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Are there known results on functional equations of the type: Given $\tau>0$ and $g$ (real numbers), find a continuous function $f$ such that $f(t)-f(t-\tau)=g$ or $f(t)+f(t-\tau)=g$ (these are distinct equations)?

For the second case, the constant function $f(t)=g/2$ works while for the first case, affine functions $f(t)=a+g/\tau t$ work as well. I am expecting many more, at least for the first case.

Mohsen Shahriari
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pluton
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1 Answers1

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Since these are inhomogeneous linear equations, the general solution is of the form $f(t) = p(t) + h(t)$ where $p(t)$ is one particular solution and $h(t)$ is the general solution of the homogeneous equation.

For the first equation, we can take $p(t) = (g/\tau) t$. The homogeneous equation is $h(t) - h(t-\tau) = 0$, which just says $h$ is periodic with period $\tau$.

For the second equation, we can take $p(t) = g/2$. The homogeneous equation is $h(t) + h(t-\tau) = 0$. Take any $s \in [0, \tau]$ and define $h$ as any continuous function on $[s, s+\tau]$ with $h(s) = h(s+\tau) = 0$. Extend to $[s, s+2\tau]$ with $h(t+\tau) = -h(t)$ for $0 \le t \le \tau$, then extend again to make it periodic on $\mathbb R$ with period $2\tau$.

Robert Israel
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  • Thank you. And the linearity of the functional equation tells us that there is no other particular solution to expect... – pluton Apr 03 '20 at 23:15
  • You wanted to say "extend to $[s,s+2\tau]$" maybe? Because otherwise, it looks like $s$ is lost or $t$ has been redefined to be $0$ at $s$. – pluton Apr 04 '20 at 17:46
  • Any solution could be the "particular solution". 2) Yes, thanks (edited).
    • – Robert Israel Apr 05 '20 at 02:25