How to integrate \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}?

[Petar Mali]

I have a problem with this equation

\frac{d^2\Omega(\alpha)}{d\alpha^2}+\frac{(1+\Phi)+\Phi e^{-\alpha}}{(1+\Phi)-\Phi e^{-\alpha}}\frac{d\Omega(\alpha)}{d\alpha}-S(S+1)\Omega(\alpha)=0

Boundary conditions are

\Omega(0)=1

[\Pi^S_{p=-S}(p-\frac{d}{d\alpha})]\Omega(\alpha)|_{\alpha=0}=0

How to to transform part before first derivative \frac{(1+\Phi)+\Phi e^{-\alpha}}{(1+\Phi)-\Phi e^{-\alpha}}?

 

[Petar Mali]

\Phi and S are constants! Thanks for your answer!

 

[gato_]

The general solution to the equation (maple obtained) is

\Omega(x)= \frac{e^{-x/2}}{-\Phi-1+\Phi e^{-x}}(C_{1}e^{x(S+1/2)}+C_{2}e^{-x(S+1/2)})

The first boundary condition puts the constraint

C_{1}+C_{2}=-1

while the second one says

\Pi_{p=-S}^{p=S} [p-[(1+2S)C_{2}+S-\Phi]] =0

which is a product with a finite number of terms. I assume S is an integer or half integer A solution is obtained by cancelling any factor of the product. You get a solution for any C2 satisfying

(1+2S)C_{2}+S-\Phi=p, -S\le p \le S

All this needs some checking

 

[Petar Mali]

Ok! But how to get solution without use of any computer programme? How to solve this with pencil and paper? What is the idea?

 

[gato_]

Well, first thing I thought was eliminating the term in the first derivative by a standart trick. If

u''(x)+A(x)u'(x)-S(S+1)u(x)=0then it turns out that u(x)=exp(-\frac{1}{2}\int{A(x')dx'})v(x) transforms the equation into

v''+[-S(S+1)-A^2/4-A'/2]v=0

In this case, we have

\int{A(x)dx}=-2ln[\frac{exp(-x/2)}{-1-\Phi+\Phi exp(-x)}]

And the resulting equation for v(x) is

v''(x)-(S+1/2)^{2}v(x)=0

resulting in the solution above. This equation has several lucky "coincidences", but the procedure to eliminate the first derivative term is standard, and very useful for non-homogeneous oscilators

 

[Petar Mali]

Well, first thing I thought was eliminating the term in the first derivative by a standart trick. If

u''(x)+A(x)u'(x)-S(S+1)u(x)=0


then it turns out that u(x)=exp(-\frac{1}{2}\int{A(x')dx'})v(x) transforms the equation into

v''+[-S(S+1)-A^2/4-A'/2]v=0

Are you sure about this?

I get equation

v''+[-S(S+1)+\frac{A^2}{4}-\frac{A}{2}-\frac{A'}{2}]v=0

 

[Petar Mali]

Well, first thing I thought was eliminating the term in the first derivative by a standart trick. If

u''(x)+A(x)u'(x)-S(S+1)u(x)=0


then it turns out that u(x)=exp(-\frac{1}{2}\int{A(x')dx'})v(x) transforms the equation into

v''+[-S(S+1)-A^2/4-A'/2]v=0

In this case, we have

\int{A(x)dx}=-2ln[\frac{exp(-x/2)}{-1-\Phi+\Phi exp(-x)}]

And the resulting equation for v(x) is

v''(x)-(S+1/2)^{2}v(x)=0

resulting in the solution above. This equation has several lucky "coincidences", but the procedure to eliminate the first derivative term is standard, and very useful for non-homogeneous oscilators

How you get this form v''(x)-(S+1/2)^{2}v(x)=0 from v''+[-S(S+1)-A^2/4-A'/2]v=0?

 

[gato_]

By substituting A

 

[Petar Mali]

A(x)=\frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}

A'(x)=[\frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}]'

A'(x)=\frac{-\Phi e^{-x}[(1+\Phi)-\Phi e^{-x}]-[(1+\Phi)+\Phi e^{-x}]\Phi e^{-x}}{[(1+\Phi)-\Phi e^{-x}]^2}=\frac{-2\Phi e^{-x}(1+\Phi)}{[(1+\Phi)-\Phi e^{-x}]^2}

\frac{-A'(x)}{2}=\frac{\Phi e^{-x}(1+\Phi)}{[(1+\Phi)-\Phi e^{-x}]^2}

\frac{-A^2}{4}=-\frac{\frac{(1+\Phi)^2}{4}+\frac{1}{2}(1+\Phi)\Phi e^{-x}+\frac{1}{4}\Phi^2 e^{-2x}}{[(1+\Phi)-\Phi e^{-x}]^2}

\frac{-A^2}{4}-\frac{A'(x)}{2}=-\frac{\frac{(1+\Phi)^2}{4}-\frac{1}{2}(1+\Phi)\Phi e^{-x}+\frac{1}{4}\Phi^2 e^{-2x}}{[(1+\Phi)-\Phi e^{-x}]^2}=-\frac{1}{4}

\upsilon''(x)-(S+\frac{1}{2})^2\upsilon(x)=0


Thanks!

 

[gato_]

You are welcome

 

[Petar Mali]

In this case, we have

\int{A(x)dx}=-2ln[\frac{exp(-x/2)}{-1-\Phi+\Phi exp(-x)}]

I got

\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=2ln[\frac{(1+\Phi)-\Phi e^{-x}}{\Phi e^{-x}}]


I have done this integral like

\int \frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=\int\frac{1+\Phi}{(1+\Phi)-\Phi e^{-x}}dx+\int\frac{\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx

and got

\int\frac{1+\Phi}{(1+\Phi)-\Phi e^{-x}}dx=ln[\frac{(1+\Phi)-\Phi e^{-x}}{\Phi e^{-x}}]

and

\int\frac{\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}dx=ln[{(1+\Phi)-\Phi e^{-x}}]