How to find the conserved quantities of $\phi^4$ model? Consider
\begin{equation}\label{1}
\partial^2_t\phi-\partial^2_x\phi=\phi -\phi^3,\: \ (x,t) \in \Bbb{R}\times \Bbb{R} \tag{1}
\end{equation}
the $\phi^4$ model, often used
in quantum field theory and other areas of physics. It can be found here, Ref. [1].
How to calculate the conserved quantities for this model?

[1] Michał Kowalczyk, Yvan Martel and Claudio Muñoz: Kink dynamics in the $\phi^4$ model: Asymptotic stability for odd perturbations in the energy space, J. Amer. Math. Soc. 30 (2017), 769-798. doi:10.1090/jams/870
 A: Define the energy
$
E(t) = \frac12\! \int_{\Bbb R}  {\phi_t}^2 + {\phi_x}^2 \,\text d x
$
as usually done for wave equations. Thus,
\begin{aligned}
\frac{\text d}{\text d t}E(t) &= \int_{\Bbb R}  {\phi_t}\phi_{tt} + {\phi_x}\phi_{xt} \,\text d x \\
&= \int_{\Bbb R}  {\phi_t}(\phi_{xx} + \phi - \phi^3) + {\phi_x}\phi_{tx} \,\text d x \\
&= \int_{\Bbb R} ({\phi_t}\phi_{x})_x + {\phi_t} (\phi - \phi^3) \,\text d x \\
&= \int_{\Bbb R}  \big(\tfrac12\phi^2 - \tfrac14\phi^4\big)_t \,\text d x \\
&= -\frac{\text d}{\text d t}D(t)
\end{aligned}
with $D(t) = \frac14\! \int_{\Bbb R} \phi^2(\phi^2-2) \,\text d x$. Therefore, the energy $E +D$ is conserved. Note that the energy found in the article is also conserved. Indeed, it is of the form $E+C$ with $C(t) = \frac14\! \int_{\Bbb R} (1-\phi^2)^2 \,\text d x$; the densities corresponding to $C$ and $D$ differ only by a constant.
A: A Lagrangian density for the system in 1+1D spacetime is
$$ {\cal L}~=~\frac{1}{2}(\partial_t\phi)^2 -\frac{1}{2}(\partial_x\phi)^2-{\cal V}, \qquad {\cal V}~= ~\frac{1}{4}\phi^4-\frac{1}{2}\phi^2. $$
We can now calculate the conserved quantities as Noether charges for the symmetries of the action functional. Poincare symmetry with Lie algebra $iso(1,1)$ is generated by 2 spacetime translations and 1 boost. The corresponding 3 Noether charges are energy, momentum & boost,
$$\begin{align} 
H~=~& \int_{\mathbb{R}} \!\mathrm{d}x~{\cal H}, \qquad 
{\cal H}~=~\frac{1}{2}(\partial_t\phi)^2 +\frac{1}{2}(\partial_x\phi)^2+{\cal V},\cr 
P~=~& \int_{\mathbb{R}} \!\mathrm{d}x~{\cal P}, \qquad 
{\cal P}~=~\partial_t\phi~\partial_x\phi,\cr
B~=~& \int_{\mathbb{R}} \!\mathrm{d}x~{\cal B}, \qquad 
{\cal B}~=~t{\cal P}-x{\cal H}.
\end{align}$$
