TWO DIMENSIONAL FLOWS Lecture Linear and Nonlinear Systems

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TWO DIMENSIONAL FLOWS Lecture Linear and Nonlinear Systems Powered By Docstoc
					TWO DIMENSIONAL
      FLOWS


Lecture 4: Linear and
 Nonlinear Systems
4. Linear and Nonlinear Systems in 2D

In higher dimensions, trajectories have more
room to manoeuvre, and hence a wider range
of behaviour is possible.

4.1 Linear systems: definitions and examples

A 2-dimensional linear system has the form
                ˙
                x = ax + by
                ˙
                y = cx + dy
where a, b, c, d are parameters. Equivalently,
in vector notation
                   ˙
                   x = Ax                   (1)
where
             a b                   x
        A=            and    x=             (2)
             c d                   y
The Linear property means that if x1 and x2
are solutions, then so is c1x1 + c2x2 for any
c1 and c2.

The solutions of x = Ax can be visualized
                  ˙
as trajectories moving on the (x, y) plane, or
phase plane.
                                        1
Example 4.1.1 m¨ + kx = 0 i.e. the simple
                 x
harmonic oscillator




                  Fig. 4.1.1

The state of the system is characterized by
          ˙
x and v = x

                  ˙
                  x = v
                           k
                  ˙
                  v = −      x
                           m
                                         ˙ ˙
i.e. for each (x, v) we obtain a vector (x, v) ⇒
vector field on the phase plane.

                                          2
As for a 1-dimensional system, we imagine a
fluid flowing steadily on the phase plane with
a local velocity given by (x, v) = (v, −ω 2x).
                           ˙ ˙




                 Fig. 4.1.2


 • Trajectory is found by placing an imag-
   inary particle or phase point at (x0, v0)
   and watching how it moves.


 • (x, v) = (0, 0) is a fixed point:
   static equilibrium!


 • Trajectories form closed orbits around (0, 0):
   oscillations!

                                        3
The phase portrait looks like...




                 Fig. 4.1.3


 • NB ω 2x2 + v 2 is constant on each ellipse.
   This is simply the energy


Example 4.1.2

            ˙
            x        a 0           x
                =
            ˙
            y        0 −1          y

                                        4
The phase portraits for these uncoupled equa-
tions are...




                   Fig. 4.1.4

Solution is
               x         x0eat
                     =
               y         y0e−t



                                       5
Some terminology...


 • x∗ = 0 is an attracting fixed point in Figs
   (a) - (c) since x(t) → x∗ as t → ∞.


 • x∗ = 0 is called Lyapunov Stable in Figs
   (a) - (d) since all trajectories that start
   sufficiently close to x∗ remain close to it
   for all time.


 • Fig. (d) shows that a fixed point can
   be Lyapunov stable but not attracting ⇒
   it is neutrally stable. It is also possible
   for a fixed point to be attracting but not
   Lyapunov stable!


 • If a fixed point is both Lyapunov sta-
   ble and attracting, we’ll call it stable, or
   sometimes asymptotically stable


 • x∗ is unstable in Fig. (e) because it is
   neither attracting nor Lyapunov stable

                                         6
4.2 Classification of Linear Systems

Consider a general 2 ×2 matrix A such that
˙
x = Ax

To solve: try

    x(t) = eλtv (v is a constant vector)
 ⇒ λeλt v = eλt Av
  ⇒ Av = λv
Hence if we obtain the eigenvectors v and
eigenvalues λ, we will have two independent
                                  a b
solutions x(t). Recall that A =         has
                                  c d
eigenvalues λ1 and λ2, where

        τ+     τ 2 − 4∆          τ−   τ 2 − 4∆
 λ1 =                     λ2 =
                2                      2

        with    τ = trace(A) = a + d
               ∆=    det(A)      = ad − bc


                                             7
 • Useful check when calculating eigenval-
   ues: λ1 + λ2 = τ and λ1λ2 = ∆


                ˙
                x          1 1    x
Example 4.2.1       =
                ˙
                y          4 −2   y


                             1
 • ⇒ λ1 = 2 with v1 =             λ1 > 0
                             1
   hence solution grows


                             1
 • ⇒ λ2 = −3 with v2 =                λ2 < 0
                             −4
   hence solution decays




                Fig. 4.2.1
                                       8
 • straight line trajectories in Fig. 4.2.1 are
   the eigenvectors v1 and v2


Example 4.2.2 Consider λ2 < λ1 < 0




                 Fig. 4.2.2


 • Both solutions decay exponentially!




