Math 142 { Mathematical Modeling
Math 142 – Mathematical ModelingUniversity of California, Los AngelesDuc VuFall 2021This is math 142 – Mathematical Modeling taught by Professor Huang. We meet weeklyon MWF from 9:00am – 9:50am for lecture. There is one textbook used for the class, which isMathematical Models by Haberman. You can find other lecture notes at my blog site. Pleaselet me know through my email if you spot any mathematical errors/typos.Contents1 Lec 1: Sep 24, 20211.1 Intro to Mathematical Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . .332 Lec 2: Sep 27, 20212.1 An Example of Modeling a Mass-Spring System . . . . . . . . . . . . . . . . . . . . .553 Lec 3: Sep 29, 20213.1 An Example (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3.2 Population Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7784 Lec 4: Oct 1, 20214.1 Population Dynamics (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4.2 Continuous Population Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9995 Lec 5: Oct 4, 2021115.1 Continuous and Discrete Population Models . . . . . . . . . . . . . . . . . . . . . . . 115.2 Discrete One-Species Model with an Age Distribution . . . . . . . . . . . . . . . . . 126 Lec 6: Oct 6, 2021136.1 Stable Age Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Lec 7: Oct 7, 2021157.1 Stable Age Distribution (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157.2 Logistic Equations with Phase Plane Solution . . . . . . . . . . . . . . . . . . . . . . 168 Lec 8: Oct 11, 2021178.1 Logistic Equation with Phase Plane Solution (Cont’d) . . . . . . . . . . . . . . . . . 178.2 SIR Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Lec 9: Oct 13, 2021189.1 SIR Model (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189.2 SIRS Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1910 Lec 10: Oct 15, 20212010.1 Solutions to System of Differential Equations . . . . . . . . . . . . . . . . . . . . . . 20
Duc Vu (Fall 2021)Contents11 Lec 11: Oct 18, 20212211.1 Solutions to System of Differential Equations (Cont’d) . . . . . . . . . . . . . . . . . 2212 Lec 12: Oct 22, 20212512.1 Asymptotic Properties of Solutions to Linear ODE System . . . . . . . . . . . . . . . 2513 Lec 13: Oct 25, 20212813.1 Asymptotic Properties (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2813.2 Introduction to Two-Species Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 2914 Lec 14: Oct 27, 20213014.1 Two-Species Models (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3014.2 Predator-Prey Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3015 Lec 15: Oct 29, 20213215.1 Predator-Prey Models (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3216 Lec 16: Nov 1, 20213416.1 Predator-Prey Models (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3417 Lec 17: Nov 3, 20213717.1 Cooperation Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3718 Lec 18: Nov 5, 20213818.1 Cooperation Model (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3819 Lec 19: Nov 8, 20214019.1 Cooperation Models (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4019.2 Stochastic Population Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4120 Lec 20: Nov 10, 20214320.1 Stochastic Population Growth (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . 4321 Lec 21: Nov 12, 20214521.1 Stochastic Population Growth (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . 4522 Lec 22: Nov 22, 20214822.1 Stochastic Population Growth (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . 4822.2 Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4923 Lec 23: Nov 24, 20215023.1 Flow (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5023.2 Diffusion Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5124 Lec 24: Nov 29, 20215224.1 Diffusion Equation (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5225 Lec 25: Dec 1, 20215425.1 Diffusion Equations (Cont’d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5425.2 Diffusion on a Bounded Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562
