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I have a Sturm-Liouville BVP that I would like to evaluate, I want the general solution for eigenvalues, and I would like to plot maybe 10 of them:

$$-y''+a^2y-b y=0 \in D$$ $$y'=0 \in \partial D$$ $$\forall |x|\le d, a<\pi /d$$

My futile attempt led to some baffling "conclusions":

Firstly, I rewrote the boundaries to $x \in [-d,d]$, this implies that $y(d)=y(-d)=0$.

Auxiliary equation was deduced: $-\lambda ^2+(a^2-b)=0$, which gives two complex roots of $\pm i \sqrt {b-a^2}$. For convenience (laziness), I will call this $\pm \alpha$.

The general solution is thus:

$$y(x)=C_1\cos(\alpha x)+C_2\sin(\alpha x)$$

Putting the probably completely wrong boundary conditions, I get two sets of equations:

$$y(d)=C_1\cos(\alpha d)+C_2\sin(\alpha d)=0$$ $$y(-d)=C_1\cos(-\alpha d)+C_2\sin(-\alpha d)=0$$

This is completely nonsense. Firstly, cosine is an even function. Since both equations give 0, they can be equated. $C_1\cos(-\alpha d)-C_1\cos(\alpha d)=0$.

I end up with this mess:

$$C_2\sin(\alpha d)=C_2\sin(-\alpha d) \Leftrightarrow \frac{\sin(\alpha d)}{\sin(-\alpha d)}=1$$ NO. Sine is an odd function, this equality is obviously false.

Some observations:

  • $y'=0$ was not used. I differentiated my result and nothing of value was obtained.
  • $\forall |x|\le d, a<\pi d$ I think rewriting the modulus caused the issue. Is this condition actually suggesting that the function lies on a unit disk?
  • $y'=0 \in \partial D$ What does this mean? First time seeing a condition written like this.
  • I looked up $y'=0 \in \partial D$, and it is called a Neumann Condition?

Any help is appreciated.

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To answer your questions in order:

  1. Yes, you did not use $y'(\pm d) = 0$ because you for some reason used $y(\pm d) = 0$ whhich is NOT implied by what you write initially.. In fact, your assumption that $y(x)$ consists of linearly independent solutions that are sines and cosines may be incorrect, as it completely depends on what $a$ and $b$ are in the auxillary equation. (What if $a^2 > b$?).

  2. There's no reason to assume that this problem is multidimensional, a closed disk in $\mathbb{R}$ is just an interval, namely in this case $[-d,d]$.

  3. $y' = 0$ on $\partial D$ means that, by the last point, $y'(-d) = 0$ and $y'(d) = 0$.

  4. Yes, this is often called the Neumann Boundary Condition. When it's just $y = 0$ on $\partial D$, then its called a Dirichlet Boundary Condition.

With that in mind, You probably need to go back and do a little bit of work here, as most of your problems started off with the assumption that you'd be dealing with sines and cosines.

EDIT: If you go back, you'll see that any time that $b -a^2 \geq 0$, you will only get trivial solutions.

If then, you assume $b - a^2 < 0$, interesting things happen! You'll find that, as you saw, $y(x) = A\sin{(\alpha x)} + B \cos{(\alpha x)}$ with $\alpha = \sqrt{a^2 - b}$. Plugging this into your BC's seems like it gets you to a contradiction $\textit{but this only happens to give trivial solutions if you want it to!}$

Plugging into the BC's we immediately see that $A = 0$ by adding the equations together. We're left with having

$$B\sin{(\alpha d)} = 0$$.

This means that either $B = 0$(boring case, we get trivial solutions again), or the far interesting thing, when $\alpha d = n\pi$, $n=1,2,...$.

This leads to the interesting relation that if you have the relationship with your constants in your original problem, that

$$a^2 - b = \left(\dfrac{n \pi}{d} \right)^2$$

Then you get solutions to your problem of the form

$$\phi_n (x) = B \cos{ \left(\dfrac{n \pi x}{d} \right)}$$

These are the corresponding eigenfunctions with eigenvalues given above.

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  • $\begingroup$ That's a lot clearer! Yes I need to wrap my mind around this stuff. Too used to tossing everything into the Fourier domain. I'll give it another shot. I also split the solution into multiple cases, figured it's just wrong; can't be that messy. $\endgroup$ – Hypergeometry Aug 31 '18 at 2:23
  • $\begingroup$ I differentiated the function and ended with two sets of equations $C_2/C_1=-C_2/C_1$ which I concluded that since $C_1$ can't be 0, $C_2$ is thus 0. I put this back into the original equation which becomes $C_1cos(\alpha x)$. Stuck again. I differentiated $y'=0$ to get $y=K$, some constant. Does this mean $y(d)=y(-d)=K$? $\endgroup$ – Hypergeometry Aug 31 '18 at 4:24
  • $\begingroup$ This leads to $cos(\alpha x) / cos(\alpha x)=-1$!!! Nice... I need cry in a corner. $\endgroup$ – Hypergeometry Aug 31 '18 at 4:29
  • $\begingroup$ Hi there, I'll edit my solution to give you a nudge in the right direction.. $\endgroup$ – DaveNine Aug 31 '18 at 6:45
  • $\begingroup$ As an extension to when you go through what I wrote, try and figure out why the condition of $a < \frac{\pi}{d}$ is there. $\endgroup$ – DaveNine Aug 31 '18 at 7:04

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