[problem] **(i)** Write the complex normal form up to order three for the restriction of the following system onto the center manifold near the zero equilibrium at $\epsilon = 0$: \begin{align*} \dot{x} &= y + \epsilon \\ \dot{y} &= -x + 2zy \\ \dot{z} &= -2z + 16\epsilon y + 8y^2 \end{align*} **(ii)** Compute the first Lyapunov coefficient. Let $U$ be a sufficiently small neighborhood of zero. How many stable and unstable periodic orbits exist in $U$ at small $\epsilon > 0$ and at small $\epsilon < 0$? [/problem]

[solution] **Part (i): Center Manifold and Normal Form** We analyze the system at the critical parameter value $\epsilon = 0$: \begin{align*} \dot{x} &= y \\ \dot{y} &= -x + 2zy \\ \dot{z} &= -2z + 8y^2 \end{align*} **Step 1: Eigenstructure and Invariant Subspaces** We first establish the linear structure to justify the Center Manifold reduction. [claim:Eigenstructure] The Jacobian matrix $A$ at the origin has eigenvalues $\sigma(A) = \{ -2, i, -i \}$. The Center subspace $E^c$ is the plane $z=0$, and the Stable subspace $E^s$ is the line $x=y=0$. [/claim] [proof] The Jacobian matrix of the system at $(0,0,0)$ is: \begin{align*} A = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -2 \end{pmatrix} \end{align*} The characteristic polynomial is: \begin{align*} \det(A - \lambda I) &= (-2-\lambda)\det\begin{pmatrix} -\lambda & 1 \\ -1 & -\lambda \end{pmatrix} \\ &= -(2+\lambda)(\lambda^2 + 1) \end{align*} The roots are $\lambda_1 = -2$ and $\lambda_{2,3} = \pm i$. 1. **Stable Eigenspace ($E^s$):** For $\lambda = -2$, $(A+2I)v = 0$ implies $v_1=v_2=0$. Thus $E^s = \text{span}\{ (0,0,1)^T \}$. 2. **Center Eigenspace ($E^c$):** For $\lambda = i$, we solve $(A-iI)v = 0$: \begin{align*} \begin{pmatrix} -i & 1 & 0 \\ -1 & -i & 0 \\ 0 & 0 & -2-i \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align*} The third row implies $v_3=0$. The first row implies $v_2 = i v_1$. A complex eigenvector is $v = (1, i, 0)^T$. The real center subspace is spanned by the real and imaginary parts of $v$: $E^c = \text{span}\{ (1,0,0)^T, (0,1,0)^T \}$. This corresponds to the plane $z=0$. [/proof] **Step 2: Center Manifold Reduction** By the Center Manifold Theorem, there exists a local invariant manifold $W^c$ tangent to $E^c$ at the origin. We can represent $W^c$ as the graph of a function $z = h(x,y)$ where $h(0,0)=0$ and $Dh(0,0)=0$. [claim:Quadratic Approximation] The function $h(x,y)$ defining the center manifold is given by: \begin{align*} h(x,y) = x^2 + 2xy + 3y^2 + O(\|(x,y)\|^3) \end{align*} [/claim] [proof] We substitute the ansatz $z = h(x,y)$ into the third equation of the system ($\dot{z}$) and use the chain rule: \begin{align*} \dot{z} = \frac{\partial h}{\partial x} \dot{x} + \frac{\partial h}{\partial y} \dot{y} \end{align*} Substituting the system dynamics for $\dot{x}, \dot{y}, \dot{z}$: \begin{align*} -2h(x,y) + 8y^2 = \frac{\partial h}{\partial x}(y) + \frac{\partial h}{\partial y}(-x + 2h(x,y)y) \end{align*} We assume a Taylor expansion for $h(x,y)$ starting at order 2: \begin{align*} h(x,y) = ax^2 + bxy + cy^2 + O(3) \end{align*} We substitute this expansion into the invariance equation. **Left Hand Side (LHS):** \begin{align*} \text{LHS} &= -2(ax^2 + bxy + cy^2 + O(3)) + 8y^2 \\ &= -2ax^2 - 2bxy + (8-2c)y^2 + O(3) \end{align*} **Right Hand Side (RHS):** Compute partial derivatives: \begin{align*} \frac{\partial h}{\partial x} = 2ax + by + O(2), \quad \frac{\partial h}{\partial y} = bx + 2cy + O(2) \end{align*} Substitute into RHS: \begin{align*} \text{RHS} &= (2ax + by + O(2))(y) + (bx + 2cy + O(2))(-x + 2h(x,y)y) \end{align*} Note that $2h(x,y)y$ is of order $O(2) \cdot O(1) = O(3)$. Thus, for quadratic matching, we ignore it. \begin{align*} \text{RHS} &= (2ax + by)y + (bx + 2cy)(-x) + O(3) \\ &= 2axy + by^2 - bx^2 - 2cxy + O(3) \\ &= -bx^2 + (2a - 2c)xy + by^2 + O(3) \end{align*} **Matching Coefficients:** We equate the coefficients of $x^2, xy, y^2$ from LHS and RHS: 1. $x^2$: $-2a = -b \implies b = 2a$. 2. $xy$: $-2b = 2a - 2c$. 3. $y^2$: $8 - 2c = b$. **Solving the Linear System:** Substitute $b=2a$ into (2): \begin{align*} -2(2a) = 2a - 2c \implies -4a = 2a - 2c \implies 2c = 6a \implies c = 3a \end{align*} Substitute $b=2a$ and $c=3a$ into (3): \begin{align*} 8 - 2(3a) = 2a \implies 8 - 6a = 2a \implies 8 = 8a \implies a = 1 \end{align*} Using $a=1$, we find $b=2(1)=2$ and $c=3(1)=3$. Thus, $h(x,y) = x^2 + 2xy + 3y^2 + O(3)$. [/proof] **Step 3: Restricted System** Substitute $z = x^2 + 2xy + 3y^2$ into the $\dot{y}$ equation: \begin{align*} \dot{y} &= -x + 2(x^2 + 2xy + 3y^2)y \\ &= -x + 2x^2y + 4xy^2 + 6y^3 + O(4) \end{align*} The restricted system on $W^c$ is: \begin{align*} \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 0 \\ 2x^2y + 4xy^2 + 6y^3 \end{pmatrix} + O(4) \end{align*} **Step 4: Complex Normal Form Calculation** We transform coordinates to diagonalize the linear part and compute the cubic coefficient. [claim:Normal Form Coefficient] Using the complex coordinate $w = x - iy$, the system takes the form $\dot{w} = iw + c_1 w|w|^2 + O(|w|^4)$ with $c_1 = \frac{5}{2} - \frac{i}{2}$. [/claim] [proof] **Transformation:** Let $w = x - iy$. The inverse transformation is $x = \frac{w+\bar{w}}{2}$ and $y = i\frac{\bar{w}-w}{2} = i\frac{-(w-\bar{w})}{2}$. Wait, let us check the sign conventions. If $w = x - iy$, then $\bar{w} = x + iy$. Sum: $w + \bar{w} = 2x \implies x = \frac{w+\bar{w}}{2}$. Diff: $\bar{w} - w = 2iy \implies y = \frac{\bar{w}-w}{2i} = -\frac{i}{2}(\bar{w}-w) = \frac{i}{2}(w-\bar{w})$. Correct. **Differentiation:** \begin{align*} \dot{w} &= \dot{x} - i\dot{y} \\ &= y - i(-x + N(x,y)) \quad \text{where } N(x,y) = 2x^2y + 4xy^2 + 6y^3 \\ &= (y + ix) - iN(x,y) \\ &= i(x - iy) - iN(x,y) \\ &= iw - iN(x,y) \end{align*} We need the coefficient of the resonant term $w^2\bar{w}$ in the expansion of $-iN(x,y)$. Let us expand $N(x,y)$ term by term in powers of $w, \bar{w}$. We denote $[w^2\bar{w}]$ as the operator that extracts the coefficient of $w^2\bar{w}$. 