Duc Vu (Fall 2021)§11Lec 1: Sep 24, 2021Lec 1: Sep 24, 2021§1.1I nt r o t o M a t h e m a t i c a l M o d e l i n gFirst, let’s examine the following questionQuestion 1.1. Why do we learn mathematical modeling?There are lots of question that math may provide some explanation so that we could understandthe question deeply.Example 1.11. How is Covid-19 spread? How can we control the spread of Covid-19?2. How to control the spreading of the forest fire and how to reduce the loss?3. How does the population of human evolve over time?So,Question 1.2. What is a mathematical model and how can we create the model?Definition 1.2 (Mathematical Model) — A mathematical model is a description of a systemusing mathematical concepts and language. The process of developing a mathematical modelis called mathematical modeling.To create a mathematical model, we1. formulate the problem: approximations and assumptions based on experiments and observations2. solve the problem that is formulated above3. interpret the mathematical results in the context of the problemLet’s now explain the three steps above in more details.1. Formulationa) State the question: If the question is vague, then make it to be precise. If the question istoo “big”, then subdivide it into several simple and manageable parts.b) Identify factors: Decide important quantities and assign some notation to the corresponding quantity. Then, we need to determine the relationship between the quantities andrepresent each relationship with an equation.2. Solve the problem above: This may entail calculations that involve algebraic equations, someODE, PDE, etc; provide some theorems or doing some simulations, etc.3. Interpretation/Evaluation: We need to translate the mathematical result in step 2 back to thereal world situations and evaluate whether the model is good or not by asking the followingquestions:a) Has the model explained the real-world observations?b) Are the answers we found accurate enough?c) Were our assumptions good?3
Duc Vu (Fall 2021)1Lec 1: Sep 24, 2021d) What are the strengths and weaknesses of our model?e) Did we make any mistake in step 2?If the answer to any of the above question is not favorable, we need to go back to step 1 andgo through all the steps again until we get some satisfying results.4
Duc Vu (Fall 2021)§22Lec 2: Sep 27, 2021Lec 2: Sep 27, 2021§2.1An Example of Modeling a Mass-Spring SystemConsider the following questionQuestion 2.1. How does the spring-mass system move/work?Formulation:a) State the question: What formula can describe how the spring-mass system work?b) Identify factors:(a) initial position x0 (called natural length)(b) the spring constant k(c) friction fc(d) mass of the object m(e) position x(f) velocity v(g) acceleration a(h) force Fx 0x 0x05
Duc Vu (Fall 2021)2Lec 2: Sep 27, 2021Now, we try to find some relations between factors we listed above. First, let’s describe ourobservations. If we contract the spring (x 0), there is some force to push the spring outward(F 0). If we stretch the spring (x 0), there is some force that restores the initial shape ofthe spring (F 0). So, we can observe thatF ·x 0The relation between F and x can be summarized by the Hooke’s LawF kx(*)Next, let’s find the relation between the force and the movement of the object (F, m, v, a) byassuming that the movement of the object only depends on the force of the spring (not onother factors). This can be summarized by Newton’s second law of motion. d vd d xd2 xF m a m(**) m m 2dtdt dtdtBy (*) and (**), we deduceF kx md2 xdt2Mathematical analysis: we need to find the solution of the ODE:mx00 kx 0To solve the ODE, we want to find the solution to the characteristic equationrk2imλ k 0 x mThus, k c2 e t m ir !r !kk (c1 c2 ) cost (c1 c2 )i sintmmr !r !kk c3 cost c4 sintmmx(t) c1 et kmi6
Duc Vu (Fall 2021)§33Lec 3: Sep 29, 2021§3.1A n E x a m p l e ( C o nt ’ d )Recall that we haverx(t) c3 cosLet θ2 Lec 3: Sep 29, 2021qkm t.ktm!r c4 sin!ktmThen,x(t) Let sin θ1 c23c3 c24qc23 c24!c4c3pcos(θ2 ) p 2sin(θ2 )c23 c24c3 c24and cos θ1 c24c3 c24with tan θ1 c3c4or θ1 arctan c3c4. So,qc23 c24 sin(θ1 θ2 )!rqk θ1 c23 c24mx(t) Evaluation of x(t) A sin (ωt θ)xAt AFrom the figure above, we know x(t) is periodic with period T 2λω 2λpmkmin x(t) Amax x(t) A,ttwhere A is the amplitude and ωt T BASince x(t) is a periodic function, this means the spring will oscillate forever. However, in practice,it is impossible. Thus, we need to modify our model by removing or adding some assumption.Now, we may consider the case that there is friction when spring oscillates.Ff cdxdtThen,md2 xdx kx c ·dt2dt7
