Eigenvalue solutions of the transcendental equation

[markymarky]

Homework Statement

Given the Sturm-Liouville system:

y'' + λy = 0 , y(0) - y'(0) = 0 , y(1) + y'(1) = 0

Show using the Rayleigh Quotient that the eigenvalues are positive.

Show that these eigenvalues are given as the solutions of the transcendental equation:

tan ( √λ ) = ( 2 √λ) / ( λ - 1 )

Homework Equations

Using the Rayleigh Quotient, I found the eigenvalues to be:

λ_n = ( y_n(1)² + y_n(0)² + ∫₀¹ y_n'²dx ) / ∫₀¹y_n²dx

Therefore, as the eigenfunction y_n cannot be zero, the eigenvalues λ_n are positive for all n.

The Attempt at a Solution

Then I have no idea how to show that these eigenvalues are solutions to the transcendental equation. I've tried putting the eigenvalues into the equation and can't figure out to simplify or to show that they solve the equation?

Thanks for any help.

 

[mighty2000]

Thank you for your post. I would like to offer some insights and guidance on how to approach this problem.

First, let's review the Rayleigh Quotient. It is a mathematical tool used to find the eigenvalues of a Sturm-Liouville system. In this case, we are given the Sturm-Liouville system y'' + λy = 0 with boundary conditions y(0) - y'(0) = 0 and y(1) + y'(1) = 0. Using the Rayleigh Quotient, we can find the eigenvalues λ that satisfy this system.

As you correctly stated, the eigenvalues can be found using the expression:

λn = ( yn(1)2 + yn(0)2 + ∫01 yn'2dx ) / ∫01yn2dx

Now, to show that these eigenvalues are positive, we need to show that the numerator of this expression is greater than zero. Let's start by looking at the integrals in the numerator and denominator.

∫01 yn'2dx is the integral of the square of the derivative of the eigenfunction yn. Since yn is a solution to the Sturm-Liouville system, it must satisfy the eigenvalue equation y'' + λy = 0. This means that yn' satisfies the equation y'' + λyn' = 0. Therefore, yn'2 is always positive or zero, and the integral will always be greater than or equal to zero.

Similarly, yn(1)2 and yn(0)2 are the squares of the eigenfunction evaluated at the boundaries 1 and 0, respectively. Since yn is a solution to the Sturm-Liouville system, it must satisfy the boundary conditions y(0) - y'(0) = 0 and y(1) + y'(1) = 0. This means that yn(1) and yn(0) must have opposite signs, and their squares will always be positive.

Therefore, the numerator of the Rayleigh Quotient will always be greater than or equal to zero. Since the denominator is also greater than zero (as yn cannot be zero), the eigenvalues λn will always be positive.

Now, to show that these eigenvalues satisfy the transcendental equation tan ( √λ ) = ( 2 √λ) / ( λ - 1 ),