Dragon Notes


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Nonlinear Dynamics & Chaos



Ordinary Differential Equations

[Th1] Existence and Uniqueness

Given \(\ \dot{x}=f(x),\quad x(0)=x_0,\)
Suppose that \(f(x) \vpl\) and \(f'(x)\) are continuous on an open interval \(R\) of the \(x\)-axis, and suppose that \(x_0\) is a point in \(R\). Then the IVP has a solution \(x(t)\) on some interval \(-\tau,\ \tau\) about \(t=0\), and the solution is unique.

[Ex1] Multiple solutions

\(\vpl\)Suppose \(\dot{x}=x^{1/3}\). Show that its solution starting from \(x_0=0\) is not unique.
[Sol] The point \(x=0\) is a fixed point, so one solution is \(x(t)=0\) for all \(t\). To find another solution, separate variables and integrate:
\(\ds \vpl \int{x^{-1/3}}dx=\int dt\)
so \(\frac{3}{2}x^{2/3}=t+C\). Imposing the IC \(x(0)=0\) yields \(C=0\). Hence, another solution is \(x(t)=\left(\frac{2}{3}t\right)^{3/2}\)
In this example, the IVP has an infinite number of solutions - owing to the slope \(f'(0)\) at the fixed point \(x^*=0\) being infinite.
[Ex2] Unique but limited solutions

\(\vpl\)Consider \(\ \dot{x}=1+x^2,\ x(0)=x_0\).
Here, \(f(x)=1+x^2\). This function is continuous and has a continuous derivative for all \(x\). Hence [T1] asserts that solutions exist and are unique for any IC \(x_0\). But the theorem does not state that solutions exist for all time; they are only guaranteed to exist in a (possibly very short) time interval around \(t=0\).
For example, consider \(x(0)=0\); separating variables,
\(\ds \vpl \int \frac{dx}{1+x^2}=\int dt\), which yields \(\ds \atan{x}=t+C\)
\(x(0)=0\) implies \(C=0\). Hence the sol. is \(x(t)=\tan{(t)}\). Note, however, that this solution exists only for \(-\pi/2 < t < \pi/2\), because \(x(t)\rightarrow \pm \infty\) as \(t\rightarrow \pm \pi/2\). Outside this interval, the IVP for \(x_0=0\) has no solution.
[Ex3] Reaching a fixed point in finite time

\(\vpl\)A particle travels on a line \(x\) with a velocity given by \(\dot{x}=-x^{c}\), where \(c\) is real & constant. Can the particle ever reach the origin in a finite time? Specifically, how long does it take for the particle to travel from \(x=1\) to \(x=0\), as a function of \(c\)?
[Sol] Separate and integrate:
\(\ds \begin{align} \int \frac{dx}{x^c} &= -\int dt \Rightarrow \frac{x^{1-c}}{1-c} = -t+C_0 \\ t &= \frac{x^{-c+1}}{c-1}+C_0,\ \t{IC:}\ x(0)=1 \\ 0 &= \frac{1^{1-c}}{c-1}+C_0 \Rightarrow C_0=\frac{1}{1-c} \\ t(c) &= \frac{1}{c-1}(0^{1-c}-1) \end{align}\)
Note that \(0^{1-c}\) explodes for \(c>1\) but not \(c\leq1\), and \((1-c)^{-1}\) for \(c=1\). Taking case-by-case,
c > 1:
\(\vpl \frac{1}{2-1}(0^{1-2}-1)=(+)(\infty)\rightarrow \infty\)
c = 1:
\(\vpl \frac{1}{1-1}(0^{1-1}-1)\rightarrow \infty\)
c < 1:
\(\vpl \frac{1}{-1-1}(0^{1+1}-1)=(-)(-)=(+)\)
\[\Rightarrow \boxed{\ds t(c)_{x:1\rightarrow 0}=\left\{ \begin{array}{c} \begin{align} &\infty,&&c \geq 1 \\ &(1-c)^{-1},&&c<1 \end{align} \end{array} \right.}\]
[Ex4]\(\hspace{2px}\) Reaching infinity in finite time

\(\vpl\)Given \(\ \dot{x}=1+x^2,\ x(0)=x_0\), observe the finite time behavior.
[Sol] \(\vpl\) From [Ex2], \(x(t)\rightarrow \pm \infty\) as \(t\rightarrow \pm \pi/2\). Hence, the system has solutions that reach infinity in finite time (a 'blow-up').

