Non-linear ode

  • Thread starter CPL.Luke
  • Start date
  • #1
441
0
hey does anybody have any idea how to solve this equation?

E/m= xd^2x/dt^2


with initial conditions x=a and dx/dt(a)=0

its non-linear and so I don't have any idea what to do with it, and maple won't give me an answer.
 
Last edited:

Answers and Replies

  • #2
Yeah, it's a tricky Emden-Fowler equation. The only thing I can think of is to multiply by x' and integrate to get

[tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{x} + C [/tex]

Now, swap for x from the original equation, i.e.,

[tex]x = \frac{E}{m x^{\prime \prime}}[/tex]

such that

[tex]\frac{1}{2}x^{\prime 2} = \frac{E}{m} \ln{\frac{E}{m x^{\prime \prime}}} + C [/tex]

rearrange to get

[tex]x^{\prime \prime} = A e^{-\frac{m x^{\prime 2}}{2 E}} [/tex]

where A is a constant. Now make the change of variable

[tex]x^{\prime} = i \sqrt{\frac{2 E}{m}} z [/tex]

to get

[tex]i \sqrt{\frac{2E}{m}} z^{\prime} e^{-z^2} = A [/tex]

but of course

[tex]\frac{ d }{dt} erf{(z)} = \sqrt{\frac{2}{\pi}} z^{\prime} e^{-z^2}[/tex]

so integrating gives you

[tex]erf{(z)} = \alpha t + \beta [/tex]

where alpha and beta are constants that may or may not be complex

So there's a solution of the form

[tex]x(t) = i\sqrt{\frac{2 E}{m}} \int{erf^{-1}{(\alpha t + \beta)} dt}[/tex]

Note: the inverse erf function is integrable -- see this Wolfram page.
 
Last edited:

Related Threads on Non-linear ode

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
3K
Top