Fourier Transform of Differential Equation

[vela]

Right. That's the equation u(k,t) satisfies since, as the others have noted, u(k,t)=F[u(x,t)].

 

[Hart]

(ik)^{3} + 2ik = \frac{\partial}{\partial t}

 

[Mapes]

(ik)^{3} + 2ik = \frac{\partial}{\partial t}

Unfortunately, \partial/\partial t alone doesn't have any meaning; it's an operator that acts on u(k,t), which can't just be divided away.

You are looking for a function whose time derivative equals the function itself multiplied by -i(k^3-2k).

 

[Hart]

well..

e^{-i(k^3-2k)t}

would mean:

\left(\frac{\partial}{\partial t}\right)e^{-i(k^3-2k)t} = -i(k^3-2k)e^{-i(k^3-2k)t}

?!?

 

[Mapes]

This is what you were hoping to demonstrate at the beginning of this thread, yes?

 

[Hart]

.. indeed it is!

 

[Hart]

.. any pointers for finding A(k) for the case where u(x,t) at t=0 is given by:

u(x,0) = U_{0}\delta (x-a) where Uo is a constant ?!?