= 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 $w=C\operatorname{sech}^{p}(Bx)$. Its derivatives satisfy
$$
\frac{w''}{w}=p^2B^2-p(p+1)B^2\operatorname{sech}^2(Bx),\qquad
w^{n-1}=C^{n-1}\operatorname{sech}^2(Bx).
$$
The static equation is $w''=aw-bw^n$. Matching the constant and $\operatorname{sech}^2$ terms gives
$$
\boxed{B=\frac{n-1}{2}\sqrt a,\qquad
C=\left[\frac{a(n+1)}{2b}\right]^{1/(n-1)},\qquad
A=\frac{2b}{a(n+1)}.}
$$
Changing 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 $A$.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2013/paper-1-tp1-scalar-pulses.png]
{title=Localized scalar pulses for several nonlinear powers, and Lorentz-contracted travelling snapshots}
Now apply the <Lorentz boost> with <rapidity> $-\beta$ to the static field. It gives
$$
\boxed{\phi(x,t)=C\operatorname{sech}^{2/(n-1)}
\left[B\cosh\beta\,(x-vt)\right],\qquad v=\tanh\beta.}
$$
For 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.
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