ODE

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ODE

• jiangyushan

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Pendulum
As a example of a system that is nonlinear,
consider the swinging pendulum shown above. When
the mass of the pendulum is small in comparison
with the mass m at the end of the pendulum, the
equation of motion of the pendulum is as follows.
Here is the angular position of the pendulum
measured relative to vertical, m is mass at the
end of the pendulum, L is the length of the
pendulum, is the coefficient of viscous
friction, and is the acceleration due to
gravity. Again, this is a second-order equation
in the form of (1.1)
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Pendulum
If we define the state vector to be
, then from above (1.1) we get the
following first-order system.
This is an autonomous system because there is no
explicit dependence on t. Furthermore, it is
nonlinear due to the presence of the
term.
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Predator-prey Ecological System
Dynamic systems occur in many fields of study.
Consider, for example, the problem of modeling
the population levels of a predator-prey pair of
species. Let denote the population level of
the prey, and let denote the population level
of the predator. Suppose and are expressed
in units of ,say, thousands. The following
simplified model of population growth is referred
to as the Lotka-Volterra system.
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Predator-prey Ecological System

• Here the parameter denotes the normalized
  growth rate of the prey when the predator is not
  present . Similarly, denotes
  the rate at which the predator population
  decrease in the absence of prey .
• The term represents the decrease in
  the prey population as a result of the actions by
  the predator,
• and the term represents the increase
  in the predator population as a result of the
  availability of prey.

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Predator-prey Ecological System

• This is a nonlinear system due to the presence
  of the product terms. As we shall see, it
  has a periodic solution in which the population
  levels of the predator and prey go through
  ecological cycles. It can be shown that amplitude
  of the cycle depends on the initial conditions,
  while the period of the cycle is

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INITIAL AND BOUNDARY PROBLEM

• By itself, a differential equation does not
  uniquely determine a solution additional side
  conditions must be imposed on the solution to
  make it unique. These side conditions prescribe
  values that the solution or its derivatives must
  have at some specified point or points. If all of
  the side conditions are specified at the same
  point, then we have an initial value problem,
  which we call it an Initial Value Problem. If the
  side conditions are specified at more than one
  point, then we have a Boundary Value Problem.

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Eulers Method

• The example of dynamic systems introduced
  earlier represent special cases of the following
  general first-order nonlinear system where
  is an n1 vector.

we restrict our consideration to systems for
which the right-hand side function is
sufficiently smooth that Equation (1.5) has a
unique solution satisfying the initial
condition, Sufficient
conditions on to ensure the
existence of a unique solution over
can be found in (Vid78).
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Eulers Method

• We are interested in estimating for
  where the are equality
  spaced over the interval .That is,
  where the step size is

Suppose the value of is known. This is
certainly true for because
To find in terms of
we multiply both sides of Equation (1.5)
by and then integrate from
. This yields the following reformulation of
(1.5) as an integral equation.
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Eulers Method
The problem with applying (1.7) directly is that
we do not know the value of for
,and without it we can not evaluate
the integral. However, if the stepsize , is
sufficiently small, we can approximate the
integrand over the interval ,by
it value at the start of the
interval.
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Eulers Method

• In this case, the integral in (1.7) simplifies to
  .If
• denotes the approximate solution
  obtained in this manner, this yields the
  following solution formula, which is called
  Eulers method.
• Eulers method has a local truncation error of
  order and the global truncation error
  is of order

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An Example for Eulers method

• To illustrate the use of Eulers method,
  consider the following simple one-dimensional
  first-order system.
• Here the constant . This is a
  one-dimensional linear system whose exact
  solution is .
  Applying Eulers method in (1.8) ,we have

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An Example for Eulers method

• This difference equation is simple enough that we
  can write a closed-form expression for the
  solution. If then
• Recall that the exact solution is a decaying
  exponential that approaches zero in the steady
  state. The Euler estimate of the solution will
  go to zero as approaches infinity only if
  or

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RUNGE-KUTTA METHODS

• The coefficients of the fourth-order Runge-Kutta
  method are chosen to ensure that its local
  truncation error is of order , and its
  global truncation error is of order .

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An Example for Runge-Kutta method

• Consider the predator-prey equations discussion
  in (1.3). For convenience, suppose the parameters
  of the system
• are . Using the
  fourth-order Runge-Kutta method to solve this
  system from to using an
  initial condition of

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Objectives

• Know how to convert a higher-order differential
  equation into an equivalent system of first-order
  equations.
• Understand the difference between initial and
  boundary conditions.
• Understand the relationship between local and
  global truncation error.
• Be able to apply the Runge-Kutta single-step
  solution methods.

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• Know how to adjust the step size to control the
  local truncation error.
• Understand how ordinary differential equation
  techniques can be used to solve practical
  engineering problems.
• Understand the relative strengths and weaknesses
  of each computational method and know which are
  most applicable for a given problem.

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Exercise Chemical Reactor