Let's solve the problem step by step: --- ## 1. **Show that \( H(x, y) = -\frac{3y^2}{2} - \frac{x^2}{2} \) is a Hamiltonian** Given system: \[ \begin{cases} x' = -3y \\ y' = x \end{cases} \] The Hamiltonian system has the form: \[ \begin{cases} x' = -\frac{\partial H}{\partial y} \\ y' = \frac{\partial H}{\partial x} \end{cases} \] Given: \[ H(x, y) = -\frac{3y^2}{2} - \frac{x^2}{2} \] Compute the partial derivatives: - \(\frac{\partial H}{\partial x} = -x\) - \(\frac{\partial H}{\partial y} = -3y\) Plug into the Hamiltonian equations: \[ x' = -\frac{\partial H}{\partial y} = -(-3y) = 3y \] \[ y' = \frac{\partial H}{\partial x} = -x \] But the original equations are \(x' = -3y\) and \(y' = x\). The signs are opposite. This means the given \(H(x, y)\) actually generates the system \[ \begin{cases} x' = 3y \\ y' = -x \end{cases} \] So to match exactly, we can use \(H(x, y)\) or \(-H(x, y)\) depending on convention, but the form is correct up to sign. --- ## 2. **Draw the Phase Plane** Rewrite the system: \[ x' = -3y \\ y' = x \] ### **Nullclines:** - \(x' = 0 \Rightarrow y = 0\) (x-axis) - \(y' = 0 \Rightarrow x = 0\) (y-axis) ### **Equilibrium Point:** Set \(x' = 0\), \(y' = 0\): Only solution is \((0, 0)\). ### **Linear Stability Analysis:** The Jacobian matrix: \[ J = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{pmatrix} = \begin{pmatrix} 0 & -3 \\ 1 & 0 \end{pmatrix} \] Eigenvalues: \[ \det(J - \lambda I) = \left| \begin{matrix} -\lambda & -3 \\ 1 & -\lambda \end{matrix} \right| = \lambda^2 + 3 = 0 \implies \lambda = \pm i\sqrt{3} \] - Purely imaginary eigenvalues \(\Rightarrow\) **center** (closed orbits). ### **Shape of Orbits:** From the Hamiltonian: \[ -\frac{3y^2}{2} - \frac{x^2}{2} = C \implies \frac{x^2}{2} + \frac{3y^2}{2} = -C \] For real solutions, \(-C > 0\), so orbits are **ellipses** centered at the origin. --- ## 3. **Flow Direction** From the system, at \((1, 0)\), \(x' = 0\), \(y' = 1\) → flow upwards. At \((0, 1)\), \(x' = -3\), \(y' = 0\) → flow left. So the motion is **clockwise** around the origin. --- ## 4. **Summary and Phase Plane Sketch** - **Nullclines:** \(x\)-axis and \(y\)-axis. - **Equilibrium:** At the origin (center). - **Orbits:** Ellipses (\(\frac{x^2}{2} + \frac{3y^2}{2} = \text{const}\)). - **Flow:** Clockwise. - **Stability:** Center; neutral stability (no spiraling, closed orbits). --- ### **Final Answer (Summary Table):** | Aspect | Explanation | |------------------------|---------------------------------------------------------| | Hamiltonian | \(H(x, y) = -\frac{3y^2}{2} - \frac{x^2}{2}\) | | Nullclines | \(x = 0\), \(y = 0\) | | Equilibrium | Origin \((0, 0)\) | | Stability | Center (purely imaginary eigenvalues) | | Phase plane | Elliptical orbits, clockwise flow | | Linearization | Jacobian: \(\begin{pmatrix} 0 & -3 \\ 1 & 0 \end{pmatrix}\), \(\lambda = \pm i\sqrt{3}\) | --- **Phase portrait sketch:** - Draw ellipses centered at (0,0) - Indicate clockwise flow with arrows - Nullclines are the axes **How did we get the shape?** - **Nullclines:** Show where derivative in each direction is zero (axes). - **Flow:** Use vector field at key points. - **Linear stability:** Center, so closed orbits. - **Hamiltonian:** Level sets are ellipses. --- **If you want a drawn phase plane, let me know!**