Engineering Mathematics Class

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Engineering Mathematics Class 10 Chapter 4
(Part2)

• Sheng-Fang Huang

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Five Types of Critical Points
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Improper Node

• The system
• The two eigenvectors are x(1) 1 1T and
  x(2) 1 -1T.

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Proper Node

• The system
• has a proper node at the origin. Its
  characteristic equation (1 ?)2 0 has the root
  ? 1. Any x ? 0 is an eigenvector, and we can
  take 1 0T and 0 1T. Hence a general
  solution is

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Fig. 82. Trajectories of the system (10) (Proper
node)
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Saddle Point

• The system
• has a saddle point at the origin. Its
  characteristic equation (1 ?)(1 ?) 0 has the
  roots ?1 1 and ?2 1. For ?1 1 an
  eigenvector 1 0T is obtained from the second
  row of (A ?I)x 0, that is, 0x1 (1 1)x2
  0. For ?2 1 the first row gives 0 1T. Hence
  a general

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Fig. 83. Trajectories of the system (11) (Saddle
point)
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Canter

• The system
• (12)
• has a center at the origin. The
  characteristic equation ?2 4 0 gives the
  eigenvalues 2i and 2i. For 2i an eigenvector
  follows from the first equation 2ix1 x2 0 of
  (A ?I)x 0, say, 1 2iT. For ? 2i that
  equation is (2i)x1 x2 0 and gives, say, 1
  2iT. Hence a complex general solution is
• (12)

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• Rewrite the given equations in the form y'1
  y2, 4y1 y'2 then the product of the left
  sides must equal the product of the right sides,

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Fig. 84. Trajectories of the system (12) (Center)
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Spiral Point

• The system
• has a spiral point at the origin, as we shall
  see. The characteristic equation is ?2 2? 2
  0. It gives the eigenvalues 1 i and 1 i.
  Corresponding eigenvectors are obtained from (1
  ?)x1 x2 0. For ? 1 i this becomes ix1
  x2 0 and we can take 1 iT as an
  eigenvector. This gives the complex general
  solution

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• Accordingly, we start again from the beginning
  and instead of that rather lengthy systematic
  calculation we use a shortcut. We multiply the
  first equation in (13) by y1, the second by y2,
  and add, obtaining
• y1y'1 y2y'2 (y12
  y22).
• We now introduce polar coordinates r, t, where
  r2 y12 y22. Differentiating this with respect
  to t gives 2rr' 2y1y'1 2y2y'2. Hence the
  previous equation can be written
• rr' r2,
• Thus, r' r, dr/r dt, lnr t c,
  r ce-t.

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Fig. 85. Trajectories of the system (13) (Spiral
point)
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4.4 Criteria for Critical Points. Stability

• We continue our discussion of homogeneous linear
  systems with constant coefficients
• (1)

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• (3)
• We also recall from Sec. 4.3 that there are
  various types of critical points, and we shall
  now see how these types are related to the
  eigenvalues. The latter are solutions ??1 and ?2
  of the characteristic equation
• (4)

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• This is a quadratic equation ?2 p? q 0
  with coefficients p, q and discriminant ? given
  by
• p a11 a22,
• q det A a11a22 a12a21,
• ? p2 4q.
• From calculus we know that the solutions of
  this equation are
• (6)

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Table 4.1 Eigenvalue Criteria for Critical Points
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Stability
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Fig. 89. Stable critical point P0 of (1) (The
trajectory initiating at P1
stays in the disk of radius e.)
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Stability
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Fig. 90. Stable and attractive critical point P0
of (1)
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Table 4.2 Stability Criteria for Critical Points
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Stability Chart
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Example 1

• In Example 1, Sec. 4.3, we have y'
  y, p
• 6, q 8, ? 4, a node by Table 4.1(a),
  which is stable and attractive by Table 4.2(a).

