Integral Equation -Fourier Transform

[Camillo]

Solve the equation :

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

[ViciousGoblin]

Sorry I didn't know this.

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

So how can we solve the integral equation proposed by Camillo? We can't use Laplace transform, can we? And how can we calculate the convolution product?

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 $L^2$).

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) .

[Paolo902]

"ViciousGoblinEnters":[quote="Paolo90"]Well, I've said that for $x>=0$ the equation becomes:
$y+e^(-x)**y=e^(-x)$.

Sorry this is wrong.
It is not true that for $x\geq 0$ $(e^{-|t|}\star y(t))(x)=(e^{-t}\star y(t))(x)$ - The convolution $f\star g$ is an integral operator, whose value $(f\star g)(x)$ at some point
$x$ depends on ALL values of $f(t)$ and $g(t)$[/quote]

Sorry I didn't know this. So how can we solve the integral equation proposed by Camillo? We can't use Laplace transform, can we? And how can we calculate the convolution product?

Thank you for your explanation.

[ViciousGoblin]

"Paolo90":Well, I've said that for $x>=0$ the equation becomes:
$y+e^(-x)**y=e^(-x)$.

Sorry this is wrong.
It is not true that for $x\geq 0$ $(e^{-|t|}\star y(t))(x)=(e^{-t}\star y(t))(x)$ - The convolution $f\star g$ is an integral operator, whose value $(f\star g)(x)$ at some point
$x$ depends on ALL values of $f(t)$ and $g(t)$

[Paolo902]

"Camillo":You didn't want to follow my indication (Fourier transform); from this fact now we have an interesting problem

That's true. It's my fault. Sorry.

[Camillo]

You didn't want to follow my indication (Fourier transform); from this fact now we have an interesting problem

[Paolo902]

"Camillo":The matter is becoming very interesting... but quite obscure.
Let see if someone clarifies the situation.

All right. We will wait. Thank you very much (I love these things, I learn a lot).

[Camillo]

The matter is becoming very interesting... but quite obscure.
Let see if someone clarifies the situation.

[Paolo902]

Well, I've said that for $x>=0$ the equation becomes:
$y+e^(-x)**y=e^(-x)$
and I'm going to prove that $y=e^(-2x)$ is solution to the equation.

$e^(-2x)+e^(-x)**e^(-2x)=e^-x$
Since the operator of convolution is commutative we can write
$e^(-2x)+e^(-2x)**e^(-x)=e^-x$
The LHS becomes
$e^(-2x)+int_0^xe^(-tau)e^(-2(x-tau))d\tau=e^(-2x)+e^(-x)-e^(-2x)=e^-x$
which is, indeed, RHS.

Now, let's consider case II, i.e. $x<0$.
$y+e^(x)**y=e^(x)$
We will prove that the constant function $1$ is solution.
$1+e^(x)**1=e^(x)$
This time the convolution product is very very easy: $int_0^xe^(tau)d\tau=e^x-1$. Therefore LHS becomes
$1+e^x-1=e^x$ which is again RHS.

What have I done? Where is (or, better, are ) the mistake(s)? Please help me: I really want to understand.
Thanks a lot.

[ViciousGoblin]

"Camillo":In fact I was just going to write the solution using Fourier transform .
I will take advantage of Kroldar input and continue :

$Y (omega) = 2/(3+omega^2) = (1/sqrt(3)) (2sqrt(3))/((sqrt(3)^2+x^2)) $ and finally $y = (1/sqrt(3))e^(-sqrt(3)|x| )$.

All contributions are well appreciated, last but not least VGE's one.

Please don't make me blush

[Paolo902]

All right, I am sorry but I didn't know those things you've said in your posts. Anyway, I haven't understood why I am allowed to use Laplace transform only when the RHS is zero for $t<0$. Indeed, I am wondering why the solutions I've found are not correct: if you evaluate the equation for the function $f$ I've defined you should find an identity.

Now I'm trying and I'll let you know. Please, if you could, explain me why Laplace Transfor doesn't work.
Thank you very much indeed.

