MATH 44041/64041 Applied Dynamical Systems, 2018

Transcription

1 MATH 4441/6441 Applied Dynamical Systems, 218 Answers to Assessed Coursework 1 1 Method based on series expansion of the matrix exponential Let the coefficient matrix of α the linear system be A, then α A 2 α α α α α α α I 2, where I 2 is the 2-by-2 identity matrix Therefore, for any integer n we have A 2n A 2 n α I 2 n α n I 2 2n I 2 and A 2n+1 A 2n A α n A 2n A with α Using the definition of matrix exponential, t n Mt expta n! An n t 2n t 2n+1 2n! A2n + 2n+1! A2n+1 separating the even and odd powers n n t 2n 2n! I 1 t 2n n+1! A n n 1 cosht + sinht α 1 α cosht+ α sinht sinht sinht cosht α sinht 1 Method based on diagonalisation with eigenvectors First, the eigenvalue λ of A satisfies the equation λ α detλi A det λ 2 α λ+α That is, the two elgenvalues are λ ± ± α ± The two eigenvectors associated with the two eigenvalues are v +, v + α α α α Let S be the 2-by-2 matrix whose columns are v + and v, ie, S [ v + v ] α α then S α α 1 +α 2 α

2 Then from A S S 1, we get expt Mt expta S S 1 exp t 1 expt 2 α α αexpt exp t+expt+exp t 2 2 α 2 expt exp t 2 +α exp t α expt exp t 2 expt exp t αexpt+exp t 2 which can be simplified to get the same matrix 1 above Method based on expansion of the solution using eigenvectors of the coefficient matrix This approach is basically the same as previous approach, where everything is presented in vectors instead of matrices but we still try to use matrices here whenever possible The starting point is that the solution can be written as xt c yt + e t v + +c e t v, where c ± are constants determined by the initial condition, and v ± are the eigenvectors of the coefficient matrix as above Then the initial condition at t becomes x c1 c1 c + v + +c v [ v + v ] S In other words, the coefficients c ± are given by c1 y c 2 S 1 x y 1 2 c 2 c 2 +α x α y Therefore, the solution can be written as xt c yt + e t v + +c e t c+ e v [ v + v ] t e t c+ e t c e t [ v + v ] e t S c e t S 1 x, y e t giving exactly the same exponential matrix Mt S e t S 1 as above Method based on explicit solutions by solving the equivalent second order ODEs From the first differential equation ẋ αx+y, we get y ẋ αx Then the second equation ẏ x αy becomes d ẋ αx dt x αẋ αx, which simplifies to ẍ α x The solution to this constant coefficient second order ODE is given by xt Acosht+Bsinht with α and constants A,B determined by the initial condition x,y Using y ẋ αx, we get y ẋ αx 1 Applying the initial condition at t, [ ] Asinht+Bcosht αacosht+bsinht x A, y B αa,

3 or A x and B y +αa/ x α+y / Finally the solution can be written as xt x cosht+ x α+y sinht [cosht+ α ] sinht x + sinhty, and yt 1 [ x sinht+ x α+y α2 sinhtx + cosht α cosht α sinht y x cosht+ x α+y Since 2 α 2 + 2, α2 2 α 2 Therefore, the solution can be written as ] xt [cosht+ α sinht x + sinhty yt sinhtx + cosht α sinht y cosht+ α sinht sinht x sinht cosht α sinht y and Mt is precisely the coefficient matrix ] sinht Remark a It is always a good idea to assign α as another variable, like above, to simplify the presentation; b In general, it is simpler to use sinht and cosh than the exponential function e ±t, because sinht vanishes at the origin and the initial condition are easier to work with 2 a Let the equation for the circle be Gx,y x 2 + y b 2 R 2 with centre at,b and radius R Therefore from Ġ 2ẋx+2ẏy b 2xx2 y 2 +4xyy b 2xx 2 +y 2 2by, we get d dt G Gx,y 2xR 2 b 2 If the circle is invariant, we must have 2xR 2 b 2 or R b The constant b can be obtained from the initial condition, that is, b 2 Gx,y x 2 +y b 2 or b x 2 +y2 /2y, and the equation of the circle is x 2 +y 2 2 x2 +y2 y y Alternative approach used to construct general invariant set The invariant circle can also be obtained by solving the first order ODE dy 2xy, where the integration constant is determined dx x 2 y 2 the initial condition Since this is a homogeneous ODE, we can use the change of variable y ux, such that the original ODE is reduced to u+x du dx 2xy x 2 y 2 2u 1 u 2, or x du dx 2u 1 u 2 u u+u3 1 u 2 This new ODE is separable, and can be written as Integrating both sides, we get dx x u+u3 1 u 2du u+u 3 lnx+c 1 u 2du lnu ln1+u2,

