Sturm-Liouville Orthogonality Proof

In summary: Thanks for your help!In summary, the problem involves a set of eigenfunctions yn(x) that satisfy the Sturm-Liouville equation #1 with boundary conditions #2, where g(x) = 0. The task is to show that the derivatives un(x) = yn'(x) are also orthogonal functions, with a weighting function w(x) to be determined. To do this, one approach is to show that the un(x) form their own Sturm-Liouville problem by substituting them into equation #1 and using the orthogonality relation. Another approach is to manipulate the equations and boundary conditions to obtain an expression for f(x), the weighting function. Both methods require careful consideration of the boundary conditions and integration
  • #1
Conor_McF
8
0

Homework Statement



A set of eigenfunctions yn(x) satisfies the Sturm-Liouville equation #1 with boundary conditions #2. The function g(x) = 0. Show that the derivatives un(x) = yn'(x) are also orthogonal functions. Determine the weighting function w(x) for these functions. What boundary conditions are required for orthogonality?

Homework Equations



#1: [tex]\frac{d}{dx}[/tex]f(x)[tex]\frac{dy}{dx}[/tex] - g(x)[tex]\frac{dy}{dx}[/tex] + [tex]\lambda[/tex]w(x)y = 0 where w(x) [tex]\geq[/tex] 0 and a [tex]\leq[/tex] x [tex]\leq[/tex] b

#2: [tex]\alpha[/tex]1y + [tex]\beta[/tex]1[tex]\frac{dy}{dx}[/tex] = 0 at x = a and [tex]\alpha[/tex]2y + [tex]\beta[/tex]2[tex]\frac{dy}{dx}[/tex] = 0 at x = b

[tex]\alpha[/tex] and [tex]\beta[/tex] are both constants and cannot both equal 0.

There is also the orthogonality relation [tex]\int y_{n}w(x)y_{m}dx = 0[/tex]


The Attempt at a Solution



I guess what I'm most confused about here is what the problem is asking me to show. By saying the set "un(x) = yn'(x) are orthogonal functions", what exactly are they orthogonal to? I'm assuming it means that [tex]\int u_{n}w(x)u_{m}dx = 0[/tex], provided that m [tex]\neq[/tex] n. Is this correct or am I way off?

Since g(x) = 0, the DE becomes [tex]\frac{d}{dx}[/tex]f(x)[tex]\frac{dy}{dx}[/tex] + [tex]\lambda[/tex]w(x)y = 0. From the boundary conditions I can tell that y(a) = - [tex]\frac{\alpha_{1}}{\beta_{1}}[/tex]u(a) and y(b) = - [tex]\frac{\alpha_{2}}{\beta_{2}}[/tex]u(b), but I don't know how this would help me at all. Can somebody please point me in the right direction?

Thanks; Conor.
 
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  • #2
might be on the wrong track, but here's a guess...

based on the wording of the question i think you need to show the u's are orthogonal, as you've written, but to some weighting function yet to be determined

how about trying to show the u's constitute their own sturm-liouville problem? existence and orthogonality properties of the solutions would show the un are orthogonal?

did this and got a weighting function ~(d/dx(f'/w)+lambda)... though not sure how to show this is >0 for all x...
 
  • #3
Yeah, that's what I thought to, but how did you show that they form their own Sturm-Liouville problem? Did you replace y(x) with u(x) in #1? When you sub the u(x)'s into eq'n #1, the last term contains [tex]\int u(x)dx[/tex], because u = y'. I'm curious to see how you got that weighting function.

I did it a little differently, but I'm not sure if I'm cheating. If you take eqn #1 in both n and m (two different indicies), multiply both by ym and ym (respectively), add the two equations and then integrate over all x you get...

[tex]\int\left[y_{m}\frac{d}{dx}f\frac{dy_{n}}{dx} + y_{n}\frac{d}{dx}f\frac{dy_{m}}{dx}\right]dx = - \left( \lambda_{n} - \lambda_{m} \right) \int y_{n}w(x)y_{m}dx = 0[/tex] (because of orthogonality of y)​

It turns out that the first and second terms in the integration are equal due to symmetry in m and n, making the integral equal to...

[tex]2\left(\frac{\beta_{1}}{\alpha_{1}}u_{n}(a)f(a)u_{m}(a) - \frac{\beta_{2}}{\alpha_{2}}u_{n}(b)f(b)u_{m}(b)\right) - 2\int u_{n}f(x)u_{m}dx = 0[/tex]​

...where I've made use of the boundary conditions and integration by parts. The additional boundary condition the problem is asking for in this case could be...

[tex]\frac{\beta_{1}}{\alpha_{1}}u_{n}(a)f(a)u_{m}(a) = \frac{\beta_{2}}{\alpha_{2}}u_{n}(b)f(b)u_{m}(b)[/tex]​

...in which case [tex]\int u_{n}f(x)u_{m}dx = 0[/tex] and the [tex]u_{n}[/tex] are orthogonal, and f(x) is the weighting function. I am concerned, however, that I did not obtain an actual expression for f(x), as the problem may have been asking. If you have time, please show me how you got your weighting function. Thanks for your help!
 
  • #4
nice idea,

don't even need to add the 2 equations, Also i would try and write the boundary conditions independent of m,n maybe this could be done by

y'(a) = [tex]\sqrt{f(a)\beta 1 \alpha 2 / (f(b)\beta 2 \alpha 1)[/tex] y'(b)


not convinced with what i did now as i don't think its in correct SL form

but method was
- sub in u =y'
- divide by w & then differentiate
- try to re-arrange into SL form
 
  • #5
Yeah good call, I guess you don't need to add them. I'll try to rewrite those boundary conditions.
 

1. What is Sturm-Liouville Orthogonality Proof?

Sturm-Liouville Orthogonality Proof is a mathematical method used to prove the orthogonality of eigenfunctions associated with a Sturm-Liouville operator. It is commonly used in the study of differential equations and has applications in fields such as physics, engineering, and economics.

2. How does Sturm-Liouville Orthogonality Proof work?

The proof involves showing that the inner product of two eigenfunctions of a Sturm-Liouville operator is equal to zero if the eigenfunctions have different eigenvalues. This is achieved by using the properties of the Sturm-Liouville operator and its associated eigenfunctions.

3. Why is Sturm-Liouville Orthogonality Proof important?

Sturm-Liouville Orthogonality Proof is important because it allows us to determine the orthogonality of eigenfunctions, which is crucial in solving many physical and engineering problems. It also provides a rigorous mathematical framework for understanding the behavior of solutions to differential equations.

4. What are some applications of Sturm-Liouville Orthogonality Proof?

Sturm-Liouville Orthogonality Proof has many applications in physics, such as in the study of heat conduction, quantum mechanics, and vibration theory. It is also used in engineering to solve boundary value problems and in economics to model dynamic systems.

5. Are there any limitations to Sturm-Liouville Orthogonality Proof?

While Sturm-Liouville Orthogonality Proof is a powerful tool, it has some limitations. It is only applicable to linear differential equations and requires the eigenfunctions to be square integrable. Additionally, it may not be applicable to certain non-self-adjoint Sturm-Liouville problems.

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