Dynamical Systems Solutions to Exercises

Transcription

1 Dynamical Systems Part 5-6 Dr G Bowtell Dynamical Systems Solutions to Exercises. Figure : Phase diagrams for i, ii and iii respectively. Only fixed point is at the origin since the equations are linear and homogeneous. i ii iii iv v vi linear auto dim no no no no yes yes no yes yes yes no yes n/a n/a Tale : For each equation the tale indicates: linear; autonomous; dimension of phase space where applicale 3. i Z dy cos y Z Z x dx Z sec y dy µ x, y /4 C y tan x dx x tan y The original equation is defined for all x and y, thus the domain equals R. ³ tan x defined for all x thus the interval of definition I,. x C

2 Dynamical Systems Part 5-6 Dr G Bowtell ii Writing the equation in the form dy tan xy cos x we see that it is a first order linear equation dx with integrating factor given y R exp tan x dx exp ln cos x cos x iii Thus y y x C y cos xx C cos x cos x x y C y x cos x The original equation is defined for all y, however due to tan x, x k. Thus the domain is given y: D {x, y R : x k / k Z} x cos x is defined for all x and must contain x, since the condition is x, y. Thus accounting for the restriction on x in the domain the interval of definition I /, /. dy y dx tan y x C x, y C y tanx 4 4 The original equation defined for all x and y thus domain equals R The interval of definition must contain zero thus the interval of definition is given y: < x 4 < 3 4 < x < I 3 4 4, 4 4. i Fixed points at y cos y y, or y k where k is an integer. To classify use either the change in sign of y cos y at each fixed point or consider the sign of y cos y at each fixed point. At y, y cos y changes sign from negative to positive with increasing y: thus y is a repellor. For the other fixed points I will use the derivative method, thus: y cos y cos y y sin y. Bearing in mind that cos y at each of these points: At ±, ± 5, ± 9..., cos y y sin y <, thus attractors At ± 3, ± 7, ±..., cos y y sin y >, thus repellors See Fig i for phase diagram. ii Fixed points at y y y, or y. Aout y, y y is non-negative with increasing y, thus y is an increasing shunt. Aout y, y y changes from positive to negative with increasing y, thus y is an attractor. See Fig ii for phase diagram.

3 Dynamical Systems Part 5-6 Dr G Bowtell 3 i ii 3 3 iii iv -3 Figure : i Repellors at and ± 4k, attractors at ± 4k 3, k,,... ii Shunt at, attractor at iii diagram not defined at k k ±, ±,... All fixed points repellors iv attractor at 3 and repellor at. iii Fixed points at tan y y k k Z. However since tan y is not defined at k the phase diagram will have gaps at these points. Aout each fixed point tan y changes from negative to positive, thus each point is a repellor. Alternatively using the derivative method tan y sec y > repellors. See Fig iii for phase diagram. iv Fixed points at y y 3 y 3y y 3 y Aout y 3, y 3y changes from positive to negative hence y 3 is an attractor. Aout y, y 3y changes from negative to positive hence y is a repellor. See Fig iv for phase diagram. 5. dy dx y y 3 y 3y dy y 3y dx Using partial fractions this gives: 4 y 3 y y 3 ln x C 4 y The initial condition x, y give C hence; y 3 ln 4x y 3 y y e 4x y e 4x 3 e 4x This is plotted as the exact solution in Fig 3 and lies etween y 3 and y.

4 Dynamical Systems Part 5-6 Dr G Bowtell 4 y 3 e 4x y y exact solution - exact solution x y 3 y e 4x Figure 3: Exact solution and fixed point solutions in heavy line. Note: linear solution aout y 3 close to exact solution aout y 3 ut as x decrease and ecomes negative the difference etween the two solutions tends to infinity. Similary for the linear solution aout y Since the denominator of the solution is never zero the interval of definition is I,. Linearisation In general, aout a fixed point a of the system dy dx y a z. Xy, the linearisation is dz dx X az, where In the example Xy y y 3 X y y thus the general linearisation aout y a is z a z. The fixed points are given y Xa a 3 a thus: z 4z aout z 3 i and z 4z aout z ii i z Ae 4x y 3Ae 4x. x, y A y 3 e 4x ii z Ae 4x y Ae 4x. x, y A y e 4x These two linearisations are plotted in Fig3, where they are seen to e respectively close to the exact solution in the neighourhood of y 3 and y. 3 λ 9 6. a Eigenvalues given y the solution of det λ 8 4 λ λ Hence the eigenvalues are λ 4 and λ 5.

