[Introduction] [Results]


The shock model simulates a shock absorber driven by an external force. It is described by the following system of differential equations:

x' = y

y' = -x -(c1 + c2(x-y)^2)*y + a*w + b*z

z' = lambda*z - omega*w - A(z^2+w^2)*z

w' = omega*z + lambda*w - A(z^2+w^2)*w

Variable x represents the position and variable y represents the velocity of the oscillator. The system can easily be rewritten in the form of a second order differential equation by substituting the first equation into the second:

x" + (c1 + c2(x-x')^2)*x' + x = a*w + b*z

The above is a familiar equation of motion for a forced-damped oscillating object. The main difference between this model and a linear one is that the damping term which multiplies the velocity x' is not entirely linear. Although there is a damping constant c1 which does provide a linear term, there is also a complicated square term of the difference of the position and velocity. We now examine the effects of this term, without taking into account the driving term:

a) The term can never be negative since the square of the difference is always positive and the constant c2 is defined to be always positive. Therefore, there is always damping, rather than 'negative damping' which would force the oscillator.

b) Ignoring the forcing term of this oscillator, it's reasonable to assume that the magnitude of the velocity when x=0 is considerably greater than the magnitude of the displacement when x'=0. Therefore, the damping is greatest when the oscillator is at a position of x=0, since then the velocity is high. When the oscillator is at maximum displacement, the damping is fairly low because, as mentioned, the magnitude of the displacement is much smaller when x'=0 than the magnitude of the velocity when x=0. It is difficult to say when the damping is at its very smallest, because of the following argument: Obviously, it has to be some distance from the x=0 position, because as described above, the square of the velocity is large. If we expand the (x-x')^2 term, we get: x^2 - 2xv + v^2, where x'=v. As we move away from the equilibrium point the velocity term has an ever smaller effect, while the displacement term 'takes over'. It is not clear at which point the minimum of that term occurs, because an explicit analytical solution of the differential equation is not possible.

We now examine the other two equations. They describe the rate of change of the forcing terms. Since the equations are not linear, it is impossible to find an analytical solution and it is best to consider the Jacobian of partial derivatives to determine the stability of the origin. This is the generalized Jacobian:

If we let z=0 and w=0 and solve for the eigenvalues, we get the following:

We can extend the theory of the determination of the stability of linear equilibrium points to linearized non-linear systems. It must be noted that if lambda=0, we cannot say anything about the stability of the fixed point.

Thus, as the following graph shows, for a negative lambda, we get exponential decay and our forcing term dies away:

For positive lambda, our forcing term solution settles down to a limit cycle:

Since part of the solution for positive lambda is a transient (assuming that the solution could be found at least in principle), we can ignore it by waiting a sufficiently long amount of time. The following are the asymptotic solutions:

It is difficult to precisely say what lambda represents physically, but we showed above it's effects.


Created by: Jason Sardell and Tom Szymkiewicz