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Consider \begin{equation}\label{1} \partial^2_t\phi-\partial^2_x\phi=\phi -\phi^3,\: \ (x,t) \in \mathbb{R}\times \mathbb{R} \hspace{30pt}(1) \end{equation} the $\phi^4$ model.

I know that $$H(x)=\tanh\left(\frac{x}{\sqrt{2}}\right),\: \forall \; x \in \mathbb{R}$$ is a time-independent solution of the $(1)$ and is called kink. Taking into account that $$sn(u,1)=\tanh(u)\: \forall \; u \in \mathbb{R}$$ where $sn$ denote the elliptical snoidal function, and that $sn$ is periodic. I can say that $H$ is periodic? In addition, the $\phi^4$ model admit explicit traveling waves of the form $\phi(x,t)=\psi_c(x-ct)$ , where $c \in \mathbb{R}$? The answer to the last question I think is 'yes', because the solution a translation of $H$ also remains a solution of $(1)$.

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  • $\begingroup$ Left hand side has Lorentz symmetry. So if $c= \pm1$, you can find travelling wave solutions (these are called solitons in the loose sense). If you want to find out if solutions travelling at other speeds exist, you can try to express the LHS in terms of derivatives $ x\pm ct$. I doubt if such solutions will exist. $\endgroup$
    – Vivek
    Commented Apr 17, 2020 at 22:34

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Can I say that $H$ is periodic?

No, you cannot. $\tanh(x)$ is pretty clearly non-periodic as a function of a real variable. Note that the period of $sn(u,m)$ is $K(m)$, with $K$ the complete elliptic integral of the first kind, but $K(1) = \tilde{\infty}$.

Does the $\phi^4$ model admit traveling wave solutions of the form $\phi(x,t)=\psi_c(x-ct)$? [...] the solution a translation of $H$ also remains a solution of (1).

A constant translation is still a solution, but not necessarily a time-dependent one. This can be checked by explicit differentiation - simply plug your proposed traveling wave into the differential equation and see what happens.

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  • $\begingroup$ If I replace $\phi(x,t)=\psi_c(x-ct)$ in the equation, I'll find $$(1-c²)\psi_c''=\psi_c-\psi_c^3.$$So what can I conclude from that? $\endgroup$
    – Guilherme
    Commented Apr 14, 2020 at 13:18
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    $\begingroup$ @GuilhermedeLoreno I think you're missing a minus sign. But if $\psi_c(x)$ also satisfies the equation, then three of the four terms in the equation you wrote cancel each other out. $\endgroup$
    – J. Murray
    Commented Apr 14, 2020 at 13:21
  • $\begingroup$ yes, there's a minus sign missing anyway. But why are they canceling if on one side we have $\psi''_c$ and the other $\psi_c$? $\endgroup$
    – Guilherme
    Commented Apr 14, 2020 at 13:26
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    $\begingroup$ @GuilhermedeLoreno If $\psi_c(x)$ satisfies the equation of motion, then $\psi_c'' = -(\psi_c + \psi_c^3)$. If we plug $\psi_c(x-vt)$ into the equation, we will find that $(1-v^2)\psi_c''=-(\psi_c+\psi_c^3)$. But given the first part, that reduces to $v^2 \psi_c'' = 0$, which is not true for non-zero $v$, which means that the traveling wave you propose is not a solution to the equation of motion. $\endgroup$
    – J. Murray
    Commented Apr 14, 2020 at 13:28
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    $\begingroup$ @GuilhermedeLoreno Not if $\psi_c(x)$ also satisfies the equation of motion, but if it doesn't, then such solutions exist. See e.g. sciencedirect.com/science/article/pii/S221137971500011X $\endgroup$
    – J. Murray
    Commented Apr 14, 2020 at 13:47

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