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Example 2 Free Motions of a Mass on a Spring

• What kind of critical point does my" cy' ky
  0 in Sec. 2.4 have?
• Solution. Division by m gives y'' (k/m)y
  (c/m)y'. To get a system, set y1 y, y2 y'.
  Then y'2 y'' (k/m)y1 (c/m)y2. Hence

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• We see that p c/m, q k/m, ? (c/m)2
  4k/m. From Tables 4.1 and 4.2, we obtain the
  following results. Note that in the last three
  cases the discriminant ? plays an essential role.
• No damping. c 0, p 0, q gt 0, a center.
• Underdamping. c2 lt 4mk, p lt 0, q gt 0, ? lt 0, a
  stable and attractive spiral point.
• Critical damping. c2 4mk, p lt 0, q gt 0, ? 0,
  a stable and attractive node.
• Overdamping. c2 gt 4mk, p lt 0, q gt 0, ? gt 0, a
  stable and attractive node.

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4.5 Qualitative Methods for nonlinear Systems

• Qualitative methods are methods of obtaining
  qualitative information on solutions without
  actually solving a system.
• These methods are particularly valuable for
  systems whose solution by analytic methods is
  difficult or impossible.

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• Phase plane method is a kind of qualitative
  method.
• In this section, we will extend phase plane
  method from linear system to nonlinear systems
• 
  y'1 f1(y1, y2)
• (1) y' f(y), thus
• y'2 f2(y1, y2)
• We assume that (1) is autonomous.
• Constant coefficients the independent variable t
  does not occur explicitly.

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• As a convenience, each time we first move the
  critical point P0 (a, b) to be considered to the
  origin (0, 0). This can be done by a translation
• which moves P0 to (0, 0). Thus we can assume
  P0 to be the origin (0, 0), and for simplicity we
  continue to write y1, y2 (instead of y1, y2).

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Linearization of Nonlinear Systems

• How to determine the kind and stability of a
  critical point P0 (0, 0) of nonlinear system?
• Linearization
• Rewrite (1) as
• 
  y'1 a11y1 a12y2 h1(y1, y2)
• (2) y' Ay h(y), thus
• 
  y'2 a21y1 a22y2 h2(y1, y2)
• A is constant (independent of t). One can prove
  that

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Example 1 Free Undamped Pendulum.

• A pendulum consisting of a body of mass m and a
  rod of length L. Determine the locations and
  types of the critical points.

Assume that the mass of the rod and air
resistance are negligible.
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Example 1 Free Undamped Pendulum.

• Solution. Step 1. Setting up the mathematical
  model.
• Let ? denote the angular displacement, measured
  counterclockwise from the equilibrium position.
  The weight of the bob is mg (g the acceleration
  of gravity).
• The mathematical model is
• mL?'' mg sin? 0.
• Dividing this by mL, we have
• (4) ?'' k sin? 0

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• Step 2. Critical points (0, 0), (2p, 0), (4p,
  0), ?? , Linearization. To obtain a system of
  ODEs, we set ? y1, ?' y2. Then from (4) we
  obtain a nonlinear system (1) of the form
• y'1 1(y1, y2) y2
• (4)
• y'2 2(y1, y2) k
  sin y1.

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• The right sides are both zero when y2 0 and
  sin y1 0. This gives infinitely many critical
  points (np, 0), where n 0, 1, 2, ??. We
  consider (0, 0). Since the Maclaurin series is
• the linearized system at (0, 0) is

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• To apply our criteria in Sec. 4.4
• p a11 a22 0
• q det A k g/L (gt 0)
• ? p2 4q 4k.
• Thus, we conclude that (0, 0) is a center, which
  is always stable. Since sin? sin y1 is periodic
  with period 2p, the critical points (np, 0), n
  2, 4, ?? , are all centers.

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• Step 3. Critical points (p, 0), (3p, 0), (5p,
  0), ?? , Linearization.
• We now consider the critical point (p, 0),
  setting ?p y1 and (?p)' ?' y2. Then in
  (4),
• the linearized system at (p, 0) is now

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• We see that
• p 0,
• q k (lt 0)
• ? 4q 4k.
• Hence, this gives a saddle point, which is always
  unstable.
• Because of periodicity, the critical points (np,
  0), n 1, 3, ??, are all saddle points.

