MA1506 Tutorial 11 Solutions including question6

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MA1506 Tutorial 11 Solutions• ‚1 31. Tr =¡1;det=¡5)SADDLE1 ¡2• ‚2 ¡2 2Tr =2, det=8, T ¡4det=¡28<0)SPIRAL SOURCEr4 0• ‚¡2 ¡4 2Tr =¡2, det=40, T ¡4det=¡36<0) SPIRAL SINKr10 0• ‚¡5 4 2Tr =¡4, det=3, T ¡4det=4>0) NODAL SINKr¡2 1• ‚5 ¡4 2Tr =4, det=3, T ¡4det=4>0) NODAL SOURCEr2 ¡1• ‚0 1 2Tr =0, det=10, T ¡4det=¡40<0) CENTREr¡10 02. Let E(t) be the number of elves, D(t) the number of Dwarves. Let B ;D be theE Ebirth and death rates per capita for Elves, similarly B ;D for Dwarves. We areD Dtold that B > B and D < D . So B ¡D > B ¡D > 0 (more birthsE D E D E E D Dthan deaths).dENow is controlled by (B ¡D )E in the usual Malthus model, but here we areE EdtdEtold that is reduced by the presence of Dwarves, so we proposedtdE=(B ¡D )E¡P DE E Edtwhere P is a constant that measures the prejudice of the elves. SimilarlyEdD=(B ¡D )D¡P ED D DdtWhere P represents the prejudice of the Dwarves. SoD0 1dE • ‚µ ¶dt (B ¡D ) ¡P EE E E@ A=¡P (B ¡D ) DD D DdDdtSo in a concrete case, the constants in the matrix should satisfy B ¡ D >E E• ‚5 ¡4B ¡D > 0, and we are told that P > P . Indeed satisfles all ofD D E D ¡1 22these. Tr = 7, det=6, T ¡4det = 49¡24 = 25 > 0 so we have a nodal source.rEigenvalues(5¡‚)(2¡‚)¡4=0)‚=1 or 6:1µ ¶ µ ¶1 1Eigenvectors are ; .1 ¡1=4So the phase plane diagram(we only care about theflrst quadrant, sinceD‚0;E‚0) isbisectedbythelineD =E. AllpointsABOVEthatlinewillmovealongtrajectoriesthat eventually hit the D axis (that is, E ...

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• Publié par: Intae
• Nombre de lectures: 123
• Langue: English

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MA1506Tutorial 11Solutions · ¸ 1 3 1.T r=−1,det =−5⇒SADDLE 1−2 · ¸ 2−2 2 T r= 2, det = 8,T−4 det=−28<0⇒SPIRAL SOURCE r 4 0 · ¸ −2−4 2 T r=−2, det = 40,T−=4 det−36<0⇒SPIRAL SINK r 10 0 · ¸ −5 4 2 T r=−4, det = 3,T−4 det= 4>0⇒NODAL SINK r −2 1 · ¸ 5−4 2 T r= 4, det = 3,T−= 44 det>0⇒NODAL SOURCE r 2−1 · ¸ 0 1 2 T r= 0, det = 10,T−=4 det−40<0⇒CENTRE r −10 0

2. LetE(t) be the number of elves,D(t) the number of Dwarves.LetB E , D E be the birth and death rates per capita for Elves, similarlyB D , D D for Dwarves.We are told thatB E > B D andD E < D D . SoB E −D E > B D −D D >0 (more births than deaths). dE Now iscontrolled by (B E −D E )Ein the usual Malthus model, but here we are dt dE told thatis reduced by the presence of Dwarves, so we propose dt dE = (B E −D E )E−P E D dt whereP E Similarlyis a constant that measures the prejudice of the elves. dD = (B D −D D )D−P D E dt WhereP D Sorepresents the prejudice of the Dwarves.   dE · ¸µ ¶ dt (B E −D E )−P E E   = dD −P D (B D −D D )D dt So in a concrete case, the constants in the matrix should satisfyB E −D E > · ¸ 5−4 B D −D D >0, and we are told thatP E > P D satisfies all of. Indeed −1 2 2 these.T r= 7, det=6,T−= 494 det−24 = 25>0 so we have a nodal source. r Eigenvalues (5−λ)(2−λ)−4 = 0⇒λor 6= 1. 1

µ ¶µ ¶ 1 1 Eigenvectors are,. 1−1/4 So the phase plane diagram (we only care about the first quadrant, since D≥0, E≥0) is

bisected by the lineD=E. Allpoints ABOVE that line will move along trajectories that eventually hit theDaxis (that is,ESo if at any time= 0).D > E, then the Elf population may increase for a while, but eventually it will reach a maximum and then collapse to zero.Rivendell is completely taken over by Dwarves, EVEN THOUGHB E > B D ANDD E < D D . Theprejudice of the Elves cancels out their other advantages and causes them to lose the competition.