Linear Stability

\(\ds \dot{\eta} \approx \eta f'(x^*)\)\(\quad \bb{1}\) [𝕯



\(\equiv\) Linearization about \(x^*\); \(f'(x^*)\neq 0\)
[P0] Shows that the perturbation \(\eta (t) \vplup \) exponentially

\(\t{grows}\ \t{ if } f'(x^*)>0\qquad \t{decays}\ \t{ if } f'(x^*)<0 \vplup\)

[Ex1]\(\hspace{1px}\) Assess the linear stability of \(\dot{x}=\Sin{x}\)

\(\vplup\)The fixed points occur where \(f(x)=\Sin{x}=0\). Thus \(x^*=k\pi , k\in \mathbb{Z}\). Then,
\(\ds \vpLup f'(x^*)=\Cos{k\pi} =\left\{ \begin{array}{c} \begin{align} 1, & \quad k\ \t{ even} \\ -1, & \quad k\ \t{ odd} \end{align} \end{array} \right. \Rightarrow\) \(x^* =\left\{ \begin{array}{c} \begin{align} \t{unstable}, & \quad k\ \t{ even} \\ \t{stable}, & \quad k\ \t{ odd} \end{align} \end{array} \right.\)
Plot shown above the problem to the right.
[Ex2]\(\hspace{1px}\) Assess the linear stability of: (1) \(\dot{x}=-x^3\), (2) \(\dot{x}=x^3\), (3) \(\dot{x}=x^2\)

See plot below 'Linear Stability'.


Bifurcations are changes in dynamics of vector fields creating or destroying fixed points, or changing their stability.
\(\vplLup\)In a

saddle-node bifurcation

, as a parameter is varied, two fixed points move toward each other, collide, and mutually annihilate.
\(\vplLup\)Consider the first-order system
where \(r\) is a parameter.
  \(\b{[r<0]}\): Two fixed points, one stable and one unstable \(\vplup\)
  \(\b{[r=0]}\): The fixed points merge into a half-stable fixed point
  \(\b{[r>0]}\): The fixed point vanishes, leaving none at all
\(\vplLup\)A saddle-node bifurcation occurs at \(r=0\), changing the vector fields.
\(\dot{x}=r+x^2\), along \(\dot{x}=r-x^2\), are normal forms of saddle-node bifurcations - equations with algebraic forms matching those of other functions with said bifurcations in vicinity of the bifurcation points.

Shown to the right is a plot of \(x^*\) vs. \(r\), called the bifurcation diagram of the system - showing the range of fixed points \(x^*\) as \(r\) is varied. Solid lines indicate stable fixed points, dashed unstable.

Transcritical bifurcations

are changes in fixed point stability.
\(\vplLup\)Consider the first-order system (and normal form)
where \(r\) is a parameter.
  \(\b{[r<0]}\): Unstable fixed point at \(x^*=r\), and stable at \(x^*=0\) \(\vplup\)
  \(\b{[r=0]}\): The fixed points merge into a half-stable fixed point
  \(\b{[r>0]}\): The origin becomes unstable, and \(x^*=r\) is stable
A transcritical bifurcation occurs at \(r=0\vplLup\), changing fixed point stability. In this case, an exchange of stabilities takes place.
Note also that as \(r\vplLup\) changes, the vertex of \(\dot{x}\) traces out a parabola of its equal and negated parent function fixed at the origin.


pitchfork bifurcations

, fixed points appear and disappear in symmetrical pairs. Two types occur:

