Homework 4.1 Solutions Math 5110/6830

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1 Homework 4. Solutios Math 5/683. a) For p + = αp γ α)p γ α)p + γ b) Let Equilibria poits satisfy: p = p = OR = γ α)p ) γ α)p + γ = α γ α)p ) γ α)p + γ α = p ) p + = p ) = The, we have equilibria poits p = ad p =. The, fp) = αp γ α)p γ α)p + γ αp γ α)p ) γ α)p + γ α γ α)p ) γ α)p + γ f p) = αγ α)p γ α)p + γ) αpγ α)p γ α)) γ α)p γ α)p + γ) = αγ α)p αγ α)p + αγ αγ α)p + αγ α)p γ α)p γ α)p + γ) = αγ αγ α)p γ α)p γ α)p + γ) At p = : f ) = α γ > whe α > γ ustable) < whe α < γ stable) At p = : f ) = So, we ca t be sure of the stability of p = from this. Do a cobweb to determie that this poit is stable whe p is ustable, etc.. a) First we have: p + = = αp + β p )p αp + β p )p + γ p ) ) α β)p + βp α β + γ)p γ β)p + γ

2 b) Fixed poits satisfy: c) Let p = p = OR = α β + γ)p ) γ β)p + γ = α β)p + β α β + γ)p ) γ β)p + γ α β)p β = α β + γ)p ) α 3β + γ)p + γ β = α β + γ)p ) α 3β + γ)p + γ β = ) α 3β + γ p ) p + γ β = α β + γ α β + γ p ) p γ β ) = α β + γ So, the equilibria poits are p = p = p 3 = γ β α β + γ α β)p ) + βp α β + γ)p ) γ β)p + γ α β)p + β α β + γ)p ) γ β)p + γ p ad p exist for all parameter values. However, p 3 eeds to be positive to exist: The, fp) = γ β α β + γ > the γ > β ad α + γ > β or γ < β ad α + γ < β α β)p + βp α β + γ)p γ β)p + γ f p) = [α β)p + β][α β + γ)p γ β)p + γ] [α β)p + βp][α β + γ)p γ β)] [α β + γ)p γ β)p + γ] For p = : For p = : = αγ + αβ + βγ)p + αγ βγ)p + βγ [α β + γ)p γ β)p + γ] f ) = β γ > whe β > γ ustable) < whe β < γ stable) f ) = β α > whe β > α ustable) < whe β < α stable)

3 For p 3 = α β+γ : ) γ β f α β + γ = αγ + αβ + βγ) [ α β + γ) α β+γ α β+γ ) ) + αγ βγ) α β+γ ) ) γ β) α β+γ + γ + βγ ] = αγ + αβ + βγ) ) α β+γ) + αγ βγ) α β+γ + βγ [ ] ) α β+γ + γ = αγ + αβ + βγ) ) α β+γ) + αγ))α β+γ) α β+γ) [ ] ) α β+γ + γ α β+γ α β+γ + βγ α β+γ) α β+γ) = αγ + αβ + βγ)γ β) + αγ βγ)γ β)α β + γ) + βγα β + γ) [ γ β) + γα β + γ)] = αγ αβ βγ)αγ beta ) αγ β ) αγ αβ βγ = αγ β For this to be stable we eed f p) < < αγ αβ βγ α < for αγ αβ βγ αγ β < Remember that for p 3 to exist, we eed γ > β AND α + γ > β OR γ < β AND α + γ < β. The, αγ βα + γ) αγ β > αγ ββ) αγ β > αγ β ) αγ β = The for γ > β which meas also meas that α > β, this poit is ustable. But for β > γ ad β > α, the this poit is stable. d) We ll start with oe of the coditios we had for p 3 to exist: γ > β AND α + γ > β. I this eviromet, the black moths are favored over the gray moths sice γ > β. However, we are also favorig the peppered moths over the gray moths sice β < α + γ < α + β so β < α Now let s look at the other coditio we could have for p 3 to exist: γ < β AND α + γ < β. I this eviromet, the grey moths are favored over the black moths sice γ < β. But they are also favored over the peppered moths sice β > α + γ > α + β so β > α We ca t tell aythig from these coditios about the selective advatages betwee the peppered ad black moths.

