# Homework Help: Sturm-Liouville Question on Orthogonality

1. Oct 7, 2012

### royblaze

1. The problem statement, all variables and given/known data
For the following diff. eqns (fcns of t)

X''m + λmXm=0
Xm (1)=0
X'm=0

X''n + λnXn=0
Xn (1)=0
X'n=0

Show that ∫XmXndt from 0 to 1 equals 0 for m≠n.

2. Relevant equations
Qualitative differential equations... no idea really what to put in this section.

3. The attempt at a solution
Using theory I am able to prove that the λ term must be positive in order to have non-trivial solutions. Using this, I am able to obtain explicit solutions for Xm and Xn respectively. However, when I attempt to take the integral I immediately am lost in how to show that their integrated product is 0.

My solutions are in this general form for both Xm and Xn, where C2 is some non-zero constant (to avoid trivial cases):

X = Cw*cos([((∏/2)+k∏)^2]x) for k = 0, 1, 2, ...

Any help would be greatly appreciated.

2. Oct 7, 2012

### voko

I cannot see how your solutions satisfy the boundary conditions.

3. Oct 7, 2012

### andrien

multiply the 1st eqn by Xn and second by Xm and then subtract .take the integral from 0 to 1 of the resulting expression and you will get
(X'mXn - X'nXm )01=(λm-λn)∫XmXn dt
if m≠n then λm≠λn
and the left side is zero because of boundary condition and the required result follows.(may be some +,- error is there but that does not matter)

4. Oct 7, 2012

### royblaze

Voko, I used the characteristic equation of the differential equation, and using Euler's formula as the general form of the homogenous equation, I solved three different cases based on the sign of λ. The negative and zero case for λ gives trivial solutions of X = 0. Only the positive case gives valuable solutions where the constant that arises is understood to be nontrivial (we say that the constant is non-zero).

This causes the trigonometric term to be the only term that can be zero. Solving for λ, we would obtain the periodic coefficient in front of the t in the cos term.

Andrien, thank you for your response, I will try it tomorrow (it is 4:20AM right now) and report back with my results.

5. Oct 7, 2012

### voko

The approach is generally correct. However, the result is not. Just evaluate your solution at x = 0 and x = 1. You must have made a mistake along the way.

6. Oct 7, 2012

### royblaze

Andrien, perhaps you can assist in my integration? I am a little confused. My sum (after subtraction as you suggested):

XnX"m + λmXmXn - XmX"n - λnXnXm

I tried to integrate term by term but I am getting very confused by the multiple instances of integration by parts. When can I stop integrating by parts? Or am I just missing a step that can be simplified with the Fundamental Thm of Calculus...?

7. Oct 8, 2012

### andrien

XnX"m- XmX"n=d/dt(XnX'm - XmX'n), which you can easily verify.then just integrate with respect to t and since this term is pure derivative it will come out with limits as I have written above and then it is all easy.

8. Oct 8, 2012

### royblaze

Thanks, I managed to get it :D