Dragon Notes

UNDER CONSTRUCTION
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Nonlinear Dynamics & Chaos:Solved Problems

$$\bb{Pr1}$$[Characteristic Equation] Solve the IVP, and draw its phase portrait:
\ds \begin{align}\dot{x}&=x+y,\\ \dot{y}&=4x-2y,\end{align}$$\hspace{15px}(x_0,y_0)=(2,-3)$$

Linear systems can contain straight-line trajectories - a trajectory starting on one of the coordinate axes stays on that axis forever, and exhibits
simple exponential growth or decay along it. Example below.
In general, we seek to find the analog of these straight-line trajectories - that is, trajectories of the form
$$\dsup \bn{x}(t)=e^{\lambda t}\bn{v}$$
$$\vplup$$where $$\bn{v}\neq\bn{0}$$ is some fixed vector, and $$\lambda$$ is a growth rate. If such solutions exist, they correspond to exponential motion along the line spanned
by the vector $$\bn{v}$$.
To determine $$\bn{v}$$ and $$\lambda$$, we substitue $$\bn{x}(t)=e^{\lambda t}\bn{v}$$ into $$\bn{\dot{x}}=A\bn{x}$$, and obtain $$\lambda Ae^{\lambda t}=e^{\lambda t}A\bn{v}$$. Canceling $$e^{\lambda t}\neq 0$$ yields
$$\dsup A\bn{v}=\lambda\bn{v}$$
$$\vplup$$which says that the desired solutions exist if $$\bn{v}$$ is an eigenvector of $$A$$ with corresponding eigenvalue $$\lambda$$. This eigensolution can be obtained from the
characteristic equation $$\t{det}(A-\lambda I)=0$$.
$$\vplup$$For the given IVP, the matrix equation is
$$\ds \mtx{\dot{x}}{\dot{y}}=\mtxx{1}{1}{4}{-2}\mtx{x}{y}$$
$$\vplup$$The corresponding eigenvalues are $$\lambda_1=2,\ \lambda_2=-3$$. Finding the corresponding eigenvectors,
$$\ds\mtxx{1-\lambda}{1}{4}{-2-\lambda}\mtx{v_1}{v_2}=\mtx{0}{0}$$
$$\Rightarrow$$
$$\ds\u{\lambda_1=2}$$
$$\ds\mtxx{1-\lambda}{1}{4}{-2-\lambda}\mtx{v_1}{v_2}=\mtx{0}{0}$$
$$\ds\Rightarrow (v_1,v_2)=(1,1)$$
$$\ds\u{\lambda_2=-3}$$
$$\mtxx{4}{1}{4}{1}\mtx{v_1}{v_2}=\mtx{0}{0}$$
$$\Rightarrow (v_1,v_2)=(1,-4)$$

$$\Rightarrow \boxed{\bn{v}_1=\mtx{1}{1},\quad \bn{v}_2=\mtx{1}{-4}}\hspace{100px}$$

$$\vplup$$
Now, we write the general solution as a linear combination of eigenvectors:

$$\ds\bn{x}(t)=c_1\mtx{1}{1}e^{2t}+c_2\mtx{1}{-4}e^{-3t}\ \ \bb{1^*}$$
Finally, we compute $$c_1$$ and $$c_2$$ to satisfy the IC $$(x_0,y_0)=(2,-3)$$. At $$t=0$$,
$$\ds \mtx{2}{-3}=c_1\mtx{1}{1}+c_2\mtx{1}{-4}$$

Solving as a system of equations, $$c_1=1,\ c_2=1$$. Substituting into $$\bb{1^*}$$,
\bxred{\begin{align}x(t)&=e^{2t}+e^{-3t}\\y(t)&=e^{2t}-4e^{-3t}\end{align}} The phase portrait of the system is shown below:
Stable manifold: line spanned by $$\bn{v}_2=(1,-4)$$ (decaying eigensolution)
Unstable manifold: line spanned by $$\bn{v}_1=(1,1)$$ (growing eigensolution)
The two exponentially growing and decaying eigensolutions render the origin a saddle point. A typical trajectory approaches the unstable manifold as $$t\rightarrow\infty$$, and the stable manifold as $$t\rightarrow -\infty$$.

