= Solution
The Fourier modes evolve according to the dispersion relation found in part (a), so
$$
\phi(Vt,t)=\int_{-\infty}^{\infty}
A(k)e^{it\psi(k)}\,dk,
\qquad
\psi(k)=(V-U)k-\frac{k^9}{9}.
$$
For $V>U$, the stationary-point equation is
$$
\psi'(k)=V-U-k^8=0.
$$
It has the two real solutions
$$
k=\pm k_0,
\qquad
k_0=(V-U)^{1/8}.
$$
Define
$$
\Omega=\psi(k_0)=\frac89k_0^9,
\qquad
|\psi''(\pm k_0)|=8k_0^7.
$$
The <one-dimensional stationary-phase formula> gives one contribution from each stationary point:
$$
\boxed{
\phi(Vt,t)\sim
\sqrt{\frac{\pi}{4tk_0^7}}
\left[
A(k_0)e^{i(\Omega t-\pi/4)}
+A(-k_0)e^{-i(\Omega t-\pi/4)}
\right].}
$$
Because the initial field is <real-valued function>[real], its <Fourier transform> has the conjugate symmetry $A(-k)=A(k)^*$. Hence the same result can be written
$$
\boxed{
\phi(Vt,t)\sim
\sqrt{\frac{\pi}{tk_0^7}}\,
\operatorname{Re}\!\left[
A(k_0)e^{i(\Omega t-\pi/4)}
\right].}
$$
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