4 or u 1+u 2 xec The constant C or e C is determined by the initial condition x,y with u y /x That is The invariant set is then given by which can be written as e C y x 2 +y 2 u x 1+u 2 y x 2 +y2 u x1+u 2 ec y, x 2 +y2 x 2 +y c 2 r 2, which is a cirlce with the y-coordinate of the centre c x 2 +y 2 /2y and the radius r c b Since r 2 x 2 +y 2, we have 2rṙ 2xẋ+2yẏ 2xx 2 y 2 +4xy 2 2xx 2 +y 2 2xr 2, or ṙ xr r 2 cosθ From θ arctany/x, ẏ 1 d y θ 1+y 2 /x 2 dtx 1 1+y 2 /x 2 x ẋy x 2 xẏ yẋ x 2 +y 2 y rsinθ Threfore, the system in polar coordinate is ṙ r 2 cosθ, θ rsinθ c The above system in polar coordinate imples that dr dθ r2 cosθ rsinθ rcosθ with the initial sinθ condition r 1,θ π/2 This is a linear ODE of r as a function of θ with integrating factor 1/ sin θ Therefore, from d r 1 dr dθ sinθ sinθdθ rcosθ sin 2 θ, we get r/sinθ r /sinθ 1 is conserved This is exactly the expression of the invariant circle x 2 +y 2 +y from a Using the relation r sinθ, the equation for θ is then θ rsinθ sin 2 θ, which is a separable ODE and can be integrated to obtain t θt θ dθ sin 2 θ cotθ cotθt cotθt, or θt arccot t π π, π/2 for t > and rt sinθt sinarccot t 1/ 1+t 2 In the original coordinate, xt rtcosθt 1 1+t 2 cosarccott π t 1+t 2, yt rtsinθt sin2 θt 1 1+t 2 Alternative method The solution can also be obtained by taking x and y be the real and imaginary parts of a complex number z That is, if we define zt xt+iyt, then ż ẋ+iẏ x 2 y 2 +2xyi x+iy 2 z 2,

5 with initial condition z x + iy i Complex valued differential equations can be solved exactly in the same way; this separable ODE can be written as dt z 2 dz That is, zt zt t z 2 dz d1/z 1 z z z 1 zt, or zt z 1 tz i 1+it i1 it 1+it1 it t i 1+t 2 Thefore, xt Re zt t 1+t 2,yt Imzt 1 1+t 2 Remark 1 The alternative approach in part a may not work for more complicated system, because the ODE for the trajectories may not be solved explicitly 2 the solution in c can be solved in many different, but there are subtle issues when one variable is eliminated For example, θ can be eliminated using the relation r sinθ in ṙ r 2 cosθ, using cosθ ± 1 sin 2 θ 1 r 2 We must choose the negative sign, to get the separable ODE ṙ r 2 1 r 2 That is, rt dr t+c r r 2 1 r 2 1 r 2 r rt 1 1 rt 2 rt The initial condition r 1 implies rt 1/ 1+t 2 Similarly, the relation x 2 +y 2 +y applying the initial condition, 1 can also be used to find the solution For instance, x y 2 +y we should choose the negative sign here, which is true at least for t close to zero, and we get the separable ODE ẏ 2xy 2y y 2 +y The solution can be obtained, by the fact that dy y 2 y+1 +C 2 +y y 3 a This is a separable ODE Integrating both sides of x 1dx 2tdt with the initialcondition x, we get y 2sds t 2 xt x x 1dx 1 xt 2 x2 x 1 x 2 xt2 xt That is xt t 2, or xt 1± 1+2t 2 The initial condition x implies the choice of the negative sign and the solution is xt 1 1+2t 2 b The equivalent integral equation is 2s t xt x + xs 1 ds 2s xs 1 ds and the Picard s iteration for the sequence of approximations {x n t} is given by Therrefore x t x, and x 2 t x 1 t x n+1 t 2s x n s 1 ds 2s x s 1 ds 2sds t 2 2s t x 1 s 1 ds 2s t t s 2 1 ds 1 s 2 +1 ds2 +1 ln1+s 2 ln1+t 2