[Camillo]

In fact I was just going to write the solution using Fourier transform .
I will take advantage of Kroldar input and continue :

$Y (omega) = 2/(3+omega^2) = (1/sqrt(3)) (2sqrt(3))/((sqrt(3)^2+x^2)) $ and finally $y = (1/sqrt(3))e^(-sqrt(3)|x| )$.

All contributions are well appreciated, last but not least VGE's one.

[Kroldar]

I add that in Fourier domain the problem is very very simple.

It results

$ccF[e^(-|t|)] = 2/(omega^2+1)$

so we have

$Y(omega) (1 + 2/(omega^2+1)) = 2/(omega^2+1)$

and so on...

[ViciousGoblin]

"Paolo90":Hi Camillo, thank you very much for proposing this integral equation. I love them also because I have recently learnt how to solve them. I beg your pardon but I won't listen to your advice: indeed, I'll use Laplace transform to solve it, and not Fourier one. Please do not get angry.

I 'd like to point out that using the Fourier or the Laplace transform is not
not a just matter of taste - it depend on WHERE you are looking for solutions. If you're
looking for solutions in $L^2(RR)$ ( or more generally in the space of tempered distributions)
you can use Fourier. In this case, since the right hand side is in $L^2(RR)$ Fourier is O.K.

You're allowed to use Laplace (at least the one sided transform) if you're looking for
one-sided solutions, i.e. solutions $y(t)$ which vanish for $t<0$. But for this you need the
right hand side to be zero for $t<0$, which is not tha case here (as Kroldar already pointed out).

[Kroldar]

We don't know if $y$ is zero for $t < 0$, so we must use the double-side Laplace transform, defined as

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

When we consider $t<0$, we have

$ccL[e^(-|t|) u(-t)] = ccL[e^t u(-t)] = int_(-oo)^(0) e^t e^(-st) dt = int_(-oo)^(0) e^((1-s)t) dt = 1/(1-s), AA Re < 1$

so it results

$Y(s) + (Y(s))/(1-s) = 1/(1-s) => Y(s) = 1/(2-s)$

There's a problem...

[Paolo902]

Hi Camillo, thank you very much for proposing this integral equation. I love them also because I have recently learnt how to solve them. I beg your pardon but I won't listen to your advice: indeed, I'll use Laplace transform to solve it, and not Fourier one. Please do not get angry.

First of all, let's consider the case $x>=0$. Then our integral equation becomes
$y+e^(-x)**y=e^(-x)$
Now, by applying Laplace operator to both RHS and LHS we get
$L(y+e^(-x)**y)(s)=L(e^(-x))(s)$
But remember that $L$ is linear; so we can write
$L(y)(s)+L(e^(-x)**y)(s)=L(e^(-x))(s)$
Now let $Y(s)$ be the Laplace transform of $y(x)$.
Using the convolution theorem ($L(f(x)**g(x))=L(f(x))L(g(x))$ (in other words, Laplace transforms the convolution operator into a simple algebraic product) we have
$Y(s)+L(e^(-x))Y(s)=L(e^(-x))$

The Laplace transform of $e^(-x)$ in $s$ is $\frac{1}{1+s}$: therefore
$Y(s)+\frac{1}{1+s}Y(s)=\frac{1}{1+s}$
which gives
$Y(s)=\frac{1}{2+s}$
By using Inverse Laplace transform we obtain $y=e^(-2x)$.

Now let's consider the case $x<0$. The equation becomes
$y+e^x**y=e^x$

In order to solve this one we do exactly the same things we have just done for case $x>=0$ (if you want I'll send you all the passages). We get
$Y(s)=1/s$
using the Inverse Laplace transform we obtain $y=1$.

We can conclude now that the solution of the integral equation is
$f(x):={[e^(-2x) \mbox{ if } x>=0],[1 \mbox{ if } x<0]:}$.

I do hope that this is correct (I have checked it, it works). Let me know.
Thank you for everything.

Looking forward to hearing from you soon.