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