Differential equation

[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

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


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

 

[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

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