# Math Help - eigenvalue problem

1. ## eigenvalue problem

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....

2. 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.

3. 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.

4. 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}}
}
$

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

6. 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

$\lambda_{1} =-16,

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

7. Thanks, but how to continue?

8. 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?

9. 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?

10. 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).

11. 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...

12. 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$

13. 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 $-16$.

14. Thank you very much