ANSWERS Final Exam Math 250b, Section 2 (Professor J. M. Cushing), 15 May 2008 PART 1


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1 ANSWERS Final Exam Math 50b, Section (Professor J. M. Cushing), 5 May 008 PART. (0 points) A bacterial population x grows exponentially according to the equation x 0 = rx, where r>0is the per unit rate of growth (per minute). We want to eliminate this population by killing the bacteria it at a rate h. However, it turns out we cannot sustain the treatment at a constant rate. Instead the treatment s potency decreases exponentially over time according to the formula h(t) =me at. Here m>0 is the initial (and maximal) potency and a>0 is the decay rate of the treatment s potency over time. (a) ( points) The differential equation for x is x 0 = rx me at. This follows immediately from the basic rate balance law x 0 = input rate output rate. (b) (5 points) Suppose the bacterial population has size x 0 > 0 at time t 0 =0when the treatment starts. Find a formula for the solution of this initial value problem. ³ ANSWER: x(t) = x 0 m a+r e rt + m a+r e at The general solution has the form x = x h + x p where x h = ce rt is the general solution of the associated homogeneous equation x 0 = rx. To calculate a particular solution x p we can use the Methods of Undetermined Coefficients and let x p = ke at. Then x 0 = ake at and rx me at = (rk m)e at so that we need to choose k so that ak = rk m. Thus k = m/(a + r) and x p = m a+r e at.thenx(0) = c + m a+r = x 0 so that c = x 0 m a+r. (c) ( points) The treatment succeeds if the bacterial population level x(t) equals 0 at some time t. Show that the treatment will fail if the initial treatment rate m is less than a critical value m, but will succeed if m>m.findaformulaform and show that it is proportional to x 0. ANSWER: The second term in the solution x(t) tends to 0 as t +. The first term is unbounded and tends to + as t + if m<(a + r)x 0. In this case, the treatment fails. If m>(a + r)x 0 then the first term tends to as t +. and as a result x(t) equals zero at some time t. Thus, the critical value for m is m =(a + r)x 0, which is a multiple of x 0.. (5 points) Atumor ssizex grows at a decreasing per unit rate according to the equation x 0 /x = +t. Suppose at time t 0 =0the size of the tumor is x 0 > 0. (a) (6 points) Find a formula for the solution of the initial value problem for t>0 and use it to show that the tumor size grows without bound (and is therefore presumably lethal). ANSWER: x(t) =x 0 (t +) The equation x 0 = +t x is linear homogeneous with a general solution x(t) =cep (t) where P (t) = R +t dt =ln(+t). Simplifying, we have x(t) =celn +t = c +t which equals c(t+).for t>0.the initial condition implies c = x 0.
2 (b) (6 points) Chemotherapy is applied with the result that tumor cells are removed at a constant rate r>0, so that x 0 = +tx r. Find a formula for the solution of this initial value problem. ANSWER: x(t) =(t +)(x 0 r ln (t +)) The solution is given by the Variation of Constants Formula Z t x(t) =x 0 e P (t) + e P (t) e P (s) q(s)ds where P (t) = R p(t)dt = R +t dt =ln(t +) and hence ep (t) = t +: Z t x(t) =x 0 (t +)+(t +) 0 s + ( r)ds = x 0 (t +) (t +)rln (t +) (c) ( points) Does your answer in (b) show that the chemotherapy treatment will cause the tumor to disappear or not? Explain. ANSWER: Yes. The equation predicts the tumor size reduces to 0 at time t = e x0/r We ask: does there exist a time t>0 such that x(t) =(t +)(x 0 r ln (t +)) = 0? In other words, can this equation be solved for t>0? Yes, as the following shows: x 0 r ln (t +)=0= ln (t +)=x 0 /r = t +=e x0/r = t = e x0/r. 0. (5 points) Consider the equation: x 0 = x e x. (a) ( points) Find all equilibria. ANSWER: x e = and x e x =0= x =0 or e x =0 In the first case, x = ± and in the second case x =. (b) ( points) Which equilibria are hyperbolic? Which are nonhyperbolic? ANSWER: hyperbolic: x e = nonhyperbolic: x e = f(x) = x e x = df ½ e.7 6= 0if x = e = dx 0 if x xe e = (c) ( points) Classify each equilibrium (sink, source or shunt). x e Classification Since df dx = e < 0 for x e =, this equilibrium xe sink is a sink (by the derivative test or the linearization principle). shunt x e =is a shunt because f(x) does not change sign as x passes through. Or, because d f(x)/dx = e x xe x + x e x equals and hence f(x) has a local maximum at x =.
