Differential equation with Fourier Transform

[skrat]

Homework Statement

Without solving the differential equation, find the differential equation that solves Fourier transformation of given differential equation for $a>0$.
a) $y^{'}+axy=0$
b) For what $a$ is the solution of part a) an eigenfunction of Fourier Transform

Homework Equations

$f(x)$ ...---... $F(f)(\xi )=\int _{-\infty }^{\infty }f(x)e^{-2\pi i x\xi } dx$

The Attempt at a Solution

I have to apologize in advance. The instructions are weird even in my language, I can't imagine how horrible they sound in English.

$F(y^{'})=2\pi i \xi F(y)$

and $F(axy)=\int _{-\infty }^{\infty }axy(x)e^{-2\pi i x\xi } dx=\frac{a}{-2\pi i}\frac{\partial }{\partial \xi}F(y)$

The differential equation is than: $2\pi i \xi F(y)-\frac{a}{-2\pi i}\frac{\partial }{\partial \xi}F(y)=0$.

That should be the answer to part a).
part b):

Here is where I have problems. For eigenfunctions $F(y)=y$.

$2\pi i \xi y-\frac{a}{2\pi i}\frac{\partial }{\partial \xi}y=0$

Therefore $y=exp(-\frac{2\pi ^2}{a}\xi ^2)$.

Now this has to be wrong... So, my question? How do I find the answer for part b).

 

[BruceW]

The differential equation is than: $2\pi i \xi F(y)-\frac{a}{-2\pi i}\frac{\partial }{\partial \xi}F(y)=0$.

That should be the answer to part a).

This is almost correct. Check that negative sign. (maybe it's just a typo).

 

[BruceW]

$2\pi i \xi y-\frac{a}{2\pi i}\frac{\partial }{\partial \xi}y=0$

I agree with this bit. (the extra negative sign has disappeared compared to before). The notation is a little bit fast & loose. But I see what you're saying.

 

[BruceW]

Therefore $y=exp(-\frac{2\pi ^2}{a}\xi ^2)$.

I thought the question told you not to try to solve the differential equation. You don't need to do this to find out what 'a' must be, for the eigenfunction. You already have
$$y^{'}+axy=0$$
And you now have the equation
$$2\pi i \xi y-\frac{a}{2\pi i}\frac{\partial }{\partial \xi}y=0$$
so what does 'a' need to be, to ensure that both equations are true?

 

[skrat]

Oh, sorry, there is a sign error, yes!

To ensure both equations $a^2=\frac{4\pi ^2\xi }{x}$.

 

[BruceW]

Well, this is where the notation was a bit too fast & loose. You had
$$2\pi i \xi F(\xi)-\frac{a}{2\pi i}\frac{\partial }{\partial \xi}F(\xi)=0$$
Where I am explicitly showing that $F(\xi )$ is a function of $\xi$. It is clear that $F$ is the Fourier transform of $y$ So you don't need to keep writing $F(y)$ to show this.
Anyway, for eigenfunctions of the Fourier transform, we have $F(x) = y(x)$ i.e. the functions are the same for every value of the same variable $x$. But what you have in the above equation is $F(\xi )$ so you first need to re-write the above equation, in terms of $x$ instead of $\xi$.

 

[skrat]

But what you have in the above equation is $F(\xi )$ so you first need to re-write the above equation, in terms of $x$ instead of $\xi$.

Ok. And how do I do that?

 

[BruceW]

simple change of variables for the equation. hehe, I think you're over-thinking it.

 

[skrat]

:D Just a bit :D

$2\pi i x F(x)-\frac{a}{2\pi i}\frac{\partial }{\partial x}F(x)=2\pi i x y(x)-\frac{a}{2\pi i}\frac{\partial }{\partial x}y(x)=0$

and taking in mind that $y^{'}+axy=0$,

than $a^2=4\pi ^2$.

 

[BruceW]

yep. looks good. and they say $a>0$ so now you have the answer!

 

[skrat]

Perfect!

Thanks a lot!