Analytical method for solving "continuous difference equations" and/or integral equations

Question

I'm trying to find a function, $G(t)$, that satisfies the relation

$$G(t+1) - G(t) = f(t)$$

for some known function, $f(t)$, for $t \in R$. I have been able to come up with solutions for a few instances of $f(t)$ by trial and error and inspection, namely:

$$\begin{align} f(t) & = m t + f_0 & & => & & G(t) = (m/2) t^2 + (f_0 - m/2) t \\ f(t) & = f_0 (1 + r)^t & & => & & G(t) = (f_0 / r) (1 + r)^t \end{align}$$

where $f_0$, $r$, and $m$ are constants. But I can't seem to identify any method which can be applied generally.

I'm having a tough time even naming the class of problem - I wrote "continuous difference equation," though that doesn't seem to be a commonly used term. My current thinking is that I might be able to write this as an integral equation in some way, since the $G(t+1) - G(t)$ is reminiscent of the Fundamental Theorem of Calculus, and the solutions I've found so far are kind of similar to the integrals of $f(t)$, though just a bit off. If anyone has seen anything like this before or knows even what to call this type of problem it would be greatly appreciated!

Answer 1

That's exactly the subject of what is called Indefinite Sum $$ \Delta G(t) = G(t + 1) - G(t) = f(t)\quad \Rightarrow \quad G(t) = \Delta ^{\, - 1} f(t) = \sum\nolimits_{\,t} {f(t)} $$ that is to find the function $G(t)$ whose finite difference is a given function $f(t)$.

The topic has many similarities with the (definite / indefinite) integrals, and the solution methods are much alike:
- basic theory and identities;
- table of basic integrals;
- lot of expertise and intuition to put up a chain combining the above.

Answer 2

A possible approach is through the Laplace transform. This gives for your equation $$ (e^{-s}-1)G(s)=f(s), $$ that is $$ G(s)=\frac{f(s)}{e^{-s}-1}. $$ This yields the solution $$ G(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}e^{st}\frac{f(s)}{e^{-s}-1}\,ds $$ where the integral should be performed as explained here.