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The full problem reads:

Show that the system with $\dot{x} = y$ and $\dot{y} = -2x-y-3x^4+y^2$ has no limit cycle in $\mathbb{R}^2$. (HINT: a function in the form $\mathrm{e}^{\beta x}$ might be useful)

I know that I need to use Dulac's Criteria here, but I can't find the correct $\beta$ to go along with the hint. Any advice is welcome!

Thanks!

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For any planar system of the form

$\dot x = X(x, y), \tag 1$

$\dot y = Y(x, y), \tag 2$

the Bendixson-Dulac criterion states that there are no periodic orbits in a region $\Omega$ where

$\nabla \cdot (\phi(X, Y)) = \nabla \cdot (\phi X, \phi Y) \ne 0, \tag 3$

where $\phi(x, y)$ is some differentiable scalar function on $\Omega$. For the present system, we may take $\Omega = \Bbb R^2$ and

$\dot x = y = X(x, y), \tag 4$

$\dot y = -2x - y - 3x^4 + y^2 = Y(x, y); \tag 5$

if we set

$\phi(x, y) = e^{\beta x}, \tag 6$

then we calculate:

$\nabla \cdot (\phi(X,Y)) = \dfrac{\partial (\phi X)}{\partial x} + \dfrac{\partial (\phi Y)}{\partial y} = \dfrac{\partial (e^{\beta x}y)}{\partial x} + \dfrac{\partial (e^{\beta x}(-2x - y - 3x^4 + y^2))}{\partial y}$ $= \beta e^{\beta x}y + (e^{\beta x}(-1 + 2y)) = e^{\beta x}(\beta y - 1 + 2y) = e^{\beta x}((\beta + 2)y - 1); \tag 7$

we observe that the function $e^{\beta x}((\beta + 2)y - 1)$ occurring on the right-hand side of (7) takes on the value zero precisely at that $y_0$ for which

$(\beta + 2)y_0 - 1 = 0, \tag 8$

that is, along the line

$y = y_0 = \dfrac{1}{\beta + 2}, \tag 9$

in the $xy$-plane. If we choose

$\beta = \epsilon - 2, \tag{10}$

for some real $\epsilon$, then since

$\epsilon = \beta + 2, \tag{11}$

we see that

$y_0(\epsilon) = \dfrac{1}{\beta + 2} = \dfrac{1}{\epsilon}, \tag{12}$

and we further observe that by choosing $\epsilon > 0$ sufficiently small, not only may we ensure that $y_0(\epsilon)$ is arbitraritly large, but also that the function

$(\beta + 2)y - 1 = \epsilon y - 1 \tag{13}$

increases with increasing $y$; thus it is positive for $y > y_0$ and negative for $y < y_0$.

Now if we had a periodic orbit, it would form the boundary of a compact subset $K \subset \Bbb R^2$ upon which the function $y$ is continuous, hence bounded above by some

$y_M = \sup \{y \mid y \in K \}; \tag{14}$

thus for suffiiently small $\beta + 2 = \epsilon > 0$ we have

$y_M < y_0; \tag{15}$

that is, on $K$,

$(\beta + 2)y - 1 < 0 \Longrightarrow e^{\beta x}((\beta + 2)y - 1) < 0; \tag{16}$

but now by Bendixson-Dulac, via (7), this contradicts the existence of a periodic trajectory of the system (1)-(2); thus in particular this system cannot have a limit cycle in $\Bbb R^2$.

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  • 1
    $\begingroup$ This is a much more rigorous answer than I was expecting. I agree in it's correctness, however, is there a simpler way to do this? Can we find a $\phi$ such that $\nabla (\phi(X,Y)) \neq 0$ as you mention in line 3 and just use that? Or is that not an option for this problem? $\endgroup$ – obewanjacobi Jul 9 '18 at 2:24
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    $\begingroup$ @obewanjacobi: Well, I figured that no matter what $\phi$ you choose you have to show it works. I went with the suggestion and tried to find the right range of $\beta$, one thing led to another, and voila! It's really not that complicated but you have to get $\beta$ in the right range. By the way, love your screen name--very clever. Cheers! $\endgroup$ – Robert Lewis Jul 9 '18 at 2:34
  • $\begingroup$ @obewanjacobi: obviously the whole trick in my method is to realize that the line where $(2 + \beta)y - 1 = 0$ can be moved out of the region surrounded by the periodic orbit! $\endgroup$ – Robert Lewis Jul 9 '18 at 2:37
  • $\begingroup$ @RobertLewis I am still trying to understand how the line can be moved out of the region surrounding the periodic orbit? $\endgroup$ – BAYMAX Jul 9 '18 at 5:00
  • $\begingroup$ @BAYMAX Well, the position of the line depends on $\beta$, a free parameter. I just chose $\beta$ so that the zero set of $\nabla \cdot e^{\beta x} (X, Y)$ does not intersect the closed orbit or its interior, which then implies this divergence is of fixed sign on that region. Of course, it is a "happy accident" that this can be so simply done. Obviously if $X, Y$ or $\phi$ was more complicated, things could get messy fast. $\endgroup$ – Robert Lewis Jul 9 '18 at 5:08
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Let us take $\beta=-2$. Then: $$ \begin{aligned} &\frac{\partial}{\partial x}\Big(\ e^{-2x}\cdot y\ \Big) + \frac{\partial}{\partial y}\Big(\ e^{-2x}\cdot (-2x-y-3x^4+y^2)\ \Big) \\ &\qquad = -2e^{-2x}\cdot y+ e^{-2x}\cdot (-1+2y)=-e^{-2x} \\ &\qquad<0\ . \end{aligned} $$ This is enough to apply Dulac's criteria.

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