Let's tackle each part step-by-step, as requested: --- ### **Given System** \[ \begin{aligned} \dot{\theta}_1 &= K \sin(\theta_2 - \theta_1) - \sin \theta_1 \\ \dot{\theta}_2 &= K \sin(\theta_1 - \theta_2) - \sin \theta_2 \end{aligned} \] where \(K \geq 0\). --- ## **(a) Find and classify all fixed points of the system.** **Fixed points:** \(\dot{\theta}_1 = \dot{\theta}_2 = 0\). So, \[ \begin{cases} K \sin(\theta_2 - \theta_1) - \sin \theta_1 = 0 \\ K \sin(\theta_1 - \theta_2) - \sin \theta_2 = 0 \\ \end{cases} \] But \(\sin(\theta_1 - \theta_2) = -\sin(\theta_2 - \theta_1)\): \[ \begin{cases} K \sin(\theta_2 - \theta_1) = \sin \theta_1 \\ -K \sin(\theta_2 - \theta_1) = \sin \theta_2 \\ \end{cases} \] Adding: \(\sin \theta_1 + \sin \theta_2 = 0\) So: \(\sin \theta_2 = -\sin \theta_1\) Thus, either: 1. \(\theta_2 = -\theta_1\) 2. \(\theta_2 = \pi - \theta_1\) **Let’s check both cases:** ### **Case 1:** \(\theta_2 = -\theta_1\) Plug into the first equation: \[ K \sin(-\theta_1 - \theta_1) = \sin \theta_1 \implies K \sin(-2\theta_1) = \sin \theta_1 \] But \(\sin(-2\theta_1) = -\sin 2\theta_1\): \[ -K \sin 2\theta_1 = \sin \theta_1 \implies K \sin 2\theta_1 + \sin \theta_1 = 0 \] \[ \sin \theta_1 (K \cdot 2 \cos \theta_1 + 1) = 0 \] So: - \(\sin \theta_1 = 0\): \(\theta_1 = 0, \pi\) - or \(2K \cos \theta_1 + 1 = 0 \implies \cos \theta_1 = -\frac{1}{2K}\) Thus: 1. \((\theta_1, \theta_2) = (0, 0)\) 2. \((\theta_1, \theta_2) = (\pi, -\pi)\) (but \(-\pi\) is equivalent to \(\pi\) modulo \(2\pi\), so effectively \((\pi, \pi)\)) 3. \((\theta_1, \theta_2) = (a, -a)\), where \(\cos a = -\frac{1}{2K}\), provided \(|2K| \geq 1 \implies K \geq \frac{1}{2}\). --- ### **Case 2:** \(\theta_2 = \pi - \theta_1\) Plug into the first equation: \[ K \sin((\pi - \theta_1) - \theta_1) = \sin \theta_1 \implies K \sin(\pi - 2\theta_1) = \sin \theta_1 \] \[ \sin(\pi - 2\theta_1) = \sin 2\theta_1 \] So: \[ K \sin 2\theta_1 = \sin \theta_1 \implies 2K \sin \theta_1 \cos \theta_1 - \sin \theta_1 = 0 \] \[ \sin \theta_1 (2K \cos \theta_1 - 1) = 0 \] So: - \(\sin \theta_1 = 0: \theta_1 = 0, \pi\) (already found above) - or \(\cos \theta_1 = \frac{1}{2K}\) So: - \((\theta_1, \theta_2) = (b, \pi - b)\), where \(\cos b = \frac{1}{2K}\), provided \(K \geq \frac{1}{2}\). --- ### **Summary Table of Fixed Points** | \(\theta_1\) | \(\theta_2\) | Condition | |--------------|------------------|--------------------------------| | 0 | 0 | all \(K\) | | \(\pi\) | \(\pi\) | all \(K\) | | \(a\) | \(-a\) | \(\cos a = -\frac{1}{2K}\), \(K \geq \frac{1}{2}\) | | \(b\) | \(\pi-b\) | \(\cos b = \frac{1}{2K}\), \(K \geq \frac{1}{2}\) | --- ### **Stability Analysis** Linearize about fixed points. Compute Jacobian \(J\): \[ J = \begin{bmatrix} \frac{\partial \dot{\theta}_1}{\partial \theta_1} & \frac{\partial \dot{\theta}_1}{\partial \theta_2} \\ \frac{\partial \dot{\theta}_2}{\partial \theta_1} & \frac{\partial \dot{\theta}_2}{\partial \theta_2} \\ \end{bmatrix} \] \[ \begin{aligned} \frac{\partial \dot{\theta}_1}{\partial \theta_1} &= -K \cos(\theta_2 - \theta_1) - \cos \theta_1 \\ \frac{\partial \dot{\theta}_1}{\partial \theta_2} &= K \cos(\theta_2 - \theta_1) \\ \frac{\partial \dot{\theta}_2}{\partial \theta_1} &= K \cos(\theta_1 - \theta_2) \\ \frac{\partial \dot{\theta}_2}{\partial \theta_2} &= -K \cos(\theta_1 - \theta_2) - \cos \theta_2 \\ \end{aligned} \] #### **For \((0,0)\):** \[ \cos(\theta_2 - \theta_1) = 1, \ \cos \theta_1 = 1, \ \cos \theta_2 = 1 \] So, \[ J_{(0,0)} = \begin{bmatrix} -K - 1 & K \\ K & -K - 1 \end{bmatrix} \] Eigenvalues of \(\lambda I - J\): \[ \begin{vmatrix} \lambda + K + 1 & -K \\ -K & \lambda + K + 1 \end{vmatrix} = (\lambda + K + 1)^2 - K^2 = 0 \] \[ (\lambda + K + 1 + K)(\lambda + K + 1 - K) = 0 \implies (\lambda + 2K + 1)(\lambda + 1) = 0 \] So, \(\lambda_1 = -1\), \(\lambda_2 = -(2K + 1) < 0\). - **Stable node for all \(K > 0\).** #### **For \((\pi, \pi)\):** \[ \cos(\theta_2 - \theta_1) = 1,\ \cos \theta_1 = -1,\ \cos \theta_2 = -1 \] So, \[ J_{(\pi, \pi)} = \begin{bmatrix} -K + 1 & K \\ K & -K + 1 \end{bmatrix} \] Eigenvalues: \[ (\lambda + K - 1)^2 - K^2 = 0 \implies (\lambda + K - 1 + K)(\lambda + K - 1 - K) = 0 \] \[ (\lambda + 2K - 1)(\lambda - 1) = 0 \implies \lambda_1 = 1, \ \lambda_2 = 2K - 1 \] - For \(K < 1/2\): one positive, one negative eigenvalue: **saddle**. - For \(K > 1/2\): both positive: **unstable node**. #### **For \((a, -a)\), \(\cos a = -\frac{1}{2K}\):** \[ \theta_1 = a, \ \theta_2 = -a \implies \cos(\theta_2 - \theta_1) = \cos(-2a) = \cos 2a \] \[ \cos \theta_1 = \cos a, \ \cos \theta_2 = \cos(-a) = \cos a \] So, \[ J = \begin{bmatrix} -K\cos 2a - \cos a & K \cos 2a \\ K \cos 2a & -K\cos 2a - \cos a \end{bmatrix} \] Eigenvalues: \[ (\lambda + K \cos 2a + \cos a)^2 - (K \cos 2a)^2 = 0 \implies (\lambda + K\cos 2a + \cos a + K\cos 2a)(\lambda + K\cos 2a + \cos a - K\cos 2a) = 0 \] \[ (\lambda + 2K\cos 2a + \cos a)(\lambda + \cos a) = 0 \] So, \(\lambda_1 = -\cos a\), \(\lambda_2 = -2K\cos 2a - \cos a\). But, \(\cos a = -\frac{1}{2K}\): \[ \lambda_1 = \frac{1}{2K} \] \[ \cos 2a = 2\cos^2 a - 1 = 2\left(\frac{1}{4K^2}\right) - 1 = \frac{1}{2K^2} - 1 \] \[ \lambda_2 = -2K\left(\frac{1}{2K^2} - 1\right) + \frac{1}{2K} = -\frac{1}{K} + 2K + \frac{1}{2K} = 2K - \frac{1}{2K} \] So, - \(\lambda_1 = \frac{1}{2K}\) (always positive for \(K > 0\)), so always **unstable**. #### **For \((b, \pi-b)\), \(\cos b = \frac{1}{2K}\):** Similar calculation shows that one eigenvalue is always positive. --- ### **Table of Fixed Points and Stability** | \((\theta_1, \theta_2)\) | Stability | |-----------------------------------|--------------------------| | (0, 0) | Stable node (all \(K\)) | | (\(\pi\), \(\pi\)) | Saddle (\(K<1/2\)), Unstable node (\(K>1/2\)) | | (a, -a), \(\cos a = -\frac{1}{2K}\) | Unstable (\(K \geq 1/2\)) | | (b, \(\pi-b\)), \(\cos b = \frac{1}{2K}\) | Unstable (\(K \geq 1/2\)) | --- ## **(b) Show that a bifurcation occurs at \(K = 1/2\). What type of bifurcation is it?** At \(K = 1/2\): - The fixed points \((a, -a)\) and \((b, \pi-b)\) are created when \(|2K| \geq 1 \implies K \geq 1/2\). - The stability of (\(\pi, \pi\)) changes at \(K = 1/2\). This is a **pitchfork bifurcation**: as \(K\) increases past 1/2, two new fixed points are created (symmetrically), and the stability of an existing fixed point (\(\pi, \pi\)) changes. --- ## **(c) Show the system is a gradient system: \(\dot{\theta}_i = -\partial V/\partial \theta_i\) for some \(V(\theta_1, \theta_2)\).** Let’s try to find \(V\): \[ \dot{\theta}_1 = K \sin(\theta_2 - \theta_1) - \sin \theta_1 = -\frac{\partial V}{\partial \theta_1} \] \[ \dot{\theta}_2 = K \