$$\bb{Pr2}$$[2D Linearization] Show that the linearization of the following system incorrectly predicts the origin to be a center for all $$a$$:
\ds \begin{align}\dot{x} &= -y+ax(x^2+y^2) \\ \dot{y} &= \ph{-}x+ ay(x^2+y^2) \end{align}

[Sol]: To linearize about $$(x^*,y^*)=(0,0)$$, we can either compute the Jacobian matrix directly from definition, or take a shortcut: for a system with a fp at the origin, $$x$$ and $$y$$ represent deviations from the fixed point, since $$u=x-x^*=x$$ and $$v=y-y^*=y$$. Hence, we can linearize by simply omitting the nonlinear terms - yielding $$\dot{x}=-y,\ \dot{y}=x$$. The Jacobian is thus
$$\ds A=\mtxx{0}{-1}{1}{0}$$
which has $$\tau = 0,\ \Delta=1>0$$, so the origin is always a center, according to the linearization.
$$\vplup$$To analyze the nonlinear system, we transfer to polar coordinates. Let $$x=r\cos{\theta},\ y=r\sin{\theta}$$. To derive a diff-eq for $$r$$, we note $$x^2+y^2=r^2$$, so $$x\dot{x}+y\dot{y}=r\dot{r}$$. Substituting this for $$\dot{x}$$ and $$\dot{y}$$ yields
\ds\begin{align} r\dot{r} &= x(-y+ax(x^2+y^2)) + y(x+ay(x^2+y^2)) \\ &= a(x^2+y^2)^2 \\ &= ar^4 \end{align}
Hence $$\dot{r}=ar^3$$. To write a diff-eq for $$\dot{\theta}$$, we use the identity $$\theta = \atan{y/x}$$ and a reference triangle:
\ds \begin{align} \tan{\theta} &= y/x \\ \frac{dy}{dt}&=\frac{dx}{dt}\tan{\theta}+x\ \t{sec}^2(\theta)\frac{d\theta}{dt} \\ x\ \t{sec}^2(\theta)\ \dot{\theta} &= \dot{y} - \tan{\theta}\ \dot{x}; \\ \t{sec}^2(\theta) &= \Frac{r}{x}^2 = 1 + y^2/x^2 \Rightarrow \\ x(1+y^2/x^2)\ \dot{\theta} &= \dot{y}-(y/x)\dot{x} \\ \dot{\theta} &= \frac{x\dot{y}-y\dot{x}}{x^2(1+y^2/x^2)} = \frac{x\dot{y}-y\dot{x}}{x^2+y^2} \Rightarrow \\ \dot{\theta} &= \frac{x\dot{y}-y\dot{x}}{r^2} \end{align}
Substituting $$\dot{x}$$ and $$\dot{y}$$, we find $$\dot{\theta}=1$$. Thus, in polar coordinates, the original system becomes,
\begin{align}\dot{r} &= ar^3 \\ \dot{\theta} &= 1 \end{align}
All trajectories rotate about the origin with constant angular velocity $$\dot{\theta}=1$$. The radial motion will depend on $$a$$ as shown:
\ds \begin{align} \b{[a<0]} & \quad r(t)\rightarrow 0\ \t{as }t\rightarrow\infty && \t{origin }=\t{ stable spiral}\\ \bn{[a=0]} &\quad r(t)=r_0 && \\ \bn{[a>0]} &\quad r(t)\rightarrow \infty\ \t{as }t\rightarrow \infty && \t{origin }=\t{ unstable spiral} \end{align}
We can see now why centers are so delicate: all trajectories are required to close perfectly after one cycle. The slightest miss converts the center into a spiral. Similarly, stars and degenerate nodes can be altered by small nonlinearities, but unlike centers, their stability doesn't change. (A stable star may change into a stable spiral, but not into an unstable spiral). This is apparent from the 2D system linear stability map: stars and degenerate nodes live squarely in the stable or unstasble region, whereas centers live on the razor's edge between stability and instability (see Stability
).
$$\vplup$$System simulations are shown below.

$$\b{a<0}$$

$$\b{a=0}$$

$$\b{a>0}$$