dx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω.

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1 Chpter 8 Stility theory We discuss properties of solutions of first order two dimensionl system, nd stility theory for specil clss of liner systems. We denote the independent vrile y t in plce of x, nd x,y denote dependent vriles. Let I R e n intervl, nd Ω R 2 e domin. Let us consider the system dx = F (t, x, y), dt dy = G(t, x, y), dt where the functions re defined on I Ω, nd re loclly Lipschitz w.r.t. vrile (x, y) Ω. Definition 8.1 (Autonomous system) A system of ODE hving the form (8.1) is clled n utonomous system if the functions F (t, x, y) nd G(t, x, y) re constnt w.r.t. vrile t. Tht is, dx = F (x, y), dt dy = G(x, y), dt Definition 8.2 A point (x 0, ) Ω is sid to e criticl point of the utonomous system (8.2) if F (x 0, ) = G(x 0, ) = 0. (8.3) A criticl point is lso clled n equilirium point, rest point. Definition 8.3 Let (, ) e solution of two-dimensionl (plnr) utonomous system (8.2). The trce of (, ) s t vries is curve in the plne. This curve is clled trjectory. Remrk 8.4 (On solutions of utonomous systems) (i) Two different solutions my represent the sme trjectory. For, (8.1) (8.2) (1) If (x 1 (t), y 1 (t)) defined on n intervl J is solution of the utonomous system (8.2), then the pir of functions (x 2 (t), y 2 (t)) defined y (x 2 (t), y 2 (t)) := (x 1 (t s), y 1 (t s)), for t s + J (8.4) is solution on intervl s + J, for every ritrry ut fixed s R. (2) However, trces of oth the solutions on the respective intervls is the sme. If the independent vrile t is interpreted s time, then we note tht the two different solutions visit every point on the trjectory with time lg s. Also see Exmple

2 (ii) The trjectories do not cross. For, (1) Suppose two trjectories γ 1 nd γ 2 cross t point (x 0, ) Ω. Let (x 1 (t), y 1 (t)) defined on intervl I 1 e solution whose trce is γ 1, nd (x 2 (t), y 2 (t)) defined on intervl I 2 e solution whose trce is γ 2. By the ssumption of crossing, there exist t 1 I 1, nd t 2 I 2 such tht (x 1 (t 1 ), y 1 (t 1 )) = (x 0, ) = (x 2 (t 2 ), y 2 (t 2 )). (8.5) Let us define pir of functions (x 3 (t), y 3 (t)) on intervl t 1 t 2 + I 2 y (x 3 (t), y 3 (t)) := (x 2 (t t 1 + t 2 ), y 2 (t t 1 + t 2 )) (8.6) It cn e esily checked, vi Chin rule, tht (x 3 (t), y 3 (t)) is solution of (8.2) on the intervl t 1 t 2 + I 2. Note tht (x 3 (t 1 ), y 3 (t 1 )) = (x 2 (t 1 t 1 + t 2 ), y 2 (t 1 t 1 + t 2 )) = (x 2 (t 2 ), y 2 (t 2 )) = (x 0, ) (8.7) Thus, (x 3 (t), y 3 (t)) nd (x 1 (t), y 1 (t)) re solutions of the sme initil vlue prolem, contrdicting the uniqueness of solutions to IVP. Therefore two trjectories do not cross ech other. (iii) The trjectories fill the domin Ω, since through every point trjectory psses. This is consequence of existence theorem. (iv) Through every point in the phse spce Ω, exctly one trjecory psses. This is consequence of uniqueness of solutions to IVPs. (v) From the lst two remrks, it follows tht the trjectories prtition the phse spce Ω. In fct, defining reltion on Ω y sying tht two points (x 1, y 1 ), (x 2, y 2 ) Ω re relted if (x 1, y 1 ) nd (x 2, y 2 ) lie on the sme trjectory, it is esy to verify tht this reltion is n equivlence reltion nd therey giving rise to prtition of Ω in terms of equivlnece clsses. Ech equivlence clss is trjectory. (vi) Note tht trjectories consisting of single point correspond to criticl points. (vii) types of trjectories: For utonomous systems with two dimensionl phse spce, three types of trjectories re possile. A trjectory consisting of single point (corresponding to equilirium solutions), nd if trjectory hs more thn one point then it could e closed curve (corresponding to periodic solutions), or curve without self-intersection. (viii) For liner utonomous systems, specil clss of systems (8.2) for which F nd G re liner in x, y, note tht sturted solutions re glol, i.e., sturted solutions re defined on the entire rel line R. Hence for liner utonomous systems, we do not mention the intervl on which given solution is defined. The ove remrk is illustrted y the following exmple. Exmple 8.5 dx dt = y, dy dt = 4x. (8.8) Note tht (x 1 (t), y 1 (t)) := (cos 2t, 2 sin 2t) is solution of (8.8). The trjectory pssing through the point (1, 0) R 2 is the ellipse x 2 + y2 4 = 1 trvelled counterclockwise. Consider the solution (x 2 (t), y 2 (t)) := (cos(t π/2), 2 sin 2(t π/2)). This solution stisfies (x 2 ( π 2 ), y 2( π 2 )) = (1, 0), nd hence hs the sme trjectory s (x 1 (t), y 1 (t)). Drw figure

