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Solve this differential equation with periodic boundary conditions: (u'[x])^3 - u'''[x] = 0
*To*: mathgroup at smc.vnet.net
*Subject*: [mg93808] Solve this differential equation with periodic boundary conditions: (u'[x])^3 - u'''[x] = 0
*From*: "Charlie Brummitt" <cbrummitt at wisc.edu>
*Date*: Wed, 26 Nov 2008 05:14:22 -0500 (EST)
Hello,
I am trying to solve the nonlinear differential equation
(u'[x])^3 - u'''[x] = 0
with periodic boundary conditions
u[0] = u[L]
u'[0] = u'[L]
(Note: The equation is (du/dx)^3 - (third derivative of u) = 0.)
I am trying the following ansatz (which clearly satisfies the boundary
conditions)
u[x] = Sum(n=1 to infinity) a_n Sin[2 pi n x / L] + b_n Cos[2 pi x / L].
When you plug this into the differential equation, it reduces to
Sum(n=1 to infinity) n^3 (-a_n Cos[2 pi n x / L] + b_n Sin[2 pi n x / L] -
(Sum(n=1 to infinity) n (a_n Cos[2 pi n x / L] - b_n Sin[2 pi n x / L]))^3 =
0 (*)
By equation the coefficients of each of the "modes," we get nonlinear
algebraic equations for the a_i's and b_i's. The question becomes: can we
solve for finitely many a_i, b_i by truncating the solution? If so, is this
solution reasonable, or do the a_i's and b_i's change appreciably if we
include more and more a_i's and b_i's?
I have tried entering the left-hand side of equation (*) into Mathematica as
a function of k, where the sums run from n = 1 to k (rather than n = 1 to
infinity). I am having trouble equation coefficients of the various modes
and solving for the a_i and b_i.
Can anyone help?
Much thanks,
Charlie Brummitt
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