Integral Equation -Fourier Transform

[Camillo]

Solve the equation :

$y+e^(-|x|)$*$y = e^(-|x| )$ . (* means convolution ).

[Paolo902]

Thanks a lot for your answer, Kroldar.

Paolo

[Kroldar]

"Paolo90":Once I have attended to a university lesson in which the teacher presented briefly Fourier Transform (and Gabor one). He said that the Fourier Transform "is a good ear, which hears perfectly but does not remember". Is it the same thing you were saying in your last post? Can we say that it's the same for Laplace transform?

I'm not sure about what that teacher meant. Maybe he wanted to underline that with Fourier transform you can only make a static analysis of a system, in fact Fourier transform forgets which were the conditions of the system when the analysis started. This is the same for bilater Laplace transform. Unilateral Laplace transform, instead, lets you study also the dynamic part of the answer of a system (transitory answer); we can make sure of that if we consider that, for $x(t)$ a.c. in the compacts of $(0,+oo)$ (and $x(t)$ of exponential order... I dunno if that's the correct translation), it results $ccL_u[x'(t)] = sX(s) - x(0)$... so unilateral Laplace transform can remember everything about the system when the analysis started ($t=0$ or a different instant, no matter the origin of times).

"Paolo90":
I promise, that's the last post (I think that 33 are more than enough.. I hope I haven't disturbed you).
Thank you.

No prob... that's not boring to answer. That's only difficult to explain this things in English because I don't know the exact translation of many technical words

"Paolo90":
P.S.: Have you got any links and/or books concerning these subjects? I mean integral, integro-differential, differential equations and functions transforms?
Thanks again.

I only studied this things on my teacher's book for the exam (a part of the exam) of Mathematical Methods.

[Paolo902]

Right, now I've understood. Thanks Camillo, I hadn't understood why you and Kroldar had told me that property of Laplace transform; now it is ok.

Kroldar, many thanks for your explanations. I just want to ask you one thing concerning what you've said in your last post: in particularly, you say that for time-invariant systems the output signal does not depend on where we put the origin of times. Once I have attended to a university lesson in which the teacher presented briefly Fourier Transform (and Gabor one). He said that the Fourier Transform "is a good ear, which hears perfectly but does not remember". Is it the same thing you were saying in your last post? Can we say that it's the same for Laplace transform?

I promise, that's the last post (I think that 33 are more than enough.. I hope I haven't disturbed you).
Thank you.

P.S.: Have you got any links and/or books concerning these subjects? I mean integral, integro-differential, differential equations and functions transforms?
Thanks again.

[Kroldar]

I see that many people identiy "L-transform" with "one side L-transform". The most general transform is double side one... but I guess why one side one is so diffused. One side L-transform is a very powerful tool which helps to study linear and time-invariant systems. Of course we can study a real system starting from a defined moment ($t=0$ or $t=3$ or $t=5.6$ and so on). For example, if we study a system from the instant $0$, double side L-transform is useless because the input-signal is $0$ for $t<0$. So we have seen that, if starting time is $t=0$, one side L-transform works well! But... if we want to study our system from $t=4$ ($4$ seconds i.e.)? No problem... in fact for time-invariant systems the output signal does not depend on where we put the origin of times, so we can apply one side L-transform and shift the output signal found (in time domain) of $4$ units of time.

[Camillo]

No, it is not correct.

Double side L transform

You have shown that $ccL(e^(-|t|)) = 2/(1-s^2)$.
Now you have to antitrasform $ 2/(3-s^2)$, knowing that ( see Kroldar and mine posts)
$ccLe^(-a|t|)= (1/a)2/(1-(s/a)^2) = a*2/(a^2-s^2)$.
Consequently, in our case, $ a=sqrt(3) $ and $y(t) = (1/sqrt(3))*e^(-sqrt(3)|t|)$.

[Paolo902]

Thank you, but I haven't understood what you (both Kroldar and Camillo) mean.
The equation
$y+e^(-|x|)**y=e^(-|x|)$
now becomes (after indicating with $Y(s)=ccL(y(x))(s)$:

$Y(s)+\frac{2}{1-s^2}Y(s)=2/(1-s^2)$
which yields
$Y(s)=2/(3-s^2)$

Now the question is how to inverse-transformate $Y(s)$. I have obtained
$y(x)=-\frac{e^(-sqrt3x)(-1+e^(2sqrt3x))}{sqrt3}=-frac{2sqrt3\sinh(xsqrt3)}{3}$

Is it correct? Thanks... I know, I'm hopeless...

[Kroldar]

"Camillo":For " standard " L tansform the following applies :

$(ccLf)((ax))(s) =(1/a)(ccLf)(s/a) ,AAa> 0 $

If the same applies( check ) for Double Side L transform then it is easy to arrive to the solution.

More in general, we have

$ccL[x(at)] = 1/|a| X(s/a), AA a in RR-{0}$


I add that my teacher denoted with $ccL$ double side L-transform and with $ccL_u$ one side L-transform.

