# Solving a non linear pde using a function

1. Oct 27, 2011

### maggie56

I have the non linear pde

du/dt = d/dx [3 u^2 - d^2u/dx^2]

the question supposes that there is a solution u(x,t) = f(x-ct) where c is constant and f(y) for y=x-ct satisfies f tends to 0, f' tends to zero and f'' tends to zero but y tends to + or - infinity.

so i have tried to reduce the above equation to an ode, i have to show that a family of solutions of the pde are given by u(x,t) = -c/2 sech^2 [ c^{1/2} /2 (x-ct)]

i find
du/dt = -cf'
du/dx = f'
d^2u/dt^2 = c^2 f''
d^2u/dx^2 = f''

but when substituting these into the pde and simplifying i get

f'''-cf'-6f'=0 so f''' = (6+c)f'

is it possible to 'cancel the derivatives' so that f''=(6+c)f and f'=(6+c) ?

im really stuck on this question
any help would really be appreciated

thanks