Consider the PDE $$u_t +4u_x +6u_{xxx} =0$$ for the function $u = u(x, t).$
(a) Determine for what wave speeds $c$ there exist bounded travelling wave solutions of $u$, and find the form of the solutions. (The solution is bounded if there exists constant $K$ such that $|u(x, t)| < K$ for all $x$ and $y$.
I am looking for some input as to whether my solution is sound.
Attempt:
We will make use of the following two facts.
(i) in the literature I have been using $c=\dfrac{\omega}{k}$
(ii) $|\sin(\alpha)|\leq1$
Given that $|u(x, t)| < K$ we want to find $|u_t +4u_x +6u_{xxx}|<k$. First see that we have the following partial derivatives \begin{align*} u_t&=A\omega\sin(kx-\omega t)\\ u_t&=-Ak\sin(kx-\omega t)\\ u_{xxx}&=Ak^3\sin(kx-\omega t) \end{align*}
For ease of notation let use set $\alpha = kx-\omega t$. Then making the proper substitutions we must show $$|A\omega\sin(\alpha)-4Ak\sin(\alpha) + 6Ak^3\sin(\alpha)|<k.$$ Now see that, \begin{align*} |A\omega\sin(\alpha)-4Ak\sin(\alpha) + 6Ak^3\sin(\alpha)|&\leq|A\omega||\sin(\alpha)|+|-4Ak||\sin(\alpha)| + |6Ak^3||\sin(\alpha)|\\ &\leq|A\omega|+|-4Ak| + |6Ak^3|\\ &\leq A\omega+4Ak+6Ak^3\\ &\leq \omega+4k+6k^3. \end{align*} now see that \begin{align*} \omega=-6k^3-4k&\Rightarrow \omega=k(-6k^2-4)\\ &\Rightarrow\dfrac{\omega}{k}=-6k^2-4\\ &\Rightarrow c=-6k^2-4 \end{align*} So, if we set $K=-6k^2-4$ we have a bound for $c$.