EXERCISES FOR SECTION 3.1

Transcription

1 174 CHAPTER 3 LINEAR SYSTEMS EXERCISES FOR SECTION 31 1 Since a > 0, Paul s making a pro t > 0 has a bene cial effect on Paul s pro ts in the future because the a term makes a positive contribution to d/ However, since b < 0, Bob s making apro t > 0 hinders Paul s abilit to make pro t because the b term contributes negativel to d/ Roughl speaking, business is good for Paul if his store is pro table and Bob s is not In fact, since d/,paul spro ts will increase whenever his store is more pro table than Bob s Even though d/ d/ for this choice of parameters, the interpretation of the equation is eactl the opposite from Bob s point of view Since d < 0, Bob s future pro ts are hurt whenever he is pro table because d < 0 But Bob s pro ts are helped whenever Paul is pro table since c > 0 Once again, since d/, Bob s pro ts will increase whenever Paul s store is more pro table than his Finall, note that both and change b identical amounts since d/ and d/ are alwas equal Since a, Paul s making a pro t > 0 has a bene cial effect on Paul s future pro ts because the a term makes a positive contribution to d/ However, since b 1, Bob s making a pro t > 0 hinders Paul s abilit to make pro t because the b term contributes negativel to d/ In some sense, Paul s pro tabilit has twice the impact on his pro ts as does Bob s pro tabilit For eample, Paul s pro ts will increase whenever his pro ts are at least one-half of Bob s pro ts since d/ Since c d 0, d/ 0 Consequentl, Bob s pro ts are not affected b the pro tabilit of either store, and hence his pro ts are constant in this model 3 Since a 1andb 0, we have d/ Hence, if Paul is making a pro t > 0, then those pro ts will increase since d/ is positive However, Bob s pro ts have no effect on Paul s pro ts Note that d/ is the standard eponential growth model Since c andd 1, pro ts from both stores have a positive effect on Bob s pro ts In some sense, Paul s pro ts have twice the impact of Bob s pro ts on d/ 4 Since a 1andb, Paul s making a pro t has a negative effect on his future pro ts However, if Bob makes a pro t, then Paul s pro ts bene t Moreover, Bob s pro tabilit has twice the impact as does Paul s In fact, since d/ +, Paul spro ts will increase if + > 0or,in other words, if Bob s pro ts are at least one-half of Paul s pro ts Since c andd 1, Bob is in the same situation as Paul His pro ts contribute negativel to d/ since d 1 However, Paul s pro tabilit has twice the positive effect Note that this model is smmetric in the sense that both Paul and Bob perceive each others pro ts in the same wa This smmetr comes from the fact that a d and b c 5 Y, p 7 Y q, r dy dy 1 Y 6 Y, Y 0 73 dy 0 3 Y 03 3π

2 31 Properties of Linear Sstems and The Linearit Principle d 3 + π d 4 9 d β d γ 10 a b c, 40 0 t t 1 t 11 a b c, 3 t t t 1 a b c t, t t

3 176 CHAPTER 3 LINEAR SYSTEMS 13 a b c, t t t 14 a If a 0, then det A ad bc bc Thus both b and c are nonzero if det A 0 b Equilibrium points 0, 0 are solutions of the simultaneous sstem of linear equations a 0 + b 0 0 c 0 + d 0 0 If a 0, the rst equation reduces to b 0 0, and since b 0, 0 0 In this case, the second equation reduces to c 0 0, so 0 0 as well Therefore, 0, 0 0, 0 is the onl equilibrium point for the sstem 15 The vector eld at a point 0, 0 is a 0 +b 0, c 0 +d 0, so in order for a point to be an equilibrium point, it must be a solution to the sstem of simultaneous linear equations a 0 + b 0 0 c 0 + d 0 0 If a 0, we know that the rst equation is satis ed if and onl if 0 b a 0 Now we see that an point that lies on this line 0 b/a 0 also satis es the second linear equation c 0 + d 0 0 In fact, if we substitute a point of this form into the second component of the vector eld, we have c 0 + d 0 c b 0 + d 0 a bc a + d 0 ad bc 0 a det A 0 a 0,