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Fig. 92. Example 1 (C will be explained in
Example 4.)
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Example 2 Linearization of the Damped Pendulum
Equation

• Now we add a damping term c?' (damping
  proportional to the angular velocity) to equation
  (4), so that it becomes
• (5) ?'' c?' k sin? 0
• where k gt 0 and c 0. Setting ? y1, ?'
  y2, as before, we obtain the nonlinear system
  (use ?'' y'2 )
• y'1 y2
• y'2 k sin y1 cy2.

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• Critical points as before, namely,
• (0, 0), (p, 0), (2p, 0), ??.
• We consider (0, 0). Linearizing sin y1 y1 as
  in Example 1, we get the linearized system at (0,
  0)
• (6)
• p
• q
• ?

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• We now consider the critical point (p, 0). We
  set ?p y1, (?p)' ?' y2 and linearize
• sin? sin (y1 p) sin
  y1 y1.
• This gives the new linearized system at (p,
  0)
• (6)

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• p a11 a22 c,
• q det A k
• ? p2 4q c2 4k.
• This gives the following results for the
  critical point at (p, 0).
• No damping.
• c 0, p 0, q lt 0, ? gt 0, a saddle point.
• Damping.
• c gt 0, p lt 0, q lt 0, ? gt 0, a saddle point.
• Since sin y1 is periodic with period 2p, the
  critical points (2p, 0), (4p, 0), ??are of the
  same type as (0, 0), and the critical points (p,
  0), (3p, 0), ??are of the same type as (p, 0),
  so that our task is finished.

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Fig. 93. Trajectories in the phase plane for the
damped pendulum in Example 2
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LotkaVolterra Population ModelExample 3
PredatorPrey Population Model

• This model concerns two species, say, rabbits and
  foxes, and the foxes prey on the rabbits.
• Step 1. Setting up the model.
• We assume the following.
• 1.Rabbits have unlimited food supply. Hence if
  there were no foxes, their number y1(t) would
  grow exponentially,
• y1' ay1
• 2.Actually, y1 is decreased because of the kill
  by foxes, say, at a rate proportional to y1y2,
  where y2(t) is the number of foxes. Hence y1'
  ay1 by1y2, where a gt 0 and b gt 0.

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• 3.If there were no rabbits, then y2(t) would
  exponentially decrease to zero, y2' ly2.
  However, y2 is increased by a rate proportional
  to the number of encounters between predator and
  prey together we have y2' ly2 ky1y2, where k
  gt 0 and l gt 0.
• This gives the LotkaVolterra system
• y1' 1(y1, y2) ay1
  by1y2
• (7)
• y2' 2(y1, y2) ky1y2
  ly2 .

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• Step 2. Critical point (0, 0), Linearization. We
  see from (7) that the critical points are the
  solutions of
• (7) 1(y1, y2) y1(a by2)
  0,
• 2(y1, y2) y2(ky1 l)
  0.
• The solutions are (y1, y2) (0, 0) and (l/k,
  a/b). We consider (0, 0). Dropping by1y2 and
  ky1y2 from (7) gives the linearized system
• Its eigenvalues are ?1 a gt 0 and ?2 l lt
  0. They have opposite signs, so that we get a
  saddle point.

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• Step 3. Critical point (l /k, a/b),
  Linearization. We set
  
• Then the critical point (l/k, a/b) corresponds
  to
• (0, 0).
• Since we obtain from
  (7)

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• Dropping the two nonlinear terms and
  , we have the linearized system
• (7)

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Fig. 94. Ecological equilibrium point and
trajectory of the linearized
LotkaVolterra system (7)
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• We see that the predators and prey have a cyclic
  variation about the critical point.
• Beginning at the right vertex, where the rabbits
  have a maximum number. Foxes are sharply
  increasing in number until they reach a maximum
  at the upper vertex.
• The number of rabbits is then sharply decreasing
  until it reaches a minimum at the left vertex,
  and so on.
• Cyclic variations of this kind have been observed
  in nature, for example, for lynx and snowshoe
  hare near the Hudson Bay, with a cycle of about
  10 years.