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Solving the Van Der Pol nonlinear differential equation using first order approximation perturbation method

Nasser M. Abbasi

2009   page compiled on June 28, 2015 at 4:15am

Abstract

The Van Der Pol differential equation

|---------------------------------|
x′′(t)− α (1−  x2(t))x ′(t)+ x(t) = 0|
-----------------------------------

is solved using perturbation with first order approximation. Two different solutions are obtained. The first solution restricted the initial conditions to be x (0)2 + x′(0)2 = 4  which resulted in the forcing function that caused resonance to be eliminated. This gave a stable solution but with initial conditions restricted to be near the origin of the phase plane space. The second solution allowed arbitrary initial conditions any where in the phase plane but the the resulting forcing function that caused resonance resulting in a solution which became unstable after some time. Phase plane plots are used to compare the two solutions.

1 First solution: Restriction on initial condition. No resonance

The Van Der Pol equation is

pict

And assuming initial conditions are x(0) = φ  and x˙(0 ) = ξ  .

If α ⋘   1  , then the equation becomes x¨0 + x0 =  0  which has an exact solution. We perturb the exact solution for ¨x0 + x0 = 0  to approximate the solution of the nonlinear equation ¨x − α(1 − x2) ˙x + x = 0  in terms of perturbation parameter α  Let the solution of the Van Der Pol equation be x(t)  and the solution to the exact equation ¨x0 + x0 = 0  be x0(t)  , then

|------------------------------------|
x(t) = x0(t) + αx1(t) + α2x2 (t) + ⋅⋅⋅
--------------------------------------

First order approximation results in

pict

Now we need to determine x0 (t)  and x1(t)  . Substituting (2) into (1) gives

pict

Collecting terms with same power of α  gives

pict

Setting terms which multiply by higher power α  to zero and since it is assumed that α  is very small therefore

|-------------(-------------------)-----|
|(¨x0 + x0) + α ¨x1 − x˙0 + x20x˙0 + x1  =  0|
-----------------------------------------

For the LHS to be zero implies that

pict

And

pict

Equation (3) is solved for x0   and (4) is solved for x1   in order to find the solution x(t)  . Solution to (3) is

x0(t) = A0 cost + B0 sin t

Assuming initial conditions for x1(t)  are zero, then initial conditions for x0(t)  can be taken to be those given for x(t)  . Solving for A0,B0   gives

|----------------------|
x0(t)-=-φ-cost-+-ξsint--

Substituting the solution just found (and its derivative) into (4) gives

pict

Using         2    1
sint cos t = 4 (sint + sin 3t)  and        2    1
costsin t = 4(cos t − cos 3t)  the above can be simplified to

pict

Using     3         3    1
ξsin t − φ cos t = 4 (− 3φ cos t − φ cos3t + 3ξ sin t − ξ sin 3t)  the above can be simplified further to

pict

The above is the equation we will now solve for x1(t)  , which has four forcing functions, hence four particular solutions. two of these particular solutions will cause a complete solution blows up as time increases (the first two in the RHS shown above). If we however restrict the initial conditions such that

(        2      3     2  )
  − φ − ξ-φ-+ φ-- + 3ξ-φ-  =  0
         2     4      4

And

(      3      2        )
  ξ − ξ- + ξφ--−  3ξφ2   = 0
      4     2     4

Then those forcing function will vanish. The above two equation result in the same restriction, which can be found to be

4 = ξ2 + φ2

By restricting ξ2 + φ2   to be exactly 4  , then the solution we obtain for x (t)
 1  from (5) will not blow up. Hence (4) can now be rewritten without the two forcing functions which caused resonance resulting in

            (         )           (          )
              ξ2 −-3φ2-             φ2-−--3ξ2
x¨1 + x1 =  ξ      4      cos3t + φ      4      sin 3t

But φ2 =  4 − ξ2   and ξ2 = 4 − φ2   , hence the above becomes after some simplification

|-----------(------)----------(-------)------|
x¨1 + x1 =  ξ ξ2 − 3 cos 3t + φ φ2 − 3  sin3t |
---------------------------------------------
(6)

The homogeneous solution to the above is

x   =  c cost + c sint
  1,h    1        2

And we guess two particular solutions x1,p1 = a1 cos3t + a2sin3t  and x1,p2 = d1cos 3t + d2sin 3t  . By substituting each of these particular solutions into (6) one at a time, and comparing coefficients, we can determine a1,a2,d1,d2   . This results in

|-----------------------|
|       ξ-(3 −-ξ2)      |
|x1,p1 =     8     cos 3t|
------------------------

