91Ó°ÊÓ

To find the nullclines, we set each equation to zero and solve for \(x\) and \(y\). First, for the \(x\)-nullcline, set \(x^{\prime} = 0\): $$ -\alpha - x + y = 0 \Rightarrow y = x + \alpha $$ The \(x\)-nullcline is a straight line with slope 1 and the \(y\)-intercept at \(\alpha\). Next, for the \(y\)-nullcline, set \(y^{\prime} = 0\): $$ -4x + y + x^{2} = 0 \Rightarrow y = 4x - x^{2} $$ The \(y\)-nullcline is a downward-facing parabola with vertex at \((2, 4)\). As \(\alpha\) increases, the \(x\)-nullcline shifts vertically upwards. The critical points, which are the intersections of the nullclines, will move accordingly.

To find the critical points, we solve the nullcline equations simultaneously: $$ x + \alpha = 4x - x^{2} $$ Rearrange the equation to find \(x\): $$ x^2 - 3x + \alpha = 0 $$ This is a quadratic equation for \(x\), and we can use the quadratic formula to find the values of \(x\) as a function of \(\alpha\): $$ x = \frac{3 \pm \sqrt{9 - 4\alpha}}{2} $$ Since for each value of x there is a corresponding y value, substitute x to find y as a function of \(\alpha\): $$ y = \frac{3 \pm \sqrt{9 - 4\alpha}}{2} + \alpha $$ Now we have the critical points \((x, y)\) as a function of \(\alpha\).

When \(\alpha=2\), we calculate the critical points by plugging it into our expressions for \(x\) and \(y\): $$ x = \frac{3 \pm \sqrt{1}}{2} = 1, 2 \\ y = \frac{3 \pm \sqrt{1}}{2} + 2 = 3, 4 $$ So, we have two critical points: \((1,3)\) and \((2,4)\). Next, we linearize the system by finding the Jacobian matrix: $$ J(x,y) = \begin{bmatrix} \frac{\partial x^{\prime}}{\partial x} & \frac{\partial x^{\prime}}{\partial y} \\ \frac{\partial y^{\prime}}{\partial x} & \frac{\partial y^{\prime}}{\partial y} \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2x - 4 & 1 \end{bmatrix} $$ Now, evaluate the Jacobian matrix at the critical points \((1,3)\) and \((2,4)\): At \((1,3)\): $$ J(1,3) = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} $$ The eigenvalues of this matrix are \(-1 \pm \sqrt{2}\), which are both negative. Hence, the critical point \((1,3)\) is a stable node. At \((2,4)\): $$ J(2,4) = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} $$ The eigenvalues of this matrix are \(-1\) and \(1\), which indicates that the critical point \((2,4)\) is a saddle point. We can now draw the phase portrait of the system with nullclines, critical points, and their classifications.

Bifurcation occurs when the discriminant of the quadratic equation for \(x\) is equal to zero: $$ 9 - 4\alpha_{0} = 0 \Rightarrow \alpha_{0} = \frac{9}{4} $$ At the bifurcation point \(\alpha_{0}=\frac{9}{4}\), there is only one critical point \((x,y)\). Plug in the value of \(\alpha\): $$ x = \frac{3}{2} \\ y = \frac{3}{2} + \frac{9}{4} = \frac{15}{4} $$ So the critical point at the bifurcation point is \((\frac{3}{2}, \frac{15}{4})\). Next, calculate the eigenvalues of the Jacobian matrix at the bifurcation point: $$ J(\frac{3}{2}, \frac{15}{4}) = \begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix} $$ The eigenvalues are \(0\) and \(2\). Since one of the eigenvalues is zero, the linear approximation fails at the bifurcation point. We can now draw the phase portrait for the system with nullclines and the critical point at the bifurcation point.

For \(\alpha>\alpha_{0}\), no critical points exist. Choose a value for \(\alpha\) greater than \(\frac{9}{4}\), such as \(\alpha=3\). With this value, we can draw the phase portrait with the nullclines but no critical points. In conclusion, this exercise demonstrates how to analyze a given system of equations, find and classify critical points with different values of the parameter \(\alpha\), and investigate the bifurcation point. The phase portraits help visualize the system's behavior for various values of \(\alpha\).