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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 $u_{xx} + (u_{x}^3)_{x}$?

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closed as off-topic by Will Jagy, Daniel Moskovich, Chris Godsil, David White, Carlo Beenakker Oct 2 '13 at 19:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Will Jagy, Daniel Moskovich, Chris Godsil, David White, Carlo Beenakker
If this question can be reworded to fit the rules in the help center, please edit the question.

    

The technique to solve this problem, as suggested by the formulation of the question, is perturbation theory. One writes the solution in the form $$ u(x,t)=\epsilon u_1(x,t)+ \epsilon^2 u_2(x,t)+\epsilon^3 u_3(x,t)+\ldots = \sum_{n=1}^\infty \epsilon^n u_n(x,t) $$ and put this into the equation and the initial and boundary conditions. In this case one gets the set of non-trivial equations $$ u_{1tt}=u_{1xx}+u_{1xxt} $$ with $u_1(0,t)=u_1(1,t)=0$, $u_1(x,0)=f(x)$ and $u_{1t}(x,0)=g(x)$. This equation is linear and can be solved with standard techniques. Then, at second order one has $$ u_{2tt}=u_{2xx}+u_{2xxt} $$ with $u_2(0,t)=u_2(1,t)=0$, $u_2(x,0)=f(x)$ and $u_{2t}(x,0)=g(x)$ and so, $u_2=0$ the trivial solution. At the third order one has a non-trivial equation $$ u_{3tt}=u_{3xx}+u_{3xxt}+(u_{1x}^3)_x $$ that has a non-null solution and the solution can be given using the Green function of the linear equation. From this, you can iterate to whatever order you want.

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