Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
add comment

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.

1 Answer

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.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.