# Expand function as series of eigenfunctions

## 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.

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
$$< f, \phi_2 > = c_2 \left( \dfrac{1-\cos\beta_2}{\beta_2} \right).$$
Then,
$$< f, \phi_2 > \phi_2 = c_2^2 \left( \dfrac{1-\cos\beta_2}{\beta_2} \right) u_2(x) = \left(\dfrac{4\beta_2}{2\beta_2 - \sin 2\beta_2} \right) \left( \dfrac{1-\cos\beta_2}{\beta_2} \right) u_2(x).$$
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,
$$\begin{array}{rl} \left(\dfrac{4\beta_2}{2\beta_2 - \sin 2\beta_2} \right) \left( \dfrac{1-\cos\beta_2}{\beta_2} \right) &= \left(\dfrac{4\beta_2}{2\beta_2 - 2 \frac{\beta_2}{\sqrt{\lambda_2}}\frac{2}{\sqrt{\lambda_2}} } \right) \left( \dfrac{1-\frac{2}{\sqrt{\lambda_2}}}{\beta_2} \right) \\ & = \left( \dfrac{4\beta_2 \lambda_2}{2\beta_2 \lambda_2 - 4\beta_2} \right) \left(\dfrac{1-\frac{2}{\sqrt{\lambda_2}}}{\beta_2} \right) \\ & = \dfrac{2 \sqrt{\lambda_2} ( \sqrt{\lambda_2} - 2 )}{(\lambda_2 - 2)\beta_2}. \end{array}$$

Therefore,
$$< f, \phi_2 > \phi_2 = \dfrac{2 \sqrt{\lambda_2} ( \sqrt{\lambda_2} - 2 )}{(\lambda_2 - 2)\beta_2} u_2(x).$$

• 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##.

Last edited:
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)}##

• Daeho Ro