Convolution: How to construct it for a given function? 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?
 A: 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!
