Solving $u_{t}+u_{xxx} = u\sin{x}$ using 1. Laplace transforms 2. KdV equation comparison I'm trying to solve: 
$$u_{t}+u_{xxx} = u\sin{x}$$
where $u=u(x,t)$, subject to some initial condition, $u(x,0) = f(x)$ 
First attempt - via Laplace transforms wrt $t$:
$$\frac{\partial^3{U(x,s)}}{\partial{x}^3}+U(x,s)\left[s-\sin{x}\right]=u(x,0)=f(x)$$
I notice from the right hand side that the left must be a function of $x$ only.  Not sure where to go from here.
Second attempt:
The Pde seems to be similar to the KdV equation (https://en.m.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation).
$$u_{t}+uu_x+\delta^2u_{xxx}=0$$
And comparing the two equations I notice that $u_x=-\sin{x}$ and $\delta^2=1$. - Can I do this here?
Hence 
$$u(x,t)=\cos{x}+h(t)$$
Substituting this back into the Pde, I get the following:
$$h_t(t)-(\sin{x})h(t)+\sin{x}(1-\cos{x})=0$$
This does not seem to have any possible solutions for $h(t)$.
Questions:


*

*Is my working correct thus far?

*Can I take $u_x = \sin{x}$? If not why not? Are there not possible solutions for $h(t)$ with this assumption?

*What methods should I try for solving? Please give examples.

