I need to solve the following:

$$(1 - \phi^2)\phi'' + \phi(\phi')^2 =0.$$

Is there any standard method I can use?

  • 1
    $\begingroup$ I don't think this is as easy as you think. $\endgroup$ – Ataraxia Jun 4 '13 at 22:14
  • $\begingroup$ It may be harder :S I just saw two exponentials function in the Mathematica solution and i though it was easy :S $\endgroup$ – José D. Jun 4 '13 at 22:18

Just a lot of pattern matching and manipulation. Rewrite the equation as

$$\frac{\phi''}{\phi'} = -\frac{\phi \, \phi'}{1-\phi^2}$$

This can be written as

$$\frac{d}{dx} \log{\phi'} = \frac12 \frac{d}{dx} \log{(1-\phi^2)}$$

This may be integrated and subsequently exponentiated to produce

$$\phi' = A \left (1-\phi^2\right)^{1/2}$$

where $A$ is a constant of integration. We may then rewrite this equation in differential form as

$$\frac{d\phi}{\left (1-\phi^2\right)^{1/2}} = A \, dx$$

which integrates to

$$\arcsin{\phi} = A x + B$$

where $B$ is another constant of integration. The solution to the above equation is then

$$\phi(x) = \sin{(A x+B)}$$

You may verify that this is indeed the solution by plugging it back into the original equation.

  • $\begingroup$ Mathematica's solution is $Cosh(C1 (t + C2))$ :S Is that right? $\endgroup$ – José D. Jun 4 '13 at 22:20
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    $\begingroup$ @Trollkemada: You tell me. I plugged my solution into the original equation and the equation was satisfied. I don't see how your Mathematica solution works, unless your constants of integration are imaginary. $\endgroup$ – Ron Gordon Jun 4 '13 at 22:22
  • $\begingroup$ You're right. I think both solutions are right. $\endgroup$ – José D. Jun 4 '13 at 22:23
  • $\begingroup$ @RonGordon However, it's easy to verify that $\mathrm{cosh}(at+b)$ is indeed a solution of the original equation. When you integrate $u'/u$, don't forget you get $\log |u|$, not simply $\log u$. $\endgroup$ – Jean-Claude Arbaut May 16 '14 at 15:48

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