6 Remark 1 Once the initial condition is specified, any ambiguity in the solution like two signs for the sqrt root can be clarified 2 In general, the approximate solution x n t at n-th step agrees with the exact solution up to and including order n That is, xt x n t o t n, as we can verify 4 a From the definition, ϕ s ϕt x,y ϕ s x e t,y e t +x 2 e t e 2t x e t e s,y e t +x 2 e t e 2t e s +x e t 2 e s e 2s x e t+s,y e t+s +x 2 e t+s e 2t+s +x 2 e 2t+s e 2t+2s x e t+s,y e t+s +x 2 e t+s e 2t+2s ϕ t+s x,y Therefore, ϕ t satisfies the semi-group property b Let x,y ϕ t x,y x e t,y e t + x 2 e t e 2t or equivalently, x xe t,y ye t x 2 1 e t ye t x 2 e 2t e t Taking the derivative of x,y ϕ t x,y with respect to t, d dt x,y d x e t,y e t +x 2 dt e t e 2t x e t, y e t +x 2 e t +2e 2t Therefore Fx,y ẋ x e t x and Gx,y ẏ y e t +x 2 e t +2e 2t ye t x 2 e 2t e t e t +xe t 2 e t +2e 2t y +x 2 Alternatively, we can apply the definition ẋ,ẏ Fx,y,Gx,y at t with x x,y y, to get Fx,y d dt x t x e t t x which agrees with the first approach Gx,y d dt y t y e t +x 2 e t +2e 2t t y +x 2, 5 a The stationarypoints satisfy y xy,x y, or y x x 2 Therefore the stationary points are, and 1, 1 y 1 x b The Jacobian matrix is Jx,y At the fixed point,, A J, 1 1 1, the two eigenvalues satisfy detλi A λ λ+1, or λ ± 1± 3i 2 This is a stable focus To find the direction of the spiral, we can take points on the positive axis, say 1,, then the local direction field is 1 1, pointing upwards Therefore, the spiral rotates in counter-clockwise direction

7 e 2 e 1 Figure 1: The local phase portrait for the system ẋ y xy,ẏ x y near, left figure and 1, 1 right figure 1 At the fixed point 1, 1, A J 1, 1 The two eigenvalues are λ ,λ 2 1 the eigenvalues of the lower or upper triangular matrix are exactly the diagonal entries The eigenvector e 1 corresponding the eigenvalue λ 1 1 satisfies A λ 1 Ie 1 e 1 2 1, that is e 1 that is e 2 2 The eigenvector e 1 2 corresponding to λ 2 1 satisfies 1 A λ 2 Ie 2 c The global phase portrait is shown in Figure 2 2 e 1 2, Remark 1 There is no need to find the complex eigenvectors associated with stable/unstable focus what is important is the direction of rotation 2 Do not introduce new intersection points of trajectories, when drawing the phase portrait If possible, check whether there are any obvious invariant sets, like the vertical line passing the saddle in this question

8 Figure 2: Global phase portrait for the system ẋ y xy,ẏ x y