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Write $T=\epsilon t$ and expandAt order ${\epsilon }^2$, the assumptions $f^{\prime \prime }(0)=0$ and the orthogonality normalization leave a homogeneous equation for $u_2$ with no forcing, hence $\boxed{u_2=0}$.

At order ${\epsilon }^3$, project the equation onto the null mode $\sin z$. Sincethe Fredholm solvability condition obtained directly from the definitions printed in the question isThe paper asks for $A/f^{\prime }(0{)}^2$ in place of $f^{\prime }(0{)}^{2}A$, but that coefficient is incompatible with $R_c=-1/f^{\prime }(0)$ and ${\epsilon }^2=R-R_c$: differentiating $1/R$ at $R_c$ gives $-1/R_c^2=-f^{\prime }(0{)}^2$. Thus the displayed target appears to contain a reciprocal typo. It would agree with the expansion only under a correspondingly rescaled definition of the small parameter.

The imposed orthogonality of every $u_j$, $j\ge 2$, removes the freedom to transfer a multiple of $\sin z$ between $A$ and the higher-order terms.

The linear $\cos x$ mode has frequency squared $k^2-1$, so $k_c(0)=1$. To resolve the small frequency near threshold, writeThe leading equationand the initial data give

At $O({\epsilon }^2)$, the equation for ${\theta }_2$ has a forcing whose $\cos x$ component isThe Fredholm solvability condition removes this secular term and yields the slow amplitude equationIt has the conserved energyThe potential has maxima atA periodic orbit launched from $A(0)=1$, $A^{\prime }(0)=0$ exists only when that turning point lies inside the two maxima, namelyAt equality the orbit is the separatrix; below it, the assumed real periodic oscillation is lost. Therefore