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The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and total energy areThe Euler-Lagrange field equation is ${\varphi }_{tt}-{\varphi }_{xx}+U^{\prime }(\varphi )=0$. Multiplication by ${\varphi }_t$ gives the local conservation of energy identityThus $E$ is conserved when the energy flux vanishes at spatial infinity. For finite energy with asymptotically constant fields, the endpoints ${\varphi }_{\pm }$ must be scalar-field vacua, where $U=0$.

Choose an antiderivative $G^{\prime }(\varphi )=\sqrt{2U(\varphi )}$. For either sign $s=\pm 1$, completing the square givesChoosing $s$ to make the boundary term nonnegative proves the Bogomolny boundThese Bogomolny equations obtained by square completion for a one-dimensional kink describe minima within a fixed endpoint sector when a saturating configuration exists. Differentiating the spatial equation gives ${\varphi }_{xx}=U^{\prime }(\varphi )$ wherever $U>0$, with smooth extension to its limiting scalar-field vacua. The absolute minimum across all sectors is a constant zero-potential scalar-field vacuum when one exists; the nontrivial minima are sector-specific kinks.

For $\beta \ne 0$, write $b=|\beta |$. The scalar-field vacua are $-b,0,b$. A static finite-energy solution has the first integralbecause differentiating its left side gives ${\varphi }_x({\varphi }_{xx}-U^{\prime })=0$, and its value is zero at infinity. In either adjacent interval, $\sqrt{2U}=|\varphi |(b^2-{\varphi }^2)$. Set $y={\varphi }^2/b^2$. The first-order equation becomes $y_x=\eta 2b^{2}y(1-y)$, with $\eta =\pm 1$, and separation gives $\log [y/(1-y)]=2\eta b^2(x-x_0)$. HenceFor $\eta =1$ the endpoints are $0\to \sigma b$; for $\eta =-1$ they are $\sigma b\to 0$. Thus there are four oriented static kink families, up to translation, namely two increasing kinks and two decreasing antikinks. Each has rest energyA static field joining $-b$ directly to $b$ would have to pass through $\varphi =0$ at finite $x$. The first integral then gives ${\varphi }_x=0$, and the Picard-Lindelöf theorem applied to the smooth second-order equation with those initial conditions forces $\varphi \equiv 0$. Therefore there is no such additional kink; this is the intermediate-vacuum obstruction to a kink. The Bogomolny classification of a rescaled phi-six kink also shows that if $\beta =0$, only the scalar-field vacuum $0$ remains and there is no nontrivial finite-energy static kink. Indeed the first integral gives a monotone field wherever it is nonzero, precluding a return to the same scalar-field vacuum.

A Lorentz boost of the $0\to b$ profile gives a moving solution with arbitrary $|v|<1$:If $K$ denotes the static profile and $\xi =\gamma (x-vt-x_0)$, then ${\varphi }_{tt}-{\varphi }_{xx}={\gamma }^2(v^2-1)K^{\prime \prime }=-K^{\prime \prime }$, proving the Euler-Lagrange field equation directly. Its energy is $\gamma b^4/4$: since $K^{\prime 2}=2U(K)$, changing variable to $\xi$ in the energy integral gives the factor $\gamma$. Reversing $\sigma$ or $\eta$ yields the other moving elementary profiles.