# How to approach this functional equation?

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: $$$$\frac{f(x+T)}{f(x)} = e^T = \text{constant}$$$$ for some period $$T$$. For instance the function $$f(x)=2^x$$ satisfies $$$$\frac{f(x+1)}{f(x)} = 2$$$$ Now, since here in Italy the growth rate of the infected is falling, the function describing the total amount of covid-19 cases since the start of the outbreak might satisfy the following equation: $$$$\frac{f(x+T)}{f(x)} = g(x)\qquad \text{with g decreasing to 1}$$$$

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

• 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 – Hagen von Eitzen Mar 27 at 18:48
• You need to read more on epidemiology models. This is a good try but to say bluntly, too naive. – timur Mar 27 at 18:55
• @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 – marco trevi Mar 27 at 19:03
• agree with timur's comment: spreading is a multiplicative diffusion process quite similar to nuclear fission – G Cab Mar 27 at 19:06

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 $$[0,T+b]$$.