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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?

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    $\begingroup$ 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
    Commented May 26, 2016 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
    Commented May 26, 2016 at 11:24
  • $\begingroup$ @scjorge, do you mean partial differential equations or ordinary differential equations $\endgroup$ Commented May 31, 2016 at 18:20
  • $\begingroup$ well, for my quiestion it is a nonhomogenous first order ordinary differential equation. $\endgroup$
    – scjorge
    Commented May 31, 2016 at 18:29

1 Answer 1

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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!

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  • $\begingroup$ 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
    Commented May 31, 2016 at 18:31
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    $\begingroup$ @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. $\endgroup$ Commented May 31, 2016 at 19:22
  • $\begingroup$ Great answer! excellent help! I will slowly edit my question and your answer and then mark it (: $\endgroup$
    – scjorge
    Commented Jun 1, 2016 at 3:21

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