                                         9
Example 4.2.3 What happens if λ1, λ2 are
complex?

Fixed point is either...




                  Fig. 4.2.3


 • If λ1, λ2 are purely imaginary, all solutions
   are periodic


 • If λ1 = λ2 we get a star node or a degen-
   erate node



                                          10
Classification of Fixed Points

λ1,2 = 1 (τ ± τ 2 − 4∆),
       2                 where
∆ = λ1λ2 and τ = λ1 + λ2




                 Fig. 4.2.4
                                 11
4.3 Phase Portraits

       ˙
Recall x = f (x), i.e.

               x˙1 = f1(x1, x2)
               x˙2 = f2(x1, x2)
where x = (x1, x2) and f (x) = (f1(x), f2(x))
(not necessarily linear now). The trajectories
x(t) wind their way through the phase plane.




The entire phase plane is filled with trajec-
tories!

4.4 Example of a phase portrait

- Shows a sample of the qualitatively different
trajectories
                                        12
                 Fig. 4.4.1


• Fixed points A, B and C satisfy f (x∗ ) = 0
  and correspond to steady states or equi-
  libria


• Closed orbit D corresponds to periodic
  solutions, i.e. x(t + T ) = x(t) for all t for
  some T > 0


• The existence and uniqueness theorem
  given for 1-dimensional systems can be
  generalized to 2-dimensional systems ...
  fortunately ⇒ different trajectories never
  intersect!

                                          13
4.5 Fixed points and Linearization

This is the same idea as for 1-dimensional
systems
                 ˙
                 x = f (x, y)
                 ˙
                 y = g(x, y)
Suppose (x∗, y ∗) is a fixed point. Expand
around (x∗, y ∗) using u = x−x∗ and v = y−y ∗.
u = x = f (x∗ + u, y ∗ + v)
˙   ˙
                        ∂f     ∂f
              ∗ ∗
       = f (x , y ) + u    +v     + O(u2, v 2, uv)
                        ∂x     ∂y
            ∂f     ∂f
        ≃ u    +v
            ∂x     ∂y
Similarly
                     ∂g     ∂g
               ˙
               v≃u      +v
                     ∂x     ∂y
Hence a small disturbance around (x∗, y ∗) evolves
as
                    ∂f    ∂f
                            
            ˙
            u                    u
                =  ∂x
                    ∂g
                          ∂y 
                          ∂g
            ˙
            v                    v
                    ∂x    ∂y
where the matrix is known as the Jacobian
matrix A at (x∗, y ∗), and is the multivariable
equivalent of f ′ (x∗) for 1-D systems.
                                          14
Example 4.5.1

                     x = −x + x3
                     ˙
                     ˙
                     y = −2y
                          ˙
Fixed points occur where x = 0 and y =   ˙
0 simultaneously. Hence x = 0 or x = ±1
and y = 0 ⇒ 3 fixed points (0, 0), (1, 0) and
(−1, 0)

Jacobian matrix A
             ˙
            ∂x    ˙
                 ∂x
                   
                            −1 + 3x2 0
    A   =  ∂x
             ˙
            ∂y
                 ∂y 
                  ˙
                 ∂y   =
                               0     −2
            ∂x   ∂y


                       −1 0
At (0, 0)       A=             ⇒ stable node
                        0 −2


                        2 0
At (±1, 0)      A=             ⇒ both are sad-
                        0 −2
dle points.


                                          15
                  Fig. 4.5.1

In general, we must obtain fixed points by
        ˙         ˙
solving x = 0 and y = 0 simultaneously.

          e.g.   ˙
                 x = x(3 − x − 2y)
                 ˙
                 y = y(2 − x − y)
yields fixed points (0, 0), (0, 2), (3, 0) and
(1, 1)

In general, A will not be diagonal at (x∗, y ∗).
Hence we must diagonalize A, i.e. find eigen-
values λ1 and λ2 and eigenvectors v1 and v2
of A
                                          16
Basically, we are doing the same here as be-
fore for 2D linear systems, since we are treat-
ing the nonlinear system as linear near (x∗, y ∗).
Knowledge of λ1 and λ2, and v1 and v2, en-
ables us to sketch the phase portrait near
(x∗, y ∗).

