Cubic NLW/NLKG

The cubic nonlinear wave and Klein-Gordon equations have been studied on R, on R^2, and on R^3.

Exact solutions

This kind of equation displays a class of solutions with a peculiar dispersion law. 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

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

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

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

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.

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

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

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

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

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

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

So, 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. This effect is known as spontaneous breaking of symmetry in physics.