4 31 Properties of Linear Sstems and The Linearit Principle 177 since we are assuming that det A 0 Hence, the line 0 b/a 0 consists entirel of equilibrium points If a 0andb 0, then the determinant condition det A ad bc 0 implies that c 0 Consequentl, the vector eld at the point 0, 0 is b 0, d 0 Since b 0, we see that we get equilibrium points if and onl if 0 0 In other words, the set of equilibrium points is eactl the -ais Finall, if a b 0, then the vector eld at the point 0, 0 is 0, c 0 + d 0 In this case, we see that a point 0, 0 is an equilibrium point if and onl if c 0 + d 0 0 Since at least one of c or d is nonzero, the set of points 0, 0 that satisf c 0 + d 0 0 is precisel a line through the origin 16 a Let v d/then/ d / q pd/ q pv Thus we obtain the sstem d v q pv In matri form, this sstem is written as d 0 1 q p v b The determinant of this matri is q Hence, if q 0, we know that the onl equilibrium point is the origin c If is constant, then v d/ is identicall zero Hence, / 0 Also, the sstem reduces to d 0 1, q p 0 which implies that / q Combining these two observations, we obtain / q 0, and if q 0, then 0 17 The rst-order sstem corresponding to this equation is a If q 0, then the sstem becomes d v q pv d v pv,

5 178 CHAPTER 3 LINEAR SYSTEMS and the equilibrium points are the solutions of the sstem of equations v 0 pv 0 Thus, the point,vis an equilibrium point if and onl if v 0 In other words, the set of all equilibria agrees with the horizontal ais in the v-plane b If p q 0, then the sstem becomes d v 0 but the equilibrium points are again the points with v 0 18 In this case, / d / 0, and the rst-order sstem reduces to d v 0 a Since / 0, we know that vt c for some constant c b Since d/ v c, we can integrate to obtain t ct + k where k is another arbitrar constant Hence, the general solution of the sstem consists of all functions of the form t, vt ct + k, c for arbitrar constants c and k c 19 Letting v d/ and w d / we can write this equation as the sstem d dw v d w d3 r qv pw 3

6 In matri notation, this sstem is where Y v w 31 Properties of Linear Sstems and The Linearit Principle 179 dy AY and A r q p 0 If there are more than the usual number of buers, then b > 0 If this level of buing means that prices will increase and that fewer buers will enter the market, then the effect on db/ should be negative Since db/ αb + βs, we epect that the αb-term will be negative if b > 0 Consequentl, α should be negative 1 If there are fewer than the usual number of buers, then b < 0 If this level of b has a negative effect on the number of sellers, we epect the γ b-term in ds/ to be negative If γ b < 0andb < 0, then we must have γ>0 If s > 0, there are more than the usual number of houses for sale and house prices should decline Declining prices should have a positive effect on the number of buers and a negative effect on the number of sellers Since db/ αb + βs, we epect the βs-term to be positive Since βs > 0if s > 0, the parameter β should be positive 3 In the model, ds/ γ b + δs Ifs > 0, then the number of sellers is greater than usual and house prices should decline Since declining prices should have a negative effect on the number of sellers, we epect the δs-term to be negative If δs < 0whens > 0, we should have δ<0 4 a Substituting Y 1 t in the left-hand side of the differential equation ields dy 1 0 e t Moreover, the right-hand side becomes Y 1 t e t 0 e t Since the two sides of the differential equation agree, Y 1 t is a solution Similarl, if we substitute Y t in the left-hand side of the differential equation, we get dy e t e t Moreover, the right-hand side is Y t 1 1 e t e t