Duc Vu (Fall 2021)§3.23Lec 3: Sep 29, 2021Po p u l a t i o n D y n a m i c sConsider the following questionQuestion 3.1. Can we predict whether a species or its population will thrive or go extinct?In order to answer it, let’s first investigate an example.Example 3.1How many people will there be in the U.S. in the next 4 years?First let’s reformulate the question in the example to be more specific:Question 3.2. Can we build a math model to predict the number of people in the U.S. in 1,2, 3, 4 year?Assumptionthe death and birth rate are constantthe counting period (of the population) is fixedthe growth of the population only depends onthe death and birth rateFactorbirth rate: bdeath rate: dthe periodinitial population: N0the distribution of the population: N (a)migration ratethe # of years from the current time: tthe # of population at time t: N (t)the growth rate: RTo study N (t) we need to consider the relation between N (t) and N (t t)N (t t) N (t) # of new birth at [t, t t] # of death at [t, t t] N (t) (b d) t · N (t) (1 (b d) t) · N (t)Thus,N (t t) (1 R t) N (t)N (1) (1 R)N0N (2) (1 R)N (1) (1 R)2 N0N (3) (1 R)N (2) (1 R)3 N0N (4) (1 R)N (3) (1 R)4 N08
Duc Vu (Fall 2021)§44Lec 4: Oct 1, 2021Lec 4: Oct 1, 2021§4.1Po p u l a t i o n D y n a m i c s ( C o nt ’ d )Example 4.1N0 300 millions, R 0.6%, t 1N (1) (1 r)N0 (1 0.6%) · 300 300 1.8 301.8 millionsN (2) (1 r)2 N0 (1 0.6%) · 300 301.8 · 100.6%N (3) (1 R)3 N0 (1 0.6%)3 · 300N (4) (1 R)4 · N0 (1 0.6%)4 · 300Consider:N (t t) (1 R · t) · N (t)where t0 0, t1 t, t2 2 t,. . . , tn n tn N (n · t) (1 R · t)N ((nt) t) . . . (1 R t) N0We have1(1 R t) tR·Rn t1· N0 (1 R t) R tRtN01Set t 0, we obtain (1 R t) R t e. Then,N (t) eRt N0 as t 0Next, let’s analyze the property of the model above:nN (n t) (1 R t) N01. 1 R t 1, then N (n t) , as n 2. 0 1 R t 1, then N (n t) 0 as n Conclusion: When 0 1 R t 1, the model is acceptable; however, when 1 R t 1(R 0),the model should be modified. Thus, we may change our assumption: the growth rate is constant(e.g., the growth rate depends on the population itself)§4.2C o nt i nu o u s Po p u l a t i o n M o d e lHave:N (t) eRt N0Let’s start from the previous lectureN (t t) N (t) R t · N (t)9
Duc Vu (Fall 2021)4Lec 4: Oct 1, 2021SoN (t t) N (t) R · N (t) tN (t t) N (t)lim lim R · N (t) R · N (t) t 0 t 0 tdN (t) R · N (t)Z dtZdN (t) RdtN (t)ln(N (t)) Rt CN (t) eC eRt N0 eRtEvaluate the continuous model N (t) eRt N01. 0 R 1: N (t) as t and N (t) as t 2. 1 R 0: N (t) 0 as t and N (t) as t Conclusion: When R 0, the model is acceptable; however, when the growth rate R 0,theindividuals (of a species) will compete each other as the resource is limited, N (t) as t .Now, let’s consider the density-dependent growth. Assumption: The growth rate is density dependent, i.e., R(t) R (N (t)) If the population is small, then the influence of the environment is small, then we hope thatthe population has exponential growth. As N (t) gets large enough, we don’t expect the growth of N (t). In other word, the growthrate R (N (t)) 0 when N (t) is large enough (since R(t) is usually assume to be smooth,R (N (t)) 0 when N (t) is large enough)dN R (N (t)) · N (t)dtFrom our assumption, R (N (t)) should be a constant when N (t) is small and R (N (t)) 0 as N (t)is large enough. So we can consider R (N (t)) of the formR (N (t)) a bN (t)Thus, the model becomesdN (a bN )NdtThis is known as the logistic model.Remark 4.2. The discrete-time population model is called Beverton-Holt model.(R0 (N (t 1)· t)N (t · t) 1 N((t 1) t)/MR(N ) R01 N ((t 1)· t)/M10
Duc Vu (Fall 2021)§55Lec 5: Oct 4, 2021Lec 5: Oct 4, 2021§5.1C o nt i nu o u s a n d D i s c r e t e Po p u l a t i o n M o d e l sRecall the continuous logistic population modeldN N (a bN )dtLet’s manipulate thisdN dtN (a bN )ZZb1 dN dtaNa(a bN )11ln N ln a bN t caaN at c̃lna bNN eat c̃ Ceata bNaN b Ce atSince N (0) N0 N0 ab C ,we haveaN (t) b b e ataN0Let’s now consider the relation between continuous logistic population and discrete-time logisticmodel for t 1. For the discrete case,(R0 N (t 1)N (t) 1 N(t 1)/MR (N (t)) R01 N (t 1)/MFor the continuous case,aN (t) b aN0 b e atThen,aN (t 1) b aN0 b e at eaNotice that ba1 b eat /aN (t)aN0 1baaat ae · b e e /a · eaN (t 1)aN01eabb eaN (t) N (t 1)a aFor the continuous model, as t , we can see that N (t) 11abwhich is a good model.
