= Solution
As $|x|\to\infty$, the operator $A$ tends to $-4\partial_x^3$. Evolve the normalized scattering eigenfunction by
$$
\partial_t\phi_k=A\phi_k+4ik^3\phi_k.
$$
The added scalar term keeps the coefficient of $e^{-ikx}$ at $-\infty$ equal to one. At $+\infty$,
$$
(-4\partial_x^3+4ik^3)e^{-ikx}=0,
\qquad
(-4\partial_x^3+4ik^3)e^{ikx}=8ik^3e^{ikx}.
$$
Consequently
$$
\dot a(k,t)=0,
\qquad
\dot b(k,t)=8ik^3b(k,t).
$$
It follows that
$$
\boxed{T(k,t)=T(k,0),
\qquad
R(k,t)=R(k,0)e^{8ik^3t}.}
$$
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