the given boundary conditions.

Next, I would look at the given information and try to understand the significance of the condition 0<β<1. Since β is a constant and not a variable in the differential equation, this condition must have some relation to the eigenvalues. From the given equation tan(λπ)=λ/β, we can see that the eigenvalues are determined by the ratio of λ to β. This means that the value of β will affect the value of the eigenvalues.

To determine the sign of the eigenvalues, we can use the Rayleigh quotient, which is given by λ2= u'(π)/u(π). From this, we can see that for the eigenvalues to be strictly positive, u'(π) and u(π) must have the same sign. This means that for 0<β<1, u(π) and u'(π) must both be positive, leading to positive eigenvalues.

For β>1, we can see that u(π) and u'(π) must have opposite signs for the eigenvalues to be negative. This means that there must be a point where u(π)=0 and u'(π)<0, which can only happen if u(0)=0 and u'(0)>0. This leads to the conclusion that for β>1, there is at least one negative eigenvalue.

To determine the exact number of negative eigenvalues, we can use the Sturm-Liouville theory, which states that the number of negative eigenvalues is equal to the number of sign changes in the eigenfunction u(x) between the boundary points. In this case, since u(0)=0 and u'(0)>0, the eigenfunction must have at least one sign change between x=0 and x=π, leading to at least one negative eigenvalue.

In summary, for 0<β<1, the eigenvalues are strictly positive, and for β>1, there is at least one negative eigenvalue and potentially more depending on the number of sign changes in the eigenfunction. This can be determined using