3 Chpter 8 : Stility theory Solving liner plnr systems with constnt coefficients Consider the system of ODE ( ( x x x y = =: A. (8.9) c d) y y) Fundmentl mtrix Definition 8.6 (Fundmentl mtrix) A mtrix vlued function Φ whose columns re solutions of the system of ODE (8.9) is clled solution mtrix. A solution mtrix Φ is clled fundmentl mtrix if the columns of Φ form fundmentl pir of solutions for the system (8.9). A fundmentl mtrix Φ is clled the stndrd fundmentl mtrix if Φ(0) is the identity mtrix. Remrk 8.7 Since the columns of solution mtrixφ re solutions of (8.9), the mtrix vlued function Φ stisfies the system of ODE Φ = AΦ. (8.10) Exercise 8.8 A solution mtrix is fundmentl mtrix if nd only if its determinnt is not zero. Exercise 8.9 Prove tht if Ψ is fundmentl mtrix then ΨC is lso fundmentl mtrix for every constnt invertile mtrix C. Prove tht ll fundmentl mtrices occur this wy. Computtion of fundmentl mtrix By definition of fundmentl mtrix, finding fundmentl pir of solutions to system of ODE (8.9) is equivlent to finding fundmentl mtrix. In view of Exercise 8.9, fundmentl mtrix is not unique ut the stndrd fundmentl mtrix is unique. (1) Oserve tht e is non-trivil solution of (8.9) if nd only if ( ) Tht is, λ is n eigenvlue nd ( ( ( 0, A = λ. (8.11) 0) ) ) is n eigenvector corresponding to λ. (2) Question Is it possile to find fundmentl pir, oth of which re of the form e? Answer Supposing tht φ 1 (t) = e λ 1t nd φ 2 (t) = e λ 2t c re two solutions of d (8.9), φ 1, φ 2 form fundmentl pir if nd only if c d 0, (8.12) since the ove determinnt is the Wronskin of φ 1, φ 2 t t = 0. Tht is, the mtrix A should hve two linerly independent eigenvectors. Note tht this is equivlent to sying tht A is digonlisle.

4 Solving liner plnr systems with constnt coefficients (3) Question Wht if the mtrix A does not hve two linerly independent eigenvectors? This cn hppen when A hs only one eigenvlue of multiplicity two. Inspired y similr sitution in the context ( of ) constnt coefficient( second ) order liner ODE, we re tempted to try φ 1 (t) = e λ1t nd φ 2 (t) = te λ1t s fundmentl pir. But note tht φ 1, φ 2 does not form fundmentl pir since Wronskin t t = 0 will e zero, lso note tht φ 2 is not even solution of the liner system (8.9). Nevertheless, we cn find solution hving the form of φ 1. Therefore, we try vrint of ove suggestion to find nother solution tht together φ 1 constitutes fundmentl pir. Let φ 2 (t) = te λ1t + e λ1t c (8.13) d Then φ 2 (t) solves the system (8.9) if nd only if ( ( c (A λ 1 I) =. (8.14) d) ) One cn esily verify tht y φ 1 (t) = e λ 1t, φ 2 (t) = te λ 1t is fundmentl pir, where, ( ( c, re linerly independent. Thus, φ ) d) 1, φ 2 defined + e λ 1t c d c re relted y the eqution (8.14). d (8.15) (4) In cse the mtrix A does not hve rel eigenvlues, then eigenvlues re complex conjugtes of ech other. In this cse, (λ, v) is n eigen pir if nd only if (λ, v) is lso n eigen pir for A. (8.16) α + iβ Denoting λ = r + iq (note q 0), v =, define γ + iδ φ 1 (t) = e rt α cos qt β sin qt, φ γ cos qt δ sin qt 2 (t) = e rt α sin qt + β cos qt. (8.17) γ sin qt + δ cos qt Verifying tht the pir of functions defined ove constitute rel-vlued fundmentl pir of solutions is left s n exercise Mtrix exponentils Inspired y the formul for solutions to liner sclr eqution y = y, given y = ce t, where y(0) = c, we would like to sy tht the system (8.9) hs solution given y ( ( x(0) = exp(ta), where =. (8.18) ) ) y(0) First of ll, we must give mening for the symol exp(a) s mtrix, which is clled the exponentl of mtrix B. Then we must verify tht the proposed formul for solution, in (8.18), is indeed solution of (8.9). We do this next. Definition 8.10 (Exponentil of mtrix) If A is 2 2 mtrix, then exponentil of A (denoted y e A, or exp(a), is defined y e A = I + k=1 A k k! (8.19)