Duc Vu (Fall 2021)§5.25Lec 5: Oct 4, 2021Discrete One-Species Model with an Age DistributionMotivation: The birth and death rates will vary a lot if state A has more young citizens than stateB.Let’s consider the period t 1 year, define variables for a population at each ageN0 (t) # individuals whose age 1N1 (t) # of individuals one year oldN2 (t) # of individuals two years old.NM (t) # of individuals M years oldwhere M is the oldest age with proper population. Supposebm birth rate for a population that is m years olddm death rate for a population that is m years oldLet’s consider the population Nm (t 1)N0 (t 1) b0 N0 (t) b1 N1 (t) . . . bM NM (t)N1 (t 1) N0 (t) d0 N0 (t) (1 d0 )N0 (t)N2 (t 1) N1 (t) d1 N1 (t) (1 d1 )N1 (t).NM (t 1) NM 1 (t) dM 1 NM 1 (t) (1 dM 1 ) NM 1 (t)In matrix notation, N0 (t) N1 (t) N (t) N2 (t) . . NM (t) Then, b0N0 (t 1) N1 (t 1) 1 d0 0 . . .NM (t 1)0 b101 d1.1 dM 1 (t 1) LN (t) – the matrix is called Leslie matrix. N12 bM N0 (t)0 N1 (t) 0 . . . . NM (t)0
Duc Vu (Fall 2021)§6§6.16Lec 6: Oct 6, 2021Lec 6: Oct 6, 2021Stable Age DistributionDefinition 6.1 (Stable Age Distribution) — A stable age distribution exists if the populations1 (t) vapproach an age distribution that is independent of time as time increases, i.e., kN (t)kN1as t whereMX kN (t)k1 Ni (t) i 0Assume that the Leslie matrix 2L 0.44and 10 100 N (0) 100Let’s track the evolution of the population age groups. We have (t 1) L · N (t)N 1 100300 (1) LN (0) 2N 0.44 0 10044 1 300644 (2) LN (1) 2N 0.44 0 44132Continue this process we obtain (3) 1420 ,N2834 3123.4,.624.8Observation: The population appears to grow over time without bound.a The ratioN1 (t 1)N1 (t)300N0 (2)644N0 (1) 3 2.1467N0 (0)100N0 (1)300N0 (3)1420N0 (4) 2.2050 2.1996N0 (2)300N0 (3)Apply the same process to N1 and we can notice that they both approach 2.2, i.e., N0 (t 1)N (t) 2.2 0N1 (t 1)N1 (t)The fraction of the population in age 0 and fraction of the population in age 0 is 1.1001N0 (1)300N0 (0) 0.872N0 (0) N1 (0)100 1002N0 (1) N1 (1)344N0 (2)N0 (3) 0.8407 0.8336 . . .N0 (2) N1 (2)N0 (3) N1 (3)With these calculations, we can see thatN0 (t)N1 (t) 0.833 0.167N0 (t) N1 (t)N0 (t) N1 (t)13N0 (t 1)N0 (t)and
Duc Vu (Fall 2021)So6Lec 6: Oct 6, 2021 N0 (t)0.833 0.167 (t)k N1 (t)kN1Recall that N0 (t 1)N0 (t)N0 (t) L 2.2N1 (t 1)N1 (t)N1 (t)Claim 6.1. 2.2 is one eigenvalue of the Leslie matrix L. 0.833is an eigenvector of the Leslie matrix L. Let’s check.Guess:0.167 det(L λI) det20.44 1λ 000λ (2 λ)( λ) 0.44 (λ 2.2)(λ 0.2)Thus, λ 2.2, λ 0.2 which verifies our claim. When λ 2.2, we can find the correspondingeigenvector as follows 212.2 0L 2.2I 0.44 00 2.2 0.21 0.44 2.2We need to find the null space of L 2.2I, i.e. 0.21x10 0.44 2.2 x20which is"5# x15x26 6x2 1x2x26 5 0.8336Thus, 1 is the corresponding eigenvector (of 2.2).0.1676From this example, we may guess in order to find the stable age distribution, we need to find themaximum eigenvalue of the Leslie matrix and then find the corresponding normalized eigenvector.Now, we will try to check our guess for the general Leslie model. (t t) LN (t)Nwith N0 (t) N1 (t) (t) N . . M (t)Nand L R(M 1) (M 1) (0) N 0 , then we have N (n · t) LN ((n 1) · t) being a non-negative. Let’s assume that N 0 . Suppose that the Leslie matrix L is diagonalizable, i.e., there are M 1 eigenvalues. . . Ln · Nλ1 λ2 . . . λM 1 and M 1 linearly independent eigenvectors v1 , . . . , vM 1 .14