1. **Term $2x^2y$:** \begin{align*} 2x^2y &= 2 \left( \frac{w+\bar{w}}{2} \right)^2 \left( \frac{i}{2}(w-\bar{w}) \right) \\ &= \frac{i}{4} (w^2 + 2w\bar{w} + \bar{w}^2)(w - \bar{w}) \end{align*} We expand to find cubic terms involving $w^2\bar{w}$: \begin{align*} (w^2)(-\bar{w}) + (2w\bar{w})(w) = -w^2\bar{w} + 2w^2\bar{w} = w^2\bar{w} \end{align*} Contribution to $N$: $\frac{i}{4} (1) = \frac{i}{4}$. 2. **Term $4xy^2$:** \begin{align*} 4xy^2 &= 4 \left( \frac{w+\bar{w}}{2} \right) \left( \frac{i}{2}(w-\bar{w}) \right)^2 \\ &= 4 \left( \frac{w+\bar{w}}{2} \right) \left( -\frac{1}{4} \right) (w^2 - 2w\bar{w} + \bar{w}^2) \\ &= -\frac{1}{2} (w + \bar{w})(w^2 - 2w\bar{w} + \bar{w}^2) \end{align*} Terms involving $w^2\bar{w}$: \begin{align*} (w)(-2w\bar{w}) + (\bar{w})(w^2) = -2w^2\bar{w} + w^2\bar{w} = -w^2\bar{w} \end{align*} Contribution to $N$: $-\frac{1}{2}(-1) = \frac{1}{2}$. 3. **Term $6y^3$:** \begin{align*} 6y^3 &= 6 \left( \frac{i}{2}(w-\bar{w}) \right)^3 \\ &= 6 \left( \frac{-i}{8} \right) (w - \bar{w})^3 \\ &= -\frac{3i}{4} (w^3 - 3w^2\bar{w} + 3w\bar{w}^2 - \bar{w}^3) \end{align*} Terms involving $w^2\bar{w}$: $-3w^2\bar{w}$. Contribution to $N$: $-\frac{3i}{4}(-3) = \frac{9i}{4}$. **Summation:** The coefficient of $w^2\bar{w}$ in $N$ is: \begin{align*} C_N = \frac{i}{4} + \frac{1}{2} + \frac{9i}{4} = \frac{1}{2} + \frac{10i}{4} = \frac{1}{2} + \frac{5i}{2} \end{align*} The equation for $w$ is $\dot{w} = iw - i (C_N w^2\bar{w} + \dots)$. Thus, the normal form coefficient $c_1$ is: \begin{align*} c_1 = -i C_N = -i \left( \frac{1}{2} + \frac{5i}{2} \right) = -\frac{i}{2} - \frac{5i^2}{2} = \frac{5}{2} - \frac{i}{2} \end{align*} [/proof] The Normal Form is: \begin{align*} \dot{w} = iw + \left( \frac{5}{2} - \frac{i}{2} \right) w|w|^2 + O(|w|^4) \end{align*} **Part (ii): Lyapunov Coefficient and Bifurcation Analysis** **1. First Lyapunov Coefficient:** \begin{align*} L_1 = \text{Re}(c_1) = \frac{5}{2} \end{align*} **2. Existence and Stability of Periodic Orbits (for $\epsilon \neq 0$)** [claim:Equilibrium Stability] For small $\epsilon \neq 0$, there exists a unique equilibrium point which is asymptotically stable. [/claim] [proof] We solve $F(X, \epsilon) = 0$: 1. $\dot{x} = y + \epsilon = 0 \implies y_0 = -\epsilon$. 