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Let \(z=x+i y, \bar{z}=x-i y\) and show that the vector fields in the complex plane defined by $$ \dot{z}=z^{k} \quad \text { and } \quad \dot{z}=\bar{z}^{k} $$ have unique critical points at \(z=0\) with indices \(k\) and \(-k\) respectively. Hint: Write \(\dot{x}=\operatorname{Re}\left(z^{k}\right) ; \dot{y}=\operatorname{Im}\left(z^{k}\right)\) and let \(z=r e^{i \theta} .\) Sketch the phase portrait near the origin in those cases with indices \(\pm 2\) and \(\pm 3\)

Short Answer

Expert verified
The vector fields \( \dot{z} = z^k \) and \( \dot{z} = \bar{z}^k \) have unique critical points at \( z=0 \) with indices \( k \) and \( -k \) respectively.

Step by step solution

01

Express in Polar Coordinates

Let the complex number be represented in polar coordinates, such that \( z = r e^{i \theta} \). Here, \( r \) is the magnitude of \( z \) and \( \theta \) is the angle.
02

Compute \( z^k \) and \( \bar{z}^k \)

Using the polar form, calculate \( z^k = (r e^{i \theta})^k = r^k e^{i k \theta} \) and \( \bar{z} = r e^{-i \theta} \) to get \( \bar{z}^k = (r e^{-i \theta})^k = r^k e^{-i k \theta} \).
03

Derive Real and Imaginary Parts

Let \( z^k = u + iv \) where \( u \) and \( v \) represent the real and imaginary parts respectively. So, \( u = \operatorname{Re}(z^k) = r^k \cos(k \theta) \) and \( v = \operatorname{Im}(z^k) = r^k \sin(k \theta) \). For \( \bar{z}^k \), \( u = r^k \cos(-k \theta) = r^k \cos(k \theta) \) and \( v = r^k \sin(-k \theta) = -r^k \sin(k \theta) \).
04

Write System of Differential Equations

Using the hint, write the differential equations. For \( \dot{z} = z^k \), we have \( \dot{x} = r^k \cos(k \theta) \) and \( \dot{y} = r^k \sin(k \theta) \). For \( \dot{z} = \bar{z}^k \), we get \( \dot{x} = r^k \cos(k \theta) \) and \( \dot{y} = -r^k \sin(k \theta) \).
05

Identify Critical Points

The critical points occur where both \( \dot{x} \) and \( \dot{y} \) are zero. Since the equations include \( r^k \) terms, the critical point \( (x, y) = (0, 0) \) is unique.
06

Compute Indices

The index of a vector field defined by \( \dot{z} = z^k \) at \( z=0 \) is \( k \). The index of the field defined by \( \dot{z} = \bar{z}^k \) at \( z=0 \) is \( -k \).
07

Sketch Phase Portraits

Sketch the phase portraits near the origin with the indices \( \pm 2 \) and \( \pm 3 \). Note that the pattern follows the number of rotations determined by the indices.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

critical points
In the context of vector fields in the complex plane, critical points are locations where the vector field is zero. These points are essential as they represent equilibrium states in a dynamic system. To find critical points for the vector fields given by \(\dot{z} = z^k\) and \(\dot{z} = \bar{z}^k\), we need to set the real and imaginary components of the vector field to zero.

By expressing the functions in terms of their real and imaginary parts, we arrive at the system of equations \(\dot{x} = r^k \cos(k \theta)\) and \(\dot{y} = r^k \sin(k \theta)\) for \(\dot{z} = z^k\). Setting these to zero, we see that the only solution is when \(r = 0\), which corresponds to \(z = 0\).Hence, \(z = 0\) is a critical point.

