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I have solved the unforced van der Pol oscillator,

$$\frac{\mathrm d^2x}{\mathrm dt^2} + \epsilon (x^2-1)\frac{\mathrm dx}{\mathrm dt}+\omega_0^2x=0$$

using the multiple time scaling method up to $O(\epsilon)$.

$$x(t,\epsilon)=X(t,\tau,\epsilon)=X_0(t,\tau)+\epsilon\,X_1(t,\tau)+\ldots$$

But I don't know how to work out the solutions using the initial conditions, $x(0,\epsilon)=1$ and $\frac{\mathrm dx(0,\epsilon)}{\mathrm dt}$=0

I can't seem to work out how $(1)$ implies $(2)$ and $(3)$.

$$X(0,0,\epsilon)=1, \frac{\partial X(0,0,\epsilon)}{\partial t}+\epsilon \frac{\partial X(0,0,\epsilon)}{\partial\tau}=0\tag1$$

$$\implies X_0(0,0)=1, \frac{\partial X_0(0,0)}{\partial t}=0\tag2$$

$$X_1(0,0)=0, \frac{\partial X_1(0,0)}{\partial t}+\frac{\partial X_0(0,0)}{\partial \tau}=0\tag3$$

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1 Answer 1

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Take the following notation: $$x(t) = \hat x(t,\tau) = \sum^{\infty}_{i=0}\varepsilon^ix_i(t,\tau) $$ Then the derivative in time expands to: $$\frac{d}{dt} = \frac{\partial}{\partial t}+\frac{\partial \tau}{\partial t}\frac{\partial}{\partial \tau} \stackrel{\tau = \varepsilon t}{=} \frac{\partial}{\partial t}+\varepsilon\frac{\partial}{\partial \tau}$$

It can be seen that if the conditions $x(0) = 1$ and $\frac{\partial x}{\partial t}|_0 = 0$ are given then for the expanded term with a small $\varepsilon$: $$x(0) = \hat x(0,0) = \sum^{\infty}_{i=0}\varepsilon^ix_i(0,0)\approx x_0(0,0)$$ And respectively for the derivative:

$$\frac{\partial x}{\partial t}|_0 = \left(\frac{\partial}{\partial t}+\varepsilon\frac{\partial}{\partial \tau}\right)\sum^{\infty}_{i=0}\varepsilon^ix_i(t,\tau) \approx \frac{\partial x_0}{\partial t}|_0$$

In words: When you perform a multiple time scales expansion with $0<\varepsilon\ll1$ (small!) you know that for Van der Pol like equations, the boundary conditions mainly affect $x_0$. All further low frequent terms ($\varepsilon x_1,\varepsilon^2 x_2,...$) have homogenous boundary conditions.

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