# Sturm-Liouville: Eigenvalues, Eigenfunctions and Square Norm

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Please help with the following problems.

1. Find the eigenvalues and eigenfunctions of the boundary- value problem

y''+lambay = 0 , y(0) = 0 , y(pi/4) = 0 ( n = pi)

2. Find the eigenvalues and eigenfunctions of the boundary- value problem

y''+(lamba+1) y = 0 , y'(0) = 0 , y'(1) = 0

3. Find the square norm of the following problem along with the eigenvalues and eigenfunctions

y''+lambay = 0 , y'(0) = 0 , y(1) + y'(1) = 0

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Sturm-Liouville Problems, Eigenvalues, Eigenfunctions and Square Norm are investigated. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.

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sturm-liouville problems

1. find the eigenvalues and eigenfunctions of the boundary- value problem

y''+λ y = 0 , y(0) = 0 , y(π /4) = 0 ( n = pie)

with

y(0) = 0, y(pi/4) = 0

Problem Solution:

I. Show that this is a special case of a Sturm-Liouville Problem:

We first let . With these substitutions,

becomes (See attached file for equation)

or (See attached file for equation)

which is the desired ODE for this problem.

Now with boundary points a=0 and b = , the coefficients in the general boundary condition equation become

With these substitutions, the BCs in the general equation can be written simply as

y(0) = 0 and y(pi/4) = 0

(See attached file for equation)

which again are the desired conditions for this problem. Thus, the current situation is indeed a special case of the general Sturm-Liouville problem.

II. Find the eigenvalues and eigenfunctions for this problem:

In trying to determine the solution to this problem, there are three possibilities - the eigenvalues might be negative, zero, or positive (note that, since the coefficient functions are real, we already know that the eigenvalues will be real). Let's try each possibility:

Case 1 - Negative Eigenvalues: For this case we try . With this substitution, the original ODE becomes

(See attached file for equation)

This is just a simple, constant coefficient, second-order ODE with characteristic equation

(See attached file for equation)

and roots

(See attached file for equation)

Thus, the general solution for the negative eigenvalue assumption is

(See attached file for equation)

Applying the first boundary condition gives

(See attached file for equation)

Similarly, the second boundary condition gives

(See attached file for equation)

Therefore, for real , , and the only solution is the trivial solution, . Thus, for this problem, letting be negative was not a good choice.

Case 2 - Zero Eigenvalues: For this case we try . With this substitution, the original ODE reduces to . This can be integrated twice to give

(See attached file for equation)

Applying the first ...

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