Nonlinear initial-boundary value problems using Taylor expansion of parameter Let $u(x,t; \epsilon)$ satisfy the nonlinear initial boundary value problem 
$$
u_{tt} = (u_{x} + u_{x}^3)_{x} + u_{xxt}, \space 0 \lt x \lt 1
$$
$$
u(0,t) = 0 \\
u(1,t) = 0 \\
u(x,0) = \epsilon f(x) \\
u_{t}(x,0) = \epsilon g(x)
$$
where $f(0) = 0 = f(1)$ and $g(0) = 0 = g(1)$. How do I go about finding the equation satisfied by $u$ if I assume that the solution depends smoothly on $\epsilon$ in the neighborhood of $0$, so that $u$ has the beginnings of a Taylor expansion in $\epsilon:$ 
$\\\\ u(x,t; \epsilon) = 0 + u_{1}(x,t)\epsilon + \frac{1}{2}u_{2}(x,t)\epsilon^2 + \frac{1}{3}u_{3}(x,t)\epsilon^3 + \space...\space + \frac{1}{n}u_{n}(x,t)\epsilon^n $
And is it possible to prove that such a solution is unique? Also, regarding notation, does $(u_{x} + u_{x}^3)_{x}$ signify the partial derivative of $(u_{x} + u_{x}^3)$ with respect to x? Is this equivalent to $u_{xx} + (u_{x}^3)_{x}$?
 A: One thing you can try. First split your equation into the linear and non-linear parts
$$ u_{tt} - u_{xx} - u_{xxt} = 3 u_x^2 u_{xx} $$
If you assume the solution has Taylor expansion with leading term of size $\epsilon$, the nonlinear term would be order $\epsilon^3$ and hence much smaller. 
So the small data ansatz can be satisfied if you solve 
$$ u^{(1)}_{tt} - u^{(1)}_{xx} - u^{(1)}_{xxt} = 0 $$
and iterate with 
$$ u^{(n)}_{tt} - u^{(n)}_{xx} - u^{(n)}_{xxt} = \Pi_n \left( 3 u^{(<n)}_{x} u^{(<n)}_{x} u^{(<n)}_{xx} \right)$$
where
$$ u^{(<n)} = \sum_{k = 1}^{n-1} u^{(k)} \epsilon^k $$
and $\Pi_n$ means truncate to just the term that contains $\epsilon^n$ and no higher orders. 
This in general is not guaranteed to converge. Luckily for you your linear part has a damping term: if you multiple the linear equation with $u_t$ and integrate by parts you get
$$ \int \partial_t (u_t)^2 + \partial_t (u_x)^2 + 2 (u_{xt})^2 ~\mathrm{d}x = 0 $$
which tells you that you have monotonic energy decay. So you have a chance to close the argument. 
Lastly, about uniqueness, applying the energy method as above to the full nonlinear equation you get
$$ \int \frac12  \partial_t (u_t)^2 + (u_x + u_x^3) u_{tx} + u_{tx}^2 \mathrm{d}x = 0 $$
or 
$$ \int \frac12 \partial_t (u_t)^2 + \frac12 \partial_t (u_x)^2 + \frac14 \partial_t (u_x)^4 + u_{tx}^2 ~\mathrm{d}x = 0 $$
Which shows that the nonlinear energy
$$ E[u](t) = \int \frac12 (u_t)^2 + \frac12 (u_x)^2 + \frac14 (u_x)^4 ~\mathrm{d}x $$
is in fact monotonically decreasing. 
Similarly, if you look at the difference between two solutions, 
$$ (u-v)_{tt} - (u - v)_{xx} - (u_x^3 - v_x^3)_x - (u-v)_{xxt} = 0 $$
You can expand
$$ u_x^3 - v_x^3 = (u_x - v_x)(u_x^2 + u_x v_x + v_x^2) $$
what is important to note is that $(u_x^2 + u_x v_x + v_x^2) \geq 0$ since
$$ u_x^2 + u_x v_x + v_x^2 = \frac12 (u_x^2 + v_x^2) + \frac12 (u_x + v_x)^2 \geq 0$$
So if we multiply by $(u-v)_t$ and integrate by parts we end up with (writing $w = u-v$)
$$ \int \frac12 \partial_t (w_t)^2 + \frac12 \partial_t (w_x)^2 + (u_x^2 + u_x v_x + v_x^2) \frac12 \partial_t (w_x)^2 + (w_{xt})^2 ~\mathrm{d}x = 0 $$
This tells us that the functional 
$$ F[u,v](t) = \int (w_t)^2 + (1 + u_x^2 + u_x v_x + v_x^2) (w_x)^2 ~\mathrm{d}x $$
which is manifestly non-negative can only decay in time. For two solutions with the same initial data (note that the functional only depends on the first derivatives of $u,v$, and not the initial acceleration), $F[u,v](0) = 0$. This means that $F[u,v](t)= 0$ for all $t > 0$ and hence necessarily $w = u-v \equiv 0$ for all $t > 0$ and therefore the forward solution is unique. 
