eigenvalue problem

• Jan 1st 2011, 08:22 AM
sinichko
eigenvalue problem
Please, help me with this one:

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

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

and i stuck here....
• Jan 1st 2011, 10:08 AM
InvisibleMan
I get the following DE:

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

Though I might have mistaken somewhere.
• Jan 1st 2011, 11:01 AM
Opalg
Quote:

Originally Posted by sinichko
Please, help me with this one:

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

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

and i stuck here....

Not sure if this is what is wanted, but you could try for a solution of the form $\displaystyle \psi(y) = y^\alpha$. That will give you two eigenvalues.
• Jan 1st 2011, 01:18 PM
zzzoak
Sorry, there is a mistake in post #2, line 3, last term.

$\displaystyle \displaystyle { \frac{d}{dy}\frac{1}{\sqrt{y^2-1}}=- \: \frac{y}{\sqrt{(y^2-1)^3}} }$
• Jan 1st 2011, 03:08 PM
Jester
Your ODE can be solved using a 4th order Darboux transformation. Do you have a second IC for $\displaystyle \psi$? Also is $\displaystyle \lambda > 0$ or $\displaystyle < 0$?
• Jan 1st 2011, 04:51 PM
sinichko
Quote:

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

$\displaystyle \lambda < 0$ and real

$\displaystyle \lambda_{1} =-16, \lambda_{2} =-16$, i don't have a second IC for $\displaystyle \psi$...
• Jan 1st 2011, 06:02 PM
sinichko
Thanks, but how to continue?
• Jan 1st 2011, 11:01 PM
sinichko
Quote:

Originally Posted by InvisibleMan
I get the following DE:

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

Though I might have mistaken somewhere.

Thanks, but how to continue?
• Jan 2nd 2011, 07:04 AM
Jester
Quote:

Originally Posted by sinichko
Thanks, but how to continue?

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

Second, are you familar with Wronskians?
• Jan 2nd 2011, 07:44 AM
Opalg
The equation $\displaystyle (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi =0$ has eigenfunctions $\displaystyle \psi(y) = y^5$ and $\displaystyle \psi(y) = y^{-4}$, with corresponding eigenvalues $\displaystyle \lambda = -25$ and $\displaystyle \lambda=-16$ (see my comment #3 above).
• Jan 2nd 2011, 09:14 AM
sinichko
Quote:

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

Second, are you familar with Wronskians?

I took it from an article: problem (4.8)

Attachment 20309

Wronskian sounds familiar...
• Jan 2nd 2011, 09:17 AM
sinichko
Quote:

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

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

$\displaystyle \lambda_{1} =-16, \lambda_{2} =-4$
• Jan 2nd 2011, 10:41 AM
Opalg
Quote:

Originally Posted by sinichko
Quote:

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

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

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

It seemed obvious to me that there should be solutions of the form $\displaystyle \psi(y) = y^\alpha$, for some values of $\displaystyle \alpha$. 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 $\displaystyle \psi(y) = y^\alpha$ in the expression $\displaystyle (y^{2}-1)\frac{d^{2}\psi}{ dy^{2}}+y\frac{d\psi}{dy}+\left ( \frac{20}{y^{2 }} + \lambda\right )\psi$, each term will be a multiple of either $\displaystyle y^\alpha$ or $\displaystyle y^{\alpha-2}$.

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

\displaystyle \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 $\displaystyle y^\alpha$ and $\displaystyle y^{\alpha-2}$ equal to 0), you get $\displaystyle \alpha(\alpha-1) = 20$, from which $\displaystyle \alpha = 5$ or $\displaystyle -4$; and $\displaystyle \lambda = -\alpha^2$, from which $\displaystyle \lambda = -25$ or $\displaystyle -16$.
• Jan 3rd 2011, 08:39 AM
sinichko
Thank you very much