5 Chpter 8 : Stility theory 71 We record elow, without proof, some properties of mtrix exponentils. Lemm 8.11 Let A e 2 2 mtrix. (1) The series in (8.19) converges. (2) e 0 = I. (3) (e A ) 1 = e A (4) exp(a + B) = exp(a) exp(b) if the mtrices A nd B stisfy AB = BA. (5) If J = P 1 AP, then exp(j) = P 1 exp(a)p. (6) d dt eta = Ae ta. Theorem 8.12 If A is 2 2 mtrix, then Φ(t) = e ta is fundmentl mtrix of the system (8.9). If (, ) is solution of the system (8.9) with (x(t 0 ), y(t 0 )) = (x 0, ), then (, ) is given y = e (t t0)a x0. (8.20) Remrk 8.13 Thnks to the ove theorem, we do not need to struggle to find fundmentl mtrix s we did erlier. We need to to tke just the exponentil of the mtrix ta nd tht would give very esily fundmentl mtrix. But it is not s simple s it seems to e. In fct, summing up the series for exponentil of mtrix is not esy t ll, even for reltively simpler mtrices. To relise this, solve the exercise following this remrk. Exercise 8.14 Find exponentil mtrix for 2 0 A 1 =, A = 1 1, A = 1 2. (8.21) 2 3 Remrk 8.15 After solving the ove exercise, one would relise tht it is esier to find fundmentl mtrices y finding fundmentl pirs of solutions insted of summing up series! There is n lternte method to clculte exponentil mtrix for ta, vi fundmentl pirs, y oserving tht exponentil mtrix is nothing ut the stndrd fundmentl mtrix. So, find fundmentl mtrix Φ, then exponentil mtrix e ta is given y e ta = [Φ(0)] 1 Φ(t). (8.22) 8.2 Stility of n equilirium point Definition 8.16 (1) A criticl point (x 0, ) of the system (8.2) is sid to e stle if for ech ɛ > 0 there exists δ > 0 such tht every solution (, ) for which there is n s such tht (x(s), y(s)) (x 0, ) < δ, exists for ll t s nd stisfies (, ) (x 0, ) < ɛ, t s. (8.23) (2) A criticl point (x 0, ) of the system (8.2) is sid to e unstle if it is not stle. (3) A criticl point (x 0, ) of the system (8.2) is sid to e symptoticlly stle if (i) the criticl point (x 0, ) is stle, nd, (ii) if there exists δ 0 > 0 (0 < δ 0 < δ) such tht (x(s), y(s)) (x 0, ) < δ 0 for some s = (, ) (x 0, ) s t. (8.24)

