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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