Non linear pde need to change it to an ode

[maggie56]

Homework Statement

my non linear pde is
du/dt = d/dx [3u² - d²u/dx² ]
The question says to let u(x,t) = f(x-ct)
Where the function f tends to 0, f' tends to 0 and f'' tends to 0 but the (x-ct) tends to positive or negative infinity.

Homework Equations

i thought the solution was to find du/dt du/dx d²u/dt² and d²u/dx²
Which gave me in order as above, -cf' , f' , c²f'' , f''

The Attempt at a Solution

But when i substitute these into the pde i get -cf' = d/dx[3f² - f"] which I am not sure actually helps

Could someone please tell me if i am using the correct method or if i have done it completely wrong

Thank you

 

[mighty2000]

for your question. It seems like you are on the right track, but there are a few things that need to be clarified.

First, when you substitute u(x,t) = f(x-ct) into the PDE, you should get:

du/dt = d/dx [3f(x-ct)^2 - d^2f(x-ct)/dx^2]

Next, you need to take the derivatives with respect to x and t. Remember that f is a function of x only, so when you take the derivative with respect to t, the x-ct term will be treated as a constant. Therefore, you should get:

du/dt = -cf'(x-ct)

d/dx [3f(x-ct)^2 - d^2f(x-ct)/dx^2] = f'(x-ct) - c^2f''(x-ct)

Substituting these into the original PDE, you should get:

-cf'(x-ct) = f'(x-ct) - c^2f''(x-ct)

This is the same equation you obtained, but it is now in terms of f and its derivatives with respect to x. From here, you can solve for f'' and then integrate to find f'.

Hope this helps! Let me know if you have any further questions.