Why $ccL_u$? Because in our language one side L-transform is called "trasformata unilatera di Laplace".

[Camillo]

For " standard " L tansform the following applies :

$(ccLf)((ax))(s) =(1/a)(ccLf)(s/a) ,AAa> 0 $

If the same applies( check ) for Double Side L transform then it is easy to arrive to the solution.

[Paolo902]

"Kroldar":[quote="Paolo90"]
$int_0^(+oo)e^(-t-st)=1/(1+s)int_0^(oo)(1+s)e^(-t(1+s))=-1/(1+s)$

Pay attention here [/quote]


What a mistake!! Oh my Gosh! I think yesterday night I was totally crazy (I've also forgotten all the differentials $dt$...). Anyhow,
$int_0^(+oo)e^(-t-st)$dt$=1/(-1-s)int_0^(oo)(-1-s)e^(t(-1-s))dt=1/(1+s)$

This time
$ccL[e^(-|t|)]=int_(-oo)^(+oo)e^(|t|)e^(-st)dt=1/(1-s)+1/(1+s)=2/(1-s^2)$

You were right, as always.

Now, last thing: I've tried to solve Camillo's equation and I've found the following result:
$ccL(y(x))(s)=2/(3-s^2)$

What about the inversion? How can I obtain $y(x)$? If it were a normal Laplce transform it would be $y(x)= -\frac{e^(-sqrt3x)(-1+e^(2sqrt3x))}{sqrt3}$; but in this case what can I do?

Thanks a lot for your help.

[Kroldar]

"Paolo90":
$int_0^(+oo)e^(-t-st)=1/(1+s)int_0^(oo)(1+s)e^(-t(1+s))=-1/(1+s)$

Pay attention here

[Paolo902]

Let's evaluate L-double side transform of $e^(-|t|)$.
$int_(-oo)^0e^(t-st)=1/(1-s)int_(-oo)^0(1-s)e^(t(1-s))=1/(1-s)$ since the exponential $to 0$ for $t->-oo$.
$int_0^(+oo)e^(-t-st)=1/(1+s)int_0^(oo)(1+s)e^(-t(1+s))=-1/(1+s)$

Now:
$int_(-oo)^(oo)e^(-|t|)e^(-st)=1/(1-s)-1/(1+s)=2s/(1-s^2)$

Well, I am quite satisfied now, even though the result is a little bit different from yours. Where is the mistake (that's the last time I asked you this thing.. I think you are fed up with me!)

Thanks again. Good night.

[Paolo902]

"Kroldar":$int_(-oo)^(+oo) e^(-|t|) e^(-st) dt = int_(-oo)^0 e^t e^(-st) dt + int_0^(+oo) e^(-t) e^(-st) dt$

You can calculate the integrals separately and add the values (integral is additive)... you can't instead calculate the solution of the whole problem in $(-oo,0)$ and $(0,+oo)$ separately and then get the sum of them and say that the function you found is the solution in $RR$.

Ahhhhhhhh!! Now it's clear! I'm a big idiot! I've understood now, thanks. I'm trying then I'll let you know my solutions.
Thanks Thanks Thanks Thanks Thanks Thanks Thanks Thanks Thanks

[Kroldar]

$int_(-oo)^(+oo) e^(-|t|) e^(-st) dt = int_(-oo)^0 e^t e^(-st) dt + int_0^(+oo) e^(-t) e^(-st) dt$

You can calculate the integrals separately and add the values (integral is additive)... you can't instead calculate the solution of the whole problem in $(-oo,0)$ and $(0,+oo)$ separately and then get the sum of them and say that the function you found is the solution in $RR$.

[Paolo902]

"Kroldar":[quote="Paolo90"]

Sorry Kroldar I hadn't seen you post. Fisrt of all thanks for it.
Secondly, what do you mena with double-side Laplace transform? I've never heard of it: how does it work? Would you mind giving to me an example (not necessarily Camillo's equation).
Thank you very much.

Please let me know how much it will be (send me the bill, please )

Thanks

Let $x(t) in L_(loc)^1$, we define double-side L-transform this way

$ccL[x(t)] = int_(-oo)^(+oo) x(t) e^(-st) dt$

Pay attention: $-oo to +oo$ and not $0 to +oo$.

Let's try to calculate double-side L-transform of $x(t) = e^(-|t|)$... uhm no... you try!!!

The result should be $2/(1-s^2)$ if I remember well...

Then you could try to solve the original problem using double-side L-transform and verify that the solution is correct.[/quote]

Ok, thanks for your help and thanks for homework you've given to me (I'm not joking).
Well, I would like to ask you some things:

$int_(-oo)^(+oo)x(t)e^(-st)dt=int_(-oo)^(+oo)e^(-|t|)e^(-st)dt=int_(-oo)^(+oo)e^(-|t|-st)dt=?$

Sorry but how can I calculate an indefinite integral of a function like $e^(-|t|-st)$? I do not like that absolute value...
I try by using the identity $|x|=x*sign(x)$. The integral becomes
$int_(-oo)^(+oo)e^(-t*signt-st)dt=int_(-oo)^(+oo)e^t*e^(s-signt)dt$.
We can integrate by parts and we obtain - I hope there aren't mistakes! - $e^(s-signt)e^t$. Now the problem is with the limits. It's quite obvius that the limit $x->oo$ is $oo$. What have I done?