7 180 CHAPTER 3 LINEAR SYSTEMS e t e t + e t e t e t Since the two sides of the differential equation also agree for this function, Y t is another solution b At t 0, Y0, 1 B the Linearit Principle, an linear combination of two solutions is also a solution Hence, we solve the given initial-value problem with a function of the form k 1 Y 1 t + k Y t where k 1 and k are constants determined b the initial value That is, we determine k 1 and k via k 1 Y k Y 0 Y0 1 We get 0 1 k 1 + k This vector equation is equivalent to the simultaneous linear equations k k 1 + k 1 From the rst equation, we have k Then from the second equation, we obtain k 1 1 Therefore, the solution to the initial-value problem is Yt Y 1 t Y t 0 e t e t e t e t e t e t Note that as alwas we can check our calculations directl B direct evaluation, we know that Y0, 1 Moreover, we can check that Yt satis es the differential equation The left-hand side of the differential equation is dy 4e t e t 4e t, and the right-hand side of the differential equation is 0 0 e t Yt e t 4e t

8 31 Properties of Linear Sstems and The Linearit Principle 181 4e t e t 4e t Since the left-hand side and the right-hand side agree, the function Yt is a solution to the differential equation, and since it assumes the given initial value, this function is the desired solution to the initial-value problem The Uniqueness Theorem sas that this function is the onl solution to the initial-value problem 5 a Note that substituting Yt into the left-hand side of the differential equation, we get dy e t + te t e t t + 1e t e t + te t 3e t te t Substituting Yt into the right-hand side, we get 1 1 te t te t + t + 1e t 1 3 t + 1e t te t 3t + 1e t e t + te t 3e t te t Since the left-hand side of the differential equation equals the right-hand side, the function Yt is a solution b At t 0, Y0 0, 1 B the Linearit Principle, an constant multiple of the solution Yt is also a solution Since the function Yt has the desired initial condition, we know that te t Yt t + e t is the desired solution B the Uniqueness Theorem, this is the onl solution with this initial condition Given the formula for Yt directl above, note that we can directl check our assertion that this function solves the initial-value problem without appealing to the Linearit Principle 6 a Substitute Y 1 t into the differential equation and compare the left-hand side to the right-hand side On the left-hand side, we have dy 1 3e 3t 3e 3t, and on the right-hand side, we have 1 e 3t e 3t e 3t 3e 3t AY 1 t 5 e 3t e 3t 5e 3t 3e 3t

9 18 CHAPTER 3 LINEAR SYSTEMS Since the two sides agree, we know that Y 1 t is a solution For Y t, dy 4e 4t 8e 4t, and 1 e 4t e 4t e 4t AY t 5 e 4t e 4t 10e 4t 4e 4t 8e 4t Since the two sides agree, the function Y t is also a solution Both Y 1 t and Y t are solutions, and we proceed to the net part of the eercise b Note that Y 1 0 1, 1 and Y 0 1, These vectors are not on the same line through the origin, so the initial conditions are linearl independent If the initial conditions are linearl independent, then the solutions must also be linearl independent Since the two solutions are linearl independent, we proceed to part c of the eercise c We must nd constants k 1 and k such that 1 1 k 1 Y k Y 0 k 1 + k 1 3 In other words, the constants k 1 and k must satisf the simultaneous sstem of linear equations k 1 + k k 1 + k 3 It follows that k 1 1andk 1 Hence, the required solution is e 3t + e 4t Y 1 t + Y t e 3t + e 4t 7 a Substitute Y 1 t into the differential equation and compare the left-hand side to the right-hand side On the left-hand side, we have dy 1 3e 3t + 8e 4t 3e 3t + 16e 4t, and on the right-hand side, we have 1 e 3t e 4t AY 1 t 5 e 3t 4e 4t 3e 3t + 8e 4t 3e 3t + 16e 4t Since the two sides agree, we know that Y 1 t is a solution For Y t, dy 6e 3t 4e 4t 6e 3t 8e 4t,

