Are there solitary waves in $\phi^4$ theory in 3+1 dimensions? In 3+1 dimensions with signature +1 -1 -1 -1, 
$$ \mathcal{L}= \frac{1}{2}\partial^\mu\phi\partial_\mu\phi -\phi^2/2 -\phi^4/4$$
field equation: $$\square\phi+\phi+\phi^3=0$$ (check this)
$$\square=\partial^2_t-\nabla^2$$
Note that it is not a Mexican hat. I guess this haven't been solved exactly before but, somebody have shown or discarded that there could be soliton solutions or at least solitary waves?
 A: The solution for the soliton in a $\phi^4$ model is given by making a field $\phi$ which depends only on x and t, and is independent of any other spatial dimensions. This is a classical one dimensional problem.
When the mass-squared parameter is negative, then the soliton appears. It is the solution to the equation
$$ \partial_x^2 \phi + \phi - \phi^3 = 0 $$
Where x is rescaled to absorb $\mu^2$, and $\phi$ is rescaled to absorb $\lambda$. The solution is gotten by using a version of conservation of energy, which works here because the above is a second order differential equation, which looks just like the motion of a particle in a potential
$$ V(\phi) = {1\over 2} \phi^2 - {1\over 4} \phi^4$$
Note that this is the inverted field potential appearing in the Lagrangian. The solution for $\phi$ has a x-conservation of x-energy, because if you call x "time", then the second order equation turns into Newton's laws for a one-dimensional motion. The conserved quantity is
$$ {1\over 2} (\partial_x \phi)^2 + V(\phi) = E$$
For the soliton solution, $\phi$ should go to the vacuum solution at $x=\pm\infty$. The two vacua are the two minima of the original potential, the places where
$$ \phi - \phi^3 = 0$$
or
$$\phi = \pm 1 $$
The potential at these field values gives the energy, because the field gradient has to go to zero at infinty. This makes the x-energy 1/4 at infinity.
The conservation of x-energy then tells you the field gradient
$$ (\partial\phi(x))^2 + \phi^2 - {1\over 2} \phi^4 = 1/2 $$
or that
$$ {1\over (\phi^2-1) } (\partial_x \phi) = \pm t+C $$
or
$$ \tanh^{-1} \phi = t+C $$
Which gives the standard $\phi^4$ domain wall soliton
$$ \phi(x) = \tanh(t+C)$$
This solution is a particle in 1d (1+1), a line in 2d (2+1), a domain wall in 3d (3+1), and in general, a d-1 dimensional object in d dimensions.
A: I would like to give you another solution to your equation that was recently published (see here). The equation
$$\partial_x^2\phi+\mu_0^2\phi+\lambda\phi^3=0$$
admits the exact solution
$$\phi(x)=\pm\sqrt{\frac{2\mu^4}{\mu_0^2 + \sqrt{\mu_0^4 + 2\lambda\mu^4}}}{\rm sn}\left(p\cdot x+\theta,\sqrt{\frac{-\mu_0^2 + \sqrt{\mu_0^4 + 2\lambda\mu^4}}{-\mu_0^2 - 
   \sqrt{\mu_0^4 + 2\lambda\mu^4}}}\right)
$$
being sn a Jacobi elliptic function, $\theta$ and $\mu$ two integration constants and provided that
$$
   p^2=\mu_0^2+\frac{\lambda\mu^4}{\mu_0^2+\sqrt{\mu_0^4+2\lambda\mu^4}}.
$$
This has the appearance of a dispersion relation and $p$ is a quasi-momentum as happens in solid state with quasi-particles in a strong interacting environment.
As pointed out before, this is not a soliton solution.