The fixed points can be classified according
to their stability as follows:


 • If Re(λ1) > 0 and Re(λ2) > 0
   ⇒ repeller (unstable node)


 • If Re(λ1) < 0 and Re(λ2) < 0
   ⇒ attractor (stable node)


 • If Re(λ1) > 0 but Re(λ2) < 0 (or vice
   versa) ⇒ saddle


 • If λ1 and λ2 are both imaginary ⇒ centre

                                           17
4.6 Example: Rabbits vs Sheep

An example of the Lotka-Volterra model of
competition between two species (e.g. rab-
bits and sheep) grazing the same food supply
(grass).


 • Each species grows to its carrying capac-
   ity in the absence of the other - logistic
   growth (rabbits faster...!)


 • When species encounter each other, the
   larger (sheep) has an advantage.


 • Conflicts occur at a rate proportional to
   the size of each population. Conflicts re-
   duce the growth rate of each species (but
   more for rabbits).


A model encapsulating these properties could
be (see above!)
            ˙
            x = x(3 − x − 2y)
            ˙
            y = y(2 − x − y)
                                       18
Fixed points at
                         3 0
(0, 0)   where    A =           ⇒ λ = 3, 2
                         0 2
                         −1 0
(0, 2)   where    A =               ⇒ λ = −1, −2
                         −2 −2
                         −1 −2                √
(1, 1)   where    A =               ⇒ λ = −1 ± 2
                         −1 −1
                         −3 −6
(3, 0)   where    A =               ⇒ λ = −3, −1
                         0 −1

(0, 0): λ = 3, 2 ⇒ unstable node (repeller)




λ = 2 ⇒ v = (0, 1)      “slow eigendirection′′
λ = 3 ⇒ v = (1, 0)      “fast eigendirection′′
General rule...

Trajectories are tangential to the slow
eigendirection (i.e. smallest |λ|) at a node
                                         19
(0, 2): λ = −1, −2 ⇒ stable node (attrac-
tor)




 • Once again... Trajectories are tangential
   to the slow eigendirection at a node


 • Here λ = −1 ⇒ v = (1, −2) is the slow
   eigendirection.

                √
(1, 1): λ = −1 ± 2 ⇒ saddle point




                                      20
(3, 0): λ = −3, −1 ⇒ stable node (attrac-
tor)




Putting these together, the phase portrait
becomes....




                Fig. 4.6.1

NB: You don’t really need to calculate the
eigenvectors to get the right shape!
                                     21
Biological interpretation...


 • In general, one species eventually drives
   the other to extinction; which species even-
   tually dominates depends on initial pop-
   ulations x0 = (x0, y0)


 • Basin of attraction of an attracting fixed
   point x∗ defined as the set of initial con-
   ditions x0 such that x → x∗ as t → ∞.


 • In this case, basin boundary is the stable
   manifold of the saddle node at (1, 1)




                  Fig. 4.6.2
                                        22
4.7 Conservative Systems

          ˙
Consider x = f (x). A conserved quantity of
this system is a real-valued continuous func-
tion E(x) that is constant on trajectories i.e.
dE/dt = 0.

Example 4.7.1 m¨ = −dV (z)/dz = F (z)
               z

                   ˙
Take x = z and y = z ⇒

                ˙
                x = y
                    1
                ˙
                y =   F (x)
                    m
        1 ˙
E(z) = 2 mz 2 +V (z) is the total energy, which
is constant
                        1
          ⇒ E(x) ≡        my 2 + V (x)
                        2
            dE(x)
                   = 0
              dt
since total energy is constant.


                                         23
              ¨
Example 6.5.2 θ + sin θ = 0




e.g. undamped simple pendulum
                  ˙
                  θ = ν
                 ˙
                 ν = − sin θ
Fixed points at (θ ∗, ν ∗) = (kπ, 0)

                  0 1
(0, 0) :   A=              ⇒ λ = ±i ⇒ centre
                  −1 0
(oscillations)

                    1
Energy E(θ, ν) = 2 ν 2 − cos θ is conserved,
since
      dE
             ˙         ˙     ¨
         = ν ν + sin θ θ = ν[θ + sin θ] = 0
      dt

                  0 1
(π, 0) :   A=            ⇒ λ = ±1 ⇒ saddle
                  1 0
                                       24
Phase portrait becomes...




                Fig. 4.7.1




                             25

				
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