Supercritical pitchfork bifurcations

, with a normal form
where \(r\) is a parameter.
  \(\b{[r<0]}\): Stable origin, \(x^*=0\vplup\)
  \(\b{[r=0]}\): Origin is still stable, but less so due to vanished
\(\hspace{80px}\)linearization (see critical slowdown)
  \(\b{[r>0]}\): Origin becomes unstable, and two new stable fixed points
\(\hspace{80px}\)emerge symmetrically at \(x^*=\pm\sqrt{r}\)
- The cubic term\(\vplup\) is stabilizing: it acts as a restoring force that pulls \(x(t)\) back toward \(x=0\)
- The system is invariant under the change of variables \(x\rightarrow -x\); i.e., \((-)\) signs cancel with \(-x\) plugged in on both sides - hence symmetry

Subcritical pitchfork bifurcations

have a normal form
where \(r\) is a parameter.
  \(\b{[r<0]}\): Stable origin, \(x^*=0\vplup\), with unstable symmetric fixed
\(\hspace{80px}\)points \(x^*=\pm\sqrt{r}\)
  \(\b{[r=0]}\): Origin is still stable, but less so due to vanished
\(\hspace{80px}\)linearization (see critical slowdown)
  \(\b{[r>0]}\): Origin becomes unstable, absorbing the unstable
\(\hspace{80px}\)symmetric fixed points
- The\(\vplup\) cubic term is destabilizing, leading to blow-up: \(x(t)\rightarrow\pm\infty\) in finite time, starting from any IC \(x_0\neq 0\)
- The system is invariant under the change of variables \(x\rightarrow -x\)

Critical slowdown

is a second-order phase transition where the system relaxes to equilibrium much slower than usual.
\(\vplLup\)Consider the system \(\dot{x}=-x^3,\ \ x(0)=10\). Solving,
\(\ds \hspace{160px}x(t)=\frac{10}{\sqrt{200t+1}}\).
As \(t\rightarrow \infty ,\ x(t)\rightarrow 0\), but the decay rate is not exponential. Now consider \(\dot{x}=-x\) w/ the same IC; the solution is
\(\ds \hspace{160px}x(t)=10e^{-t}\).
The system decays to zero as before, but now exponentially. See plots of the two shown, along for the combined system of the two.

To see why this occurs, observe the rates behavior near origin: the rate of \(\dot{x}=-x^3\) drops to zero very rapidly near origin compared with that of \(\dot{x}=-x\), and as can be seen by their difference function (shown), is virtually relatively negligible very close to the origin.

[Ex1]\(\hspace{1px}\) Show that the first-order system \(\dot{x}=x(1-x^2)-a(1-e^{-bx})\) undergoes a transcritical bifurcation at \(x=0\) when parameters \(a,b\) satisfy a certain equation (to be determined). Find an approximate formula for the fixed point that bifurcates from \(x=0\), assuming that the parameters are close to the bifurcation curve.

[Sol] \(x^*=0\) is a fixed point for all (a,b). This makes transcritical bifurcation plausible, if bifurcating at all. For small \(x\),
\(\ds \begin{align} 1-e^{-bx} &= 1-[1-bx+\sfrac{1}{2}b^2x^2+O(x^3)] \\ &= bx-\sfrac{1}{2}b^2x^2+O(x^3) \\ \Rightarrow \dot{x} &= x-a(bx-\sfrac{1}{2}b^2x^2)+O(x^3) \\ &= (1-ab)x+(\sfrac{1}{2}ab^2)x^2+O(x^3) \end{align}\)
Hence a transcritical bifurcation occurs when \(ab=1\); this is the equatoin for the bifurcation curve. The nonzero fixed point is given by the solution of \(1-ab+(\frac{1}{2}ab^2)x\approx 0\), i.e. \(\ds x^* \approx 2(ab-1)/(ab^2)\).


[P0] Oscillations are impossible [first-order systems]

    All trajectories either approach a fixed point or diverge to \(\pm \infty\); this is their only possible behavior:
      ➢ Trajectories must either increase or decrease monotonically, or remain constant
      ➢ The phase point never reverses direction
    The approach to equilibrium (a fixed point) is always monotonic - overshoot and damped oscillations can never occur in a first-order system.\(\vplup\)
    For the same reason, undamped oscillations are impossible. Hence, there are no periodic solutions to \(\dot{x} = f(x)\).

Dragon Notes,   Est. 2018     About

By OverLordGoldDragon