San José State University

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Thayer Watkins
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Perturbation Analysis

## General Principles

Often the mathematical solutions to a set of equations representing a particular problem are functions of the parameters of the problem and it is enlightening and useful to determine how changes in the parameters affect the solutions. Ususally the changes in the paramters are small and are called perturbations. One of the techniques for determining the effect of perturbations is an asymptotic expansion of a solution with respect to a perturbation parameter. While such asymptotic expansions are invoked with respect to small values of a parameter it has been found that important information can be found for asymptotic expansions with respect to a large parameter.

Dimensional parameters cannot be characterized as large or small. For example, the 40,000 km circumference of the Earth is large on the scale of the physical dimensions of humans but it is infinitesimal on the scale of galactic dimensions. On the other hand, a poppy seed is small on the scale of human dimensions but gigantic on the scale of molecular dimensions. In order to carry out valid analysis of perturbations in a problem it has to be reduced to a nondimensional form.

### An Introductory Example

In order to gain insights into the nature of perturbation analysis one should start with a problem simple enough to have a complete analytical solution. Consider a simplified version of the shock absorption mechanism for an automobile. The car body is connected to the car frame by means of a spring and what is usually called a shock absorber. The force on the car body is the net sum of the force due to the spring, which depends upon the position of the car body, and the absorber, which depends upon the motion of the car body. Let x be the vertical distance of the car body above a reference point. Furthermore let the reference point be such that when x=0 the force due to the spring is zero. In other words, x=0 is the equilibrium position for the car body. The spring force is then -kx, where k is a constant which indicates the stiffness of the spring. The negative sign indicates that when x is positive the spring force is downward. The force due to the shock absorber is -s(dx/dt), where s is a constant. The negative sign indicates that if the motion is upward the force is directed in the opposite direction, downward.

By Newton's law of motion the force on the body of the car is equal to the mass of the body m times the acceleration of the body, d2x/dt2. Thus,

#### m(d2x/dt2) = -s(dx/dt) - kx.

The additional information needed to determine a solution for the position x(t) as a function of time is the initial conditions, which are taken to be x(0)=x0 and dx/dt=0 at t=0.

The position x has the dimensions of length and dx/dt has of course the dimensions of length per unit time. The spring constant parameter k has the dimensions of force per unit length, which is the same as mass times per time2. The shock absorber parameter constant s has dimensions of force times time per unit length, which is the same as mass per unit time.

Although it would be appropriate to first transform the problem into a nondimensional form it is instructive for this simple problem to find an analytic solution first to see what is involved in the nondimensionalization of a problem.

The equation for the problem,

#### m(d2x/dt2) + s(dx/dt) + kx = 0,

is a second order linear homogeneous differential equation. It has solutions of the exponential form, exp(λt). The parameter λ must be such that

#### mλ2 + sλs(dx/dt) - k = 0.

The solutions to the above quadratic equation are:

#### λ1 = (-s + √(s2-4mk))/(2m) and λ2 = (-s - √(s2-4mk))/(2m)

The general solution to the diffential equation is then

#### x(t) = c1exp(λt) + c2exp(λ2t)

The complete solution to the problem is found by requiring c1 and c2 to be such that the intitial conditions for x(t) are satisfied; i.e.,

#### c1 + c2 = x0and λ1c1 + λ2c2 = 0.

The second equation implies that c2 = - c112). For convenience let (λ12) = μ. Then the values of c1 and c2 are:

#### c1 = x0/(1-μ) c2 = -x0μ/(1-μ)

The complete solution is then

#### x(t) = x0(exp(λ1t) - μexp(λ2t))/(1-μ) or, equivalently x(t) = x0(λ2exp(λ1t) - λ1exp(λ2t))/(λ2-λ1)

This result can be put into a more conveniet form that is related to the procedure of of nondimensionalization. The equations for the lambdas can be put into the form:

#### λ1 = (-β + √(β2-α)) and λ2 = (-β - √(β2-α))

where β=s/2m and α=k/2m. The dimension of β is per unit time and that α is per unit time2. The reciprocal of β is a natural unit of time for the system. The character of the solution depends upon the sign of (β2-α). If this quantity is negative the solution is oscillatory. For convenience let (β2-α)= γ2.

Since λ1=(-β+γ) and λ2=(-β-γ) so (λ21)=-2γ the solution can be put into the form

#### x(t)/x0 = (-λ2exp(λ1t) + λ1exp(λ2t))/(2γ) = ((β+γ)exp((-β+γ)t) + (-β+γ)exp((-β-γ)t))/(2γ) = exp(-βt)((β+γ)exp(γt)+(-β+γ)exp(-γt))/(2γ) = exp(-βt)(((β/γ)sinh(γt) + cosh(γt))

If γ2 is negative it is convenient to let γ=iδ so the last formula above becomes:

#### x(t)/x0 = exp(-βt)(((β/δ)sin(δt) + cos(δt))

The natural unit of length for the problem is x0. There are two natural units of time, the reciprocals of β and of γ or δ. Division of x(t) by x0 gives a dimensionless variable z(t). The variable βt gives a dimensionless relative time u so the complete solution can be expressed as:

#### z(u) = exp(-u)[(sin(εu)/ε) + cos(εu)]

where ε = δ/β is the frequency of the oscillations. This function satisfies the initial conditions that z(0)=1 and z'(0)=0. For γ2 positive the solution would be:

with ε = γ/β.