Duc Vu (Fall 2021)§7§7.17Lec 7: Oct 7, 2021Lec 7: Oct 7, 2021S t a b l e A g e D i s t r i b u t i o n ( C o nt ’ d ) ((n 1) · t) . . . Ln N 0 . Suppose that (0) N 0 , then we have N (n · t) LNAssume that Nthe Leslie matrix L is diagonalizable, i.e., there are M 1 eigenvalues λ1 . . . λM 1 and M 1linearly indep. eigenvectors v1 , . . . , vM 1 .L V DV 1where λ1 D λ2. , . V v1. . . vM 1 λM 1Since v1 , v2 , . . . , vM 1 are linearly independent, { v1 , . . . , vM 1 } is a basis for RM 1 . Then, thereexists c1 , c2 , . . . , cM 1 s.t.M 1X N0 ci vii 1Thus, (n · t) Ln N 0NM 1Xn L!ci vii 1 M 1Xci (Ln vi )i 1 M 1Xci λni vii 1 c1 v1 M 1X cii 2If λ1 λi for i 2, then λi λ1 λiλ1 n vi 1 which meansλiλ1n 0 as n for i 2Therefore, we have nM 1Xλi1 N(n· t) c v c vi c1 v11 1inλ1λ1i 2 (n · t) by c1 λn v1 .as n . Thus, for large value of n, we can approximate N1The process to find “stable age distribution”:1. Find the maximum eigenvalue of the Leslie matrix Ldet(L λI) 02. λ1 λi 3. Find one corresponding eigenvector vi associated to λ14. Normalize v1 : v1k v1 k15
Duc Vu (Fall 2021)§7.27Lec 7: Oct 7, 2021Logistic Equations with Phase Plane SolutionDefinition 7.1 (Phase Plane) — A phase plane is a visual display of certain characteristics ofcertain kinds of differential equations. A coordinate plane with axes being the values of twovariables.Logistic Equation:dN N · (a bN )dtNotice that this is an autonomous differential equation. One important thing for autonomous DE isthe stability of the equilibrium points.N (a bN ) 0 N 0,N abWe can observe that the equilibrium point N (t) ab is stable and N (t) 0 is unstable. Now, let’sshow the stability of equilibrium points from an analytical aspect. We will first analyze the solutionin the neighborhood of N ab . Let’s consider the Taylor’s expansion of f (N ) N (a bN ) atN ab .f (N ) N · (a bN ) a d ff (N ) abdNN ab (N b ) a 0 ( a) N ( b) N b a a · N bd2 f (N )dN 2 a 2b1a 2N ab · 2 (N b )Therefore, dNa N · (a bN ) ( a) N dtbanear the neighborhood of N b . dNa a N dtbLet y N ab dydt dNdtdy ay y Ce atdtaN Ce atbaN (t) Ce atbas t , we have N (t) ab . Thus, N (t) abis stable.16
Duc Vu (Fall 2021)§8§8.18Lec 8: Oct 11, 2021Lec 8: Oct 11, 2021L o g i s t i c E q u a t i o n w i t h P h a s e P l a n e S o l u t i o n ( C o nt ’ d )We’d like to illustrate N (t) ab ε · N1 (t) by assuming thatabis stable from perturbation analysis point of view. Let N (t) ab εN1 (t) into the original DE: εN1 (t) Let’s substitute N (t) abdN N (a bN )dt d ad εN1 (t) ε N1 (t)dt b dta εN1 (t) (a (a εbN1 (t)) ba εbN1 (t) ε2 bN12 (t)b aεN1 (t) ε2 bN12 (t)dN1 (t) aN1 (t) εbN12 (t)dt aN1 (t)Thus, N1 (t) Ce at 0 as t and N (t) §8.2abas t . So, N (t) abis stable.SIR ModelThe SIR model was first used by Kermack and McKendrick in 1947. Now this model is popularlyused to study the spread of infectious disease such as measles, Covid 19, etc. It consists of threeparts: S: the number of susceptible individuals I: the number of infected individuals R: the number of recovered individualsThe process of the spread of the infectious disease is at the beginning where all the individuals aresusceptible. The some of them become infectious and then become recovered individuals.SIRWe assume that the total populationN S I Ris fixed. Let β be the contact rate (individuals who come into contact with each other). Let γ bethe recovery rate for the infected individuals.17
