Convolution (Possibly using Fourier transform)

[Incand]

Homework Statement

Find a function $u$ such that
$\int_{-\infty}^\infty u(x-y)e^{-|y|}dy=e^{-x^4}$.

Homework Equations

Not really sure how to approach this but here's a few of the formulas I tried to use.
Fourier transform of convolution
$\mathscr{F} (f*g)(x) \to \hat f(\xi ) \hat g(\xi )$.
$\mathscr{F} e^{-a|x|} \to 2a(\xi² +a^2)^{-1}$
Plancherel's Theorem
If $f,g \in L^2$ then
$\langle \hat f,\hat g \rangle = 2\pi \langle f, g \rangle$.

The Attempt at a Solution

The left side is the convolution $u*(e^{-|x|}$ Taking the Fourier transform I have
$\hat u2(\xi^2+1)^{-1}= \mathscr{F}(e^{-x^4})$.
Which I don't know the Fourier transform of nor how to calculate the inverse.

I also tried to somehow look at the integral as a scalar product and use Plancherel's formula but that didn't work at all. Any ideas on how I should approach the problem? I don't even know if I should use the Fourier transform for this, the problem was just grouped with other exercises on Fourier transforms.

 

[RUber]

The Fourier transform of $e^{-x^4}$ is often written as $\int_{-\infty}^{\infty} e^{-x^4}e^{-2\pi i x \xi } dx$ and the inverse operation for $\hat u(\xi)$ would be
$u(x) = \int_{-\infty}^\infty \hat u(x) e^{2\pi i x \xi } d\xi$.
But how does this help you?
You have
$\hat u(\xi) \mathcal{F}(e^{-|x|}) = \int_{-\infty}^{\infty} e^{-x^4}e^{-2\pi i x \xi } dx$
Which might be written as
$\frac{2}{\xi ^2 + 1} \left( \int_{-\infty}^{\infty} u(x)e^{-2\pi i x \xi } dx\right) = \int_{-\infty}^{\infty} e^{-x^4}e^{-2\pi i x \xi } dx$
Since the integral is not a function of xi, you can pass that variable inside the integral of u.
Option 1:
$\int_{-\infty}^{\infty} \frac{2}{\xi ^2 + 1} u(x)e^{-2\pi i x \xi } dx = \int_{-\infty}^{\infty} e^{-x^4}e^{-2\pi i x \xi } dx$
Try to make the left side match the right side. (this might be challenging since u(x) would be defined as a function of xi as well.

Option 2: move the $\displaystyle \frac{1}{\xi ^2 + 1}$ to the other side to get:
$2 \int_{-\infty}^{\infty} u(x)e^{-2\pi i x \xi } dx = \int_{-\infty}^{\infty}( \xi ^2 + 1) e^{-x^4}e^{-2\pi i x \xi } dx$
Then take the inverse Fourier transform of both sides to see what happens.

 

[Incand]

I'm going to use $\hat f(\xi ) = \int f(x)e^{-i\xi x} dx$ and $f(x) = \frac{1}{2\pi}\int \hat f(\xi)e^{i\xi x}d \xi$ since I'm more used to these definitions (I just end up with the $2\pi$ in other places like in the plancherel theorem above.)
I tried option 2 so I have
$2 u(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \left( \int_{-\infty}^\infty (\xi^2+1)e^{-x^4}e^{-i\xi x} dx \right) e^{i \xi x} d\xi$
Which looks promising until I discover I can't change order of integration since I have a divergent inner integral if the $e^{i\xi x}$ terms cancel each other. But I could write it as
$4\pi u(x) = \iint_{\mathbf R^2} (\xi^2+1)e^{-x^4}dxd\xi$. I tried revisiting my multivariable calculus book but apparently since the integrand doesn't change sign I'm indeed allowed to treat it as two iterated integrals and therefore the integral is divergent? Did I mess up?

For option 1 I don't really see how to go about it I still end up with a very hard integral It looks like. I mean if $u(x) = \frac{\xi^2+1}{2}e^{-x^4}$ that solves it but I don't how getting those $\xi$ into $x$ is anything but the problem I already have.

 

[RUber]

I'm going to use $\hat f(\xi ) = \int f(x)e^{-i\xi x} dx$ and $f(x) = \frac{1}{2\pi}\int \hat f(\xi)e^{i\xi x}d \xi$ since I'm more used to these definitions (I just end up with the $2\pi$ in other places like in the plancherel theorem above.)
I tried option 2 so I have
$2 u(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \left( \int_{-\infty}^\infty (\xi^2+1)e^{-x^4}e^{-i\xi x} dx \right) e^{i \xi x} d\xi$
Which looks promising until I discover I can't change order of integration since I have a divergent inner integral if the $e^{i\xi x}$ terms cancel each other.

The x inside the first integral is not the same as the x outside the integral. You can change the first x (dx) to z to make it clear.

 

[Incand]

The x inside the first integral is not the same as the x outside the integral. You can change the first x (dx) to z to make it clear.

Alright but I still get a divergent integral here anyway. However I think I have another idea on how too solve it by using differentiation.

Differentiating a convolution we have $(f*g)' = f'*g=f*g')$ so in our case we have
$u(x)*e^{-|x|} = e^{-x^4}$ Differentiate
$-u(x)*\text{sgn} (x) e^{-|x|} = -4x^3e^{-x^4}$ and again (The derivative of $\text{sgn}(x)$ at $x=0$ being a dirac spike actually works too our advantage!)
$u(x)*(-2\delta(x) +e^{-|x|} = (-12x^2+16x^6)e^{-x^4}$.
Using the linearity of Convolution we have
$2u(x)*\delta(x) = (1+12e^2-16e^6)e^{-x^4} \Longrightarrow u(x) = \frac{(1+12e^2-16e^6)e^{-x^4}}{2}$
Which is the correct answer! In a way I cheated coming up with this since I looked at the answer(which you didn't have access too) which looked very much like derivatives so I tried that. But really thanks for your help, always helps the thinking having someone to discuss it with!
I also misled us both thinking I should use the Fourier transform however I believe integral equations of this type (renewal equations) are quite often solved using the Fourier or Laplace transforms!

 

[RUber]

Nice solve.