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Let $\zeta$ be the classical Riemann zeta function.

We define a differential equation on $\mathbb{R}^{2} \setminus \{1\}$ by $\dot Z= \zeta(Z)$. From a foliation point of view this vector field can be counted as a smooth vector field on whole $\mathbb{R}^{2}$ with the following equivalent formulation(They have the same trajectories).

$$\dot Z= \lVert z-1\rVert^2 \zeta(Z)$$

Then the field has a saddle point at $1$.

Are there some researches about this dynamical system?Are there closed orbits for this equation?The latter is equivalent to ask: "Are there zeroes of the Riemann Zeta function whose Taylor expansion (after translation to the origin and real rescalling ) is in the form $"iz+...."$. Every zero of a holomorphic map with this linear part is necessarily a center, a singularity surrounded by a band of closed orbits.

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    $\begingroup$ if you are talking about dynamical system of zeta function all research are occurs over Ruelle $ \zeta$ -function where .$ \zeta(Z,s)$ is non zero and analytic for $|z|<e^{p(-re(s)f}$ where p is pressure and if you meant the geometrical part then you shoud talk about selberg zeta function here there are a closed orbit or closed geodisc flows , i think you should explain more and specify your problem $\endgroup$
    – zeraoulia rafik
    Commented Feb 5, 2017 at 23:35
  • $\begingroup$ What is Z(1) then? $\endgroup$
    – Fan Zheng
    Commented Feb 6, 2017 at 1:11
  • $\begingroup$ @FanZheng thank you I revise the question $\endgroup$
    – Ali Taghavi
    Commented Feb 6, 2017 at 3:30
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    $\begingroup$ For the link between $\zeta(s)$ and dynamical systems, I'd look at $\theta(x,z) =\sum_{n=-\infty}^\infty e^{-\pi n^2 x}e^{2i \pi x z}$ the kernel of the heat equation, whose Mellin transform is $\int_0^\infty x^{s-1} (\theta(x,0)-1)dx = 2 \Gamma(s) \pi^{-s}\zeta(2s)$ $\endgroup$
    – reuns
    Commented Feb 6, 2017 at 13:06
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    $\begingroup$ A trivial matter of LaTeX: for norms, using \parallel is ill-suited, as they are binary operators (which affects spacing). I took the liberty to replace them by the corresponding parentheses, \lVert and \rVert. $\endgroup$
    – Benoît Kloeckner
    Commented Feb 13, 2017 at 9:29

2 Answers 2

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This is not an answer, but rather a long comment. Software like Mathematica can plot trajectories of this dynamical system, by (numerically) solving the system of ODEs. Here are trajectories that start on a circle of radius $1/10$ centered around the lowest nontrivial zero of $\zeta(s)$ (sorry, I can't make myself call the variable $Z$.) enter image description here

The picture near the zero is exactly what you expect when you think of approximating $\zeta(s)$ by the linear approximation: the ODE in $s=\sigma+it$ is very nearly $\sigma^\prime=\sigma$ and $t^\prime=t$, at least, up to the scaling and rotation of the picture by $\zeta^\prime$ evaluated at the zero. Further from the zero, higher order terms come in to play.

Here are trajectories that start on a (larger) circle of radius $1/2$ centered on the pole at $s=1$. This picture is also what you should expects when you think about the Laurent expansion of $\zeta(s)$ at $s=1$

enter image description here

Update (Things I should have remembered from complex analysis in my original posting): The function $f(s)=(s-1)\zeta(s)$ is holomorphic in $\mathbb C\backslash \{1\}$. Let $u(\sigma,t)$ and $v(\sigma,t)$ be the real and imaginary parts, as functions of $s=\sigma+i t$ (NB: $t$ is not time in this notation.) From the point of view of complex analysis, it makes more sense to look instead at the Polya vector field $(u,-v)$ , which is conservative. Indeed, with $$ F(s)=\int_{s_0}^s f(w)\, dw=U(\sigma,t)+iV(\sigma,t) $$ we have $\triangledown U=(U_x,U_y)=(U_x, -V_x)=(u,-v)$ The trajectories for this vector field are the level curves of $V$.

I don't believe there are any closed orbits.

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    $\begingroup$ Does not the Hadamard product formula imply that the picture is qualitatively the same at all zeros? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 17, 2017 at 19:32
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    $\begingroup$ The zero is a source (spiral out) if $\text{Re}\zeta^\prime(\rho)>0$, and is a sink (spiral in) if $\text{Re}\zeta^\prime(\rho)<0$. This latter happens less often; the first example is zero number 127. See oeis.org/A153815 $\endgroup$
    – Stopple
    Commented Feb 20, 2017 at 20:52
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A colleague gives me this article: Does the Riemann zeta function satisfy a differential equation?

where the next formula appear $$\zeta'(z)=\zeta(z)\sum\limits_{n=1}^\infty{\ln p_n\over1-p_n^z}.$$ So, your original problem symplifies to the system $$\dot{y}=F(z)y^2$$ $$\dot{z}=y$$ (of course $F(z)=\sum\limits_{n=1}^\infty{\ln p_n\over1-p_n^z}$).

This system have been studied for $F$s in many spaces. But I can't tell what kind of function is it.

I hope it helps. About closed orbits, I really don't know (but I don't think they are).

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