Thanks. I beg your pardon, I'm really taking advantage of your patience... sorry.
Thank you very much.

[Paolo902]

"Camillo":$L^2 (RR)$ is a vector space ( look on Google for the exact meaning ) of the square integrable functions i.e. such that
$int_(RR)|f(x)|^2 dx <+oo$. Look on Google for $L2 space $.

Thanks a lot, Camillo. Don't worry, last summer I studied a bit of Linear Algebra and I remember the concept of Vector space. Now I'm trying to trasform following Kroldar instructions. I let you know.

[Camillo]

$L^2 (RR)$ is a vector space ( look on Google for the exact meaning ) of the square integrable functions i.e. such that
$int_(RR)|f(x)|^2 dx <+oo$. Look on Google for $L2 space $.

[Kroldar]

"Paolo90":

Sorry Kroldar I hadn't seen you post. Fisrt of all thanks for it.
Secondly, what do you mena with double-side Laplace transform? I've never heard of it: how does it work? Would you mind giving to me an example (not necessarily Camillo's equation).
Thank you very much.

Please let me know how much it will be (send me the bill, please )

Thanks

Let $x(t) in L_(loc)^1$, we define double-side L-transform this way

$ccL[x(t)] = int_(-oo)^(+oo) x(t) e^(-st) dt$

Pay attention: $-oo to +oo$ and not $0 to +oo$.

Let's try to calculate double-side L-transform of $x(t) = e^(-|t|)$... uhm no... you try!!!

The result should be $2/(1-s^2)$ if I remember well...

Then you could try to solve the original problem using double-side L-transform and verify that the solution is correct.

[Paolo902]

"Kroldar":You can also use Laplace transform, but you must use double-side Laplace transform for all $t$.

You can't solve the whole problem first in $(0,+oo)$ and then $(-oo,0)$ and finally add the solutions.

Sorry Kroldar I hadn't seen your post. Fisrt of all thanks for it.
Secondly, what do you mean with double-side Laplace transform? I've never heard of it: how does it work? Would you mind giving to me an example (not necessarily Camillo's equation)?
Thank you very much.

Please let me know how much it will be (send me the bill, please )

Thanks

[Paolo902]

First of all, VCE, thank you very much for your wonderful explanation.

"ViciousGoblinEnters":

Sorry I didn't know this.

So you have something to learn - but hey, don't hurry.

Yes, I've a lot to learn (but, you're right, I shouldn't hurry... I'm 17... ). I'm happy that you help me learn. I apologize for disturbing you but I really want to understand deeply. Well, let's see my doubts.

Well that's a long story - the short answer is : use Fourier transform, as the valiant guys here already told you
(why Fourier? - because the involved function $e^{-|t|}$ is in the space of square integrable functions $L^2(RR)$
and the Fourier transforms works well in $L^2$).

All right. I do trust you ( ) but I would like to ask you one thing: what do you mean with $L^2(RR)$? I've already seen it but I've never found a good definition. Would you mind explaining it to me or - if you prefer - giving to me some links where I can find it? (well, if you prefer we could also continue this conversation by email, in italian: for me it's the same).
What do you mean exactly with space of funtions? I mean why is it called "space" and not for example "set"?

Laplace transform works well if you have to deal with functions $f(t)$ which vanish for $t<0$ (but in turn you can allow
$f$ to grow exponentially at $+\infty$). If you were given the analogous problem:

$y(x)+((H(t)e^{-t})\star y(t))(x)=H(t)e^{-x}$ ($H(t)=1$ for $t>0$ and $H(t)=0$ for $t<0$, $H$ is the Heavyside function)

you could find a solution $y$ such that $y(t)=0$ for $t<0$ USING THE LAPLACE TRANSFORM
(while Fourier could not be used)

I recall that $g\star g)(x):=\int_{RR}f(t)g(x-t) dt$, let's say whenever the integral makes sense (also this point is a subtle one and should be more precise) .

That is more clear (if I'm not wrong the Heavyside function is the integral of Dirac's delta). Well, I didn't remember this definition of convolution; are there any differences with "mine"? I mean which are the differences between
$(f**g)(x)=int_0^xf(tau)g(tau-y)dy$
and $(f**g)(x)=int_-oo^(+oo)f(t)g(x-t) dt$?

I hope I have not disturbed you. Please help me, and forgive me - I know that my knowledge is extremely asystematic...

Thanks a lot.

Paolo

[Kroldar]

You can also use Laplace transform, but you must use double-side Laplace transform for all $t$.

You can't solve the whole problem first in $(0,+oo)$ and then $(-oo,0)$ and finally add the solutions.