# A Primer on Ordinary Differential Equations

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```					EMEM 440: Numerical Methods                                                           20031

A Primer on Ordinary Differential Equations
Many mathematical models of physical systems result in equations governing the
derivative of some dependent variable. To solve such a differential equation we must
find the functional form of the variable that, when properly differentiated, satisfies the
given relationship. The solution process is effectively an integration, although we can
rarely integrate the equation directly. As an example, consider the rack & pinion device
shown below in Figure 1, whose motion is governed by (dots indicate time derivatives):

( J  mR 2 ) x  ( R 2 b) x  (  R 2 k ) x   R 2  f (t )                            (1)

Where:
m = mass of the rack;
b = friction coefficient;
 = torsional spring constant of axle;
k = linear spring constant;
R = radius of pinion gear;
J = polar moment of inertia of pinion gear;
f(t) = applied (input) forcing function.                      Figure 1

Time t is the only independent variable in (1), and the highest derivative is second-order.
The resulting equation is therefore a second-order, ordinary differential equation
(ODE) for rack position x(t). To be precise, it is a second-order, linear, non-
homogeneous, ordinary differential equation with constant coefficients. No differential
equation is complete for solution without the specification of conditions to fix arbitrary
coefficients that occur. These conditions may either be of the initial value type, or of the
boundary value type. In an initial value problem, the conditions are all specified at the
same value of the independent variable. In a boundary value problem, the conditions
are specified at different values of the independent variable that tend to “bound” the
domain of the problem. When numerical solutions are concerned, each type is
approached differently.

When dealing with higher-order ODE’s, it is very common to write the single equation as
a system of coupled first-order equations instead. This “expansion” process is essential
for numerical solution. Consider the example above, which need not be expanded for
theoretical purposes, but must be expanded for numerical solution. If we choose v = dx/dt
as an intermediate variable, then we can re-write (1) as:

v  ( J  mR 2 ) 1 [  ( R 2 b)v  (  R 2 k ) x  R 2  f (t )]                      (2)

Note that (2) is now only a first-order equation. However, it is coupled to our original
definition; that is, it cannot be solved by itself without considering:

xv                                                                                      (3)

John D. Wellin                                                                      03/05/11
EMEM 440: Numerical Methods                                                                20031

Therefore the combination of (2) and (3) constitutes a coupled system of first-order
ODE’s. Each needs an initial condition for solution, which in turn are easily derived
from any initial conditions specified for the single equation (1). In this example, our
intermediate variable is conveniently interpreted as the velocity, but in general the
intermediate(s) need not have any specific bearing upon the problem. In fact, the choice
of intermediates is essentially arbitrary and unlimited.

As another example, consider the mechanical system shown in Figure 2. The position x1
is governed by:

3MB2 K21  x1  ( 3 M  3 B1 B2 K21 ) x1
1

 ( 3 B1  3 B2 (K1  K2 )K21 )  x1  3 K1  x1  f a (t )
1                                   1
(4)

Where:
M = mass;
B1, B2 = friction coefficients;                                               Figure 2
KK = spring constants;
L = armature length;
fa(t) = applied forcing function.

To solve (4) numerically, we must expand it
to a coupled system of three first-order
equations. The simplest approach results in
the following, where v(t) and a(t) are the
chosen intermediates:

x1  v                                                                                       (5)

va                                                                                          (6)
a   K 2 (3 MB2 ) 1  ( 3 M  3 B1 B2 K 21 )  a

1

 ( 3 B1  3 B2 ( K1  K 2 )K 21 )  v  3 K1  x1  f a (t )
1                                     1
        (7)

The numerical solution therefore results in a simultaneous succession of values for all
three dependent variables.

John D. Wellin                                                                          03/05/11

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