10 and 1 AY t 5 31 Properties of Linear Sstems and The Linearit Principle 183 e 3t + e 4t e 3t + e 4t 6e 3t 4e 4t 6e 3t 8e 4t Since the two sides agree, the function Y t is also a solution Both Y 1 t and Y t are solutions, and we proceed to the net part of the eercise b Note that Y 1 0 1, 3 and Y 0 3, 4 These vectors are not on the same line through the origin, so the initial conditions are linearl independent If the initial conditions are linearl independent, then the solutions must also be linearl independent Since the two solutions are linearl independent, we proceed to part c of the eercise c We must nd constants k 1 and k such that 1 3 k 1 Y k Y 0 k 1 + k In other words, the constants k 1 and k must satisf the simultaneous sstem of linear equations k 1 + 3k 3k 1 + 4k 3 It follows that k 1 1/5 andk 3/5 Hence, the required solution is 1 5 Y 1t + 3 e 3t 5 Y + e 4t t e 3t + e 4t 8 a First we substitute Y 1 t into the differential equation The left-hand side becomes dy 1 cos 3t 3sin3t e t + e t sin 3t 3cos3t cos3t 3sin3t e t, 3cos3t sin3t and the right-hand side is 3 AY 1 t 3 cos3t 3sin3t Y 1 t e t 3cos3t sin3t Since the two sides of the differential equation agree, the function Y 1 t is a solution Using Y t,wehave dy sin 3t 3cos3t e t + e t cos 3t 3sin3t sin3t 3cos3t e t, 3sin3t cos3t

11 184 CHAPTER 3 LINEAR SYSTEMS and 3 sin3t 3cos3t AY t Y t e t 3 3sin3t cos3t The two sides of the differential equation agree Hence, Y t is also a solution Since both Y 1 t and Y t are solutions, we proceed to part b b Note that Y 1 0 1, 0 and Y 0 0, 1, and these vectors are not on the same line through the origin Hence, Y 1 t and Y t are linearl independent, and we proceed to part c of the eercise c To nd the solution with the initial condition Y0, 3, wemust nd constants k 1 and k so that k 1 Y k Y 0 3 We have k 1 andk 3, and the solution with initial condition, 3 is cos3t 3sin3t Yt e t sin3t + 3cos3t 9 a First, we check to see if Y 1 t is a solution The left-hand side of the differential equation is dy 1 e t + 36e 3t e t + 1e 3t, and the right-hand side is 3 AY 1 t 1 0 e t + 1e 3t e t + 4e 3t e t 36e 3t e t + 1e 3t Consequentl, Y 1 t is a solution However, dy e t e t, and 3 e t 4e t AY t 1 0 e t e t Consequentl, the function Y t is not a solution In this case, we are not able to solve the given initial-value problem, so we stop here 30 a This holds in all dimensions In two dimensions the computation is a b AkY k c d a b k c d k

12 31 Properties of Linear Sstems and The Linearit Principle 185 ak + bk ck + dk a + b k kay c + d b To verif the rst half of the Linearit Principle, we suppose that Y 1 t 1 t, 1 t is a solution to the sstem d a + b d c + d and that k is an an constant In order to verif that the function Y t ky 1 t is also a solution, we need to substitute Y t into both sides of the differential equation and check for equalit In other words, after we write Y t in scalar notation as Y t t, t, we must show that d a + b d c + d given that we know that d 1 a 1 + b 1 d 1 c 1 + d 1 Since t k 1 t and t k 1 t, we can multipl both sides of b k to obtain d 1 a 1 + b 1 d 1 c 1 + d 1 k d 1 ka 1 + b 1 k d 1 kc 1 + d 1 However, using standard algebraic properties and the rules of differentiation, this sstem is equivalent to dk 1 ak 1 + bk 1 dk 1 ck 1 + dk 1,