Duc Vu (Fall 2021)§99Lec 9: Oct 13, 2021Lec 9: Oct 13, 2021§9.1S I R M o d e l ( C o nt ’ d )SIR model without vital dynamics We assume that the course of the infection is short. The birth and death can be ignored. The total number N can be treated as a constant.Observation: The more interactions between the people in S and I the more individuals in S will“transfer” to I.dS β · S · I/N(1)dtThe change of I will involve two parts: S I which will increase I, and I R which will decrees IdI β · S · I/n γ · IdtdR γIdt(2)(3)Let’s combine the three equations. βSIdS dt NβSIdIdt N γI dRdt γIwith S I R N being a constant. Thus, to understand the model, we only need to understand(βSIdSdt NβSIdIdt N γILet’s normalize S, I, R first by settings ds1 dtNdi1 dI dtN dtIRS, i , r NnN dS1 βSI βsidtNN 1 βSI γI βsi γiNNand we know r 1 i s.Remark 9.1. s [0, 1], i [0, 1], r [0, 1].Next, let’s analyze the new model(dsdtdidt βsi βsi γi (βs γ)iObserve that1.dsdt βsi 0 s 2.didtdidt (βs γ)i 0 i 0, s 0, s βγ .γβ.When18didt 0, we know that s γβ.Similarly, when
Duc Vu (Fall 2021)9Lec 9: Oct 13, 2021Let’s draw the graph for s, i, r together.sr1itSIR Model with Vital Dynamics:For this model, the disease will last for a long period. It is not reasonable to ignore the birth anddeath rate. It is not a reasonable assumption that S I R N where N is a constant. For thiscase, let’s introduce new parameters birth rate b and death rate d. βSIdS bN dSdtNdIβSI γI dIdtNdR γI dRdt§9.2SIRS ModelSIRS Model without Vital Dynamics:SIRSIRS βSIdS dt N αRβSIdIdt N γI dRdt γI αRand S I R N fixedSIRS with Vital Dynamics: Similar to SIR with vital dynamics, we need to take the birth anddeath rate into account. βSIdS dt N αR bN dSβSIdIand N (t) S I R not fixeddt N γI dI dRdt γI αR dRIntro to Two-Species Models: There are several different relations: competition, predator and prey,symbiosis, mutualism.19
Duc Vu (Fall 2021)§1010Lec 10: Oct 15, 2021Lec 10: Oct 15, 2021§10.1S o l u t i o n s t o S y s t e m o f D i f f e r e nt i a l E q u a t i o n sTheorem 10.1If (λ, v ) is an eigen pair of M , then eλt v is a solution ofd y (t)dt M y (t).Proof. Set y (t) eλt v . Then we havedd λt d y (t) e v ( eλt ) v λeλt vdtdtdt(1)andM y (t) M eλt v eλt M v eλt (λ v ) λeλt v(2)Combining (1) and (2) we have y (t) eλt v is a solution ofFrom the above theorem, we could find n solutions eλ1 tdy (t)dt v1 ,. . . , e M y (t).λn t vn .Question 10.1. Are these n solutions linearly independent?PnIf i 1 ci vi 0 where ci 0 in which i 1, . . . , n, then v1 , . . . , vn are linearly independent.PnKnow:vi 0 and M vi λi vi . We want to show ci 0 for all i. Let’s use mathematicali 1 ci induction to show this. When n 1, c1 v1 0 c1 0 because v1 6 0 Assume that the statement is correct when n k. We want to show now that the statement also applies for the case n k 1. Havek 1Xci M vi i 1k 1Xci λi vi 0(3)i 1Idea: get rid of one term so that we could use the induction assumption.k 1Xci vi 0 i 1k 1Xci λk 1 vi 0i 1So (3) (4),kXci (λi λk 1 ) vi 0i 1ci (λi λk 1 ) 0Thus, ci 0 since λi are distinct.k 1Xci vi ck 1 vk 1 0 ck 1 0i 1Thus, the statement is true for n k 1.20(4)
Duc Vu (Fall 2021)10Lec 10: Oct 15, 2021Theorem 10.2Ifhas n distinct eigenvalues λ1 , . . . , λn with the corresponding eigenvectors v1 , . . . , vn then Meλ1 t v1 , . . . , eλn t vn are linearly independent.Proof. Left as exercise.Example 10.3Solve the following ODE:(dxdtdydt 2x 3y x 2yLet’s rewrite the ODE into the matrix vector form. x(t)2 Y (t) , M y(t)1 3 2Now, let’s find the eigenvalues and the corresponding eigenvectors of M . 2 λ 3det (M λI) det1 2 λ λ2 1 0So, λ1,2 1. For λ1 1, 3 31 1 x y 1 v1 1(M I) v1 3 For λ 1, using the same process we obtain v2 1Therefore, t 1t 3 Y (t) c1 e c2 e11is the general solution for (t)dYdt (t). MY21
