Question

The coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ x^{\prime}=\left(\begin{array}{cc}{4} & {\alpha} \\ {8} & {-6}\end{array}\right) \mathbf{x} $$

Short answer

Answer: The phase portraits slightly below and above the critical values of α both exhibit a saddle point.

Step-by-step solution

Calculate the eigenvalues in terms of \(\alpha\)

Given the system, $$ x^{\prime}=\left(\begin{array}{cc}{4} & {\alpha} \\ {8} & {-6}\end{array}\right) \mathbf{x}, $$ we need to find the eigenvalues of the coefficient matrix. To find the eigenvalues, we need to solve the characteristic equation: $$ \text{det}(A - \lambda I) = 0, $$ where \(A\) is the coefficient matrix, \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix. In this case, we have: $$ \text{det}\left(\begin{array}{cc}{4-\lambda} & {\alpha} \\ {8} & {-6-\lambda}\end{array}\right) = (4-\lambda)(-6-\lambda) - 8\alpha = 0. $$ Expanding the equation, we have a quadratic equation in terms of \(\lambda\): $$ \lambda^2 - 2\lambda - 24 - 8\alpha = 0. $$ Now we can calculate the eigenvalues in terms of the parameter \(\alpha\). We'll use the quadratic formula, which gives: $$ \lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$ In our case, \(a=1\), \(b=-2\), and \(c=-24-8\alpha\). Applying the quadratic formula, we get: $$ \lambda_{1,2}(\alpha) = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-24 - 8\alpha)}}{2} = 1 \pm \sqrt{25 + 8\alpha}. $$ So, the eigenvalues of the matrix are: $$ \lambda_1(\alpha) = 1 + \sqrt{25 + 8\alpha}, $$ $$ \lambda_2(\alpha) = 1 - \sqrt{25 + 8\alpha}. $$

Determine critical values of \(\alpha\)

The qualitative nature of the phase portrait changes based on the signs and values of the real parts of the eigenvalues. We need to find the critical values of \(\alpha\) where these changes occur. Let's analyze the possible cases: Case a: Both eigenvalues are real and have the same sign (either positive or negative) Case b: Both eigenvalues are real and have opposite signs Case c: Both eigenvalues are complex with positive real parts Case d: Both eigenvalues are complex with negative real parts Case e: Both eigenvalues are complex with zero real parts We need to find the values of \(\alpha\) that will cause a transition between these cases. First, let's find the values of \(\alpha\) that will make either of the eigenvalues complex. This occurs when the discriminant of the quadratic equation is negative. In our case, the discriminant is \(25+8\alpha\). Thus, we need to find the \(\alpha\) for which: $$ 25+8\alpha < 0 \Rightarrow \alpha < -\frac{25}{8}. $$ Now let's consider the case when both eigenvalues are real. We know that the real parts of both eigenvalues are 1, so we don't need to analyze case c, d, or e further. Now we need to analyze cases a and b. If the eigenvalues have the same sign (either positive or negative), it means that their sum is either positive or negative: $$ \lambda_1 + \lambda_2 = 2. $$ Since 2 is positive, we know that both eigenvalues must be positive for a critical value. Therefore, we have case a: both eigenvalues are real and positive. The critical value for \(\alpha\) is when both eigenvalues are real and positive, which occurs when: $$ -\frac{25}{8} \leq \alpha. $$

Draw phase portraits for different values of \(\alpha\)

Next, we will draw phase portraits for values of \(\alpha\) slightly below and above the critical value that we found in Step 2. For this, we can choose \(\alpha_1 = -4\) (below critical value) and \(\alpha_2 = 0\) (above critical value). For \(\alpha_1 = -4\), the eigenvalues are: $$ \lambda_1(-4) = 1 + \sqrt{25 + 8(-4)} = 1 + 3 = 4, $$ $$ \lambda_2(-4) = 1 - \sqrt{25 + 8(-4)} = 1 - 3 = -2. $$ In this case, we have one positive and one negative eigenvalue, which implies a saddle point. For \(\alpha_2 = 0\), the eigenvalues are: $$ \lambda_1(0) = 1 + \sqrt{25 + 8(0)} = 1 + 5 = 6, $$ $$ \lambda_2(0) = 1 - \sqrt{25 + 8(0)} = 1 - 5 = -4. $$ In this case, we still have one positive and one negative eigenvalue, which again implies a saddle point. Therefore, the phase portraits slightly below and above the critical values of \(\alpha\) both exhibit a saddle point.