# Solving ODE equation for classical field [closed]

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, stafusaJan 25 at 12:51

• This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

• 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. – SAMCRO Jan 23 at 22:52
• 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. – 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!

• Saying "use a CAS" is a terrible answer because it promotes laziness when the math seems tractable & has an analytic solution. – Kyle Kanos Jan 24 at 2:16
• @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). – Tal Sheaffer Jan 24 at 15:12
• 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. – Kyle Kanos Jan 24 at 16:00
• 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 – SAMCRO Jan 25 at 2:40