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Sturm-Liouville problem

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Poirot

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Feb 15, 2012
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consider $y''+2y'+ty=0$ on $0<x<1$ such that $y(0)=y(1)=0$
Find the corresponding Sturm-Liouville operator and formulate the Sturm-Liouville problem. Hence, define the inner product for which the eigenfunctions are orthogonal.

I have $L=-e^-2x. d/dx(e^2x.d/dx)$ (how to do fractions in code?)

I'm not sure what is meant by formulate the problem but perhaps Ly=ty.

I don't know how to choose an inner product.

 

Ackbach

Indicium Physicus
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Jan 26, 2012
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Question: is $t$ the variable of differentiation, or is it the eigenvalue?

You can do fractions with the \frac{numerator}{denominator} command.
 
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Poirot

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Feb 15, 2012
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't' is the eigenvalue
 

Ackbach

Indicium Physicus
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Jan 26, 2012
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So, assuming you meant
$$L=-\frac{1}{e^{2x}}\,\frac{d}{dx}\left[e^{2x}\,\frac{d}{dx}\right],$$
I would agree that's the Sturm-Liouville operator.
In comparing it with the standard form of the Sturm-Liouville operator, what are $p, q,$ and $w$?
 
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Poirot

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Feb 15, 2012
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$$p(x)=e^2x$$, $$q(x)=0$$,$$r(x)= -e^2x$$. What does formulating the problem entail? By your hint on the other thread, I might guess that I should define the inner product to be $$<f,g>= \int{ \frac{1}{r}*f*g$$.
 

Ackbach

Indicium Physicus
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Jan 26, 2012
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I don't know what your $r$ is - is that your weight function? Incidentally, you can write more than one thing in an exponent, in $\LaTeX$, if you enclose it in curly braces {} thus: $e^{2x}$. Compare with $e^2x$.
 
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Poirot

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Feb 15, 2012
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yes, my r is your w.
 

Ackbach

Indicium Physicus
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Jan 26, 2012
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yes, my r is your w.
So, in writing your original DE in Sturm-Liouville form, you have now discovered what the weight function is. If you look at my big post in your other thread, note how the weight function shows up in the inner product w.r.t. which the eigenfunctions of the Sturm-Liouville operator are orthogonal. It's not quite what you have in your post # 5 of this thread.
 
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Poirot

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Feb 15, 2012
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So did formulating the problem simply mean writing in sturm-liouville form?
 

Ackbach

Indicium Physicus
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Jan 26, 2012
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So did formulating the problem simply mean writing in sturm-liouville form?
Correct. Then, to solve the problem, you simply need to recognize what the weight function is, and write down the corresponding inner product.
 

Poirot

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Feb 15, 2012
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ok so <f,g> should be the integral of 'rgf' between 0 and 1 but $$r=-e^{-2x}$$ is negative. Should I take the mod?
 

Ackbach

Indicium Physicus
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Jan 26, 2012
4,191
Yeah, you can't have a negative weight function, because then $\langle f|f\rangle$ might be negative, which is not allowed in an inner product. You should not take the mod, however. What I would do is absorb the minus sign into the eigenvalue. That is, define $\lambda=-t$ as the eigenvalue for the Sturm-Liouville operator.