Physical interpretation of convolution obtained from the solution of a linear first order DE: The bank application example.

85 Views Asked by At

Let's image the linear first order differential equation \begin{gather} y'(t) = ay(t) + q(t) \end{gather} that reflects the growth rate of a money bank application, where in the $t=0$ the amount of money $y(0)$ is applied and after this time, money is continuously applied according the function $q(t)$. It is known that the solution for this differential equation is \begin{gather*} y(t) = y(0)e^{at}+\int_0^te^{a(t-s)}q(s)ds. \end{gather*} The first term $y(0)e^{at}$ of the solution encodes the new amount of money derived from the incidence of some interest above the original amount of money $y(0)$ and the second term $\int_0^te^{a(t-s)}q(s)ds$ encodes the new amount of money derived from the incidence of the same interest above all the amount of money applied continuously according to $q(t)$.

Well, apart to the solution interpretation above, I do not fell that I fully understood the integral term of the solution, i.e. the convolution. My problem is that I not succeeding in create an imaginary function from which I would sum all the infinitesimal chunks of areas $e^{a(t-s)}q(s)ds$ from time $0$ until time $t$ to obtain the amount of new money suc derived from the incidence of some interest above the all amount of money applied continuously according to $q(t)$. So, what such a function looks like?