eigenvalue problem

**sinichko** · January 1st 2011, 08:22 AM

Please, help me with this one:

$\psi ''(z)+(\frac{20}{cosh^{2}z}+\lambda )\psi(z)=0$
$\psi '(0)=0$
I try it, and got the following:

$y=coshz$
$\frac{d\psi }{dz}=\frac{dx}{dy}sinhz$
$(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{dx}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0$

and i stuck here....

**InvisibleMan** · January 1st 2011, 10:08 AM

I get the following DE:

$\frac{dy}{dz} = sinh z$
$\frac{d\psi}{dy}=\frac{d\psi}{dz} \frac{ dz}{dy}= \frac{\psi '(z)}{sinh z}= \frac{\psi '(z)}{\sqrt{y^2-1}}$
$\frac{d^2\psi}{dy^2}= \frac{\psi ''(z)}{y^2-1} -\frac{\psi '(z)}{(y^2-1)^2} y$
Plugging back I get:
$(y^2-1)\psi ''(y) + \frac{y \psi '(y)}{\sqrt{y^2-1}} + (\frac{20}{y^2}+\lambda)\psi (y)$

Though I might have mistaken somewhere.

**Opalg** · January 1st 2011, 11:01 AM

Originally Posted by sinichko

Please, help me with this one:

$\psi ''(z)+(\frac{20}{cosh^{2}z}+\lambda )\psi(z)=0$
$\psi '(0)=0$
I try it, and got the following:

$y=coshz$
$\frac{d\psi }{dz}=\frac{dx}{dy}sinhz$
$(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{dx}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0$

and i stuck here....

Not sure if this is what is wanted, but you could try for a solution of the form $\psi(y) = y^\alpha$ . That will give you two eigenvalues.

**zzzoak** · January 1st 2011, 01:18 PM

Sorry, there is a mistake in post #2, line 3, last term.

$ \displaystyle { \frac{d}{dy}\frac{1}{\sqrt{y^2-1}}=- \: \frac{y}{\sqrt{(y^2-1)^3}} } $

**Danny** · January 1st 2011, 03:08 PM

Your ODE can be solved using a 4th order Darboux transformation. Do you have a second IC for $\psi$ ? Also is $\lambda > 0$ or $< 0$ ?

**sinichko** · January 1st 2011, 04:51 PM

Originally Posted by Danny

Your ODE can be solved using a 4th order Darboux transformation. Do you have a second IC for $\psi$ ? Also is $\lambda > 0$ or $< 0$ ?

$\lambda < 0$ and real
also i know the answer:

$\lambda_{1} =-16, \lambda_{2} =-16$ , i don't have a second IC for $\psi$ ...

**sinichko** · January 1st 2011, 06:02 PM

Thanks, but how to continue?

**sinichko** · January 1st 2011, 11:01 PM

Originally Posted by InvisibleMan

I get the following DE:

$\frac{dy}{dz} = sinh z$
$\frac{d\psi}{dy}=\frac{d\psi}{dz} \frac{ dz}{dy}= \frac{\psi '(z)}{sinh z}= \frac{\psi '(z)}{\sqrt{y^2-1}}$
$\frac{d^2\psi}{dy^2}= \frac{\psi ''(z)}{y^2-1} -\frac{\psi '(z)}{(y^2-1)^2} y$
Plugging back I get:
$(y^2-1)\psi ''(y) + \frac{y \psi '(y)}{\sqrt{y^2-1}} + (\frac{20}{y^2}+\lambda)\psi (y)$

Though I might have mistaken somewhere.

Thanks, but how to continue?

**Danny** · January 2nd 2011, 07:04 AM

Originally Posted by sinichko

Thanks, but how to continue?

Frist, where did this problem come from and how did you know that $\lambda = -16$ . Are there other values?

Second, are you familar with Wronskians?

**Opalg** · January 2nd 2011, 07:44 AM

The equation $(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0$ has eigenfunctions $\psi(y) = y^5$ and $\psi(y) = y^{-4}$ , with corresponding eigenvalues $\lambda = -25$ and $\lambda=-16$ (see my comment #3 above).

**sinichko** · January 2nd 2011, 09:14 AM

Originally Posted by Danny

Frist, where did this problem come from and how did you know that $\lambda = -16$ . Are there other values?

Second, are you familar with Wronskians?

I took it from an article: problem (4.8)

Wronskian sounds familiar...

**sinichko** · January 2nd 2011, 09:17 AM

Originally Posted by Opalg

The equation $(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0$ has eigenfunctions $\psi(y) = y^5$ and $\psi(y) = y^{-4}$ , with corresponding eigenvalues $\lambda = -25$ and $\lambda=-16$ (see my comment #3 above).

thanks, can you write please first steps of your solution, in the article he got a little different answer...

$\lambda_{1} =-16, \lambda_{2} =-4$

**Opalg** · January 2nd 2011, 10:41 AM

Originally Posted by sinichko

Originally Posted by Opalg

The equation $(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0$ has eigenfunctions $\psi(y) = y^5$ and $\psi(y) = y^{-4}$ , with corresponding eigenvalues $\lambda = -25$ and $\lambda=-16$ (see my comment #3 above).

thanks, can you write please first steps of your solution, in the article he got a little different answer...

$\lambda_{1} =-16, \lambda_{2} =-4$

It seemed obvious to me that there should be solutions of the form $\psi(y) = y^\alpha$ , for some values of $\alpha$ . The reason is that when you differentiate a power of y, it reduces the power by 1 (and when you differentiate twice, the power goes down by 2). So when you put $\psi(y) = y^\alpha$ in the expression $(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi$ , each term will be a multiple of either $y^\alpha$ or $y^{\alpha-2}$ .

In fact, if $\psi(y) = y^\alpha$ then $\frac{d\psi}{dy} = \alpha y^{\alpha-1}$ and $\frac{d^2\psi}{dy^2} = \alpha(\alpha-1) y^{\alpha-2}$ . Thus

$\begin{aligned}(y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi &= \alpha(\alpha-1)(y^{2}-1)y^{\alpha-2}+ \alpha y^{\alpha}+\left ( \frac{20}{y^{2 }} + \lambda\right )y^{\alpha} \\ &= \bigl(\alpha(\alpha-1) + \alpha + \lambda\bigr)y^{\alpha} + \bigl(-\alpha(\alpha-1) + 20\bigr)y^{\alpha-2}.\end{aligned}$

If that is 0 for all y then (putting the coefficients of $y^\alpha$ and $y^{\alpha-2}$ equal to 0), you get $\alpha(\alpha-1) = 20$ , from which $\alpha = 5$ or $-4$ ; and $\lambda = -\alpha^2$ , from which $\lambda = -25$ or $-16$ .

**sinichko** · January 3rd 2011, 08:39 AM

Thank you very much

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