Duc Vu (Fall 2021)§1111Lec 11: Oct 18, 2021Lec 11: Oct 18, 2021§11.1S o l u t i o n s t o S y s t e m o f D i f f e r e nt i a l E q u a t i o n s( C o nt ’ d )Example 11.1 (Cont’d of the last example from last lecture)Suppose that the initial conditions are x(0) 8 and y(0) 4. Find the explicit solution forthe DE. Recall t 1t 3 Y (t) c1 e c2 e11is the general solution. So, 138c1 e 0 c2 e0 114 1 3 c18 1 1 c24 1 c11 382 c21 142Question 11.1. If there are some complex eigenvalues for the real matrix M , how can we find the (t)?general real solutions for dYdt(t) M YExample 11.2Find the real solution for the ODE(Notice thatdxdtdydt x(t) y(t) x(t) y(t) (t) x(t) ,Yy(t)M 11 11First, let’s find the eigenvalues and their corresponding eigenvectors of M . 1 1λ 0det (M λI) det 1 10 λ λ2 2λ 2 0So, λ 1 i. iis a corresponding eigenvector.1 i For λ 1 i, we haveis a corresponding eigenvector.1 For λ 1 i, we haveThus, (1 i)t i(1 i)t i Y (t) c1 e c2 e11is the general solution for (t)dYdt (t). MY22
Duc Vu (Fall 2021)11Lec 11: Oct 18, 2021Question 11.2. How do we transform the general solution to general real solution?Recall thateai cos(a) i sin(a),a RSo, (1 i)t i(1 i)t i Y (t) c1 e c2 e11 i i c1 et eti c2 et e ti11 i itt c1 e (cos(t) i sin(t)) c2 e (cos( t) i sin( t))11 t (cos(t) i sin(t)) it (cos( t) i sin( t)) ( i) c1 e c2 ecos(t) i sin(t)cos( t) i sin( t) t sin(t) cos(t)it sin(t) cos(t)i c1 e c2 ecos(t) sin(t)icos(t) sin(t)i t sin(t)t cos(t) (c1 c2 )e (c1 c2 )iecos(t)sin(t)Because c1 and c2 are arbitrary numbers we could choose c1 c2 1 and c1 c2 0 or c1 c2 0and (c1 c2 )i 1. t sin(t)t cos(t)e, ecos(t)sin(t) (t). The general real solutions can beare two linearly independent real solutions of dYdt(t) M Yrepresented by (t) c 1 et sin(t) c 2 et cos(t)Ycos(t)sin(t)where c 1 , c 2 R.Method II: Exponential MethodWhen n 1, we have ODEdx mx x(t) emt x0dtis the solution ofdxdt mx. Recall thatemt eM t X(mt)jj 0 Xj 0j! X tj M j(M t)j j!j!j 123
Duc Vu (Fall 2021)To get a clearer look at e11MtLec 11: Oct 18, 2021 2, let’s consider the case that M is diagonal, e.g., M 0eM t jXt Mjj! jj 2 0 tX0 3 j!j 0 j 0j 2 tX0 3j j!j 0"P (2t)jj 0j 0 P 0 2te0j 00e3t#0j!(3t)jj! If M is diagonalizable, how can we compute eM t ?M JDJ 1 jXt MjeM t j!j 0 jXt (JDJ 1 )jj 0 jXj!t JDj J 1j!j 0 j jXtD J 1 J j!j 0 JeDt J 124 0.3
Duc Vu (Fall 2021)§1212Lec 12: Oct 22, 2021Lec 12: Oct 22, 2021§12.1Asymptotic Properties of Solutions to Linear ODESystemConsider:(dxdtdydt ax by cx dyThen, aM c b,d x(t) Y (t) y(t)So, a λdet (M λI) detcbd λ (a λ)(d λ) bc λ2 (a d)λ ad bcSet p a d, q ad bc. Then,det (M λI) λ2 pλ q 0 p2 4qThus the eigenvalues distribution of the matrix M are as follows1. 0, the eigenvalues are real and distinct (node or saddle)2. 0, repeated real eigenvalues (improper node)3. 0, the eigenvalues are complex (spiral)First, let’s consider the case where we have two real roots: 0.a) positive real roots p 0, q 0 (t) c1 eλ1 t v1 c2 eλ2 t v2YSince λ1 , λ2 0 eλ1 t 25
