Differential equations are used in the modeling of the dynamics of physi­cal systems. If space dependence is not included in the equations that describe a process, then ordinary differential equations are used, and the model is said to be a lumped-parameter model. If the space dependence is included in the equations, then partial differential equations are required, and the model is said to be a distributed-parameter model. Of course a distributed process may be handled by using a number of interconnected lumped models for portions of the system. This is similar to a finite difference approximation for solving differential equations and is usually called the nodal approach.

Another important distinction between mathematical models for a system is whether they are linear or nonlinear. A linear model is one in which the dependent variables and their derivatives appear only to the first power and not as factors in products of dependent variables. A model that violates this is a nonlinear model.

We consider the question of linearity versus nonlinearity early in this book because the testing methods presented in subsequent sections require linearity of the processes being studied. The user must appreciate the implica­tions of this requirement in regard to restrictions on test conditions and in regard to limitations on interpretation of the results.

Example 2.1.1. The following differential equations are linear:

(1) dx/dt = 2x + 4

(2) d2x/dt2 = 3(dx/dt) — 2x + 61

(3) dx/dt = 3 tx

The first two are constant-coefficient linear equations and the third is a variable-coefficient linear equation. The following equations are nonlinear:

(1) dx/dt = x + (1/x)

(2) dx/dt = x2 + 2

(3) (dx/dt)2 = 2x

(4) dx/dt = x + ex ■

Simple, well-defined techniques are available for analyzing linear systems. On the other hand, it is very difficult to analyze nonlinear systems. This motivates the attempt to approximate nonlinear models with linear models that are valid over some range of operation for the system being modeled. This process, called linearization, proceeds as follows: [1] [2] [3] [4]

1. Identify combinations of terms that identify the equilibrium condi­tion. Since the time derivatives are all zero at equilibrium, these combina­tions of terms are identically zero. The remaining terms make up the linearized model.

As example is useful for illustrating this procedure.

Example 2.1.2. Linearize the following coupled set of differential equations.

dx^/dt = 3xt + xlx2, dx2/dt = x22 + exp(x,)

Step 1. Let

xl = x10 + 5xl and x2 = x20 + <5×2

Step 2.

d bxjdt = 3(x10 + bxj + (x10 + bxj(x20 + <5×2)

d bxjdt = (x20 + <5×2)2 + exp(x10 + bxj or

d bxjdt = 3×10 + x10x20 + 3 <5x! + x10 <5×2 + x20 bxi + bxi bx2 d bxjdt = xj0 + 2×20 <5×2 + (<5×2)2 + (exp x10)(exp bxj Step 3. Substitute

exp(<5x!) = 1 + bx^ + {bx^/H + ■ ■ ■

to obtain

d bxjdt = 3×10 + x10x20 + 3 bxi + x10 <5×2 + x20 bxl + bxi bx2

d bxjdt = x|0 + 2x20bx2 + (<5×2)2 + (expx10)[l + <5xj + (<5x i)2/2! + • ■ •]

Step 4. Eliminate all terms containing products of deviations to obtain

d bxjdt = 3×10 + x10x20 + 3 bxi + x10 <5×2 + x20 bxl

d bxjdt = x20 + (exp x10) + 2×20 <5×2 + (exp x10) «Sxj

Step 5. The combinations of terms that define equilibrium may be found by setting the derivatives equal to zero in the original equations.

(dxjdt)о = 3×10 + x10x20 = 0, (dxjdt)0 = x + exp x10 = 0

The final linearized equations are

d bxjdt = (3 + x20) bxl + x10 <5×2, d bxjdt = (exp x10) <5xj + 2×20 <5×2

The important things to note are that systematic procedures are available for developing linearized models for nonlinear systems, but that the validity

of these linearized models is assured only for some “small” region around an equilibrium point. ■

It is also often convenient to write linear equations with the dependent variables expressed as deviations from equilibrium. In this case, the equa­tions for Sx are identical with the equations for x, but the initial conditions may be different. If the transient starts from an equilibrium point, the initial values for the Sx variables are all zero.

Example 2.1.3. Express the following coupled set of differential equations in terms of deviations from equilibrium:

dxl/dt = Зхі + 2×2, dx2/dt = Xi + 4×2

Because of the linearity of the equations, we may write

d Sxl/dt = 3 <5x! + 2 <5×2, d Sx2/dt = 8x^ + 4 <5×2 with all initial conditions equal to zero. ■

An important property of linear systems is the property of superposition. This property may be stated as follows:

If the output of a system is y, when the input is then the output у when N inputs are applied simultaneously is

N

У=Ї. Уі (2.1.1)

t= 1