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\begin{document}


{\bf Question}

Consider van der Pol's equation

$$\begin{array}{l}u''+u=\varepsilon(1-u^2)u',\ \varepsilon \to
0^+\\ u(0)=2,\ u'(0)=0\end{array}$$

Show that substitution of the regular ansatz

$$u(x;\varepsilon)=u_0(x)+\varepsilon u_1(x)+O(\varepsilon^2)$$

leads to a perturbation solution for finite $x$,

$$u(x;\varepsilon)=2\cos x+\varepsilon\left(\df{3}{4}\sin
x-\df{\sin 3x}{2}\right)+O(\varepsilon^2)$$


\vspace{.25in}

{\bf Answer}

$u''+u=\varepsilon(1-u^2)e'\ \ \varepsilon\to 0^+\ \ u(0)=2,\
u'(0)=0$.

Substitute $u(x;\varepsilon)=u_0(x)+\varepsilon
u_1(x)+O(\varepsilon^3)$ ????

$u_0''+\varepsilon {u_1}''+u_0+\varepsilon
u_1+O(\varepsilon^2)=\varepsilon(1-u_0^2-\varepsilon^2u_1-2u_0u_1\varepsilon+O(\varepsilon^2))
\times (u_0'+\varepsilon u_1'+O(\varepsilon^2))$

Collect terms of $O(\varepsilon^0)$ and $O(\varepsilon^1)$

$O(\varepsilon^0): {u_0}''+u_0=0 \Rightarrow u_0=A \sin x + B \cos
x$

Boundary conditions: $u_0(0)=2,\ u_0'(0)=0$

$\Rightarrow A=0,\ B=2 \Rightarrow u_0=2\cos x$

$O(\varepsilon): {u_1};;+u_1=u_0'(1-u_0^2)$

$u_0=2\cos x,\ u_0'=-2\sin x \Rightarrow RHS=-2sin x(1-4\cos^2 x)$

\ \ But $4\sin^3 x-3\sin x=-\sin 3x \left\{\begin{array}{l}=-2\sin
x(1-4+4\sin^2 x)\\ =-2\sin x(-3+4\sin^2 x)\\ =-2(4\sin^3 x-3\sin
x) \end{array}\right.$

Therefore ${u_1}''+u_1=+2\sin 3x\ \ (A)$

$u_1=u_1^{CF}+u_1^{PI} \rightarrow \begin{array}{l} u_1^{CF}:
{u_1^{CF}}''+u_1^{CF}=0\\ u_1^{PI}: u_1^{PI}=C\sin 3x+D\cos 3x
\end{array}$

${}$

By substituting in (A):

$\ -9C\sin 3x-9D\cos 3x+C\sin 3x+D\cos 3x=2\sin 3x$

$\Rightarrow -8C\sin 3x-8D\cos 3x=2\sin 3x$

$\Rightarrow C=-\df{1}{4},\ D=0$

and $u_1^{CF}=E\sin x+D\cos x$

Therefore $u_1=E\sin x+D\cos x-\df{1}{4}\sin 3x$

${}$

Boundary conditions:

$u_1(0)=u_1'(0)=0$ (from perturbation of boundary conditions)
$\Rightarrow D=0$

$u_1'(0)=E\cos 0-\df{3}{4}\cos 0 \Rightarrow E=\df{3}{4}$

Therefore $u(x;\varepsilon) \sim 2 \cos
x+\varepsilon\left(\df{3}{4}\sin x-\df{\sin 3x}{2}\right)$


\end{document}