13 186 CHAPTER 3 LINEAR SYSTEMS which is the same as the desired equalit d a + b d c + d To verif the second half of the Linearit Principle, we suppose that Y 1 t 1 t, 1 t and Y t t, t are solutions to the sstem d a + b d c + d To verif that the function Y 3 t Y 1 t+y t is also a solution, we need to substitute Y 3 t into both sides of the differential equation and check for equalit In other words, after we write Y 3 t in scalar notation as Y 3 t 3 t, 3 t, we must show that d 3 a 3 + b 3 d 3 c 3 + d 3 given that we know that d 1 a 1 + b 1 d 1 c 1 + d 1 and d a + b d c + d Adding the two given sstems together ields the sstem which can be rewritten as d 1 d 1 + d + d a 1 + b 1 + a + b c 1 + d 1 + c + d, d 1 + a b 1 + d 1 + c d 1 + But this last sstem of equalities is the desired equalit that indicates that Y 3 t is also a solution

14 31 Properties of Linear Sstems and The Linearit Principle a If 1, 1 0, 0, then 1, 1 and, are on the same line through the origin because 1, 1 is the origin So 1, 1 and, are linearl dependent b If 1, 1 λ, for some λ, then 1, 1 and, are on the same line through the origin To see wh, suppose that 0andλ 0 The λ 0 case was handled in part a above In this case, 1 0 as well Then the slope of the line through the origin and 1, 1 is 1 / 1, and the slope of the line through the origin and, is / However, because 1, 1 λ,,wehave 1 λ 1 λ Since these two lines have the same slope and both contain the origin, the are the same line The special case where 0 reduces to considering vertical lines through the origin c If 1 1 0, then 1 1 Once again, this condition implies that 1, 1 and, are on the same line through the origin For eample, suppose that 1 0, then But we alread know that 1, 1 so we have, 1, 1 1 B part b above where λ / 1, the two vectors are linearl dependent If 1 0but 1 0, it follows that 1 0, and thus 0 Thus, both 1, 1 and, are on the vertical line through the origin Finall, if 1 0and 1 0, we can use part a to show that the two vectors are linearl dependent 3 If 1 1 is nonzero, then 1 1 If both 1 0and 0, we can divide both sides b 1, and we obtain 1, 1 and therefore, the slope of the line through the origin and, is not the same as the slope of the line through the origin and 1, 1 If 1 0, then 0 In this case, the line through the origin and 1, 1 is vertical, and the line through the origin and, is not vertical 33 The initial position of Y 1 t is Y 1 0 1, 1 B the Linearit Principle, we know that ky 1 t is also a solution of the sstem for an constant k Hence, for an initial condition of the form k, k, the solution is ky 1 t a ThecurveY 1 t e t, e t is the solution with this initial condition b We cannot nd the solution for this initial condition using onl Y 1 t c The constant function 0Y 1 t 0, 0 represented b the equilibrium point at the origin is the solution with this initial condition d The curve 3Y 1 t 3e t, 3e t is the solution with this initial condition

15 188 CHAPTER 3 LINEAR SYSTEMS 34 a If Yt t, t /, thent t and t t / Then d/ 1, and d/ t So Yt satis es the differential equation b For Yt,wehavet t,andt t In this case, we need onl consider d/ to see that the function is not a solution to the sstem 35 a Using the Product Rule we compute dw d 1 d + 1 d d 1 1 b Since 1 t, 1 t and t, t are solutions, we know that d 1 a 1 + b 1 d 1 c 1 + d 1 and that d a + b d c + d Substituting these equations into the epression for dw/, we obtain dw a 1 + b c + d a + b 1 c 1 + d 1 After we collect terms, we have dw a + dw c This equation is a homogeneous, linear, rst-order equation as such it is also separable see Sections 11, 1, and 18 Therefore, we know that the general solution is W t Ce a+dt where C is an constant but note that C W 0 d From Eercises 31 and 3, we know that Y 1 t and Y t are linearl independent if and onl if W t 0 But, W t Ce a+dt,sow t 0 if and onl if C W 0 0 Hence, W t 0iszeroforsomet if and onl if C W 0 0 EXERCISES FOR SECTION 3 1 a The characteristic polnomial is 3 λ λ 0, and therefore the eigenvalues are λ 1 andλ 3