3
$\begingroup$

While working on my thesis my advisor handed me an unfinished paper which states the following:

First, define the operators \begin{align*} A_i &:= -\operatorname{div}(\sigma_i\nabla) \\ A_e &:= -\operatorname{div}(\sigma_e\nabla) \\ C &:= A_i + A_e \\ G &:\approx C^{-1} \\ R &:= \mathrm{Id} - CG \end{align*} where the $\sigma$'s are tensors for an internal- and an external-"influence/action".
Second, consider the function $$ \xi(t) := R \left[A_i u(t) - C v(t) + \epsilon^{-1}r(t) \right] \quad \text{on} \quad [0,T]\subset \mathbb{R}$$ and the differential equation $$ \dot r (t) = \xi(t) -\epsilon^{-1}r(t) \quad \text{also on} \quad [0,T]\subset \mathbb{R}$$ where $u$, $v$, $r$, and $\xi$ are elements of the Sobolev space $$ W^{1,2}\left(0,T,H^1(\Omega),H^1(\Omega)^*\right) := \left\{ \varphi \, \Big| \, \varphi \in L^2\left(0,T,H^1(\Omega)\right),\, \dot\varphi \in L^2\left(0,T,H^1(\Omega)^*\right) \right\}.$$ Thus, we can write \begin{equation} r(t) = \int_{\tau = 0} ^t \exp\left( -\tfrac{t-\tau}{\epsilon} \right) \xi (\tau)\, d\tau \quad \text{f.a.a} \quad t \in [0,T].\hspace{80pt} (1) \end{equation} Further, using $$ \delta_t(x) := \begin{cases} + \infty & \text{if } x = t \\ 0 & \text{otherwise,} \end{cases} \quad\quad\quad \int_{-\infty}^{+\infty} \delta_t(x) \, dx = 1 $$ we can write $$ \dot{r} (t) = \big( \xi * \left( \delta _t - \mu \right) \big) (t). $$

I can see that with $$ \mu (t) := \left\{ \begin{array}{ll} \epsilon^{-1}\exp\left(-\tfrac{t}{\epsilon} \right) & \mbox{for } t \geq 0 \\ 0 & \mbox{otherwise} \end{array}\right.$$ $(1)$ is a convolution $$r (t) = (\xi * \mu)(t)=\int_{-\infty}^{+\infty}\xi(\tau)\mu(t-\tau)\,d\tau$$ But, why does $r = \xi * \mu $ ? How is this calculated? or how is such a function $\mu$ constructed?
Is there a standard method to express any given function as a convolution? Or a differential equation as a convolution?

$\endgroup$
  • 1
    $\begingroup$ Just to be clear about your question. Given functions $f$ and $g$ you want a method to build $\mu$ such that $f = g\star \mu$ ? $\endgroup$ – C. Dubussy May 26 '16 at 10:14
  • $\begingroup$ yes. my question could be simplifyed that way: given functions $f=r$ and $g=\xi$, such that $g=c \cdot f$, $c \in \mathbb{R}$ constant, how would a function $\mu$ be constructed such that $f = g * \mu$ $\endgroup$ – scjorge May 26 '16 at 11:24
  • $\begingroup$ @scjorge, do you mean partial differential equations or ordinary differential equations $\endgroup$ – moonshine May 31 '16 at 18:20
  • $\begingroup$ well, for my quiestion it is a nonhomogenous first order ordinary differential equation. $\endgroup$ – scjorge May 31 '16 at 18:29
2
+50
$\begingroup$

First: How to construct $\mu$.

I think the function $\mu$ would be constructed by "deconvolution". This is needed in a lot of experimental situations, because in many measurement processes the measuring instrument acts like a convolution filter. The most standard way of doing the deconvolution is taking the Fourier transforms of both $r(t)$ and $(\xi*\mu)(t)$, let's say $$R(\omega)=\int_{-\infty}^{\infty}r(t)e^{-2\pi i\omega t}dt$$ $$\Xi(\omega)=\int_{-\infty}^{\infty}\xi(t)e^{-2\pi i\omega t}dt$$ $$M(\omega)=\int_{-\infty}^{\infty}\mu(t)e^{-2\pi i\omega t}dt$$

Then you have $$R(\omega)=\Xi(\omega)M(\omega)$$ So $$M(\omega)=\frac{R(\omega)}{\Xi(\omega)}$$ And in consequence, you can obtain $\mu(t)$ by taking the inverse Fourier transform, i.e. $$\mu(t)=\int_{-\infty}^{\infty}M(\omega)e^{2\pi i\omega t}d\omega$$ Or, more explicitly $$\mu(t)=\int_{-\infty}^{\infty}\frac{\int_{-\infty}^{\infty}r(t)e^{-2\pi i\omega t}dt}{\int_{-\infty}^{\infty}\xi(t)e^{-2\pi i\omega t}dt}e^{2\pi i\omega t}d\omega$$ The best about this method is that there are very good algorithms to calculate the direct and inverse Fourier transform quickly (see Fast Fourier Transform).

Second: Can differential equations be expressed as convolutions?

Yes if they are linear with constant coefficients. Let's say you have an ordinary differential equation like: $$a_nD^nx(t)+a_{n-1}D^{n-1}x(t)+...+a_0=f(t)$$

You can use the property of the Fourier transform $$\mathcal{F}[D^{n}x]=(2\pi i\omega)^nX(\omega)$$ To make $$(a_n(2\pi i\omega)^n+a_{n-1}(2\pi i\omega)^{n-1}+...+a_0)X(\omega)=F(\omega)$$ And simply dividing by the polynomial $$X(\omega)=\frac{F(\omega)}{a_n(2\pi i\omega)^n+a_{n-1}(2\pi i\omega)^{n-1}+...+a_0}=\frac{F(\omega)}{P(\omega)}$$ Now the solution $x(t)$ can be written as a convolution $$x(t)=\left(\frac{1}{p}*f\right)(t)$$ Here $p(t)=\mathcal{F}^{-1}[P(\omega)]$. If you have a partial differential equation instead, you'd use the multidimensional Fourier transform $$\mathcal{F}[f(\vec{r})]=F(\vec{k})=\int_{\vec{k}\in \mathbb{R}^m}f(\vec{r})e^{-2\pi i\vec{k}\cdot\vec{r}}d^mk$$ With the property $$\mathcal{F}\left[\frac{\partial f(\vec{r})}{\partial x_j}\right]=(2\pi ik_j)F(\vec{k})$$ And you should be able to express the solution as a spatial convolution. Also, if the time were restricted, like $t>0$ for instance, the Laplace transform must be used in place of the Fourier's for the time dimension.

Bottom line (advertising): Fourier (and Laplace) transforms are awesome, use them after every meal!

$\endgroup$
  • $\begingroup$ interesting answer, thanx! I didnt yet know about this. I've gotta do some check-up and calculations to know if this (or an edited version of it) is the right answer. $\endgroup$ – scjorge May 31 '16 at 18:31
  • 1
    $\begingroup$ @scjorge thank you back, I put something about expressing differential equations as convolutions via Fourier transforms. This is a huge method in PDEs and ODEs, I just gave a tablespoon. $\endgroup$ – moonshine May 31 '16 at 19:22
  • $\begingroup$ Great answer! excellent help! I will slowly edit my question and your answer and then mark it (: $\endgroup$ – scjorge Jun 1 '16 at 3:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.