Solving a Sturm-Liouville problem

Question

I want to solve the Sturm-Liouville problem:

$\mathcal{A}u=-u''$

with I.C:

\begin{cases} u(0)=0\\ u'(L)=0 \end{cases}

where $\mathcal{A}$ is the Sturm-Liuville operator defined on $\mathscr{D}_A:\{u\in C^2([0,L])$.

By reading this post from SE , I see that it is practically the same as a PDE. So using the characteristic equation, I obtain:

$u''+Au=0$

$m=\pm\sqrt{-4A}/2=\pm iA$

$u(x)=Be^{iAx}+Ce^{-iAx}$

We rewrite to :

\begin{gather} u(L)= \begin{cases} B\cos Ax\\ iB\sin Ax \end{cases} \end{gather}

Since the first conI seedition yields a existence of u at the origin, we can disregard from the cosine solution. We get thus:

\begin{equation} u(x)=iB\sin Ax \end{equation}

Using first IC, we have $u(x)=\sin Ax$

Then using the second IC:

\begin{equation} \begin{array} fu'(x)=A\cos Ax\\ u'(L)=A\cos A L\\ 0=A\cos A x\\ \cos AL =0\\ A=\big(\frac{\pi}{2}+n\pi\big)\frac{1}{L} \end{array} \end{equation}

This gives

\begin{equation} u(x)=\sin \bigg(\big(\frac{\pi}{2}+n\pi\big)\frac{x}{L}\bigg) \end{equation}

What is wrong?

Thanks!

How do you conclude $A=\pi n$ from the first initial condition? $u(x) = iB \sin(A x)$ already fulfills $u(0)=0$ for any $B, A$, no need to fix $B$ or $A$. The second initial condition will aid you to fix $A$ and not $L$ which is part of the data in the problem. $B$ on the other hand is free up to this point. — FeedbackLooper, Mar 21 '22 at 16:39
@FeedbackLooper, I can't "find L".... is this meant to be as this? Thanks — Luthier415Hz, Mar 21 '22 at 17:11
$L$ disappears when you evaluate the function at $L$. However, $u$ should be a function of another variable, say $x$, not $L$. $L$ is a constant parameter of the problem. — FeedbackLooper, Mar 21 '22 at 17:13
Yes, I misinterpreted L as a the variable. But x is still vanishing. — Luthier415Hz, Mar 21 '22 at 17:23
When you use the second initial condition, you need to use $x=L$. So you solve for $A$ in terms of $L$ and plug that in the expresion for $u(x)$. — FeedbackLooper, Mar 21 '22 at 17:26

Udi Fogiel · Answer 1 · 2022-03-21T18:04:57.650

1

After you have got $u'(x)=A\cos (Ax)$ you used the second I.C wrong. we know that $u'(L)=0$ so that would give

$$ 0=u'(L)=A\cos(AL)\,\Rightarrow\,A=\frac{1}{L}\cdot\left(\frac{\pi}{2}+\pi\cdot n\right) $$

for $n\in\mathbb{Z}$. It implies that $u(x)=\sin\left(\frac{1}{L}\left(\frac{\pi}{2}+\pi\cdot n\right)x\right)$

edited Mar 21 '22 at 18:04

answered Mar 21 '22 at 17:32

Udi Fogiel

1,624

This I struggle with all the time. Isn't cosx=0 x=n pi/2 ? Why adding $\pi n$ when it is already given in $n\pi/2$? – Luthier415Hz Mar 21 '22 at 17:55
1

No, for example $\cos(\pi)=-1$ and $\cos(0)=1$. If you know that $\sin(x)=0$ implies $x=\pi\cdot k$ for $k\in\mathbb{Z}$, remember that $\cos(x)$ is just a shift of $\frac{\pi}{2}$ units of $\sin(x)$ to the left. – Udi Fogiel Mar 21 '22 at 17:58
OK, so I add npi to what ever value that gives 0, in this case. – Luthier415Hz Mar 21 '22 at 17:59
1

Yes, Exactly :) – Udi Fogiel Mar 21 '22 at 18:00
One thing, should it be AsinAx? I couldn't find a reason to keep A in front of it, so I got sinAx – Luthier415Hz Mar 21 '22 at 18:02
1

No, it is my mistake. I will edit the answer. – Udi Fogiel Mar 21 '22 at 18:04

score 1 · Accepted Answer · answered Mar 29 '22 at 14:33

Your Sturm-Liouville problem is the eigenvalue problem $$ -u''(x)=\lambda u(x),\;\;\; 0 \le x \le L $$ subject to the following endpoint constraints: $$ u(0)=0,\;\; u'(L)=0. $$ The problem is to expand a function $f\in L^2[0,L]$ in terms of the eigenfunction solutions of the above. In your case, you have a regular Sturm-Liouville problem, and the expansion will be a discrete expansion in terms of the eigenfunctions $$ e_n(x)=\sin((n+1/2)\pi x/L),\;\; n=1,2,3,\cdots. $$ It is easy to verify that $e_n(0)=0$ is satisfied for all $n=1,2,3,\cdots$, and the right endpoint restriction is also satisfied: $$ e_n'(L)=\cos((n+1/2)\pi)\{(n+1/2)\pi/L\}=0,\;\; n=1,2,3,\cdots. $$ The Fourier expansion of $f\in L^2[0,L]$ in terms of these eigenfunctions is given by $$ f(x) \sim \sum_{n=1}^{\infty}\frac{\int_{0}^{L}f(y)e_n(y)dy}{\int_{0}^{L}e_n^2(y)dy}e_n(x). $$ This Sturm-Liouville eigenfunction expansion is guaranteed to converge to $f$ in the norm of $L^2[0,L]$ even though the series expansion always converges to $0$ at $x=0$

Is my answer sufficient, though it is more simple than yours? Thanks DISINTEGRATION BY PARTS! — Luthier415Hz, Mar 29 '22 at 15:34
if you follow my profile to my homepage, you will see it there under the contact infromation. Here I can't post it. — Luthier415Hz, Mar 30 '22 at 07:53

Solving a Sturm-Liouville problem

2 Answers2