Using Fourier Transform to Solve ODE with Initial Conditions

[ondine4]

Hi, let's take this ode:
y''(t) = f(t),y(0)=0, y'(0)=0.
using the FT it becomes:
-w^2 Y(w) = F(w)
Y(w)=( -1/w^2 )F(w)
so i can say that -1/w^2 is the Fourier transorm of the green's function(let's call it G(w)).
then
y(t) = g(t) * f(t)
where
g(t) = F^-1 (G(w)) (inverse Fourier transorm)
how can i solve the integral to find g(t)?
if f(t)=0 for t<0 and f(t)=1 for t>=0 how can i say that y(t)= 1/2t^2?

 

[Delta2]

The inverse Fourier transform of $-\frac{1}{w^2}$ is $t sgn(t)$ where $sgn(t)$ the sign function https://en.wikipedia.org/wiki/Sign_function. You can plug this and compute the convolution integral.

 

[ondine4]

First of all thank you for the answer :)
My problem is that i don't know why the inverse Fourier tranform of -1/w^2 is tsgn(t).
using the definition of inverse Fourier transform I have to calculate an integral from -infinity to +infinity of a function that has some issue in w=0.
I can use Jordan's lemma and the residue theorem but w=0 lies on the integration's curve and it's a second order pole.. how can i overcome this problem?

 

[Delta2]

Don't use the definition of inverse Fourier transform. Use the properties of Fourier transform. We know that the Fourier transform of sgn(t) is $\frac{1}{iw}$ (you can prove this by the definition of Fourier transform). Therefore by a known property of Fourier transform (property 107 here https://en.wikipedia.org/wiki/Fourier_transform#Tables_of_important_Fourier_transforms)

it follows that the Fourier transform of $t~sgn(t)$ will be

$i\frac{d}{dw}(\frac{1}{iw})=-\frac{1}{w^2}$

 

[ondine4]

Thank you very much, this is a very simple and smart method to calculate some Fourier transforms :) i will keep that in mind.

 

[jasonRF]

If you are using the Fourier transform to solve this kind of problem then you need to be a little more careful. The solution of
<br /> -\omega^2 Y(\omega) = F(\omega)<br />
is
<br /> Y(\omega) = - \frac{1}{\omega^2} F(\omega) + a \, \delta(\omega) + b \, \delta^\prime(\omega),<br />
where $a$ and $b$ are arbitrary constants that you determine by using your initial conditions.

By the way, the Laplace transform is a more straightforward way of solving these kind of problems, in my opinion.

Jason

 

[ondine4]

Thank you :) yes i know that i have to add the solution of the associated homogeneus equation to find the complete solution..I wanted to use ft to solve this ode because i haven't studied laplace transform yet, thanks for the advice!