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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?

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Your field equation is correct up to a choice of sign for the metric $\eta_{\mu \nu}.$ –  Gerben Oct 25 '11 at 17:50
    
oh yeah, I always use signature +1 -1 -1 -1 –  Anthonny Oct 25 '11 at 18:04
    
You need the opposite sign mass term, then there are solitons. –  Ron Maimon Oct 25 '11 at 20:22
    
Are you sure? I have a problem with this because I need that the asymptotic behaviour to be like $e^{-r}/r$ and if it were like you are saying it cant be possible. Or not? –  Anthonny Oct 25 '11 at 20:49
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Sorry--- domain line--- its a wall in time. I wick rotated unconsciously. –  Ron Maimon Oct 26 '11 at 3:34
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2 Answers

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.

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Your answer is fine. You have just to change the sign before the quartic term to agree with the OP's question. –  Jon Dec 6 '11 at 14:02
    
@Jon: No, the sign change is in the quadratic term. The negative mass is required for a soliton solution. –  Ron Maimon Dec 6 '11 at 17:10
    
Yes, but the original question was if the equation with a plus sign admits solitons. In agreement with your assertion, I display the exact solution in this case that is not a soliton. My answer and yours are complimentary and fully answer the OP's question. –  Jon Dec 6 '11 at 17:40
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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.

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What does this answer have to do with the question? –  Ron Maimon Dec 6 '11 at 17:14
    
Dear @Jon. Don't forget to disclose if you are the author of the linked paper, cf. Physics.SE policy. –  Qmechanic Mar 6 '12 at 11:15
    
Dear @Qmechanic, as you can see this is not a secret. What really matters is if an argument is correct or not. –  Jon Mar 6 '12 at 20:49
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