The system is
$$
\dot{x} = f(x,y)=y \\
\dot{y}= g(x,y)=-ay – b \sin(x)
$$
with $a$ and $b$ real numbers.I am asked to find stable and unstable manifolds using Taylor approximation of order 3 around the equilibrium point $(\pi,0)$.
To find the stable manifold, $y=h(x)$, I use the condition
$$Dh(x)f(x,h(x)) -g(x,h(x))=0 \;\;\;(I)$$
The Taylor approximation of order 3 near $(\pi,0)$ is:
$$
h(x) \approx c_1x^2+c_2x^3
$$
From condition $(I)$, I have:
$$
(2c_1x+3c_2x^2)(c_1x^2+c_2x^3) = -a(c_1x^2+c_2x^3)-b\sin(x)
$$
Using Taylor approximation for $sin(x)$ I get:
$$
(2c_1x+3c_2x^2)(c_1x^2+c_2x^3) = -a(c_1x^2+c_2x^3)-b\left(x -\frac{1}{6}x^3\right)
$$
Equating coeffcients, I get $c_1=0$ and $c_2=-\frac{b}{6a}$ (but also $-bx=0$, which doesn't make any sense!?)
Ignoring $-bx=0$, The solutions found for $c_i$ give
$$
h(x) = -\frac{b}{6a}x^3
$$
Is this the stable manifold? How can I check if that is the case?
As for the unstable manifold, I don't even know how to start. Is there a similar approach to find it?
Best Answer
Making some local adjustments we have
$$ h(x) = \sum_{k=1}^n c_k (x-\pi)^{2k-1}\\ \sin(x)= x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^6) $$
and then for $n=3, a = \frac 18, b = 1$
$$ h_x(x)h(x)+a h(x)+b\left(x-\frac{x^3}{6}+\frac{x^5}{120}\right)=0 $$
giving the relationships
$$ \left\{ \begin{array}{rcl} \pi c_1^2+4 \pi ^3 c_2 c_1+6 \pi ^5 c_3 c_1+a \pi c_1+3 \pi ^5 c_2^2+5 \pi ^9 c_3^2+a \pi ^3 c_2+8 \pi ^7 c_2 c_3-a \pi ^5 c_3 & = & 0\\ c_1^2+a c_1+12 \pi ^2 c_2 c_1+30 \pi ^4 c_3 c_1+15 \pi ^4 c_2^2+45 \pi ^8 c_3^2+b+3 a \pi ^2 c_2+56 \pi ^6 c_2 c_3+5 a \pi ^4 c_3 & = & 0\\ 30 \pi ^3 c_2^2+12 \pi c_1 c_2+168 \pi ^5 c_3 c_2+3 a \pi c_2+180 \pi ^7 c_3^2+60 \pi ^3 c_1 c_3+10 a \pi ^3 c_3 & = & 0\\ \end{array} \right. $$
Solving for $c_k$ we obtain
$$ \left[ \begin{array}{ccc} c_1 & c_2 & c_3\\ -2.75782 & 0.57593 & -0.0296066 \\ -1.05874 & 0.0394895 & -0.000483529 \\ 0.91031 & -0.0391548 & 0.00025463 \\ \end{array} \right] $$
discarding the first solution we follow with
$$ \left\{ \begin{array}{l} h_1(x) = -0.000483529 (x-\pi )^5+0.0394895 (x-\pi )^3-1.05874 (x-\pi ) \\ h_2(x) = 0.00025463 (x-\pi )^5-0.0391548 (x-\pi )^3+0.91031 (x-\pi ) \\ \end{array} \right. $$
Follows a stream plot and $h_1,h_2$ in red and blue respectively.
NOTE
In this case the linear approximation for the stable and unstable modes is easily obtained from the jacobian eigenvectors which at $(0, \pi)$ have the values
$$ \left\{ \begin{array}{l} \vec v_1 = (\frac{a-\sqrt{a^2+4 b}}{2 b} , 1) \\ \vec v_2 = (\frac{a+\sqrt{a^2+4 b}}{2 b} , 1) \\ \end{array} \right. $$