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Away from forcing, set $F=0$. For a linear sinusoidal disturbance $A=\mathrm{Re}[\hat{A}{e}^{i(kx-\omega t)}]$, the reduced equation givesSince $\sigma ,\beta ,{\alpha }_1>0$, the phase velocity $c_p=\omega /k=\sigma k^2/(\beta {\alpha }_1)$ is positive for every nonzero wavenumber. The group velocity isThus both the linear phase and wave packet travel downstream; the sign of $k$ does not introduce an upstream branch.

Now seek a permanent nonlinear wave with $\xi =x+ct$, for which the propagation velocity is $-c$. Substitution gives $\sigma A^{(3)}-\beta {\alpha }_{1}c{A}^{\prime }-{\alpha }_{2}A{A}^{\prime }=0$. Integration with vanishing far-field data givesMultiplying by $A^{\prime }$ and integrating again, with zero integration constant, givesHere ${\alpha }_2>0$, because ${\gamma }_0>{\gamma }_1>0$ and $R<R+1$. A smooth nonzero wave approaching zero requires $c>0$: for $c<0$ the right-hand side is negative at sufficiently small nonzero $A$. At $c=0$ the allowed positive branch has no nonzero turning point, so it cannot form a smooth localized pulse.

When $c>0$, the nonzero turning point is $A_m=-3\beta {\alpha }_{1}c/{\alpha }_2$. The energy curve is positive between $A_m$ and zero and negative below $A_m$, giving a homoclinic negative pulse. The positive branch has no turning point and cannot return to zero. ThereforeAn explicit solution, also confirming smooth exponential tails, isThis establishes the opposite propagation directions in the annular displacement equation: linear waves move downstream while its localized nonlinear depression waves move upstream.