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: enter image description here

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?


1 Answer 1


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

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


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

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

enter image description here

  • $\begingroup$ 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. $\endgroup$
    – Dasherman
    Dec 16, 2018 at 14:10
  • 1
    $\begingroup$ @Dasherman It will be zero in the numerator and in the denominator, take the limit when $u\to 0$ $\endgroup$
    – caverac
    Dec 16, 2018 at 14:17
  • $\begingroup$ 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 $a_3u^3$? $\endgroup$
    – Dasherman
    Dec 16, 2018 at 14:27
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    $\begingroup$ @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 $u^4$, $u^5$, $\cdots$ $\endgroup$
    – caverac
    Dec 16, 2018 at 14:35
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    $\begingroup$ @Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$ $\endgroup$
    – caverac
    Dec 20, 2018 at 13:26

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