Extra Credit Solutions Math 181, Fall 018 Instrutor: Dr. Doreen De Leon 1. In eah problem below, the given sstem is an almost linear sstem at eah of its equilibrium points. For eah, (i Find the (real equilibrium points. (ii Find the orresponding linearized sstem z = Az at eah equilibrium point. (iii What, if anthing, an be inferred about the stabilit of the equilibrium points. (a (i We need to lve x = x + 9 9 = x 0 = x + 9 9 = x + 0 = 9 0 = x. Sine x = 0, we have 9 = 9, and = ±1. Therefore, the equilibrium points are (0, 1 and (0, 1. (ii The linearized sstem has oeffiient matrix A = [ fx (x e, e f (x e, e g x (x e, e g (x e, e ] = So, for (0, 1, the linearized sstem is [ ] 0 18 z = z, 1 0 and for (0, 1, the linearized sstem is [ ] 0 18 z = z. 1 0 [ ] xe 18 3. 1 0 (iii For eah equilibrium point e, analze the stabilit of the linearized sstem and then determine (if possible the stabilit of e. (0, 1 Find the eigenvalues of A. 0 = det(a λi = λ 18 1 λ = λ + 18,
and λ 1, = ±3 i. Sine z e = 0 is stable, but not asmptotiall stable, no onlusion an be drawn from this method about e = (0, 1. (0, 1 Find the eigenvalues of A. 0 = det(a λi = λ 18 1 λ = λ 18, (b and λ 1, = ±3. Sine λ = 3 > 0, z e = 0 is unstable. Therefore, e = (0, 1 is unstable. x = x 1 = (x + 4(x 1 (i We need to lve 0 = x 1 = x = 1 (1 0 = (x + 4(x 1 = x = 1 or x + 4 = 0 ( If x = 1, then from (1, = 1. If x + 4 = 0, then x = 4. From (1, we have 4( = 1, or = 1 4, whih has no real lution. Therefore, there is onl one equilibrium point, e = (1, 1. (ii The oeffiient matrix for the linearized sstem is [ ] [ ] fx (x A = e, e f (x e, e = e x e. g x (x e, e g (x e, e x e + 4 e 1 4x e 4 Therefore, the linearized sstem for (1, 1 is [ ] 1 1 z = z. 5 0 (iii For eah equilibrium point e, analze the stabilit of the linearized sstem and then determine (if possible the stabilit of e. To do this, find the eigenvalues of A. 0 = det(a λi = 1 λ 1 5 λ = λ(1 λ 5 = λ λ 5. Therefore, λ 1, = 1 ± 1 = 1 ± 1.
Sine λ = 1 1 + > 0, z = 0 is unstable, and therefore, e = (1, 1 is unstable.. For eah of the following homogeneous linear sstems of differential equations (i Classif the equilibrium point e = 0. (ii Sketh the phase portrait. Make sure to show the diretion along trajetories. [ ] 1 6 (a = 1 4 (i 0 = det(a λi = 1 λ 6 1 4 λ = (1 λ( 4 λ 6 = λ + 3λ + = (λ + 1(λ +. So, λ 1 = and λ 1 = 1. Sine both eigenvalues are negative, e = 0 is an asmptotiall stable (improper node. (ii So, the lution takes the form x(t = 1 ( λ1 d = 1 ( ( 4 1 = 1 e t + 3 e t. (t = 1 e λ 1t + e λ t = 1 e t + e t. ( e λ1t λ d + e λ t e t + ( 1 ( 4 1 Next, find the slopes of the onstant lines: 1 = 0: x = λ d = 1 3 = 0: Asmptotiall: x = λ 1 d = 1 e t
t : t : x 1 e t ± 1 e t ± x 1. x 3 e t 0 e t 0 x 1 3. The sketh of the phase portrait is on a different page. [ ] 1 1 (b = 1 1 (i 0 = det(a λi = 1 λ 1 1 1 λ = ( 1 λ( 1 λ + 1 = λ + λ +. So, λ 1, = 1 ± i. Sine the eigenvalues are omplex onjugate with nonzero real part, the equilibrium point is a spiral, and sine Reλ 1 = Reλ = 1 < 0, e = 0 is asmptotiall stable. (ii To do the phase portrait, we need to determine how r and θ behave dθ dt = x (t x (t = x + = x x + x x + = x + = 1 < 0. x + x( x ( x + x + (sine x = x +, = x Therefore, θ is dereasing. Sine Reλ 1 = Reλ = 1 < 0, r = x + is dereasing. The phase portrait reated b Maple is below.
[ ] 7 4 ( = 7 (i 0 = det(a λi = 7 λ 4 7 λ = ( 7 λ( 7 λ + 48 = λ 1 = (λ + 1(λ 1. So, λ 1 = 1 and λ = 1. Sine λ = 1 > 0, the equilibrium lution is unstable. Sine λ 1 < 0 < λ, it is a saddle point. (ii So, the lution takes the form x(t = 1 ( λ1 d = 1 ( 1 ( 7 = 3 1 e t + 4 e t. (t = 1 e λ 1t + e λ t = 1 e t + e t. ( e λ1t λ d + e λ t e t + ( 1 ( 7 Next, find the slopes of the onstant lines: 1 = 0: x = λ d = 1 4 = 0: x = λ 1 d = 1 3 e t
Asmptotiall: t : t : x 3 1 e t ± 1 e t ± x 1 3. x 4 e t ± e t ± x 1 4. The phase portrait is on a separate page. 3. Find the eigenvalues and eigenfuntions of The harateristi equation is There are three possibilities: λ = 0 λ < 0 λ > 0 + λ = 0, (0 = 0, (π = 0. r + λ = 0 = r = ± λ. We will look at eah ase to determine if there are eigenvalues satisfing that ondition. λ = 0 In this ase, r 1 = r = 0, and the general lution is (x = 1 + x. Appling the boundar onditions, we see that 1 is arbitrar and = 0, λ = 0 is an eigenvalue with orresponding eigenfuntion (x = 1. Therefore, λ = 0 is not an eigenvalue. λ < 0 In this ase, λ > 0, and the general lution is (x = 1 osh( λx + sinh( λx.
Next, we appl the boundar onditions to find 1 and. Sine (x = 1 λ sinh( λx + λ osh( λx, 0 = λ = = 0 0 = 1 osh(π λx + sinh(π λx 0 = 1 osh(π λ = 1 = 0. So, (x = 0 is the unique lution. Therefore, there are no eigenvalues λ < 0. λ > 0 In this ase, λ < 0, the general lution is (x = 1 os( λx + sin( λx. Sine (x = λ sin( λx + λ os( λx, 0 = λ = = 0 0 = 1 sin(π λ, whih is true if either 1 = 0 or sin(π λ = 0. If 1 = 0, then (x = 0 and we do not have an eigenvalue, assume 1 0. Then Therefore, the eigenvalues are sin(π λ = 0 π λ = nπ (n = 1,, 3,... ( n n λ = = 4. λ 0 = 0, with orresponding eigenfuntions 0 (x = 1, λ n = n, n = 1,, 3,..., 4 n (x = os ( n x, n = 1,, 3,....