Solving a non linear pde using a function

[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

 

[mighty2000]

Thank you for your question. I understand that you are trying to reduce the given PDE to an ODE and find a family of solutions. Your approach seems to be correct, but there are a few errors in your calculations.

Firstly, when you substitute the expressions for du/dt and d^2u/dx^2 into the PDE, you should not get f'''-cf'-6f'=0. The correct result should be f'''-cf'-6f''=0. This is because d^2u/dt^2 = c^2 f'' and d^2u/dx^2 = f''. Therefore, the term involving f'' should have a coefficient of -6, not 0.

Secondly, when you try to solve the resulting ODE f'''-cf'-6f''=0, you cannot simply cancel the derivatives. This is because the derivatives are with respect to different variables (t and x). In order to solve this ODE, you will need to use a method such as variation of parameters or the method of undetermined coefficients.

I hope this helps you in your solution. If you are still stuck, I would suggest consulting with your professor or a colleague for further assistance. Good luck!