Cubic NLW/NLKG

The cubic nonlinear wave and Klein-Gordon equations have been studied on ${\mathbb {R} }$, on ${\mathbb {R} }^{2}$, and on ${\mathbb {R} }^{3}$.

Exact solutions

Technique

The technique to solve a non-linear equation

$-\Box \phi +V'(\phi )=0$

relies on Lorentz invariance of the solutions. We consider the reference frame where the solutions depend only on time variable. This reduces the above PDE to an ordinary differential equation as

$\partial _{t}^{2}\phi +V'(\phi )=0.$

Then, if we are able to solve this equation, we can get back an exact solution to the equation we started from with the identity

$\,\phi (x')=\phi (\Lambda x)\!$

being this the way the scalar field changes under the effect of a Lorentz transformation $\,\Lambda \!$ and being here $\,x_{\mu }=(t,0)\!$. Indeed, one notes that the solutions in this frame, $\,\phi (t,0)\!$, are exact solutions of the given PDE.

Solutions

This kind of equation displays a class of solutions with a peculiar dispersion relation. To show explicitly this, let us consider the massless equation

$-\Box \phi +\lambda \phi ^{3}=0$

being $\,\lambda >0\!$. An exact solution of this equation is given by

$\phi (x)=\pm \mu \left({\frac {2}{\lambda }}\right)^{1 \over 4}{\rm {sn}}(p\cdot x+\theta ,i),$

being $\,{\rm {sn\!}}$ a Jacobi elliptic function and $\,\mu ,\theta \!$ two integration constants, and the following dispersion relation holds

$p^{2}=\mu ^{2}\left({\frac {\lambda }{2}}\right)^{1 \over 2}.$

We see that we started with an equation without a mass term but the exact solution describes a wave with a dispersion relation proper to a massive solution. This can be seen as the superposition of an infinite number of massive linear waves through a Fourier series of the Jacobi function, that is

$\phi (t,0)=\pm \mu \left({\frac {2}{\lambda }}\right)^{1 \over 4}\sum _{n=0}^{\infty }(-1)^{n}{\frac {2\pi }{K(i)}}{\frac {e^{-\left(n+{1 \over 2}\right)\pi }}{1+e^{-(2n+1)\pi }}}\sin \left((2n+1){\frac {\pi }{2K(i)}}\left({\frac {\lambda }{2}}\right)^{1 \over 4}\mu t\right)$

being $K(i)$ an elliptic integral. We recognize the "spectrum"

$m_{n}=(2n+1){\frac {\pi }{2K(i)}}\left({\frac {\lambda }{2}}\right)^{1 \over 4}\mu .$

But a meaning as a mass spectrum can only be given within a quantum field theory FraB2006.

Similarly, when there is a mass term as in

$-\Box \phi +\mu _{0}^{2}\phi +\lambda \phi ^{3}=0$

the exact solution is given by

$\phi (x)=\pm {\sqrt {\frac {2\mu ^{4}}{\mu _{0}^{2}+{\sqrt {\mu _{0}^{4}+2\lambda \mu ^{4}}}}}}{\rm {sn}}\left(p\cdot x+\theta ,{\sqrt {\frac {-\mu _{0}^{2}+{\sqrt {\mu _{0}^{4}+2\lambda \mu ^{4}}}}{-\mu _{0}^{2}-{\sqrt {\mu _{0}^{4}+2\lambda \mu ^{4}}}}}}\right)$

being now the dispersion relation

$p^{2}=\mu _{0}^{2}+{\frac {\lambda \mu ^{4}}{\mu _{0}^{2}+{\sqrt {\mu _{0}^{4}+2\lambda \mu ^{4}}}}}.$

Finally, we can write down the exact solution for the case

$-\Box \phi -\mu _{0}^{2}\phi +\lambda \phi ^{3}=0$

that is given by

$\phi (x)=\pm v\cdot {\rm {dn}}(p\cdot x+\theta ,i),$

being $v={\sqrt {\frac {2\mu _{0}^{2}}{3\lambda }}}$ and the following dispersion relation holds

$p^{2}={\frac {\lambda v^{2}}{2}}.$

These wave solutions are interesting as, notwithstanding we started with an equation with a wrong mass sign, the dispersion relation has the right one. Besides, Jacobi function $\,{\rm {dn}}\!$ has no real zeros and so the field is never zero but moves around a given constant value that is initially chosen. This effect is known as spontaneous breaking of symmetry in physics.