I would like to solve the following homogeneous, ODE:

$$\left[\frac{d^2}{dt^2} + m^2\right]\phi(t) + \frac{1}{6}\lambda \phi^3(t)=0.$$

I know the solution is

$$\phi(t) = \frac{z(t)}{1-\frac{\lambda}{48m^2}z^2(t)}$$ for $z(t)=z_{0}e^{i\omega t}$ in the limit $\lambda\rightarrow 0$.

I suppose the question I am asking without the physics of finding classical solution for a field is how to solve"

$$\phi''(t) + a\phi(t)+b \phi^3(t)=0.$$


closed as off-topic by Gert, ZeroTheHero, Buzz, Kyle Kanos, stafusa Jan 25 at 12:51

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Use the energy integral: After multiplication by $ d\phi/dt$ your equation becomes $$ \frac{d}{dt}\left\{ \frac 12\left( \frac{ d \phi}{dt}\right)^2 +\frac 12 m^2 \phi^2 +\frac \lambda{4!} \phi^4\right\}= 0. $$ So $$ \frac 12\left( \frac{ d \phi}{dt}\right)^2 +\frac 12 m^2 \phi^2 +\frac \lambda{4!} \phi^4= \kappa $$ for some constant $\kappa$. Choose a $\kappa$ and separate variables $$ dt= \frac {d\phi}{\sqrt{2(\kappa - \frac 12 m^2 \phi^2 +\frac \lambda{4!} \phi^4})}. $$ In general the $\phi$ integration will give an elliptic function (except in the case $\kappa=0$ which can reduce to the well-known lump solution if $m^2<0$).


A standard way to solve such nonlinear differential equation is by using the perturbation theory. Assuming the nonlinear term is weak, i.e., b<<1.

Step 1: assume the solution $\phi$ is also a function of b, i.e., $\phi(t,b)$.

Step 2: If $b$ is small, one can expand $\phi$ use Taylor expansion about $b=0$, i.e., $\phi(t,b)=\phi(t,0)+b\partial\phi(t,0)/\partial b+b^2/2\partial^2\phi(t,0)/\partial^2 b+...$

Step 3: set $b=0$, one obtain a standard linear ODE, we can solve it exactly, and obtain the so-called unperturbed solution, $\phi(t,0)$.

Step 4: Differentiate the nonlinear ODE on both side with respect to b, then set $b=0$ to solve a new differential equation for $\partial\phi(t,0)/\partial b$.

Then one can further differentiate both side again to higher orders to obtain higher order solution.

  • $\begingroup$ what I meant was that in principle the first equation can be rewritten as the final one, which is an ODE (albeit a non-standard one). Also from the solution they have $\lambda$ which is the coefficient of the non-linear term so I don't think one can ignore it. $\endgroup$ – SAMCRO Jan 23 at 22:52
  • $\begingroup$ right, the pertubative theory does not ignore it, but write the solution in terms of series of $\lambda$, the small parameter. I have added some details. $\endgroup$ – Zecheng Gan Jan 23 at 23:09

The simplest way is to throw it into Mathematica.

For a more clever way, notice that if I rewrite $\phi(t)=x(t)$ then this looks like a the eq' of motion of a particle in some potential well - hence there should be some way to use conservation of energy.

A quick way to do that is to multiply by $\phi'(t)$: $$ \phi'(t)\phi''(t) + a\phi'(t)\phi(t)+b \phi'(t)\phi^3(t)=0 $$ $$ \frac{d}{dt}\left(\frac{\phi'(t)^2}{2} + a\frac{\phi(t)^2}{2}+b\frac{ \phi^4(t)}{4}\right)=0 $$ $$ \Rightarrow\frac{\phi'(t)^2}{2} + a\frac{\phi(t)^2}{2}+b\frac{ \phi^4(t)}{4}=E $$ With $E$ ('the energy') an integration constant. Now this is a first order equation, and can be rewritten as:

$$ \intop \frac{d\phi}{\pm \sqrt{ 2E-a \phi^2-b \frac{\phi^4}{2} }}=\intop dt$$

And now throw it into Mathematica!

  • 2
    $\begingroup$ Saying "use a CAS" is a terrible answer because it promotes laziness when the math seems tractable & has an analytic solution. $\endgroup$ – Kyle Kanos Jan 24 at 2:16
  • $\begingroup$ @KyleKanos is that a guideline here on stackexchange? The question was 'how to solve' and sometimes in real life the lazy option is the best and fastest. There's something to learn from using a constant of motion but not too much to learn from carrying out that final integral (I think). $\endgroup$ – Tal Sheaffer Jan 24 at 15:12
  • $\begingroup$ for sure, this is primarily my advice/opinion on the matter. I suspect the 2 upvotes on my comment suggest I'm not alone in this thinking. That said, just saying "use CAS" alone would constitute a non-answer in my book & deletion would be imminent. Lazy options generally are fastest, but I contest that it's best because you slowly kill your ability to do any sort of math by resorting to the "easy" way out. $\endgroup$ – Kyle Kanos Jan 24 at 16:00
  • $\begingroup$ Hi All, many thanks for your input, I am going to try both methods and see which gives a sensible solution. I am being practical and want to reproduce the results of the following paper: arxiv.org/abs/hep-ph/9209203 $\endgroup$ – SAMCRO Jan 25 at 2:40

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