ODE - Fourier transform

[Camillo]

Solve the equation :

$ y''+2xy'+2y = 0 $.

[Kroldar]

"ViciousGoblinEnters":
In my opinion giving the Laplace transform $g=\mathcal L(f)$ means giving both the domain of $g$ and the actual expression of $g$.

Yes that's true, I was not accurate. We say that $x(t) in L_(loc)^1$ is L-transformable in $s in CC$ if the function $x(t) e^(-st)$ is integrable in $RR$. Of course, we have to make sure that there exists $s$...

[ViciousGoblin]

"Kroldar":[quote="Camillo"]It can be noted that only one family of solutions have been found, while the ODE is of the second order.
It means that the second family of solutions is $ !in L^1(RR)$.

The same result I found could be found by using bilateral Laplace transform. The domain of bilateral Laplace transform is not $L^1$, but $L_(loc)^1$... should we conclude that the other family of solutions is neither in $L_(loc)^1$?
Very interesting problem.[/quote]

Hmmm, I have just read the above sentence (and not the whole thead I'm to busy in these days sorry).

I think that saing that the domain of the two-sided Laplace transform is $L^1_{loc}$ is somewhat misleading.
Actually given $f$ in $L^1_{loc}$ you can try to define its transform as a map $g: Omega\to CC$, where $\Omega$ is an open subset of $CC$ of the form $]\alpha,\beta[\times i RR$. In general $]\alpha,\beta[$ might be empty so that the
domain of $g=\mathcal L(f)$ is empty. For instance $f(x)=e^{x^2}$ is in $L^1_{loc}$ but has no Laplace transform of any sort.
In my opinion giving the Laplace transform $g=\mathcal L(f)$ means giving both the domain of $g$ and the actual expression of $g$.

[Kroldar]

"Camillo":It can be noted that only one family of solutions have been found, while the ODE is of the second order.
It means that the second family of solutions is $ !in L^1(RR)$.

The same result I found could be found by using bilateral Laplace transform. The domain of bilateral Laplace transform is not $L^1$, but $L_(loc)^1$... should we conclude that the other family of solutions is neither in $L_(loc)^1$?
Very interesting problem.

[Camillo]

Who wants to find all solutions of the ODE ?

[Kroldar]

"Camillo":Why in the exponent you have $-x^2/sqrt(pi) $ , instead of just $-x^2$ ?

You're right... I made a mistake

[Camillo]

Why in the exponent you have $-x^2/sqrt(pi) $ , instead of just $-x^2$ ?

It can be noted that only one family of solutions have been found, while the ODE is of the second order.
It means that the second family of solutions is $ !in L^1(RR)$.

[Kroldar]

Let's apply Fourier transform...

$ccF[y''+2xy'+2y] = 0$

$- omega^2 Y(omega) - 2Y(omega) - 2omega Y'(omega) + 2Y(omega) = 0$

$- omega^2 Y(omega) - 2omega Y'(omega) = 0$

$Y'(omega) = - omega/2 Y(omega)$

The solution is

$y(x) = ce^(-x^2) , AA c in RR$

[Camillo]

Go on, which solution do you get ?

[Kroldar]

If we apply Fourier transform to both sides of the equation, we obtain a separable differential equation.