91Ó°ÊÓ

Given the differential equations \(\dot{r} = -r\) and \(\dot{\theta} = \frac{1}{\ln r}\), we can first solve the r equation. It's a first-order linear differential equation: 1. Separate variables: \(\frac{dr}{dt} = -r\). 2. Integrate both sides: \(\int \frac{1}{r} dr = \int -1 dt\). 3. Solve the integrals: \(\ln|r| = -t + C\). 4. Take the exponent of both sides: \(|r| = e^{-t+C}\). 5. Since \(|r| = r_0\) when \(t = 0\), we can find the constant C: \(e^C = r_0\). 6. The general solution of r(t) is \(r(t) = r_0 e^{-t}\). Now, we will solve the θ(t) equation, which is a first-order linear differential equation: 1. Substitute the result of r(t) into the equation: \(\frac{d\theta}{dt} = \frac{1}{\ln (r_0 e^{-t})}\). 2. Simplify the equation: \(\frac{d\theta}{dt} = \frac{1}{\ln r_0 - t}\). 3. Separate variables: \(d\theta = \frac{1}{\ln r_0 - t} dt\). 4. Integrate both sides: \(\int d\theta = \int \frac{1}{\ln r_0 - t} dt\). 5. Solve the integral on the right side using substitution method (let \(u = \ln r_0 - t\)) and find the general solution for θ(t): \(\theta(t) = -\ln(\ln r_0 - t) + \theta_0\).

With the general solutions of r(t) and θ(t) found in Step 1, we can determine their behavior as t → ∞: a) The r(t) function: As t → ∞, the exponential term \(e^{-t}\) approaches 0, so \(r(t) \rightarrow 0\). b) The θ(t) function: As t → ∞, the term \(\ln r_0 - t\) approaches \(-\infty\). Therefore, we have \(\ln(-\infty) = -\infty\). Thus, \(|\theta(t)| \rightarrow \infty\). As r(t) approaches 0 and θ(t) approaches infinity, the system becomes a stable spiral.

To change the system from polar coordinates to Cartesian coordinates, we use the following transformations: \(x = r \cos \theta\) and \(y = r \sin \theta\) Taking their time derivatives, we will get: \(\dot{x} = \frac{dx}{dt} = \dot{r}\cos\theta - r\dot{\theta}\sin\theta\) and \(\dot{y} = \frac{dy}{dt} = \dot{r}\sin\theta + r\dot{\theta}\cos\theta\) Substitute the differential equations of \(\dot{r}\) and \(\dot{\theta}\) into these equations: \(\dot{x} = -r\cos\theta - r\frac{1}{\ln r}\sin\theta\) and \(\dot{y} = -r\sin\theta + r\frac{1}{\ln r}\cos\theta\)

The linearized system involves finding the Jacobian matrix evaluated at the origin (r = 0), and then transforming it into the Cartesian coordinate system. Since we're taking the Jacobian at r = 0, the exponential term doesn't affect the system. Thus, the linearized system becomes: \(\dot{x} = -x\), and \(\dot{y} = -y\) The eigenvalues of the Jacobian matrix are both negative, making this a stable star in the linearized system.