Solving a Non-Linear Equation: Need Help!

  • Context: Graduate 
  • Thread starter Thread starter CPL.Luke
  • Start date Start date
  • Tags Tags
    Non-linear
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
CPL.Luke
Messages
440
Reaction score
0
hey does anybody have any idea how to solve this equation?

E/m= xd^2x/dt^2with 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:
Physics news on Phys.org
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: