 Solution

Use the amplitude placement in the original PDF: $w=[A{\cosh }^2(Bx){]}^{1/(1-n)}$. Introduce the actual peak $C=A^{-1/(n-1)}$ and $p=2/(n-1)$, so $w=C{\mathrm{sech}}^p(Bx)$. Its derivatives satisfyThe static equation is $w^{\prime \prime }=aw-b{w}^n$. Matching the constant and ${\mathrm{sech}}^2$ terms givesChanging the sign of $B$ leaves the profile unchanged. If $A$ is instead defined as an outside prefactor, as in the held TeX transcription, its value is $C$, not the PDF's $A$. This notation discrepancy does not change the physical profile.

The nonconstant positive power-law nonlinear Klein-Gordon pulse is even, smooth, has peak $C$ at zero and decays as $e^{-\sqrt{a}|x|}$. The requested sketches below compare several powers with illustrative $a=b=1$. For odd $n$, its negative is also a solution. A separate constant branch occurs for $B=0$: the positive equilibrium has $C=(a/b{)}^{1/(n-1)}$, or $A=b/a$ in the PDF convention. The zero solution is another equilibrium, obtained as a limiting profile rather than finite positive $A$.

Now apply the Lorentz boost with rapidity $-\beta$ to the static field. It givesFor positive $\beta$, the centre moves to the right at $0<v<1$, with speed one the wave-speed unit. The peak stays $C$ and the width contracts by $1/\cosh \beta$. The shape translates without changing in time. This establishes a travelling solitary pulse; it does not by itself prove stability or elastic multi-pulse scattering.