# Pursuit solutions to the Rock-paper-scissors flow and delay differential equations

The Rock-paper-scissors flow is the following reaction-diffusion system

$$r_t = \Delta r + rs-rp,$$

$$p_t = \Delta p + pr-ps,$$

$$s_t = \Delta s + sp-sr.$$

We can assume $r,p,s\geq 0$, $r+p+s$ is constant, and $1=\int_M (r+p+s) dV$.

Are there travelling wave solutions in one spacial variable of the form

$$r(x,t)=r(x-3,t)=p(x-2,t)=s(x-1,t)=u(x-vt)?$$

Here $u(y)\geq0$ is some function and $v>0$ is some constant. Such a solution would represent distributions of rock-paper-scissors which pursue each other around a circle.

It suffices to find a non-trivial, three periodic solution to the following non-linear delay differential equation

$$u''(y)+vu'(y)+u(y)(u(y-2)-u(y-1))=0.$$

Most of the references I've found on DDEs are very numerical in flavor or cover DDEs in rather specific forms.

• The problem is that non-trivial solutions do exist in general (which rules out any trivial arguments) but their non-negativity is a big issue. So far I haven't managed to find anything non-negative. I prefer $2\pi$-periodic functions with delays $2\pi/3$ and $4\pi/3$ and the same equation. It seems to me like we have enough free parameters to rescale the period.. Dec 2 '17 at 16:34
• A stupid question. $u>0$ is an understandable condition. But why do you insist on $v>0$? Dec 2 '17 at 18:08
• 1) You obviously cannot do that: the non-linear term changes! It is an issue, and a big one at that. 2) That "convenience" is quite a headache too: so far I have a whole family of (sign changing) solutions with $v<0$ but $v>0$ necessarily pushes us to high frequency ranges since then the multiplier and the derivative term work in the same direction on low frequencies, and handling that requires working in higher dimension than $4$, which I prefer to abstain from on my old dying laptop. Dec 3 '17 at 18:12
• My mistake re adding constants, result of answering this while travelling. $v$ being positive is really just convenience: these solutions are travelling waves, the sign of $v$ determines the direction of travel. Dec 3 '17 at 18:32
• OK, will try to find a positive solution with arbitrary $v$ then :-) Give me some time and check this thread now and then. Have a good trip! Dec 3 '17 at 18:39

Anyway, write $a(x)=u(x), b(x)=u(x-1), c(x)=u(x-2)$. Then $a''+va'=a(b-c)$, etc. If $a,b,c>0$, we can divide and write $$\frac{a''}a+v\frac{a'}a=b-c\,$$ i.e. $$(\log a)''+\frac{(a')^2}{a^2}+v(\log a)'=b-c\,,$$ etc. Adding these three equations up and integrating over the period, we conclude that $$\int\left(\frac{(a')^2}{a^2}+\frac{(b')^2}{b^2}+\frac{(c')^2}{c^2}\right)=0\,,$$ so $a,b,c=\operatorname{const}$.
• @JessBoling Not in the finite volume setting. You can apply the same argument directly to the original system and conclude that $\int_M \log(prs)$ is decreasing. And in the infinite volume setting the solutions are spreading to infinity due to the heat evolution for $p+r+s$. So, nothing periodic is anywhere in sight (unless you accept sign changing solutions) Dec 11 '17 at 0:12
• @JessBoling Yep, and the constants will cycle over the curve conserving the sum and the product. All that we need to show for that is that we cannot send one of the components to $0$ as time tends to infinity. Do you know if that can happen? Dec 11 '17 at 7:08
• In compact settings the maximum principle applied to the evolution of $\log⁡(rps)$ would rule that out, so long time convergence to the ODE is the way of the game. Dec 12 '17 at 21:34