For the vector field given by \(\dot{z} = \bar{z}^k\), the corresponding system of equations is \(\dot{x} = r^k \cos(k \theta)\) and \(\dot{y} = -r^k \sin(k \theta)\). Again, setting these to zero, we find the critical point is at \(z = 0\).
  • Z = 0 is the critical point for \(\dot{z} = z^k\) with index k.
  • Z = 0 is the critical point for \(\dot{z} = \bar{z}^k\) with index -k.
polar coordinates
Using polar coordinates simplifies the analysis of vector fields in the complex plane. This coordinate system represents a complex number \(z\) as \(r e^{i \theta}\), where \(r\) is the magnitude and \(\theta\) is the angle (or argument) of \(z\). This form often makes it easier to handle complex arithmetic.

To compute \(z^k\) in polar coordinates, we get \(z = r e^{i \theta}\). Thus, \(z^k = (r e^{i \theta})^k = r^k e^{i k \theta}\). Similarly, for the conjugate \(\bar{z} = r e^{-i \theta}\), \(\bar{z}^k = (r e^{-i \theta})^k = r^k e^{-i k \theta}\).
  • Using polar coordinates, complex multiplication becomes simpler.
  • The expressions \(z^k\) and \(\bar{z}^k\) yield cosine and sine terms that represent real and imaginary components respectively.
phase portraits
Phase portraits graphically represent the trajectories of a dynamical system in the phase plane. For complex vector fields defined by \(\dot{z} = z^k\) and \(\dot{z} = \bar{z}^k\), phase portraits help illustrate the behavior near the critical points.

To sketch these, we look at the indices \(\pm k\) near the origin:
  • For \(\dot{z} = z^k\) with indices \(\pm 2\), the trajectories exhibit a double rotation around the origin.
  • For \(\pm 3\), a triple rotation pattern emerges.


Generally, the number of rotations (or loops) in the phase portrait correlates with the index value \(k\), indicating how the system evolves over time.
indices of vector fields
The index of a vector field at a critical point characterizes the topological structure of the field around that point. It essentially counts how many times the vector field rotates while making a closed loop around the critical point.

For \(\dot{z} = z^k\), the index at the critical point \(z = 0\) is \(k\). It indicates that as we trace a path around the origin, the vector field completes \(k\) full rotations.

Conversely, for \(\dot{z} = \bar{z}^k\), the index at \(z = 0\) is \(-k\). Here, the field undergoes \(-k\) rotations around the origin. This negative value shows that the rotation direction is opposite to that of \(\dot{z} = z^k\).
  • The index gives insight into the stability and behavior of critical points.
  • It helps in predicting patterns in the phase portraits.

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Most popular questions from this chapter

(a) Suppose that the linearization of (1) about a periodic orbit \(\Gamma\) of (1) has a fundamental matrix solution given by $$ \Phi(t)=\left[\begin{array}{ccccc} \cos t & -\sin t & 0 & 0 & 0 \\ \sin t & \cos t & 0 & 0 & 0 \\ 0 & 0 & \frac{4 \mathrm{e}^{t}-\mathrm{e}^{-2 t}}{3} & \frac{2\left(\mathrm{e}^{-2 t}-\mathrm{e}^{t}\right)}{3} & 0 \\ 0 & 0 & \frac{2\left(\mathrm{e}^{t}-\mathrm{e}^{-2 t}\right)}{3} & \frac{4 \mathrm{e}^{-2 t}-\mathrm{e}^{t}}{3} & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] $$ Find the characteristic exponents of the periodic orbit \(\Gamma\) and the dimensions of \(W^{s}(\Gamma), W^{u}(\Gamma)\) and \(W^{c}(\Gamma)\) (b) Same thing for $$ \Phi(t)=\left[\begin{array}{ccccc} e^{-3 t} \cos 2 t & -e^{-3 t} \sin 2 t & 0 & 0 & 0 \\ e^{-3 t} \sin 2 t & \mathrm{e}^{-3 t} \cos 2 t & 0 & 0 & 0 \\ 0 & 0 & \mathrm{e}^{3 t} & 0 & 0 \\ 0 & 0 & t \mathrm{e}^{3 t} & \mathrm{e}^{3 t} & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] $$