6 Phse spce picture Convention We re interested in the ehviour of solutions s t. Therefore, for us lwys t 0, nd hence we indicte directions on trjectories only for incresing t. Remrk 8.17 In other words, criticl point is stle if every trjectory tht comes within δ distnce from the criticl point t some time, stys within n ɛ distnce t ll lter times. Further if the trjectory pproches the criticl point, then the criticl point is symptoticlly stle. 8.3 Phse spce picture Before we strt discussing the nture of trjectories for liner systems, we need to understnd the following questions. (1) Wht re ll 2 2 rel mtrices? (2) Under liner trnsformtion, wht will e the imges of point, closed curve, curve? Let us nlyse some specil cses of liner utonomous system (8.9): Cse 1: Invertile digonl mtrix A 1 = λ 0, with λ 0 µ. (8.25) 0 µ (1) If (, ) is solution of the system (8.9) corresponding to A 1, with (x(0), y(0)) = (x 0, ), then (, ) is given y = e ta 1 x0 e 0 x0. = 0 e µt (8.26) (2) To drw trjectories in the plne, we hve to drw the unique trjectory pssing through ech point. (i) The criticl point (0, 0) is clled non-degenerte since it is isolted (well-seprted from other criticl points, if ny; in fct none!) (ii) Let us fix (x 0, ) = (0, 0) The trjectory through origin consists of only one point, which is origin. (iii) Let us consider the cse of x 0 = 0, ut 0. Then solution is given y (, ) = (0, e µt ). Similrly, in the cse where = 0, ut x 0 0, the solution is given y (, ) = (e x 0, 0). (iv) Let us fix (x 0, ) with x 0 0. In this cse, we cn write µ λ =, (8.27) x 0 hence trjectory through (x 0, ) is given y µ λ x y =, with x 0 x x 0 > 0, y > 0. (8.28) (3) Now drw the trjectories nd indicte the progress of the curve s t increses. The nture of the trjectories vry nd depend on the nture of λ, µ. There re five distinct scenrios: λ = µ < 0, λ < 0 < µ, λ < µ < 0, 0 < µ < λ, 0 < λ = µ. (8.29)

7 Chpter 8 : Stility theory 73 Cse 1B: Non-invertile digonl mtrix λ 0 A 2 =, 0 µ with λµ = 0. (8.30) (1) The criticl point (0, 0) is clled degenerte since there re infinitely mny criticl points of the system in ny neighourhood of the origin. (criticl points re not well-seprted from ech other) In fct, ll points on x xis (respectively y xis) re criticl points if λ = 0 (respectively, µ = 0). (2) If (, ) is solution of the system (8.9) corresponding to A 2, with (x(0), y(0)) = (x 0, ), then (, ) is given y = e ta 2 x0 e 0 x0 = 0 e µt (8.31) (3) To drw trjectories in the plne, we hve to drw the unique trjectory pssing through ech point. (i) Let us fix (x 0, ) = (0, 0) The trjectory through origin consists of only one point, which is origin. (ii) Let us fix (x 0, ) to e ny other criticl point. The trjectory through ny criticl point consists of only one point, which is the criticl point itself. (iii) Let us fix (x 0, ) different from ny criticl point. Then solution is given y (, ) = (e x 0, e µt ). (4) Now drw the trjectories nd indicte the progress of the curve s t increses. The nture of the trjectories vry nd depend on the nture of λ, µ. There re three distinct scenrios: λ < µ = 0, 0 = µ < λ, 0 = µ = λ. (8.32) Cse 2: Mtrix hving the form λi + A 3 = λ 1, λ R (8.33) 0 λ (1) The exponentil mtrix is given y e ta 3 ( ) 0 1 = exp(i) exp t 0 0 e 0 1 t e te = 0 e = e (8.34) (2) If (, ) is solution of the system (8.9) corresponding to A 3, with (x(0), y(0)) = (x 0, ), then (, ) is given y = e ta 3 x0 = e te x0 0 e (8.35) (3) To drw trjectories in the plne, we hve to drw the unique trjectory pssing through ech point.

8 Phse spce picture (i) If λ 0, the criticl point (0, 0) is clled non-degenerte since it is isolted (well-seprted from other criticl points, if ny; in fct none!). If λ = 0, then criticl point (0, 0) is clled degenerte since there re infinitely mny criticl points of the system in ny neighourhood of the origin. (criticl points re not well-seprted from ech other); in this cse, ll points on x xis re criticl points. (ii) The trjectory through ny criticl point consists of only one point, which is the criticl point itself. (iii) Let us fix (x 0, ) different from criticl point. Let us ssume tht x 0 = 0 nd 0. Then = t, nd = log. (8.36) Thus trjectories re given y ( y λx = y log ), with y > 0. (8.37) (iv) In the generl cse of oth x 0 0 nd 0, solution (, ) stisfies = [x 0 + t ], (8.38) nd thus psses through y xis t time t = x 0. nd thus trjectory is lredy descried in the previous point. Recll tht trjectory does not chnge y introducing dely in the time t which solution visits point, nd even hve the sme ehviour s t. (v) The other cse is = 0, ut x 0 0. Then solution is given y (, ) = (e x 0, 0). (4) Now drw the trjectories nd indicte the progress of the curve s t. The nture of the trjectories vry nd depend on the nture of λ. There re three distinct scenrios: 0 < λ, λ < 0, λ = 0. (8.39) Cse 3: Mtrix hving the form A 4 =, λ R (8.40) (1) The exponentil mtrix is given y ( ) e ta 4 0 = exp(ti) exp t (8.41) Denoting J =, we must evlute exp(tj). By evluting vrious powers of 1 0 tj nd sustituting them in the definition of exponentil, it cn e seen tht exp(tj) = (1 (t)2 2! + (t)4 4! (t)6 6! + ) I+ ( t 1! (t)3 3! + (t)5 5! (t)7 7! + ) J (8.42)

