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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: xv (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 K21 x1 ( 3 M 3 B1 B2 K21 ) x1 1 ( 3 B1 3 B2 (K1 K2 )K21 ) x1 3 K1 x1 f a (t ) 1 1 (4) Where: M = mass; B1, B2 = friction coefficients; Figure 2 KK = 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) va (6) a K 2 (3 MB2 ) 1 ( 3 M 3 B1 B2 K 21 ) a 1 ( 3 B1 3 B2 ( K1 K 2 )K 21 ) 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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posted: | 3/5/2011 |

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