Show that $$ \begin{aligned} &\dot{x}=y \\ &\dot{y}=x+x^{2} \end{aligned} $$ is a Hamiltonian system with $$ H(x, y)=y^{2} / 2-x^{2} / 2-x^{3} / 3 $$ and that this system is symmetric with respect to the \(x\)-axis. Show that the origin is a saddle for this system and that \((-1,0)\) is a center. Sketch the homoclinic orbit given by $$ y^{2}=x^{2}+\frac{2}{3} x^{3} $$ Also, sketch all four trajectories given by this equation and sketch the phase portrait for this system.

Consider the nonlinear system $$ \begin{aligned} &\dot{x}=-2 y+a x\left(4-4 x^{2}-y^{2}+z\right) \\ &\dot{y}=8 x+a y\left(4-4 x^{2}-y^{2}+z\right) \\ &\dot{z}=z\left(x-a^{2}\right) \end{aligned} $$ where \(a\) is a parameter. Show that \(\gamma(t)=(\cos 4 t, 2 \sin 4 t, 0)^{T}\) is a periodic solution of this system of period \(T=\pi / 2\). Determine the linearization of this system about the periodic orbit \(\gamma(t)\), $$ \dot{\mathbf{x}}=A(t) \mathbf{x} $$ and show that is the fundamental matrix for this nonautonomous linear system which satisfies \(\Phi(0)=I\), where \(\alpha(t)\) and \(\beta(t)\) are \(\pi / 2\)-periodic functions which satisfy the nonhomogeneous linear system $$ \begin{gathered} \left(\begin{array}{l} \dot{\alpha} \\ \dot{\beta} \end{array}\right)=\left[\begin{array}{cc} a(a-4)-\cos 4 t-4 a \cos 8 t & -2-2 a \sin 8 t \\ 8-8 a \sin 8 t & a(a-4)-\cos 4 t+4 a \cos 8 t \end{array}\right] \\ \cdot\left(\begin{array}{c} \alpha \\ \beta \end{array}\right)+\left(\begin{array}{c} a \cos 4 t \\ 2 a \sin 4 t \end{array}\right) \end{gathered} $$ and the initial conditions \(\alpha(0)=\beta(0)=0\). (This latter system can be solved using Theorem 1 and Remark 1 in Section \(1.10\) of Chapter 1 and the result of Problem 4 in Section \(2.2\) of Chapter \(2 ;\) however, this is not necessary for our purposes.) Write \(\Phi(t)\) in the form of equation (3), show that the characteristic exponents \(\lambda_{1}=-8 a\) and \(\lambda_{2}=-a^{2}\), determine the characteristic multipliers of \(\gamma(t)\), and determine the dimensions of the stable and unstable manifolds of \(\gamma(t)\) for \(a>0\) and for \(a<0\)

Let \(F\) satisfy the hypotheses of Lienard's Theorem. Show that $$ \ddot{z}+F(\dot{z})+z=0 $$ has a unique, asymptotically stable, periodic solution.

Consider the system $$ \begin{aligned} &\dot{x}=-y+x\left(1-x^{2}-y^{2}\right)\left(4-x^{2}-y^{2}\right) \\ &\dot{y}=x+y\left(1-x^{2}-y^{2}\right)\left(4-x^{2}-y^{2}\right) \\ &\dot{z}=z \end{aligned} $$ Show that there are two periodic orbits \(\Gamma_{1}\) and \(\Gamma_{2}\) in the \(x, y\) plane represented by \(\gamma_{1}(t)=(\cos t, \sin t)^{T}\) and \(\gamma_{2}(t)=(2 \cos t, 2 \sin t)^{T}\), and determine their stability. Show that there are two invariant cylinders for this system given by \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=4\); and describe the invariant manifolds \(W^{\prime}\left(\Gamma_{j}\right)\) and \(W^{u}\left(\Gamma_{j}\right)\) for \(j=1,2\).

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