Question

The Van der Pol equation is y′′ −αy′ (1−y2+y=0 where α is a non-negative parameter. 1. Introduce the variable u=[ u1u2 ]T . Write the equation as a first order system u′ =F(u) 2. Show that u=0 is the only equilibrium point. 3. Show that u=0 is a center when α=0 and a repellor when α>0.

Answer 1

The Van der Pol equation is transformed into a first-order system by defining a new variable vector u. The only equilibrium point is found to be u = 0. Depending on the value of α, u = 0 acts as a center when α = 0, and as a repellor when α > 0.

Step-by-step explanation:

The Van der Pol equation is a second-order differential equation that can be expressed as a first-order system by introducing a variable vector u = [ u1 u2 ]T. To write the equation as a first-order system, we let u1 = y and u2 = yʼ, which gives us the system:

• u1ʼ = u2
• u2ʼ = αu2(1 - u12) - u1

To find the equilibrium points, we set the derivatives to zero (uʼ = 0), obtaining u2 = 0 and αu2(1 - u12) - u1 = 0. With u2 = 0, the second equation simplifies to -u1 = 0, hence u = 0 is the only equilibrium point.

When α = 0, the system reduces to a simple harmonic oscillator, and u = 0 is a center. When α > 0, the nonlinearity introduces a damping term which makes u = 0 a repellor, meaning that solutions will move away from the equilibrium point over time.