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Project, EGME 511 (Advanced Mechanical Vibration)
Analysis of Van Der Pol differential equation

Nasser M. Abbasi

June 28, 2015

Contents

1 Introduction
2 Stability
3 Phase diagram
4 Phase diagram

This report in PDF

1 Introduction

Van der Pol differential equation is given by

         (      )
x′′(t) − c 1 − x2  x′(t) + kx (t) = 0

In this analysis, we will consider the case only for positive c,k  . We will analyze the stability of this equation and generate a phase diagram.

2 Stability

The first step in examining stability of a non-linear differential equation is to convert it to state space by introducing 2 state variables.

        }                                  }                              }
 x1 = x                  x′1 = x′                         x ′1 = x2
       ′   →    ′    ′′          2  ′        →     ′          2
x2 = x         x2 = x  = c (1 − x  )x − kx        x 2 = c(1 − x1)x2 − kx1

Therefore

(   )    (                   )    (         )
  x′1               x2               g(x1,x2)
   ′   =          2            =
  x2       c(1 − x1) x2 − kx1       f (x1,x2)

Equilibrium points are found by solving (   )    (  )
  x′1      0
  x′  =   0
   2 , hence from the above, we see that x2 = 0  and from         2
c (1 − x 1)x2 − kx1 = 0   we conclude that x1 =  0  as well. Hence

      (  )
        0
xeq =   0

The system matrix is now found. First we note that -∂g      ∂g-     ∂f-
∂x1 = 0, ∂x2 = 1, ∂x1 = − 2cx2x1 − k  , and ∂∂xf = − cx21 + c
  2  , hence

     (         )    (                       )
       ∂∂gx-  ∂∂gx-            0           1
A  =   ∂f1  ∂f2  =                      2
       ∂x1  ∂x2       − 2cx2x1 − k  − cx 1 + c

Hence A  at xeq  becomes

     (       )
        0  1
A =
       − k  c

Now we find the characteristic equation

pict

Hence              √ --------        √ --------
λ1,2 = −b2 ± 12  b2 − 4ac = c ± 12  c2 − 4k  , therefore

            √ ------
λ1,2 = c ± 1- c2 − k
           2

If  2
c  > k  then both roots are on the RHS, hence system is unstable (equilibrium point is a repelling point).

If c2 < k  then we have λ1,2 = c ± jβ  , and we have spiral out equilibrium point, unstable.

3 Phase diagram

We need to obtain a relation between x2   and x1   . From the differential equation

         (      )
x′′(t) − c 1 − x2  x′(t) + kx (t) = 0

rewrite in state space variables, we obtain

pict

Hence the above is in the form dx2=  f (x ,x )
dx1       1  2  , therefore the isoclines lines can be found by setting

f (x1, x2) = ξ

Where ξ  is a constant. Hence we obtain the parameterize equation to use to plot the gradient lines as

    c-(1 −-x21)x2-−-kx1
ξ =         x2

4 Phase diagram

To generate the phase diagram1 , a program was written which allows one to adjust the initial conditions and the parameters k  and c  and observe the effect on the shape of the limit cycle. We see that starting from different initial conditions, the solution trajectory always ends up in a limit cycle.

The following is a screen shot of the program written for this project.

PIC