Expand function as series of eigenfunctions

[Incand]

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.

 

[Daeho Ro]

First of all, there are several typos for the expression of f(x), the coefficients have to be inverse of them.

You mostly well done, so, I will show an example for n=2. In this case, c_2^2 = \frac{4\beta_2}{(2\beta_2 - \sin(2\beta_2))} and
&lt; f, \phi_2 &gt; = c_2 \left( \dfrac{1-\cos\beta_2}{\beta_2} \right).
Then,
&lt; f, \phi_2 &gt; \phi_2 = c_2^2 \left( \dfrac{1-\cos\beta_2}{\beta_2} \right) u_2(x) = \left(\dfrac{4\beta_2}{2\beta_2 - \sin 2\beta_2} \right) \left( \dfrac{1-\cos\beta_2}{\beta_2} \right) u_2(x).
From the fact that
\tan \beta_n = \dfrac{\beta_n}{2},
the sine and cosine function can be obtained by
\sin \beta_n = (-1)^n \dfrac{\beta_n}{\sqrt{\lambda_n}}, \quad \text{and} \quad \cos\beta_n = (-1)^n \dfrac{2}{\sqrt{\lambda_n}}.

By using these things,
<br /> \begin{array}{rl}<br /> \left(\dfrac{4\beta_2}{2\beta_2 - \sin 2\beta_2} \right) \left( \dfrac{1-\cos\beta_2}{\beta_2} \right) &amp;= \left(\dfrac{4\beta_2}{2\beta_2 - 2 \frac{\beta_2}{\sqrt{\lambda_2}}\frac{2}{\sqrt{\lambda_2}} } \right) \left( \dfrac{1-\frac{2}{\sqrt{\lambda_2}}}{\beta_2} \right) \\<br /> &amp; = \left( \dfrac{4\beta_2 \lambda_2}{2\beta_2 \lambda_2 - 4\beta_2} \right) \left(\dfrac{1-\frac{2}{\sqrt{\lambda_2}}}{\beta_2} \right) \\<br /> &amp; = \dfrac{2 \sqrt{\lambda_2} ( \sqrt{\lambda_2} - 2 )}{(\lambda_2 - 2)\beta_2}.<br /> \end{array}<br />

Therefore,
&lt; f, \phi_2 &gt; \phi_2 = \dfrac{2 \sqrt{\lambda_2} ( \sqrt{\lambda_2} - 2 )}{(\lambda_2 - 2)\beta_2} u_2(x).

 

[Incand]

Thanks a lot! Don't know why I inverted the coefficients again when I already inverted to get them in the first place.

From the fact that
\tan \beta_n = \dfrac{\beta_n}{2},
the sine and cosine function can be obtained by
\sin \beta_n = (-1)^n \dfrac{\beta_n}{\sqrt{\lambda_n}}, \quad \text{and} \quad \cos\beta_n = (-1)^n \dfrac{2}{\sqrt{\lambda_n}}.

I'm a little unsure how you get the $\sin$ and $\cos$ functions. I guess you set $\sin \beta = A\beta_n$ and $\cos \beta = B\beta_n$ and then set $A,B$ so that $\sin^2\beta + \cos^2\beta = 1$.
The $(-1)^n$ (Actually I think it should be $(-1)^{n+1}$) you get from $\tan \beta = \frac{\beta}{2}$ only having a solution each in intervals of length $\pi$ the first would be $(\pi/2,3\pi/2), (3\pi/2, 5\pi/5), \dots)$. Is this correct or is there something I'm missing here?For the case with $n=1$ i get from
$\tanh \beta_1= \frac{\beta_1}{2}$ that
$\sinh \beta_1 = \frac{\beta_1}{\sqrt{\lambda_1}}$ and $\cosh \beta_1 = \frac{2i}{\sqrt{\lambda_1}}$.
Again I'm not entirely sure about determining these. I just made them satisfy the hyperbolic one, but I got the feeling If I'm not careful here I may get the expressions (luckily it works out in this case).

Edit: Think I got them wrong, edited the post a bit.

So the coefficient is given by
$\frac{4\beta_1}{\sinh 2\beta_1 -2\beta_1}\frac{\cosh \beta_1-1}{\beta_1} = \frac{2(\cosh \beta_1 -1)}{\sinh \beta_1 \cosh b_2 - \beta_1} = \frac{2(2i-\sqrt{\lambda_1})}{\beta_1(2i-\sqrt{\lambda_1)}}$ which doesn't look to good.

So I think I mess up when I calculate $\sinh$ and $\cosh$.

 

[Daeho Ro]

The sign problem, you are correct. It should start from the negative sign first.

The \sin \beta and \cos \beta just obtained by the properties of trigonometric functions. Let us consider the right angled triangle with width 2 and height \beta. Then, the length of incline is \sqrt{4 + \beta^2} = \sqrt{\lambda}. It gives the result what I used.

If you are still not comfortable, then you might get the same result from
1 + \tan^2 \beta = \dfrac{4 + \beta^2}{4} = \dfrac{1}{\cos^2 \beta}.

 

[Incand]

The sign problem, you are correct. It should start from the negative sign first.

The \sin \beta and \cos \beta just obtained by the properties of trigonometric functions. Let us consider the right angled triangle with width 2 and height \beta. Then, the length of incline is \sqrt{4 + \beta^2} = \sqrt{\lambda}. It gives the result what I used.

Thanks!
I also see where I messed up, last part of my post is all wrong. I had forgotten the hyperbolic one and used an incorrect version.

Going to post the rest of the solution incase anyone comes upon this thread in the future
$\sinh \beta = \frac{\beta_1 }{\sqrt{\lambda_1}}$ and $\cosh \beta = \frac{2}{\sqrt{\lambda_1}}$
Then we have
$\frac{4\beta_1}{\sinh 2\beta_1 - 2\beta_1}\frac{\cosh \beta-1}{\beta_1} = \frac{2(\cosh \beta_1-1)}{\sinh \beta_1 \cosh b_1 -\beta_1} = \frac{2(\frac{2}{\sqrt{\lambda}}-1)}{\beta_1(\frac{2}{\lambda_1}-1)} = \frac{2\sqrt\lambda_1(\sqrt{\lambda_1}-2)}{\beta_1(\lambda_1-2)}$