1. Feb 6, 2012

### fauboca

I am trying to show that there are 3 nonzero steady states of
$$\frac{du}{dt}=ru\left(1-\frac{u}{q}\right)-\left(1-\exp\left(-\frac{u^2}{\varepsilon}\right)\right)=0$$

I have tried using Mathematica and Mathematica couldn't solve it.
I tried some algebra and that wasn't going anywhere so I am at a loss here.

Then in order to determine if the model has hysteresis, I have to make the substitution
$$r=\frac{R}{\sqrt{\varepsilon}}$$
and
$$u=U\sqrt{\varepsilon}.$$

And show that there is a nose at R=0.638.

After I make the substitutions, what do I do to show the nose at R=0.638?

Thanks.

2. Feb 6, 2012

### fauboca

I forgot to mention that I am trying to demonstrate that there are 3 nonzero steady states if r and q lie in a domain in r,q space given approximately by rq > 4.

Not to sure how this helps because I haven't been able to do much with it and $0<\varepsilon\ll 1$

3. Feb 7, 2012

### epenguin

Those substitutions are what you would do in a problem like this in order to work with expressions that have the minimum number of constants which makes calculations and qualitative insights easier.

Then I think you get for steady states,:

downward curving parabola = 1 - e-U2

so to speak.

The RHS is a sigmoid kind of curve that increases from 0 and levels off at 1.

So to get three intersections between these two curves you just have to make the parabola not increase too steeply so that it will cut the sigmoid ascending so to speak and not from the top, not descending - R not too big.

You choose q though so that the maximum of the parabola is above the maximum of the sigmoid curve. The curve levels off at height 1 and is pretty close to that over an infinite range. For the maximum of the parabola to equal 1 is exactly the condition you were asked to find I think, and that is not the exact condition for a transition to multistationarity (going from one intersection to three) but close to one, as they say.

This is a typical nonlinear modelling kind of problem - finding the parameter zones where the qualitatively nonlinear stuff happens.

Would be interesting if you come back with figs and behaviour simulations if you get results. (Even if I suppose the noses are only nodes .)

4. Feb 7, 2012

### fauboca

I let epsilon = .1, r = 2, and q = 3 which shows that the model will exhibit 3 nonzero steady states for rq > 3.

I am still confused on to do the hysteresis part though.

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5. Feb 7, 2012

### epenguin

OK the idea is in that pic. They ask you to give the (approximate) condition for the onset of this multistability though, which I've already indicated.

If you look at what u will do according to your equation, just using a sketch of your total function you will see that for any starting u (call u0) u will go to one of two stable steady-state u values and it will do this according to which side of the middle (unstable) steady state u0 is.

Hysteresis I guess comes in within some wider problem which includes the above. Suppose you are able to externally to control one of those parameters, r, q, or ε then because there are two different steady states as you vary it up from low you come to one of them, but as you vary it down from high you come to another. Try that with sketches or computer calculations.

6. Feb 7, 2012

### fauboca

The problem is I don't get how I am supposed to get R = .638 after making these substitutions. What am I doing after I make the substitutions? Taking the derivative? Solving for R? What?

7. Feb 7, 2012

### epenguin

Did you get the rq > 4 bit?

I can't answer the other question, perhaps someone else can - I don't know what a nose is (in this context).

8. Feb 7, 2012

### fauboca

I understand why rq had to be greater than 4.

9. Feb 10, 2012

### fauboca

How can I find the u-r space?

10. Feb 10, 2012

### epenguin

You ask some questions I do not know the meaning of. I do not know what is a nose. I have to guess what u-r space means. I am a bit doubtful that R = 0.638 by itself can correspond to anything particular without specifying other constants. Perhaps if you reproduced the entire problem or book page we might see farther.

11. Feb 10, 2012

### fauboca

The predation P(N) on a population N(t) is very fast and a model for the prey N(t)
satisfies

see post 1 for model

where R, K, P and A are positive constants. By an appropriate nondimensionalisation
show that the equation is equivalent to

see post 1 for nondimensionalized model

where r and q are positive parameters. Demonstrate that there are three possible
nonzero steady states if r and q lie in a domain in r, q space given approximately by
rq > 4. Could this model exhibit hysteresis?

