#### Reduction of Higher Order Equations to Systems

Reduction of Higher Order Equations to Systems:

The motion of a pendulum:

Consider the motion of an perfect pendulum that consists of a mass m attached to an arm of length l.

If we ignore friction after that Newton’s laws of motion tell us

m¨θ =−{mg/l}sin θ,

where θ is the angle of displacement.

A pendulum

If we as well incorporate moving friction and sinusoidal forcing then the equation takes the form:

mθ¨+ γθ? +mg/lsin θ = A sint.

Here γ is the coefficient of friction moreover A and are the amplitude as well as frequency of the forcing.

Typically this equation would be rewritten by dividing through by m to produce:

θ¨+ cθ? + ω sin θ = a sint

where c = γ/m. ω = g/l and a = A/m.

This is a second order ODE for the reason that the second derivative with respect to time t is the highest derivative. It is nonlinear for the reason that it has the term sin θ and which is a nonlinear function of the dependent variable θ. A resolution of the equation would be a function θ(t). To get a specific solution we require side conditions. For the reason that it is second order 2 conditions are needed and the usual conditions are initial conditions

θ(0) = θ0 and θ?(0) = v0.

Converting a universal higher order equation:

The entire of the standard methods for solving ordinary differential equations are intended for first order equations. Therefore it is inconvenient to resolve higher order equations numerically. Nevertheless mainly higher-order differential equations that take place in applications can be converted to a system of first order equations and that is what is usually done in practice.

Presume that an n-th order equation can be solved for the n-th derivative that is it can be written in the form:

Afterwards it can be converted to a first-order system by this standard change of variables:

y1 = x
y2 = x?
...

The resulting first-order system is the subsequent:

y?1 = x? = y2
y?2 = x¨ = y3
...
y?n = x(n) = f(t, y1, y2, . . . , yn).

In vector form this is simply y?= f (t, y) with fi(t, y) = yi+1 for i < n and fn(t, y) = f(t, y1, y2, . . . , yn).

For the illustration of the pendulum the change of variables has the form:

y1 = θ
y2 = θ?,

And the resultant equations are:

y?1 = y2
y?2 = −cy2 − ω sin(y1) + a sin(t).

In vector form this is:

The preliminary conditions are converted to:

As stated above the major reason we wish to change a higher order equation into a system of equations is that this form is convenient for solving the equation numerically. Most universal software for solving ODEs (including Mat lab) require that the ODE be input in the form of a first-order system. Additionally there is a conceptual reason to make the change. In a system illustrate by a higher order equation knowing the position isn’t enough to know what the system is doing. In the case of a subsequent order equation such as the pendulum one must know both the angle and the angular velocity to know what the pendulum is really doing. We call the pair (θ, θ?) the state of the system. In general in applications the vector y is the state of the system described by the differential equation.

Using Matlab to solve a system of ODE’s:

In Mat lab there are several commands that are able to be used to solve an initial value problem fora system of differential equations. All of these correspond to different solving methods. Thestandard one to utilize is ode45 which uses the algorithm ‘Runge-Kutta 4 5’. We will study aboutthis algorithm later.

To realize ode45 for a system we have to input the vector function f that defines the system. For the pendulum system we will presume that all the constants are 1 except c = .1 then we are able to input the right hand side as:

> f = inline(’[y(2);-.1*y(2)-sin(y(1))+sin(t)]’,’t’,’y’)

The command ode45 is afterwards used as follows:

> [T Y] = ode45(f,[0 20],[1;-1.5]);

Here [0 20] is the time span you desire to consider and [1;-1.5] is the preliminary value of the vectory. The output T contains times as well as Y contains values of the vector y at those times Try:

>size(T)
>T(1:10)
>size(Y)
>Y(1:10,:)

Ever since the first coordinate of the vector is the position (angle) we are largely interested in its values

>theta = Y(:,1);
>plot(T,theta)

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