5 Dynamical Systems Part 5-6 Dr G Bowtell 5 The eigenvector corresponding to λ 4 is given y: a 9 9 a a a take E The eigenvector corresponding to λ 5 is given y: a 8 9 a a a a a take E a 4 Thus the matrix P is given y P E E and J 5 5 λ 4 Eigenvalues given y the solution of det λ 7 λ λ 5 Hence the eigenvalues are λ i and λ i. The eigenvector corresponding to λ i is given y: 5 i 4 a 6 i 4 a 7 i 6 i 6 ia 4 a take E i 6 E i 6 i 6 Taking the real and imaginary part of E gives: 4 e ReE e 6 ImE 4 Thus the matrix P is given y P e e and J 6 3 λ c Eigenvalues given y the solution of det λ λ Hence a single eigenvalue λ The eigenvector corresponding to λ is given y: 3 a a a a a a take E a The Jordan vector J is given y: 3 J E a

6 Dynamical Systems Part 5-6 Dr G Bowtell 6 a a Thus the matrix P is given y P E J and J take J 7. a Writing Qi in the standard form and using Q6a we have: 3 9 ẋ x. λ λ 5 saddle point E E x α e 4t β e 5t In Fig 4a the eigenlines corresponding to E and E are trajectories and since E corresponds to the positive eigenvalue points move out from the origin along this line as indicated y the arrows. a x E x E x x c x vertical cut horizontal cut x... E. Figure 4: On all diagrams arrows show direction of increasing t and the fixed point is at the origin a Eigenlines shown, E unstale and E stale. Unstale saddle point. stale clockwise focus c unstale improper inflected node. Asymptotically trajectories are parallel to eigenline E NB: These three diagrams compare well with the ones otained y drawing line segments in question Writing Qii in the standard form and using Q6 we have: 5 4 ẋ x. λ 7 i λ i stale focus e 4 6 e x r e {cos t t 4 φ 6 sin t φ The trajectories spiral aout the fixed point at the origin and are drawn into the origin as time increases. The orientation of the trajectories is not ovious at this point; it may spiral clockwise or anti-clockwise. Thus considering the trajectories that cut the positive x axis, that is to say set x and take x >, and looking at the first equation ẋ 5x 4x we see that ẋ >. Thus x is increasing and the trajectories move from left to right in this region. The arrows are entered onto the trajectories accordingly, Fig 4, to give a clockwise rotation. }

7 Dynamical Systems Part 5-6 Dr G Bowtell 7 c Writing Qi in the standard form and using Q6a we have: 3 ẋ E J 8. a x. λ only unstale improper node x e t at ae t Since there is only one eigenline the trajectories are asymptotically parallel to this line. In detail, since the node is unstale, as t the trajectories ecome parallel to E and tend to infinity; as t the trajectories enter the fixed point at the origin parallel to E. The vector J plays no ovious part in constructing the diagram. Of more use are the points where the the trajectories are horizontal and where they are vertical. These points occur where ẋ and ẋ respectively. From the equations we have therefore that the trajectories are horizontal when x x and vertical when 3x x. These two straight lines are shown on the diagram Fig 4c as dotted lines. The direction of the phase paths can e checked using the method of part, namely looking at the sign of ẋ as the curves cut the positive x -axis. a A x ẋ 6 3 x In general: x a z ẋ ż and Ax Aa Az Az Az ż Az 7 λ 39 The Eigenvalues are given y the solution of det λ 6 3 λ 4λ 3 Hence the eigenvalues are λ 3i and λ 3i. The eigenvector corresponding to λ 3i is given y: 7 3i 39 a 5 3i 39 a 6 3 3i 6 5 3i 5 3ia 39 a take E 5 3i E 5 3i 5 3i The fixed point is a stale focus since the eigenvalues are complex with negative real part. Taking the real and imaginary part of E gives: 39 e ReE e 5 ImE 3 x z a r e {cos 3t t 39 φ 5 sin 3t φ 3 } 65 8 In Fig 5a the orientation of the focus is otained y considering the curves as they cut the positive z -axis vertical dashed axis in fig. Thus with z and z > and y considering ż 7z 39z we deduce that ż >. Thus as the curves cut the positive z -axis z increases, hence the rotation is clockwise.

8 Dynamical Systems Part 5-6 Dr G Bowtell 8 ẋ 6 5 x x 6 a A λ 6 Eigenvalues given y the solution of det λ 5 λ 5λ 6 Hence the eigenvalues are λ 3 and λ. The eigenvector corresponding to λ 3 is given y: 3 6 a a a take E The eigenvector corresponding to λ is given y: 6 a a a take E 3 Since the eigenvalues are real different and oth positive the fixed point is an unstale node. The solution is given y: x z a α e 3t 3 β e t 4 c ẋ 4 9 x x a A λ 9 Eigenvalues given y the solution of det λ λ Hence there is only a single eigenvalues λ The eigenvector corresponding to λ is given y: 3 9 a a 3 3 a 3 3 take E 3 Construct the J vector to form the new ais: a 3 A λij E a a take J

9 Dynamical Systems Part 5-6 Dr G Bowtell 9 c x z vertical cut z 4z / E z x Figure 5: Diagrams for Q8ac respectively; Stale focus, unstale node and stale improper inflected node Since the repeated eigenvalue is negative the fixed point is a stale inflected improper node. The solution is given y: x z a αt βe t 3 αe t 5 9. For the linear system ẋ Ax the origin is a fixed point. Since det A cos θ sin θ the fixed point is simple. Eigenvalues given y: deta λi cos θ λ sin θ λ cos θ ±i sin θ λ cos θ ± i sin θ a star node A λi, thus sin θ θ or λ cos θ To e stale cos θ λ < θ As a ut to e unstale cos θ λ > θ c Centre if λ purely imaginary. Thus cos θ θ / d Stale focus if sin θ and cos θ <. Thus / < θ < e Unstale focus if sin θ and cos θ >. Thus < θ < /. In general for the system ẋ X x, x ẋ X x, x the Jacoian is given y J X x X X x X x x Thus: x x a J x x x cos x cos x sin x sin x

