= Solution
The <Euler-Lagrange field equation> is $F'(\phi)-\partial_t^2\phi+\partial_x^2\phi=0$. Therefore
$$
\boxed{\partial_t^2\phi-\partial_x^2\phi=f(\phi).}
$$
Put $c=\cosh\beta$, $s=\sinh\beta$, $X=cx+st$ and $T=ct+sx$. The <chain rule> gives $\partial_t=s\partial_X+c\partial_T$ and $\partial_x=c\partial_X+s\partial_T$. Cross terms cancel and $c^2-s^2=1$, so
$$
(\partial_t^2-\partial_x^2)\phi(X,T)
=(\partial_T^2-\partial_X^2)\phi(X,T)=f(\phi(X,T)).
$$
Thus \b[the boosted <scalar field> is another solution] for any constant <rapidity>. The nonlinear term in the <nonlinear Klein-Gordon equation> is unchanged because it is a local function of the scalar itself.
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