Differential equation

  • Thread starter Petar Mali
  • Start date
  • #1
290
0

Main Question or Discussion Point

I have a problem with this equation

[tex]\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[/tex]

Boundary conditions are

[tex]\Omega(0)=1[/tex]

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

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

Answers and Replies

  • #2
290
0
[tex]\Phi[/tex] and [tex]S[/tex] are constants! Thanks for your answer!
 
  • #3
116
0
The general solution to the equation (maple obtained) is

[tex]\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)})[/tex]

The first boundary condition puts the constraint

[tex]C_{1}+C_{2}=-1[/tex]

while the second one says

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

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

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

All this needs some checking
 
Last edited:
  • #4
290
0
Ok! But how to get solution without use of any computer programme? How to solve this with pencil and paper? What is the idea?
 
  • #5
116
0
Well, first thing I thought was eliminating the term in the first derivative by a standart trick. If

[tex] u''(x)+A(x)u'(x)-S(S+1)u(x)=0[/tex]


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

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

In this case, we have

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

And the resulting equation for v(x) is

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

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
 
  • #6
290
0
Well, first thing I thought was eliminating the term in the first derivative by a standart trick. If

[tex] u''(x)+A(x)u'(x)-S(S+1)u(x)=0[/tex]


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

[tex] v''+[-S(S+1)-A^2/4-A'/2]v=0[/tex]
Are you sure about this?

I get equation

[tex]v''+[-S(S+1)+\frac{A^2}{4}-\frac{A}{2}-\frac{A'}{2}]v=0[/tex]
 
  • #7
290
0
Well, first thing I thought was eliminating the term in the first derivative by a standart trick. If

[tex] u''(x)+A(x)u'(x)-S(S+1)u(x)=0[/tex]


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

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

In this case, we have

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

And the resulting equation for v(x) is

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

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 [tex]v''(x)-(S+1/2)^{2}v(x)=0[/tex] from [tex] v''+[-S(S+1)-A^2/4-A'/2]v=0[/tex]?
 
  • #8
116
0
By substituting A
 
  • #9
290
0
[tex]A(x)=\frac{(1+\Phi)+\Phi e^{-x}}{(1+\Phi)-\Phi e^{-x}}[/tex]

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

[tex]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}[/tex]

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

[tex]\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}[/tex]

[tex]\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}[/tex]

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


Thanks!
 
  • #10
116
0
You are welcome
 
  • #11
290
0
In this case, we have

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

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


I have done this integral like

[tex]\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[/tex]

and got

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

and

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

Related Threads for: Differential equation

Replies
7
Views
4K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
2
Views
2K
Top