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I am trying (for fun!) to study the ongoing COVID-19 pandemic and have the following question. we know that an exponential function satisfies the following functional equation: \begin{equation} \frac{f(x+T)}{f(x)} = e^T = \text{constant} \end{equation} for some period $T$. For instance the function $f(x)=2^x$ satisfies \begin{equation} \frac{f(x+1)}{f(x)} = 2 \end{equation} Now, since here in Italy the growth rate of the infected is falling, the function describing the total amount of covid-19 cases since the start of the outbreak might satisfy the following equation: \begin{equation} \frac{f(x+T)}{f(x)} = g(x)\qquad \text{with $g$ decreasing to 1} \end{equation}

My question is this: how to solve this, is the solution existing, is it unique? Any hint on how I could approach this?

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    $\begingroup$ The solution is not unique. Already the exponential solution was not unique: If $f$ is a solution then so is $c\cdot f$ for any constant. Apart from that, $f$ can behave arbitrarily on an initial interval $[0,T)$, say $\endgroup$ – Hagen von Eitzen Mar 27 at 18:48
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    $\begingroup$ You need to read more on epidemiology models. This is a good try but to say bluntly, too naive. $\endgroup$ – timur Mar 27 at 18:55
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    $\begingroup$ @timur Ok I am not really interested in having a good model for the epidemic (I know about SIR models etc) but I thought that this mathematical problem was interesting $\endgroup$ – marco trevi Mar 27 at 19:03
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    $\begingroup$ agree with timur's comment: spreading is a multiplicative diffusion process quite similar to nuclear fission $\endgroup$ – G Cab Mar 27 at 19:06
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Let $f(x)$ be given for $x\in[0,T)$, and let $g(x)$ be given for $x\in[0,b]$, for some $T>0$ and $b\geq T$. Then the equation $$ f(x+T)=g(x)f(x) , $$ has a unique solution for $x\in[0,b]$. In other words, the solution $f$ is unique as a function defined on $[0,T+b]$.

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