I meant r-q space not u-r in the previous post.

12. Feb 10, 2012

### epenguin

You said in #8 you had the answer to the r-q question! If not look at the graph you gave and think of the effect of varying r.

I suppose one parameter in your model represents predator numbers or intensity, don't know which. I suppose that is a parameter you could think of varying, e.g by introducing predators or culling them, perhaps other parameters. After you have varied it would things always go back to the same steady state you had before? I think that is the question they call 'hysteresis'.

Edit: And I don't see anything about noses and R=0.638 there.

Last edited: Feb 10, 2012
13. Feb 10, 2012

### fauboca

Some school that also uses Mathematical Biology by J.D. Murray posted a hint on how to do the problem and that was the hint.

14. Feb 11, 2012

### epenguin

I should have said that the rq > 4 is only part of the (approximate) description of the r-q space for 3 steady states. You should be able to see from the sort of fig you already showed that with further increase of r something else happens. If you give a qualitative description and if possible a numerically calculated diagram and set of curves with some fixed ε, say 1, then r = R, and fixed q that illustrates it you would be doing quite well.

Last edited: Feb 11, 2012
15. Feb 14, 2012

### epenguin

How did you get on?
I looked at it some more.

I found it more difficult that I thought. Perhaps this was because I also tried some algebra or math that like you, it didn't go anywhere.

One shouldn't usually waste too much time with difficult or refined math in biomath. Use exact solutions if they can be got without too much trouble or if available. Otherwise it is a vanity as you're mostly after broad qualitative pictures.

I can't afford more time on this now and got tired of it, so as much as I got is illustrated (for ε = 1 which is equivalent to making the vertical axis R) Horizontal axis is q.

I have not completely understood this system and give conclusions as far as I have got. The main space that gives three stationary points is that on the right part of this fig between the two curves - as q goes to infinity the bottom curve goes to 0 and the top curve goes to the R=0.638 you mentioned. I do not know a way to calculate this number mathematically and think it must have been calculated numerically, but id does correspond to something rational. I did manage to put mathematically a limit for R not be greater than for three sp's: 1/√2 = 0.707 which is not ridiculously far. The bottom curve on the right is the xy=4 we mentioned. I did also mention that this was an approximation; a more accurate curve is the outside curve on the left which is quite near the xy=4 curve, and as you see the difference on the right is insignificant.

The curves cross twice. I think there is a very thin sliver of conditions - the thin crescent between the curves, where there are three s.p.s too but this is no doubt of no practical significance. I think in all the rest of the space there is only one s.p.

Perhaps someone else could treat it better. You do however in biomath often just cobble rough and ready treatments together. It gives qualitative insights and also helps find parameter ranges for the interesting behaviours instead of exploring enormous spaces with computer to find them.

Last edited: Feb 14, 2012
16. Feb 15, 2012

### fauboca

What were your plots in mathematica to generate that graph? I would like to replicate it.

Thanks.

17. Feb 15, 2012

### epenguin

I think the forum rules and ethos rightly don't let me give you the formulae, at least without having seen more effort on your part. However one we already have.

You need first to work out a way to for how would you get that R = 0.638?

Nothing fancy. Crude. Just with graphs and a ruler for instance.

18. Feb 15, 2012

### fauboca

I have said before I don't know how to parametrize the curve to obtain the r-q space. So I don't know how you think I will be able to do that now.

19. Feb 15, 2012

### epenguin

You have given a fig. where there are three steady states. As you vary R and q in what ways does that fig. change? What situations does transition from 3 to 1 look like diagramatically? What conditions could you say will surely give you 1 only even if they are not the exact transition?

20. Feb 15, 2012

### fauboca

How can it ever have 1 steady state? It should have 2 since the parabola opens down. By increasing r >= 2.5 and q >= 2.5, there will only be 2 steady states.