Duc Vu (Fall 2021)12Lec 12: Oct 22, 2021Example 12.1ConsiderM 41 12thendet(M λI) λ2 6λ 7 0 λ 3 2 0 (t) c1 e(3 2)t 1 1Y c2 e(3 2)t2 1 2 1b) Two negative real solutions: p 0, q 0. (t) c1 eλ1 t v1 c2 eλ2 t v2YSince λ1 , λ2 0 eλ1 t 0, eλ2 t 0 as t . So the equilibrium solution is stable.c) λ1 0 and λ2 0 and so q 0 (t) c1 eλ1 t v1 c2 eλ2 t v2YSince λ1 0 eλ1 t 0 as t and λ2 0 eλ2 t as t . v2 v1d) One root is 0: q 0 and another root is positive: p 0. Let’s assume that λ1 0, λ2 0 (t) c1 v1 c2 eλ2 t v2Y26
Duc Vu (Fall 2021)12 v1 v2e) One root is 0: q 0, and another root is negative: p 0 (t) c1 v1 c2 eλ2 t v2Y v1 v227Lec 12: Oct 22, 2021
Duc Vu (Fall 2021)§1313Lec 13: Oct 25, 2021Lec 13: Oct 25, 2021§13.1A s y m p t o t i c P r o p e r t i e s ( C o nt ’ d )2. Real and equal: 0 (t) (c1 c2 t)eλt a) Both are positive, p 0, the equilibriu
This is math 142 { Mathematical Modeling taught by Professor Huang. We meet weekly on MWF from 9:00am { 9:50am for lecture. There is one textbook used for the class, which is Mathematical Models by Haberman. You can nd other lecture notes at myblog site. Please let me know through myemailif you spot any mathematical errors/typos. Contents
Aunty Acid Breaks the Internet, 19 Aunty Acid Laugh ’til You Leak!, 142 Aunty Acid’s Getting Older, 19, 142 Aunty Acid’s Guide to Life, 142 Aunty Acid’s Guide to Love, 142 Aunty Acid’s Office Manual, 142 Aunty Acid With Age Comes Wisdom, 142 Autograph Book of L.A., The, 151 A
So, I say mathematical modeling is a way of life. Keyword: Mathematical modelling, Mathematical thinking style, Applied 1. Introduction: Applied Mathematical modeling welcomes contributions on research related to the mathematical modeling of e
2.1 Mathematical modeling In mathematical modeling, students elicit a mathematical solution for a problem that is formulated in mathematical terms but is embedded within meaningful, real-world context (Damlamian et al., 2013). Mathematical model
Math Modeling Process- Modified from the SIAM- Moody's Mega Math Challenge website) 1. Pose the Problem Statement: Is it real-world and does it require math modeling? What mathematical questions come to mind? 2. Make Assumptions, Define, and Simplify: What assumptions do you make? What are the constraints that help you define and simplify the .
Math 5/4, Math 6/5, Math 7/6, Math 8/7, and Algebra 1/2 Math 5/4, Math 6/5, Math 7/6, Math 8/7, and Algebra ½ form a series of courses to move students from primary grades to algebra. Each course contains a series of daily lessons covering all areas of general math. Each lesson
Explanation of Transition to IFRS 137 Independent Auditors'Report to the Members of AstraZeneca PLC (Company) 139 Company Balance Sheet 140 Accounting Policies (Company) 141 Notes to the Financial Statements (Company) 142 1. Fixed asset investments 142 2. Other debtors 142 3. Non-trade creditors 142 4. Loans 142 5. Reserves 143 6 .
14 D Unit 5.1 Geometric Relationships - Forms and Shapes 15 C Unit 6.4 Modeling - Mathematical 16 B Unit 6.5 Modeling - Computer 17 A Unit 6.1 Modeling - Conceptual 18 D Unit 6.5 Modeling - Computer 19 C Unit 6.5 Modeling - Computer 20 B Unit 6.1 Modeling - Conceptual 21 D Unit 6.3 Modeling - Physical 22 A Unit 6.5 Modeling - Computer
Software development is a source of security vulnerabilities. Software-developing organizations therefore need to pay at-tention to security and apply secure development practices. However, managing software development is a challenge in itself even without the added complexity of security work. Agile methodologies like Scrum are commonly .
