Integro differential equation

[Camillo]

Solve the following equation :

$u'(x)-u(x)+2u$*$e^x = 0 $ ; $ x> 0 $ ( * means convolution.)

Feel free to use the method you prefer

[Kroldar]

"Paolo90":
Now let's think about Camillo's question: I'm not sure we can solve it by means of Fourier Tr.(because the impropre integral is between $-oo$ and $+oo$).

In fact, we can't.

"Paolo90":
Does it exist a unilateral version of Fourier transform?

I don't think so; however the Fourier integral in $(0,+oo)$ would be useless for studying linear and time-invariant systems.

"Paolo90":
Another problem: is the function $e^x$ transformable (does this word exist? I think I've actually invented a neologism... )? I mean is it $\in L^1(RR)$?

$e^x !in L^1$ and is not transformable, as Camillo said. But Fourier-transform theory can be extended out of $L^1$ ($L^2$ for example or the space of tempered distributions $ccS'$) and in $ccS'$ we see that some functions are not in $L^1$ but can be Fourier-transformed.

"Paolo90":
P.S. As tomorrow I'm leaving, I do not know if today I have free time to spend here. Anyway, I'll be back on July, 7th. Until then,

Have nice holidays then.

[Camillo]

Nice solution !

$e^x !in L^1(RR) $ and is not Fourier transformable.

Have a nice stay next week and enjoy all new things you will learn

[Paolo902]

Ohhhhhhhh, I can't believe it!! Uao! I'm very very happy!

Anyhow, as I promised, here you are the resolution.

Let us consider the integro-differential equation $u'(x)-u(x)+2u(x)**e^x=0$. We apply $ccL_u[f(x)]$ to LHS (and RHS...); after calling $L_u[u(x)](s)=U(s)$, we should remember that $L_u[f'(x)](s)=s\ccL[f(x)](s)-f(0)$: so, we obtain
$sU(s)-u(0)-U(s)+2U(s)ccL_u(e^x)(s)=0$

Solving this equation for $U(s)$ we get
$U(s)=\frac{u(0)(s-1)}{s^2-2s+3}=\frac{c(s-1)}{s^2-2s+3}$ where we suppose $u(0)=c$ (a generic constant value $\in RR$).

The anti transformation yields $u(x)$:
$u(x)=U(0)e^xcos(2sqrtx)$ which is, indeed, defined only in $(0,+oo)$ (or if we prefer $[0, +oo)$).

Well, I'm very happy for this (as I have already said it was the first integro-differential equation I've ever solved!).

Now let's think about Camillo's question: I'm not sure we can solve it by means of Fourier Tr.(because the impropre integral is between $-oo$ and $+oo$).- Does it exist a unilateral version of Fourier transform? Another problem: is the function $e^x$ transformable (does this word exist? I think I've actually invented a neologism... )? I mean is it $\in L^1(RR)$?

I hope I haven't disturbed you. Thanks a lot for your help.

P.S. As tomorrow I'm leaving, I do not know if today I have free time to spend here. Anyway, I'll be back on July, 7th. Until then,

take care. Thank you.

Paolo

[Kroldar]

"Camillo":
Additional question :Is it possible to solve the same equation by means of Fourier transform ?

We'll wait for Paolo's answer then

[Camillo]

Yes, it is correct
Additional question :Is it possible to solve the same equation by means of Fourier transform ?

[Paolo902]

"Camillo":We can just use $ u(0)$as initial value and express the solution in function of it.

Well, let's try again. Since we are looking for solution(s) in $(0,+oo)$ I have used $ccL_u[f(x)]$, I mean what in our language is called "trasformata unilatera di Laplace", as Kroldar has recently explained to me.

By using it, I've found this solution: $u(x)=u(0)e^xcos(2sqrtx)$. Is it correct? Before sending you all the resolution I would like to know if it is correct. Please let me know and - in case it is wrong - please forgive me (and remember that this is the first integro-differential equation I solve in my life.. )

Thanks a lot to everybody.

[Camillo]

We can just use $ u(0)$as initial value and express the solution in function of it.

[Kroldar]

Don't you give us initial conditions ($u(0)$)?

[Camillo]

"Kroldar":Have we to solve the problem only in $(0,+oo)$?

Yes ; give also reasons for the method adopted; are other methods feasible ?

[Kroldar]

Have we to solve the problem only in $(0,+oo)$?

[Camillo]

Of course

[Paolo902]

"Camillo": Feel free to use the method you prefer

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