Similarly,

|-----------------------|
|       φ (3 − φ2)      |
|x1,p2 = ---------- sin 3t|
-------------8-----------

Hence the solution to (6) becomes

pict

Now applying initial conditions, which are x1 (0) = 0  and ˙x1(0) = 0  , to (7) and determining c1   and      c2   results in

            2                (      )               2                  2
x1 (t) = ξ(ξ--−-3)cos t + 3φ  φ2 − 3  sin t + ξ(3 −-ξ-)cos 3t + φ(3-−-φ--)sin3t
             8            8                      8                 8

Therefore we now have the final perturbation solution, which is

pict

Where x(0) = φ  and x˙(0) = ξ  and the above solution is valid under the restriction that

x2 (0) + ˙x2(0) = 4

We notice that the above restriction implies that both x(0)  and ˙x(0)  have to be less than 4  to avoid getting a quantity which is complex. This implies that his solution is valid near the center of the phase plane only. In addition, it is valid only for small α  . To illustrate this solution, We show the phase plot which results from the above solution, and also plot the solution x (t)  . Selecting x (0 ) = 1.5  and         √ --------
x˙(0) =   4 − 1.52 = 1.3229

PIC

2 Second solution. No restriction on initial conditions. Resonance present in the solution

The same initial steps are repeated as in the first solution, up the equation to solve for x  (t)
 1

pict

There exists four particular solutions relating to the above four forcing functions

pict

For each of the above, we guess a particular solution, and determine the solution parameters. We guess x1,p1 = t (c1 sin t + c2 cost)  ,  x1,p2 = t(c1sint + c2cos t)  , x1,p3 = c1 sin 3t + c2cos3t  , and x1,p4 = c1 sin 3t + c2cos3t  . By substituting each of the above into (8) one at a time and comparing coefficients, we arrive at the following solutions

pict
pict

Now applying initial conditions which are x (0) = 0
 1  and ˙x (0) = 0
 1  and determining c
 1   and c
 2   results in

pict

We can now write the final solution for x(t)  as

pict

To illustrate this solution, we show the phase plot which results from the above solution, and also plot the solution x(t)  . We select x(0) = 1.5  and x˙(0) = 1.3229  . We note that the same initial conditions are used as in the earlier solution to compare both solutions.

PIC

We see clearly the effect of including resonance particular solutions. The limit cycle boundary is increasing with time.

The effect becomes more clear if we plot the solution x(t)  itself, and compare it to the solution earlier with no resonance. Below, We show x(t)  from both solutions. We clearly see the effect of including the resonance terms. This is the solution with no resonance terms

PIC

This is the solution with resonance terms

PIC

In the second solution (with resonance), there is no restriction on initial conditions. Hence we can for example look at the solution starting from any x(0)  and ˙x(0)  . This is the phase plot for x (0) = 5  and x˙(0) = 2  . This would not be possible using the first solution, due to the restriction on x (0)2   +  ˙x (0)2   having to be equal to 4

PIC

3 Conclusion

Van Der Pol was solved using perturbation for first order approximation. It was solved using two methods. In the first method, the solution was restricted to not include forcing functions which leads to resonance.

In the second method, no such restriction was made. In both methods, this problem was solved for general initial conditions (i.e. both x (0)  and x˙(0)  can both be nonzero.

In the first method, it was found that the we must restrict the initial conditions to be such that     2
x (0 )          2
+ ˙x (0) = 4  . This is the condition which resulted in the resonance terms vanishing from the solution. This leads to a solution which generated a limit cycle which did not blow up as the solution is run for longer time. In other words, once the solution enters a limit cycle, it remains in the limit cycle.

In the second method, no restriction on the initial conditions was made. This allowed the solution to start from any state. However, the limit cycle would grow with time, and the solution will suffer from fluttering due to the presence of the resonance terms. However, even though the solution was not stable in the long term, this second approach allowed one to examine the solution for a shorter time but with the flexibility of choosing any initial conditions.

It was also observed that increasing the value of the perturbation parameter α  gradually resulted in an inaccurate solution as would be expected, as the solutions derived here all assumed a very small value of α1   1α = 0.01 was used in generating the solutions and plots shown  .

For future research, it would be interesting to consider the effect of higher order approximation on the solutions and compare with accurate numerical solutions.