3 (d) ( points) Draw the phase line portrait. > < < x. (0 points) Draw a bifurcation diagram for the equation x 0 = p +(x ) (x +).Locate and identify the type of all bifurcations that occur. Plot the quartic polynomial p = (x ) (x +) with roots at ± and then rotate and reflect the graph. The result appears below. There are two saddle node bifurcations at p =0and another saddlenode bifurcation at p =. x  p  PART. (5 points) The following twodimensional system of differential equations is a model for a predatorprey system, where x is the density of the prey and y thedensityofthepredator. The parameter α>0 represents addition of predators at a constant rate. dx dt = x( y), dy dt = y( +x)+α. () (a) ( points) The point P :(x, y) =( α, ) is an equilibrium (fixed point) of the above system. Find any other equilibria. ANSWER: (x, y) =(0,α) The equilibrium equations are x( y) = 0 y( +x)+α = 0. The first equation implies x =0or y =. In the first case x =0, the second equation implies y( +0)+α =0or y = α. Therefore(x, y) =(0,α) is an equilibrium. In the second case y =, the second equation implies ( +x)+α =0or x = α, which gives the equilibrium P. (b) (8 points) Let α =/ and classify the equilibrium P (e.g. saddle,...). Show all your work. ANSWER: P :(x, y) =(/, ) is a stable spiral The Jacobian y x J(x, y) = y +x
4 evaluated at the equilibrium (x, y) =(/, ) J(/, ) = µ 0 / / has characteristic polynomial is λ + λ +.Sincetherootsλ = 8 ± i i of this polynomial are complex and have negative real part 8, the phase portrait of the linearization is a stable spiral. The Linearization Principle implies that the equilibrium P is also a stable spiral. (c) (8 points) Show that when α =0(i.e. if no predators are added to the system), the linearization at the fixed point P, whose coordinates are now (x =, y =), has a center. In this case (do not try to show this), system () has closed orbits (or trajectories) around P. The Jacobian evaluated at the equilibrium (x, y) =(, ) 0 J(, ) = 0 has characteristic polynomial is λ +. Since the roots λ = ±i of this polynomial are complex with zero real part, the phase portrait of the linearization is a center. (d) (5 points) Basedon(b)above,whichofthestatementsbelowbestdescribeswhathappens when predators are added at the constant rate α =/? Circle your answer. There is no need to explain your reasoning. ANSWER: Both species survive. Since the phase portrait near the equilibrium (x, y) =(/, ) is a stable spiral, nearby orbits (x(t),y(t)), that is to say, as t both species equilibrate at positive (and hence nonzero) values.. (5 points) Find a fundamental solution matrix Φ(t) for the system ˆx 0 = Aˆx where 0 A = and use it to write down the general solution. µ e t e ANSWER: Φ(t) = t x e t e t, y = c e t +c e t c e t +c e t The characteristic polynomial: λ λ 6 has roots λ = and. Associated eigenvectors are λ = ˆv = and λ = ŵ =
5 . (0 points) Calculate the eigenvalues and eigenvectors of the matrices below and use them to identify the type of phase plane portrait associated with the linear system ˆx 0 = Aˆx. Alsosketch the phase portrait (being sure to include orientation arrows and several typical orbits, including any orbits of any relevant eigensolutions. (Show your work.) (a) ( points) A = y x 5 TYPE: stable node The roots of the characteristic µ polynomial λ +λ + of A are λ =,. Associated eigenvectors are respectively ˆv = and ŵ =. Both eigenvalues λ are negative, which implies the portrait is an stable node. Orbits approach the origin tangentially to the direction of ˆv, since this vector is associated with the eigenvalue of least magnitude. (b) ( points) A = µ 0 TYPE: Saddle The roots of the characteristic polynomial λ + λ µ of the coefficient µ matrix are λ =and. Corresponding eigenvectors are, respectively, ˆv = and ŵ =. The eigenvalues are of opposite sign so the phase portrait is a saddle. 5
6 (c) ( points) Consider the orbit associated with the solution (x(t),y(t)) of the system in (b) with initial condition (x 0,y 0 )=(, ). Sketch the graph of x(t) and the graph of y(t). (Plotboth on the same axes below.) The orbit is the hyperbolic shaped orbit in part (b) running from the th to the st quadrants. Along that orbit x decreases to nearly after which it increases to approximately. y strictly increases, from to a little more than y x t. (0 points) The differential equation for the frictionless massspring system studied in class is mx 00 + kx =0where m>0 denotes the mass and k>0denotes the spring constant. By dividing by m we can rewrite this equation as x 00 +ω 0x =0where ω 0 = p k/m. The general solution of this equation x(t) =c sin ω 0 t + c cos ω 0 t is periodic with frequency ω 0, which is called the natural frequency of the massspring system. If the system is forced with a sinusoidal motion of frequency ω, the differential equation becomes x 00 + ω 0x =sinωt. Resonance is said to occur in this system if there are unbounded solutions. (a) (8 points) Find a formula for the general solution of this equation when the natural frequency ω 0 =. (Use the Method of Undetermined Coefficients twice, once when ω 6= and once when ω =.) ½ c sin t + c ANSWER: x(t) = cos t + ω sin ωt if ω 6= c sin t + c cos t t cos kt if ω = The general solution has the form x(t) =x h (t) +x p (t) where x h (t) =c sin t + c cos t. We use the Method of Undetermined coefficients to find a particular solution of the nonhomogeneous equation of the form x p (t) =k sin ωt + k cos ωt provided ω 6= (because in that case sin ωt and cos ωt are solutions of the homogeneous equation). If ω = wetake,accordingtotherulesofthe Method of Undetermined Coefficients, x p (t) =k t sin t + k t cos t In the fist case ω 6= a calculation shows x 00 p + x p = ω k sin ωt + ω k cos ωt. Therefore, in order to solve the nonhomogeneous equation we must take ω k =and ω k =0. Because ω 6=these equations tells us that k = ω and k =0. In the second case ω =a calculation shows x 00 p + k x p =k cos t k sin t. Therefore, in order to solve the nonhomogeneous equation we must take k =0and k = /. 6
7 (b) ( points) When the natural frequency of the massspring system ω 0 =, determine for which values of the external forcing frequency ω resonance will occur and for which values of ω 0 resonance does not occur. What does this mean physically about the massspring system? (Explain your answers.) ANSWER: Resonance occurs if ω =and does not occur if ω 6=. The general solution in (a) when ω 6= is a sum of bounded functions and is therefore bounded, no matter what the values of the arbitrary constants c,c are. On the other hand, if ω =the general solution in (a) has an unbounded term (namely, t cos kt) and is therefore unbounded (no matter what the values of the arbitrary constants c,c are). Thus, if a massspring system is sinusoidally forced with an external frequency equal to the natural frequency of the system, unbounded oscillations will occur that eventually lead to a break down of the system. Other external frequency do not lead to a collapse of the system. 7
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