3. LetS(t) be the number of Spartans,P(t) the number of Persians.We have dS =−P dt dP =−11,111,111.1S dt   dS · ¸· ¸ dt 0−1 S   so = dP −11,111,111. 1 0P dt 2 Eigenvaluesλ−11,111,111.1 = 0⇒λ=±3333.3333 · ¸· ¸ ±3333.3333−1 1 Eigenvectors =0 −11,111,111.1±3333.3333α ⇒α=±3333.3333.

Since we are interested only in the first quadrant, take +.We have a SADDLE.

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The trajectory that passes through the origin is a straight line parallel to the eigenvector µ ¶ 1 . Thatstraight 3333.3333 line has equationP= 3333.3333S. We want to start at the point on this straight line that hasPSo the= 1000000. 1000000 initial value ofShas to be= 300. 3333.3333 So Leonidas should take 299 soldiers, assuming that he intends to fight (and die) himself.

4. (1)Use the rule: Rate of change of the amount ofU F 6 = (concentration in) (flow rate in) - (concentra- tion out) (flow rate out), where “concentration” is the mass of uranium hexafluoride per unit volume of water−you can understand this equation by looking at the units: lbs lbsgallons =× sec gallonssec Letx A be the mass ofU F 6 in the first tank, so the concentration in that tank is x A /let100. Similarlyx B be the mass ofU F 6 Then from thein the second tank. given data we have 2 6 x˙ A = x B −x A 100 100 and 6 6 x˙ B =x A −x B . 100 100 Notice that the 4 gallons/min of pure water flowing into the first tank does not appear here−it contains noU F 6 ; it is just there so that the amounts of water in the tanks remain constant.The initial conditions arex A (0) = 25,x B (0) = 0. In matrix form, we have · ¸· ¸· ¸ 1 x˙ A − 6 2x A =. ˙x B 100 6−6x B √ −36 + 2 The eigenvalues of the coefficient matrix can be found asλ 1 = and 100 √ · ¸ · ¸ −6−2 3 1 1 √ √ λ 2 = .and . The corresponding eigenvectors are 100 3−3 Hence · ¸· ¸· ¸ x A λ 1 t 1 λ 2 t 1 √ √ =c 1 e+c 2 e . x B 3−3 Substituting the initial values, we havec 1 = 12.5,c 2 = 12.5, so we getx A (t) = √ √ λ 1 t λ 2 t λ 1 t λ 2 t 12.5(e+e),x B (t) = 12.5( 3 e−3e). 3

b) Inthe java applet,xcorresponds tox A andytox B . Youhave to click on a point on thexaxis because we are told thatx B [that is,y] is zero initially.You should see the red curve going steadily down, but the green curve goes up at first and then down. √ 25 √ c) Yes.The two curves intersect whent 0 =ln3)(2 +≈19, (which is in fact 3 the maximum point ofx B ). Beforet 0 , the uranium hexafluoride in tankAis more than that in tankB; after that, it’s always less. 2 d) HereTrace =−12/100 and det = 24/100, so Trace−4 det = (144-96)/10000 which is positive; hence we have a nodal sink.That is also clear from the java applet.

dL 5. =B L L−D L L dt (B L =birth rate per capita,D L =death rate per capita.) dL1 ButB L =uZso =uZL−D L L. (Unitsofu= .) dt year Similarly dZ =B Z Z−D Z Z dt =B Z Z−sLZ dL dZ At equilibrium,= =0, so dt dt D L uZ−DL= 0⇒Z= u B Z B Z −sL= 0⇒L= s 0.05 0.20 ⇒L12= =.5,Z= =250. 0.004 0.0008 SubstituteL= 12.5 +x,Z= 250 +y, get dL dx = =0.0008(250 +y)(12.5 +x)−0.2(12.5 +x) dt dt = 0.01y+ 0.0008xy ≈0.01y

Hence

dZ dy = =0.05(250 +y)−0.004(12.5 +x)(250 +y) dt dt =−x−0.004xy≈ −x   dx · ¸µ ¶ dt 0 0.01x   ≈ dy −1 0y dt T r= 0,det = 0.01 4

2 T r−4 det<0⇒centre. So near equilibrium, the Lion and Zebra populations should fluctuate up and down but never vary much from their equilibrium values.The equilibrium is stable.

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Question 6 From the last section of Chapter 7, we know that there is a line in the M - G plane which separates victory from defeat.The discussion there shows that if the orcs get no reinforcements, then this line [originally M = (2/3)G] gets moved vertically.We need to move this line up so that it passes through the point (15, 11), since that corresponds to the marginal situation in which the Gondor army is destroyed; any reinforcement rate above this will result in victory for Gondor, since the initial point will then lie below this line.[We measure numbers of soldiers in thousands; recall that initially there are 15 thousand men and 11 thousand orcs.]So we are interested in a line of the form M−A= (2/3)G, where A is a constant to be determined by the fact that the line has to pass through (15,11). ClearlyA = 1, so [again from the last part of Chapter 7] we have  ! −1 0 ~ −B F=, 1 ~ whereFClearlyis the vector which gives the number of reinforcements.  ! ! ! ! 0−1−3/4 03/4 ~ F=−B=−=, 1−1 01 0 so the minimum number of reinforcements needed is 751 per day [since in fact 750 would lead to defeat for Gondor!].