4 For the log term dyamics, we kow that the W allele will either disappear p = ), take over p = ), or go to a coexistece p 3. Recall the coditios for each of these to be stable. For p = to be stable we eeded γ > β. So, the W allele is lost if the black moths are favored over the grey moths. Ad, for p = to be stable we eeded α > β. So, the W allele is fixed if the peppered moths are favored over the grey moths. Both of these are true as i the case before. So, how could this happe? We have stable equilibria! Remember that we still have a ustable equilibria ibetwee the two stable oes. Mathematically this better be true!! We always eed to alterate the stability of the poits. This meas that if we start to the left of p 3, the the W allele will be lost. But, if we start to the right of p 3, the the W allele wis everythig! However, if both p ad p + are ustable ie. for β > γ & β > α), the the coexistece equilibria p 3 is stable. Therefore we get some peppered moths ad some of the other kids depedig o if γ > α or vice versa. e) We are ow give that α < β < γ with the black ww) moths at equilibrium. By our previous aalysis, p = is stable, p = is ustable ad p 3 does t exist. However, the peppered moths W W ) are makig their comeback & ow have much higher levels. What happes? Well, sice p = is stable, the the peppered moths will evetually die out agai. Poor peppered moths. f) For β < α < γ, we agai have three equilibria poits. Recall that if we re above p 3, the the peppered moths populatio will survive. Otherwise, they will be lost agai.

5 Homework 4. Solutios Math 5/683. a) with a R =.5, a J =.7, p R =., ad p J =.5: & b) with a R =.5, a J =.7, p R =.7, ad p J =.9: 4 x 9 4 x 9 & x 9 c) with a R =., a J =., p R =., ad p J =.: & d) with a R =.5, a J =.8, p R =., ad p J =.5: &

6 Matlab code: a=[.5;.7;.;.5]; b=[.5;.7;.7;.9]; c=[;;.;-.]; d=[.5;.8;.;.5]; A=[a b c d]; N=7; for i=:4 ar=a,i); aj=a,i); pr=a3,i); pj=a4,i); R=zeros,N); J=zeros,N); R)=; J)=; for =:N- R+)=ar*R)+pr*J); J+)=aj*J)+pj*R); figurei) subplot,,) plot[:n-)], R, m:.,[:n-)],j, k:h ); xlabel ); ylabel & ); leged,,); hold off; subplot,,) x=r; y=j; plotx,y, k, Liewidth,3); xlabel ); ylabel ); ed ed. a) Cobweb diagram: x x b) The poit, ) is ustable, ad the ozero equilibria is stable.

7 c) Solutio: x Matlab code for both part a) ad c): N=5; x=zeros,n); x)=.3; figure); for =:4 x+)=.8*-x))*x); axis square; hold off; plot[:n-)],x, k:d, Liewidth,); xlabel ); ylabel x_ ); ed % for the cobwebbig: t=:.:; figure) plott,.8*t.*-t)), b, Liewidth,3); hold o; axis square; fplot *y,[ ], Liewidth,3, k ); lie[x) x)],[ x)], Color, c, Liewidth,); plotx),x), kh, Liewidth,); for =:3 lie[x) x+)],[x+) x+)], Color, c, Liewidth,) lie[x+) x+)],[x+) x+)], Color, c, Liewidth,) plotx+), x+), kh, Liewidth,); xlabel x_ );ylabel x_{+} ); ed lie[x4) x4+)],[x4+) x4+)], Color, c, Liewidth,) d) The fixed poits are: x = x = Kr ) r Stability of fixed poits: let fx) = r x ) x K

8 The, f x) = r x ) rx K K = r x ) K Ad, f Kr ) r f ) ) = r = r ) r ) r = r r ) = r So, x = is stable for < r <, ad ustable otherwise. Ad, x = Kr ) r is stable for < r < 3, ad ustable otherwise. We ca also tell from this that there will be cyclig period doublig) that begis at r = 3. Your clue to kow this should have bee the fact that you ca t have ustable or stable) equilibria poits ext to each other. They always have to alterate stability ie. if oe is stable, the other is ustable, ad vice versa). Bifurcatio Diagram:

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