10 Dynamical Systems Part 5-6 Dr G Bowtell x x J sin x e x x x x cos x sin x x sin x x cos x cos x c J x x sin x x sin x x cos x cos x. In each case it is clear that X, and X,, thus the origin is a fixed points for each of the systems in Q9. Evaluating each of the aove J at the origin gives the three linearisations as follows: a ż z Since det J the fixed point is not simple, hence no classification. ż z Since det J the fixed point is simple. The eigenvalues of J are ± thus the origin is a saddle point. c ż z Since det J the fixed point is simple. J is already in the canonical form for an unstale improper node. single eigenvalue λ 3. The Eigenvalues are given y the solution of det 6 5 λ Hence the eigenvalues are λ 3i and λ 3i. The eigenvector corresponding to λ 3i is given y: 3i 3 a 6 5 3i 3 3ia 3 a 3 3 3i i take E λ 4λ 3 3 3i i i a E i Thus P 3 and J 3 ẏ ẏ 3 3 y y ẏ y 3y i ẏ 3y y ii Sustitute y r cos θ and y r sin θ into i and ii to give: ṙ cos θ r θ sin θ r cos θ 3r sin θ iii ṙ sin θ r θ cos θ 3r cos θ r sin θ iv cos θ iii sin θ iv ṙ r r r e t

11 Dynamical Systems Part 5-6 Dr G Bowtell sin θ iii cos θ iv r θ 3r θ 3 θ 3t θ Thus: y r e t cos 3t θ and y r e t sin 3t θ The solution to ẋ Ax is given y: x r e t {cos 3t θ sin 3t θ } where r and θ are aritrary constants. 3. Fixed points at x cos x and x cos x. In the given region this give five fixed points:,,,,, The Jacoian is given y cos x x J sin x x sin x cos x Thus the linearisation aout the fixed points are:, ż z ±, ± ż z ±, ż z Clearly, is an unstale star node. At ±, ± the eigenvalues are given y det λ saddle points At ±, the eigenvalues are given y det λ saddle points λ λ λ λ ± hence λ λ ± hence From the original equations dx x cos x dx x cos x Clearly x x satisfies this equation with each side equal to. Similarly x x satisfies the equation with each side equal to. The x -axis is given y x. Sustituting this into the original equations gives ẋ x for the first equation and trivially for the second equation. ẋ x x Ce t thus if C > this generates the positive x -axis and if C < the negative x -axis. Thus the x -axis is a trajectory. Similarly the x -axis is trajectory. We have estalished that oth the axes and the lines x ±x are trajectories. These along with the fixed points are exact trajectories and can e added with certainty onto the phase diagram. The details from the linearisations are then used to construct a reasonale approximation to the gloal

12 Dynamical Systems Part 5-6 Dr G Bowtell phase diagram. In some cases it is not always clear how the linear diagrams link together to give a gloal diagram. Rememer the linear phase diagrams are only approximations to the true picture and only remain good near their respective fixed points. The saddles at each of the four fixed points can e positioned using the eigenvectors of the linearised system aout each point. The usual calculation leads to the eigenvectors at the two points ±, ± eing E and E relative to the axes through the points. As we see in each case the eigenvector E lies along the trajectory of the non-linear system given y x x. However the eigenvector E is not a trajectory of the non-linear system aout either fixed point and can only e used as a guide when drawing the saddle. These are shown in Fig 6 as dotted lines. Similar remarks are also applicale at the other two fixed point, namely ±,. The completed diagram is shown in Fig 6. The arrows can e put on the diagram fairly easily y noting for example that the origin is unstale and all trajectories leave this point with increasing t Figure 6: Q3. The x and x axes are trajectories; the two lines x ±x are trajectories; Saddles at the four fixed points ±, and ±, ; star node at the origin 4. Fixed points at x x x x x x. For all points on the hyperola x /x, x x thus all points are fixed points. x x J x x det J x x, det thus the origin is simple det J 3x x 4x x on the hyperola x x det J 3 4

13 Dynamical Systems Part 5-6 Dr G Bowtell 3 Thus all points on the hyperola are non-simple fixed points. Setting x in Eq: The second equation is satisfied and the first implies ẋ x x Ae t, which gives the whole of the x axis for different values of A. Similarly the x axis is a trajectory. From Eq Solving: dx x x x dx x x x x x x x dx dx x, x x Ax x x This gives all straight lines through the origin, omitting the origin and points on the hyperola. x x Figure 7: Note the apparent change of direction of the trajectories as they approach the hyperola of fixed points. 5. No solution given.