= Solution
Allow a freely chosen center $x_0$, set $u=k(x-x_0)$ and $z=e^u$. Direct differentiation of the proposed <Sine-Gordon kink> gives
$$
\phi'=\frac{4kz}{b(1+z^2)}=\frac{2k}{b}\operatorname{sech}u,
\qquad
\phi''=\frac{4k^2z(1-z^2)}{b(1+z^2)^2}.
$$
The angle identity gives
$$
\sin(b\phi)=\sin(4\arctan z)=\frac{4z(1-z^2)}{(1+z^2)^2}.
$$
Since $k^2/b=a$, these formulas show $\boxed{\phi''=a\sin(b\phi)}$, precisely the static <equation of motion>.
For $x\to-\infty$ and $x\to+\infty$, the <field> approaches $0$ and $2\pi/b$ respectively. These are neighboring minima of the <scalar potential> when $ab>0$, and $\phi'$ vanishes exponentially at either end. The <energy> is therefore finite and the <topological charge> $Q=b[\phi(+\infty)-\phi(-\infty)]/(2\pi)$ is one. Reversing $x-x_0$ gives the corresponding antikink. If $ab<0$, the printed real exponential profile is unavailable; its formal imaginary scale is not a real finite-energy solution of the stated form.
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