2. $\dot{z} = -2z + 16\epsilon y + 8y^2 = 0$. Substitute $y_0 = -\epsilon$: \begin{align*} -2z_0 + 16\epsilon(-\epsilon) + 8(-\epsilon)^2 &= 0 \\ -2z_0 - 16\epsilon^2 + 8\epsilon^2 &= 0 \\ -2z_0 = 8\epsilon^2 \implies z_0 = -4\epsilon^2 \end{align*} 3. $\dot{y} = -x + 2zy = 0$. Substitute $y_0, z_0$: \begin{align*} -x_0 + 2(-4\epsilon^2)(-\epsilon) = 0 \implies x_0 = 8\epsilon^3 \end{align*} The Jacobian matrix at $(8\epsilon^3, -\epsilon, -4\epsilon^2)$ is: \begin{align*} J &= \begin{pmatrix} 0 & 1 & 0 \\ -1 & 2z_0 & 2y_0 \\ 0 & 16\epsilon + 16y_0 & -2 \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 & 0 \\ -1 & -8\epsilon^2 & -2\epsilon \\ 0 & 16\epsilon + 16(-\epsilon) & -2 \end{pmatrix} \\ &= \begin{pmatrix} 0 & 1 & 0 \\ -1 & -8\epsilon^2 & -2\epsilon \\ 0 & 0 & -2 \end{pmatrix} \end{align*} This matrix is block triangular. The eigenvalues are the diagonal block entries: $\lambda_3 = -2$ and the eigenvalues of: \begin{align*} M = \begin{pmatrix} 0 & 1 \\ -1 & -8\epsilon^2 \end{pmatrix} \end{align*} The characteristic equation for $M$ is: \begin{align*} \lambda^2 - \text{tr}(M)\lambda + \det(M) = 0 \implies \lambda^2 + 8\epsilon^2\lambda + 1 = 0 \end{align*} The roots are: \begin{align*} \lambda_{1,2} = \frac{-8\epsilon^2 \pm \sqrt{64\epsilon^4 - 4}}{2} \end{align*} For small $\epsilon$ ($64\epsilon^4 < 4$), the term under the square root is negative. \begin{align*} \lambda_{1,2} = -4\epsilon^2 \pm i \frac{\sqrt{4 - 64\epsilon^4}}{2} \end{align*} The real part is $\text{Re}(\lambda_{1,2}) = -4\epsilon^2$. Since $-4\epsilon^2 < 0$ for all $\epsilon \neq 0$, the equilibrium is stable. [/proof] **Periodic Orbit Calculation:** The Hopf Bifurcation Theorem usually assumes $\frac{d}{d\epsilon}\text{Re}(\lambda) \neq 0$. Here, $\text{Re}(\lambda) = -4\epsilon^2$, so the derivative at 0 is 0. This is a degenerate Hopf bifurcation. However, we can still analyze the limit cycles using the radial normal form: \begin{align*} \dot{r} = \text{Re}(\lambda(\epsilon)) r + L_1 r^3 + O(r^5) \end{align*} Substitute the calculated values: \begin{align*} \dot{r} = -4\epsilon^2 r + \frac{5}{2} r^3 \end{align*} We define a periodic orbit as a solution with constant radius $r > 0$ ($\dot{r}=0$): \begin{align*} -4\epsilon^2 r + \frac{5}{2} r^3 &= 0 \\ r \left( \frac{5}{2} r^2 - 4\epsilon^2 \right) &= 0 \end{align*} Since $r \neq 0$: \begin{align*} \frac{5}{2} r^2 = 4\epsilon^2 \implies r^2 = \frac{8}{5} \epsilon^2 \implies r = \sqrt{1.6} |\epsilon| \end{align*} This solution exists for all $\epsilon \neq 0$. **Stability of the Orbit:** We test the sign of $\dot{r}$ for $r$ slightly larger than the equilibrium radius $r_* = \sqrt{1.6}|\epsilon|$. If $r > r_*$, then $r^2 > \frac{8}{5}\epsilon^2$, so $\frac{5}{2}r^2 - 4\epsilon^2 > 0$. Thus $\dot{r} > 0$, meaning trajectories spiral away from the cycle. The periodic orbit is **unstable**. **Conclusion:** * **Case $\epsilon > 0$:** The equilibrium is stable. There is **1** unstable periodic orbit. * **Case $\epsilon < 0$:** The equilibrium is stable. There is **1** unstable periodic orbit. [/solution]