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## Homework Statement

Determine all eigenvalues and eigenfunctions for the Sturm-Liouville problem

\begin{cases}

-e^{-4x}\frac{d}{dx} \left(e^{4x}\frac{d}{dx}\right) = \lambda u, \; \; 0 < x <1\\

u(0)=0, \; \; u'(1)=0

\end{cases}

Expand the function ##e^{-2x}## as a series of eigenfunctions.

## Homework Equations

Orthonormal series expansion

##f(x) = \sum_1^\infty \langle f, \phi_n \rangle \phi_n##

where ##\phi_n## is the orthonormal eigenfunctions.

## The Attempt at a Solution

I managed to find the eigenvalues and eigenfunctions (correct according to the answer to the exercise) but I'm having trouble expanding the function as a series.

We have from the first part of the exercise (##\lambda## is eigenvalues and ##u## eigenfunctions)

##\lambda_1 = 4-\beta_1^2## where ##\beta_1## is the positive root of ##\tanh \beta = \frac{\beta}{2}##, ##u_1(x) = e^{-2x}\sinh (\beta_1 x)##

##\lambda_n = 4+\beta_n^2## where ##\beta_n##, ##n=2,3,\dots## are the positive roots of ##\tan \beta = \frac{\beta}{2}##, ##u_n(x) = e^{-2x}\sin (\beta_n x)##.

Starting with the part I'm having trouble with. Normalize the eigenfunctions. We have the weight function ##w = e^{4x}## so the norm is given by

##\int_0^1 |f(x)|^2w(x)dx##. We get

##c_1^2\int_0^1 \sinh^2 \beta x dx = c^2 \frac{\sinh (2\beta) -2\beta}{4\beta}## so we have ##c_1^2 = \frac{4\beta}{\sinh (2\beta) -2\beta}##

##c_2^2 \int_0^1 sin^2 \beta x dx = c_1^2 \frac{(2\beta-sin(2\beta))}{4\beta}## so we have

##c_2^2 = \frac{4\beta}{(2\beta-sin(2\beta))}##.

Calculating the coefficients we get

##\int_0^1 \sin (\beta x)dx = \frac{1-\cos \beta}{\beta}## and

##\int_0^1 \sinh (\beta x)dx = \frac{\cosh \beta -1}{\beta}##.

So our series should be given by

##f(x) = \frac{1}{c_1^2} \frac{\cosh \beta -1}{\beta} u_1(x) +\frac{1}{c_2^2}\sum_2^\infty \frac{1-\cos \beta}{\beta}u_n(x) = \frac{(sinh(2b)-2\beta)(\cosh \beta -1)}{4\beta^2}u_1(x)+\sum_2^\infty \frac{2\beta-(\sin (2\beta) )(1-\cos \beta)}{4\beta^2 }u_n(x) ## (Note how we get ##c_1^2## and ##c_2^2## since ##u_n## refers to the old non normalized eigenfunctions and we also get one from the coefficient calculations before.)

I'm pretty sure I made a mistake here already but I keep ending up at the same thing when I redo the exercise. The answer should be

##e^{-2x}=\sum_1^\infty \frac{2\sqrt{\lambda_n}\left[ \sqrt{\lambda_n}+2(-1)^n\right]}{\beta_n(\lambda_n-2)}u_n(x)## which I can't get close to however much algebra I apply.

I suspect I made some conceptual misunderstanding since I can't even get close to the expression in the answer.

Sorry for a rather lenghty post. Thanks a lot to anyone with the patience to take a look at it.