9 Chpter 8 : Stility theory 75 On noting tht the series expnsions in the ove eqution correspond to well-known functions, the ove eqution reduces to cos t sin t exp(tj) = cos ti + sin tj = (8.43) sin t cos t Thus exponentil mtrix ecomes ( ) 0 e ta4 = exp(ti) exp t 0 = e t cos t sin t sin t cos t (8.44) (2) If (, ) is solution of the system (8.9) corresponding to A 3, with (x(0), y(0)) = (x 0, ), then (, ) is given y = e ta 4 x0 = e t cos t sin t x0 (8.45) sin t cos t Tht is, = We cn write the ove eqution in the form = where α = tn 1 ( y0 x 0 ). Also, note e t (x 0 cos t sin t) e t. (8.46) (x 0 sin t + cos t) ( (x0, ) e t cos(t α) (x 0, ) e t sin(t α) ), (8.47) = tn(t α) (8.48) (3) To drw trjectories in the plne, we hve to drw the unique trjectory pssing through ech point. (i) The criticl point (0, 0) is clled non-degenerte since it is isolted (well-seprted from other criticl points, if ny; in fct none!). (ii) The trjectory through origin, eing criticl point, consists of only one point, which is the criticl point itself. (iii) Oserve tht () 2 + () 2 = e 2t (x y 2 0) (8.49) There re two points to e noted from here: (i). If = 0, then solution lies on circle of rdius (x 0, ). (ii). If 0, depending on the sign of, we hve (, ) goes to zero or infinity. In fct y switching to polr coordintes, x = r cos θ, y = r sin θ, r = x 2 + y 2, Trjectories re given y θ = t α. (8.50) r = () 2 + () 2 = e t (x θ+α y2 0 ) = e( (x ) y2 0 ), (8.51) which is nothing ut where c = e ( α ) (x y2 0 ). These curves re spirls. r = ce ( )θ (8.52)

10 Phse spce picture (4) Now drw the trjectories nd indicte the progress of the curve s t. The nture of the trjectories vry nd depend on the nture of,. There re four distinct scenrios: < 0 <, < 0 <, = 0& > 0, = 0& < 0. (8.53) Exercise 8.18 Discuss the phse spce picture of liner utonomous systems corresponding to ech of the mtrices A 1 =, A =, A =, A =, A 5 =, A =, A =, A =, A 9 =, A =, A =, A =, A 13 =, A =. 1 0 Exercise 8.19 Comment on the stility of origin in ech of the ove systems. Phse spce picture for liner utonomous systems corresponding to generl mtrix So fr, we discussed phse spce portrit for systems descried y mtrix of vrious specil forms. The discussion for generl mtrix cse cn e done once we hve nswers to the two questions posed t the eginning of this Section 8.3. The nswers re (1) The specil mtrices we considered re ll the 2 2 rel mtrices. It mens tht ny 2 2 rel mtrix is similr to one of the three types of specil mtrices we considered. (2) Under liner trnsformtion, imges of point, closed curve, nd curve re point, closed curve, nd curve respectively! Liner trnsformtion stnds for line goes to line. Note tht circle will ecome n ellipse under liner trnsformtion, in generl. Let P 1 AP = J. Introducing new set of dependent vriles (v, w) defined y ( v = P w) 1 x. (8.54) y It is esy to verify tht x x x solves y y = A y if nd only if ( v w) solves v = J w v w (8.55) We discussed the phse spce portrit of (v, w) since J is one of the specil mtrices for which we nlysed the phse spce picture. Now returning to the originl dependent vriles (x, y), strting from solution (v(t), w(t)) corresponding to mtrix J, solution (, ) corresponding to mtrix A is given y = P v(t). (8.56) w(t) The trjectories in xy plne re nothing ut imges of trjectories in vw plne corresponding to specil mtrix J, under liner trnsformtion defined y the mtrix P. Note tht the qulittive ehviour of criticl point remins the sme for ll systems tht correspond to mtrices which re similr. This finishes the discussion of trjectories for liner plnr utonomous system.

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