Convolutions, relatively speaking, are fairly straightforward for simple systems (from an applied perspective), but I cannot, at all, find the intuition behind the Laplace identity for convolutions. That is:
$$ \mathcal{L}\{f\star g\}=\mathcal{L}\{f\}\cdot \mathcal{L}\{g\} $$
How is it possible to make sense of this? Perhaps there is some probability theory example that sheds light on it?
Thanks in advance.
On
Well convolutions are really DEFINED to make that identity true. Their use comes into play when you want to separate the elements of the equations that relate to the system itself, those that relate to the initial conditions of the system, and those that relate to external forces. For example, consider
$$ ay''+by'+cy = g(x) $$
Transforming,
$$a\big (s^2Y(s)-sy(0)-y'(0)\big ) + b\big (sY(s) - y(0) \big ) + cY(s) = G(s) $$
Now let
$$H(s)=\frac 1{as^2+bs+c}$$ $$C(s) = sy(0)+y'(0)+y(0)$$
Then $Y(s) = G(s)H(s) + C(s)H(s) $ and using our definition of convolution,
$$y(x) = g(x) ⋆ h(x) + c(x) ⋆ h(x) $$
If the convolution is easier to calculate than the inverse transform, then the convolution theorem can be used to simplify things and separate the aspects of the equations logically.
I'll try to review this post when I'm not so tired in the morning.
If you multiply two power series $\sum_{n=0}^{\infty}a_{n}z^{n}$ and $\sum_{n=0}^{\infty}b_{n}z^{n}$, then a type of convolution also appears because of trying to gather like powers $$ \sum_{n=0}^{\infty}a_{n}z^{n}\sum_{n=0}^{\infty}b_{n}z^{n} = \sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_{k}b_{n-k}\right)z^{n} $$ You can think of a Laplace transform as a continuous version of a power series for the purposes of this discussion built from powers $e^{-ns}=(e^{-s})^{n}$. $$ \int_{0}^{\infty}a(n)e^{-n s}\,dn \int_{0}^{\infty}b(n)e^{-ns}\,dn = \int_{0}^{\infty}\left(\int_{0}^{n}a(k)b(n-k)\,dk\right)e^{-ns}\,dn. $$ The Laplace transform identity then gives $$ \mathcal{L}\{ a\} \mathcal{L}\{ b\} = \mathcal{L}\{ a\star b\}. $$ The proof of the convolution identity follows the pattern of dealing with power series by changing the order of integration (instead of summation) in order to collect like powers. So the analogy is strong. I'm not sure if this is the type of intuition you're looking for or not.
Hardy Space Mapping: Suppose $f \in L^{2}[0,\infty)$. Then the Laplace transform $\mathcal{L}\{f\}(s)$ of $f$ is holomorphic in the right half-plane, and is uniformly bounded on $\Re s \ge \delta > 0$. Furthermore, $G_{u}(v)=\mathcal{L}\{f\}(u+iv)$ is in $L^{2}(\mathbb{R})$ for each $u > 0$. This is because $G_{u}(v)$ is really the Fourier transform (to within a constant) of $e^{-ut}f(t)$ for fixed $u$. So, by Parsevel's equality, $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}|G_{u}(v)|^{2}\,dv=\int_{0}^{\infty}e^{-2ut}|f(t)|^{2}\,dt \le \|f\|^{2}_{L^{2}[0,\infty)} $$ So that means $G$ is in the hardy space $H^{2}(\Pi^{+})$ where $\Pi^{+}$ is the right half-plane. Furthermore, $$ \frac{1}{2\pi}\|\mathcal{L}\{f\}\|_{H^{2}}=\|f\|_{L^{2}}. $$ And the boundary function of this Hardy class function is $L^{2}-\lim_{u\downarrow 0}G_{u}=\mathcal{F}f$ where $\mathcal{F}$ is the Fourier transform of the function which is $0$ on $(-\infty,0)$ and is $f$ on $[0,\infty)$. On the other hand, every $f \in H^{2}$ has an $L^{2}$ boundary function $f_{0}$, which allows the Cauchy integral representation $$ \begin{align} f(s) & = -\frac{1}{2\pi}\int_{-\infty}^{+\infty}\frac{f_{0}(iu)}{iu-s}\,du \\ & =-\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{0}(iu)\int_{0}^{\infty}e^{t(iu-z)}\,dt\,du \\ & =\int_{0}^{\infty}\left(-\frac{1}{2\pi}\int_{-\infty}^{\infty}f_{0}(iu) e^{itu}\,du\right)e^{-ts}\,dt. \end{align} $$ So the correspondence between $H^{2}(\Pi^{+})$ is $L^{2}[0,\infty)$ is fully unitary.
So $f \in H^{2}(\Pi^{+})$ iff there exists $h \in L^{2}[0,\infty$ such that $$ f(z) = \int_{0}^{\infty}e^{-zt}h(t)\,dt. $$ If $k \in L^{1}[0,\infty)$, then $$ K(z) = \int_{0}^{\infty}e^{-zt}k(t)\,dt $$ is a bounded holomorphic function on the right half-plane. And $K(z)$ is a valid multiplier on $H^{2}$ that is equivalent to a convolution operator: $$ F(z)f(z) = \int_{0}^{\infty}\left( e^{-zt}\int_{0}^{t}k(u)h(t-u)\,du\right)\,dt $$ This is part of the functional calculus.