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In my following question, I am referring to an article dealing with reaction-diffusion equations. I'll give screenshots of the parts of the text I am dealing with.

The setup is the following:

enter image description here

enter image description here

My question is how to see these described "jumps" within the dynamics: How do one see these jumps from (3.2)? For example: How can one see that, starting with initial values $(u_0,v_0)$ with $v_0\geq S(u_0)$, the singular flow jumps instantaneously to the point $(u_1,v_0)$ where $v_0=f(u_1), u_1\leq u_{\text{min}}$? Or how does one see from (3.2) that from point $(u_{\text{max}},v_{\text{max}})$ we jump instantaneously to the point $(u_2,v_{\text{max}})$ where $v_{\text{max}}=f(u_2), u_2>0$?

I think, here, the case $\varepsilon=0$ is considered. Thus, (3.2) is $$ 0=F(u,v)=u(u-a)(1-u)-v,~~\frac{dv}{dt}=G(u,v). $$ Now, $0=F(u,v)=u(u-a)(1-u)-v$ means that $v=u(u-a)(1-u)=f(u)$, so the dynamics only takes place on $f(u)$. But thats all I can say.

Rhjg
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1 Answers1

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First of all take $\varepsilon=0$ (and don't think in terms of the limit behavior in $\varepsilon$).

The reason for the jumps is that the first equation now takes the form $F(u,v)=0$. This causes that if we don't start in the solution of this equation, we must jumpt to them (this is basically the last sentence in your question).

John B
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  • But why do we jump for example if $a\leq u < u_{\text{max}}$ and $v=f(u)$; then, we are on $f(u)$ but the text says we jump (to the left). Moreover, how to see in general the direction of the jumps (left or right)? – Rhjg Jan 28 '16 at 13:31
  • Since $\varepsilon>0$, you take $\varepsilon=0^+$ and so you have $u'=+\infty$ if $F(u,v)>0$ and $u'=-\infty$ if $F(u,v)<0$ (being "somewhat" rigorous...). This indicates the direction of the jumps. What you ask follows already from the specific hypotheses on the functions. – John B Jan 28 '16 at 13:37
  • But I thought we are considering $\varepsilon=0$? And if I am on $f(u)$ then $u'=0$. Sorry, that confuses me a bit. – Rhjg Jan 28 '16 at 13:40
  • How come? If $\varepsilon=0$, then $0=\varepsilon u'=F$. – John B Jan 28 '16 at 14:05
  • No matter what $\varepsilon$ is, whether $0$ or $>0$, on $f(u)$ we have $u'=0$. So I do not understand why on some parts of $f(u)$ we are jumping to the left. What confuses me also is whether I should think of $\varepsilon=0$ or of the limit process $\varepsilon\to 0$... the description of the phase portrait seems to be for the limit process rather than for $\varepsilon=0$. But again, its unclear for me why we jump, esspecially if we are on $f(u)$ since jumping changes $u$ and we have $u'=0$ on $f(u)$. – Rhjg Jan 28 '16 at 14:09
  • Your first claim is wrong, but because of this (it is not a computation): you have to think of it as a limit, although the solutions are not obtained as a limit (in the work of Greenberg et al, and in others in the area). Simply you define what they call a "singular flow" using the rules that I gave first comment above. – John B Jan 28 '16 at 14:14
  • If we have $\varepsilon >0$ and start on $f(u)$, i.e. in $(u,f(u))$. Why is then $\varepsilon u'=F(u,v)\neq 0$? – Rhjg Jan 28 '16 at 14:20
  • $F(u,f(u))=0$. That's why you don't jump then (when $\varepsilon=0$). – John B Jan 28 '16 at 14:23
  • Hm, I seem to have some blocked thoughts. I simply do not understand. What is the singular flow? . . . We always have $F(u,f(u))=0$ by assumption. Why do we jump then? – Rhjg Jan 28 '16 at 14:31
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    Sorry, I have suggest that you read again the Greenberg paper and my comments. It is impossible to be more detailed without using too much time. It also depends on whatever you now of the theory. To study parabolic pde's assumes already a lot, plus some habit of dealing with the material. So, any detailed explanation would depend on what you already know. – John B Jan 28 '16 at 17:19
  • Ok, maybe I can explain my main problem now. I think I got that we consider $\varepsilon=0^+$ so as a limit when $\varepsilon$ gets smaller and smaller. But there is one thing I do not understand. Sorry: Assume we have $\varepsilon >0$ and start at $(u,v)$ with $a<u<u_{max}$ and $v=f(u)$. Then we are on $f(u)$ and we have $F(u,f(u))=0$. When we now let $\varepsilon\to 0$ why do we jump: We are on $f(u)$ and for $\varepsilon\to 0$, we therefore should move on $f(u)$ instead of jumping! . . This is my main problem right now. Maybe you can answer this. – Rhjg Jan 28 '16 at 17:25
  • Maybe I can answer by myself. If we have $\varepsilon >0$ and have $a<u<u_{max}, v=f(u)$ we have $\varepsilon u'=0$, i.e. $u'=0$ (so we do not move in direction u) but $v'>0$, so me move upwards a bit. So when $\varepsilon\to 0$, we can define to junp to the left in order to be on $f(u)$. Dont know if this is the correct reason. – Rhjg Jan 28 '16 at 18:20
  • If I understand correctly your description, all seems to be as you say. – John B Jan 28 '16 at 23:52