Systems: Complex Eigenvalues - Differential Equations

1. A system $\mathbf{x}' = A\mathbf{x}$ has eigenvalues $\lambda = \sigma \pm i\omega$ where $\sigma < 0$ and $\omega > 0$. If the distance from any trajectory to the origin decreases by a factor of $e$ while the trajectory rotates through angle $\theta$, what is the relationship between $\sigma$, $\omega$, and $\theta$?
   Answer: $\sigma = -\frac{\omega}{\theta}$ The solution has form $\mathbf{x}(t) = e^{\sigma t}(\mathbf{u}\cos(\omega t) + \mathbf{v}\sin(\omega t))$. The distance from origin is proportional to $e^{\sigma t}$, and the angular displacement is $\omega t$. If the trajectory rotates through angle $\theta$, then $\omega t = \theta$, so $t = \frac{\theta}{\omega}$. During this time, the distance changes by factor $e^{\sigma t} = e^{\sigma \theta/\omega}$. For the distance to decrease by factor $e$ (i.e., become $\frac{1}{e}$ times the original), we need $e^{\sigma \theta/\omega} = \frac{1}{e} = e^{-1}$. Therefore $\frac{\sigma \theta}{\omega} = -1$, which gives $\sigma = -\frac{\omega}{\theta}$.
2. Consider the system $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$ where $a + d = 0$ and $ad - bc = 25$. If the solution satisfying $x(0) = 3, y(0) = 4$ has the property that $x^2(t) + y^2(t)$ is constant for all $t$, which of the following must be true about the coefficient matrix?
   Answer: $a = 0, d = 0$, and $bc = -25$ with $b$ and $c$ having opposite signs For $x^2(t) + y^2(t)$ to be constant, we need $\frac{d}{dt}(x^2 + y^2) = 2x x' + 2y y' = 0$ for all solutions. Substituting the system: $2x(ax + by) + 2y(cx + dy) = 2ax^2 + 2(b+c)xy + 2dy^2 = 0$. Since $a + d = 0$, we have $d = -a$, so $2ax^2 + 2(b+c)xy - 2ay^2 = 0$. For this to hold for all $x, y$, we need $a = 0$ (so $d = 0$) and $b + c = 0$. With $a = d = 0$ and $ad - bc = 25$, we get $-bc = 25$, so $bc = -25$. Since $b + c = 0$, we have $c = -b$, so $b(-b) = -b^2 = -25$, giving $b^2 = 25$. Thus $b = \pm 5$ and $c = \mp 5$, meaning $b$ and $c$ have opposite signs.
3. A system $\mathbf{x}' = A\mathbf{x}$ has complex eigenvalues $3 \pm 4i$. The trajectory starting at $(1, 1)$ first returns to the positive $x$-axis (where $y = 0$ and $x > 0$) at time $t^*$. What is the $x$-coordinate of this return point?
   Answer: $\sqrt{2} e^{3\pi/2}$ The eigenvalues $3 \pm 4i$ give solutions of the form $\mathbf{x}(t) = e^{3t}(c_1\mathbf{u}\cos(4t) + c_2\mathbf{v}\sin(4t))$ where $\mathbf{u}$ and $\mathbf{v}$ are the real and imaginary parts of an eigenvector. Starting at $(1,1)$, the trajectory spirals outward. The trajectory returns to the positive $x$-axis when it has completed one revolution, which occurs when $4t = 2\pi$, so $t^* = \frac{\pi}{2}$. At this time, $y = 0$ and the $x$-coordinate is the initial distance from origin times $e^{3t^*}$. The initial distance is $\sqrt{1^2 + 1^2} = \sqrt{2}$. Therefore, the $x$-coordinate at return is $\sqrt{2} \cdot e^{3\pi/2}$.
4. For the system $\mathbf{x}' = A\mathbf{x}$ where $A$ has complex eigenvalues $\lambda = -1 \pm 3i$, two students propose different fundamental matrix solutions. Student 1 claims $\Phi_1(t) = e^{-t}\begin{pmatrix} \cos(3t) & \sin(3t) \\ 2\cos(3t) + \sin(3t) & 2\sin(3t) - \cos(3t) \end{pmatrix}$. Student 2 claims $\Phi_2(t) = e^{-t}\begin{pmatrix} \cos(3t) & -\sin(3t) \\ 2\cos(3t) + \sin(3t) & 2\sin(3t) + \cos(3t) \end{pmatrix}$. Which statement is most accurate?
   Answer: Neither can be determined valid without verifying $\Phi'(t) = A\Phi(t)$ and checking linear independence A fundamental matrix $\Phi(t)$ must satisfy two conditions: (1) $\Phi'(t) = A\Phi(t)$ and (2) $\det(\Phi(t)) \neq 0$ for all $t$. While both proposed matrices have $\det(\Phi(0)) = -1 \neq 0$, this alone doesn't guarantee they satisfy the differential equation. The specific form of the matrix $A$ determines which solution is correct. We cannot conclude validity without explicitly checking that the derivative equals $A$ times the matrix. The sign of the determinant and patterns in off-diagonal terms don't determine validity by themselves - only the fundamental differential equation relationship matters.

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