 A: This is a very partial answer. Maybe you should solve the problem exactly as Airy's equation is solved. Using that 
$$
u\sin x = \frac u{2i}(e^{ix}-e^{-ix}),
$$
take Fourier transform to the equation to obtain
$$
\partial_t \hat u(t,\xi) +(i\xi)^3 \hat u(t,\xi) =\frac 1{2i}(\hat u(t,\xi-1)-\hat u(t,\xi+1)),
$$
which simplifies to
$$
\partial_t \hat u(t,\xi) = i\xi^3 \hat u(t,\xi)  + \frac 1{2i}(\hat u(t,\xi-1)-\hat u(t,\xi+1)).
$$
Unfortunately this ODE seems not easy to solve because of the delay ($\pm 1$) in $\xi$. I'm completely ignorant on how to solve these type of equations, but I assume that something can be said about its general solution if we know at least one particular solution. Unfortunately, the standard method (characteristic roots) seems not to offer any clue about it.
A: The classic solving method for linear PDE's is Fourier substitution:
$$u(x,t) = X(x)T(t),$$
which gives
$$\dfrac{X'''}X - \sin x = -\ \dfrac{T'}T = \lambda,\quad \lambda = const,$$
$$\begin{cases}
X''' - (\sin x +\lambda)X = 0\\
T' + \lambda T = 0.
\end{cases}\qquad(1)$$
System $(1)$ seems easier than similar in the "first attempt" of OP, and that is the half of the answer to the first question.
The second half of the answer is: KdV equation is immediately integrable, and this is its main difference from OP equation. So any attempts to reduce the initial equation to KdV are unpromising
The answer to the second question is: $u_x = \sin(x)$ gives (or not gives) a particular solution of the problem immediately, without a transition to KdV.
The third answer is really hard. My attempt is the next.
Structure of the solutions.
Substituting
$$X=\Phi\Psi$$
in $(1)$ transforms it to the form of
$$\Phi'''\Psi + 3\Phi''\Psi' + 3\Phi'\Psi''+ \Phi\left(\Psi''' - (\sin x+\lambda)\cdot\Psi\right) = 0$$
which can be presented as the system
$$\begin{cases}
\Psi''' - (\sin x+\lambda)\cdot\Psi = 0\\
\Phi'''\Psi + 3\Phi''\Psi' + 3\Phi'\Psi'' = 0.
\end{cases}\qquad(2)$$
Easy to see that ODE's $(2.1)$ and $(1.1)$ are identical. This means that any particular solution $X_0$ of $(1.1)$ can be obtaned both as self-contained solution and as multiplyer $\Psi$ in the (2). So the arbitrary quantity of solutions 
$$X_1 = \Phi_1X_0,\quad X_2 = \Phi_2X_1\dots$$
can be obtained throw the $(2)$ for $\Psi = X_0.$ 
That requires solving equation $(2.2)$ which are a homogenius linear ODE of the second order with attention to $\Phi'.$
Periodic solution.
Seems a promising to find the periodical solution of $(2.1)$ in the form of the Fourier series
$$X(x,\vec a(\lambda),\vec b(\lambda)) = a_0 + \sum\limits_{n=1}^\infty\left(a_n\cos nx + b_n\sin nx\right),\qquad(3).$$
Coefficients are actually functions of $\lambda,$ but for convenience of calculations this functional dependence is not further specified.
So
$$ - a_0(\sin x +\lambda) - \sum\limits_{n=1}^\infty\left(a_n\cos nx + b_n\sin nx\right)\sin x \\
 - \sum\limits_{n=1}^\infty (\lambda a_n + n^3 b_n)\cos nx + \sum\limits_{n=1}^\infty (n^3 a_n -\lambda b_n)\sin nx = 0.$$
Taking in account the trigonometric identities
$$2\cos nx\sin x = \sin(n+1)x - \sin(n-1)x,\\
2\sin nx\sin x = \cos(n-1)x - \cos(n+1)x,\\
2\cos x\sin x = \sin 2x,\\
2\sin^2 x = 1 - \cos 2x,$$
and giving similar terms at 
$$1,\quad \cos x,\quad \sin x,\quad \cos 2x,\quad \sin 2x,\quad \cos 3x,\quad \sin 3x,\quad\dots,$$
one can get the linear algebraic system for $a_n,\ b_n:$
$$\begin{cases}
-2a_0\lambda + b_1 = 0\\
-b_2 - 2(\lambda a_1 + b_1) = 0\\
- 2a_0 + a_2 + 2(a_1 - \lambda b_1) = 0\\
b_1 - b_3 - 2(\lambda a_2 + 8b_2) = 0\\
-a_1 + a_3 + 2(8a_2 - \lambda b_2) = 0\\
b_2 - b_4 - 2(\lambda a_3 + 27b_3) = 0\\
-a_2 + a_4 + 2(27a_3 - \lambda b_3) = 0\\
\dots.
\end{cases}$$
$$\begin{cases}
b_1 = 2\lambda a_0\\
b_2 = -2(\lambda a_1+b_1)\\
a_2 = 2a_0 - 2(a_1 - \lambda b_1)\\
b_3 = b_1 - 2(\lambda a_2 + 8b_2)\\
a_3 = a_1 - 2(8a_2 - \lambda b_2)\\
b_4 = b_2 - 2(\lambda a_3 + 27b_3)\\
a_4 = a_2 - 2(27a_3 - \lambda b_3)\\
\dots.
\end{cases}$$
This means that, for any $\lambda,$ coefficients $a_0$ and $a_1$ can be considered as arbitrary constants of the according solution. 
Common solution
Formally, the common solution can be written in the form of
$$u(x,t) = \int\limits_{\Omega(\lambda)} X(x,\lambda, a_0, a_1)e^{-\lambda t}\,d\lambda.$$
At the same time, at least, there are questions about the convergence of series, the existence of non-periodic solutions. And this does not make it possible to consider the problem as solved.
The signal spectrum is continuous, so the solution requires integral representation. The region of integration with respect to $\lambda$ also depends on the type of conditions in $x$ (initial or boundary?), which are absent in the statement of the problem.
A: If the function $f(x)$ is smooth then there is an elegant method, based on semigroup, that may be helpful. Rewrite equation as
$$\frac{\partial u}{\partial t} = \left[ -\partial^3_x + \sin(x) \right] u(t,x) := L(x,\partial_x ) u(t,x)$$
Now the solution can be written as
$$u(t,x) = e^{tL(x,\partial_x)} .u(0,x) = e^{tL(x,\partial_x)} .f(x)$$
In particular if $f(x)$ is a nice function so that the action of operator, expanded in Taylor series, is cyclic then you can find a closed form for the solution.
$\textbf{Extending Arigato's method:}$
The shift in Arigato's last equation can also be handled provided that the function possesses a series solution. In particular this kind of shift is the property of Bernoulli Polynomial, i.e. $B_n(1-x) = (-1)^n B_b (x)$. See symmetry section of  https://en.wikipedia.org/wiki/Bernoulli_polynomials
So the only need you need is to construct an orthogonal set from Bernoulli polynomial using Gram-Schmidt orthogonalization and then rewrite an expansion to solve Arigato's final equation.
$\textbf{Solution for $f(x) = \cos(x)$}$
I try to explain problem for this special case. Fortunately your choice (in comments) ($f(x)=\cos(x)$) is a smooth function. 
Using Taylor series of exponential function, the solution is
$$u(t,x)  = e^{tL(x,\partial_x)} \cos(x) \\
= \sum_{n=0}^{\infty} \frac{(tL)^n}{n!} \cos(x)=\sum_{n=0}^{\infty} \frac{t^n }{n!} [\sin(x)-\partial_x^3  ]^n \cos(x) \\
=\cos(x) + t[\sin(x)-\partial_x^3  ] \cos(x) + \frac{t^2}{2!} [\sin(x)-\partial_x^3  ]\times [\sin(x)-\partial_x^3  ] \cos(x) +...$$
Although the function cosine is a good function for this situation, it is still a cumbersome task to find a routine between terms in the series. But As I saw the comments and answers, the solution seems not to be straightforward. 
