# Finding approximation to stable manifold of saddle point

I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).

Consider the system $$\dot{x} = x+e^{-y}, \dot{y} = -y$$. This system has a single fixed point, $$(-1, 0)$$. This is a saddle point. The unstable manifold is $$y=0$$, but the stable manifold is some non-linear curve. Let $$(x, y)$$ be a point on the stable manifold close to $$(-1, > 0)$$ and define $$u = x + 1$$. Write the stable manifold as $$y=a_1u +a_2u^2 + O(u^3)$$. To determine the coefficients, derive two expressions for $$dy/du$$ and equate them.

I have, as a first try for an equation, simply differentiated $$y$$ wrt $$u$$: $$\frac{dy}{du} = a_1 + 2a_2u + O(u^2)$$, where I'm a bit uncertain about the $$O(u^2)$$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.

I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:

Furthermore, I don't see why they have $$\frac{dy}{du} = \frac{\dot{y}}{\dot{u}}$$ or how they calculated the Taylor approximation for $$\dot{u}$$.

Basically, I'm a bit lost, could someone perhaps give me some hints?

Remember that

$$\frac{{\rm d}y}{{\rm d}u} = \frac{{\rm d}y / {\rm d}t}{{\rm d}u / {\rm d}t} \tag{1}$$

If you replace

$$\frac{{\rm d}y}{{\rm d}t} = -y \tag{2}$$

and

$$\frac{{\rm d}u}{{\rm d}t} = \frac{{\rm d}x}{{\rm d}t} = u - 1 + \left(1 - y + \frac{y^2}{2} - \cdots \right) \tag{3}$$

In (1) you will get

$$\frac{{\rm d}y}{{\rm d}u} = -\frac{y}{u - y + y^2/2 - y^3/6 + \cdots} \tag{4}$$

Now you replace your expression for $$y$$: $$y(u) = a_1u + a_2 u^2 + \cdots$$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $$y$$ in (4) and Taylor expanding it around $$u = 0$$, the result is

$$\begin{eqnarray} \frac{{\rm d}y}{{\rm d}u} &=& \frac{a_1}{a_1 - 1} + \frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + \frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + \cdots \\ &=& a_1 + 2a_2 u + 3a_3 u^3 + \cdots \end{eqnarray}$$

You solve for coefficients, and you should get

$$a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9$$

The black line below is the result

• Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $\frac{a_1}{a_1-1}$ you find. Dec 16, 2018 at 14:10
• @Dasherman It will be zero in the numerator and in the denominator, take the limit when $u\to 0$ Dec 16, 2018 at 14:17
• Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$? Dec 16, 2018 at 14:27
• @Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $\cdots$ Dec 16, 2018 at 14:35
• @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$ Dec 20, 2018 at 13:26