91Ó°ÊÓ

To find the fixed points of the system, set \(\dot{u} = 0\) and \(\dot{v} = 0\). This gives us the following system of equations: 1. \(a(1-u)-u v^2 = 0\) 2. \(u v^2 -(a+k) v = 0\) Now we need to solve these equations simultaneously for \(u\) and \(v\).

From equation (2), we have two cases: - Case 1: \(v = 0\) If \(v=0\), equation (1) becomes: \(a(1-u) = 0\) Solving for \(u\), we get: \(u = 1\) So one fixed point is \((u, v) = (1, 0)\). - Case 2: \(uv^2 = (a+k)v\) Divide both sides by \(v\): \(uv = a + k\) Now substitute this expression for \(uv\) in equation (1): \(a(1-u)-uv^2 = a(1-u)-(a+k)v = 0\) Rearrange and factor: \((1-u)(a-v) = 0\) This gives two possibilities: \(u = 1\) or \(v = a\). If \(u = 1\), we already found this fixed point in Case 1. So, we're left with \(v=a\). Substituting \(v=a\) in the expression for \(uv\), we get: \(ua = a + k \Rightarrow u = 1 + \frac{k}{a}\) Thus, the second fixed point is \(\left(1+\frac{k}{a}, a\right)\).

We'll now linearize the given system of equations around the fixed points. The Jacobian matrix for the system is: \( J(u,v) = \begin{bmatrix} -\frac{\partial \dot{u}}{\partial u} & -\frac{\partial \dot{u}}{\partial v} \\ \frac{\partial \dot{v}}{\partial u} & \frac{\partial \dot{v}}{\partial v} \end{bmatrix}=\begin{bmatrix} -a-2uv^2 & -2uv \\ v^2 & 2uv-a-k \end{bmatrix} \) Now, compute the Jacobian at each fixed point: - Fixed Point 1: \((1,0)\) \( J(1,0) = \begin{bmatrix} -a & 0 \\ 0 & -a-k \end{bmatrix} \) - Fixed Point 2: \(\left(1+\frac{k}{a}, a\right)\) \( J\left(1+\frac{k}{a}, a\right) = \begin{bmatrix} -a+2k & -2(1+\frac{k}{a})a \\ a^2 & 0 \end{bmatrix} \)

Next, we analyze the eigenvalues for the Jacobian matrices calculated in Step 3 and find the parameter values at which saddle-node bifurcations occur. - For Fixed Point 1: The eigenvalues of \(J(1,0)\) are simply the diagonal elements, \(-a\) and \(-a-k\). As we are looking for a saddle-node bifurcation, one eigenvalue must be positive and one must be negative. In this case, both eigenvalues are negative, so there is no saddle-node bifurcation at \((1, 0)\). - For Fixed Point 2: The eigenvalues of \(J\left(1+\frac{k}{a}, a\right)\) are given by the characteristic equation: \( \det(J-\lambda I) = \begin{vmatrix} -a+2k-\lambda & -2(1+\frac{k}{a})a \\ a^2 & -\lambda \end{vmatrix} =(-\lambda)(-\lambda -2k+a)-2a^3(1+\frac{k}{a}) =0 \) Simplifying the characteristic equation, we have: \(\lambda^2 - (a-2k)\lambda - 2a^2 (a+k) = 0\) Now, apply the quadratic formula to find the eigenvalues: \(\lambda = \frac{-(a-2k) \pm \sqrt{((a-2k)^2 - 4(-2a^2 (a+k)))}}{2}\) For a saddle-node bifurcation, one eigenvalue must be positive, and one must be negative. This occurs when the discriminant of the characteristic equation is zero: \(((a-2k)^2 - 4(-2a^2 (a+k))) = 0\) Solving the equation for \(k\), we get: \(k=-a \pm \frac{1}{2} \sqrt{a}\) This shows that saddle-node bifurcations occur at \(k=-a \pm \frac{1}{2} \sqrt{a}\), as required.