### **Step 1: Rewrite the System as a First-Order System** Given: \[ \begin{cases} x'' = -ax - by \\ y'' = -bx - cy \end{cases} \] Introduce new variables using the hint: \( x' = z \), \( y' = w \). So, \[ \begin{cases} x' = z \\ z' = -ax - by \\ y' = w \\ w' = -bx - cy \end{cases} \] **Explanation:** The original system contains second-order differential equations. Introducing new variables (\(z\) for \(x'\) and \(w\) for \(y'\)) converts the system into a set of first-order equations. This is a standard step in analyzing dynamical systems, especially for Hamiltonian formulation. --- ### **Step 2: Hamiltonian Formulation** To express the system as Hamiltonian, look for a function \(H(x, y, z, w)\) such that: \[ \begin{cases} x' = \frac{\partial H}{\partial z} \\ z' = -\frac{\partial H}{\partial x} \\ y' = \frac{\partial H}{\partial w} \\ w' = -\frac{\partial H}{\partial y} \end{cases} \] Propose a Hamiltonian function: \[ H(x, y, z, w) = \frac{1}{2}(z^2 + w^2) + \frac{a}{2}x^2 + bxy + \frac{c}{2}y^2 \] Check derivatives: - \(\frac{\partial H}{\partial z} = z \rightarrow x' = z\) - \(\frac{\partial H}{\partial x} = ax + by \rightarrow z' = -ax - by\) - \(\frac{\partial H}{\partial w} = w \rightarrow y' = w\) - \(\frac{\partial H}{\partial y} = bx + cy \rightarrow w' = -bx - cy\) **Explanation:** The Hamiltonian function \(H\) includes kinetic energy terms (\(\frac{1}{2}(z^2 + w^2)\)) and potential energy terms (\(\frac{a}{2}x^2 + bxy + \frac{c}{2}y^2\)), matching the structure of the equations. The partial derivatives confirm that the system is indeed Hamiltonian. --- ### **Step 3: Stability of the Critical Point** For stability, analyze the equilibrium at \((x, y, z, w) = (0, 0, 0, 0)\). Linearize the system and find the eigenvalues of the matrix: \[ A = \begin{pmatrix} 0 & 0 & 1 & 0 \\ -a & -b & 0 & 0 \\ 0 & 0 & 0 & 1 \\ -b & -c & 0 & 0 \end{pmatrix} \] The characteristic equation for eigenvalues \(\lambda\) is: \[ \det(A - \lambda I) = 0 \] This expands to: \[ \left| \begin{matrix} -\lambda & 0 & 1 & 0 \\ -a & -b-\lambda & 0 & 0 \\ 0 & 0 & -\lambda & 1 \\ -b & -c & 0 & -\lambda \end{matrix} \right| = 0 \] This factors into two identical quadratic equations: \[ \lambda^4 + (a + c)\lambda^2 + (ac - b^2) = 0 \] For stability, all eigenvalues must be purely imaginary or have negative real parts. This is true if: - \(a + c > 0\) (sum of roots negative, so no positive real part) - \(ac - b^2 > 0\) (discriminant negative, so no real roots) **Explanation:** Stability requires that the energy is minimized at the origin, which is guaranteed if the quadratic form in the potential is positive definite. The derived conditions ensure that the critical point is a center (stable equilibrium). --- ### **Final Answer Summary** - The reduced system is Hamiltonian with the Hamiltonian function: \[ H(x, y, z, w) = \frac{1}{2}(z^2 + w^2) + \frac{a}{2}x^2 + bxy + \frac{c}{2}y^2 \] - Sufficient conditions for stability at the critical point are: \[ a + c > 0, \quad ac - b^2 > 0 \] --- #### **Parameter Definitions Recap** - **Relevancy:** The solution directly addresses the posed questions. - **Completeness:** Each calculation and logical step is shown. - **Accuracy:** The mathematical formulation and stability analysis are correct. - **Clarity:** Each step includes explanation for better understanding. - **Structure:** Steps are logically ordered for natural progression. - **Voice:** The explanation uses clear and human language.