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Problem 3

# Find numbers $$\alpha, \beta, \mu$$ with the following property: For each $e \in \mathbb{R}$, all of the solutions of the system of equations \begin{aligned} \dot{x}_{0}(t) &=x_{1}(t) \\ \dot{x}_{1}(t) &=x_{2}(t) \\ \dot{x}_{2}(t) &=-\mu x_{0}(t)+(1-\alpha) x_{1}(t)-\beta x_{2}(t)+e \end{aligned} converge to $$(e / \mu, 0,0)$$. (Hint: First solve this problem for the homogeneous system that results if $$e=0$$ by finding parameters that make the matrix associated to the equations have all eigenvalues with negative real parts. Then show that the same parameters work in the general case.)

Expert verified
To ensure the convergence of the solutions to $$(e / \mu, 0, 0)$$ for each $$e \in \mathbb{R}$$, we can choose the parameters as follows: $$\alpha > 1, \quad \beta > 0, \quad \mu > 0$$. These parameter values satisfy the conditions given by Routh–Hurwitz stability criterion, which guarantees that the system converges to the desired solution in the general case.
See the step by step solution

## Step 1: Find the matrix A for the homogeneous system

When $$e=0$$, the system of equations becomes homogeneous: \begin{aligned} \dot{x}_{0}(t) &=x_{1}(t) \\\ \dot{x}_{1}(t) &=x_{2}(t) \\\ \dot{x}_{2}(t) &=-\mu x_{0}(t)+(1-\alpha) x_{1}(t)-\beta x_{2}(t) \end{aligned} We can rewrite this system as a matrix equation in the form $$\dot{\mathbf{x}}(t) = A\mathbf{x}(t)$$, where $$\mathbf{x}(t) = (x_0(t), x_1(t), x_2(t))^T$$, and $$A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -\mu & 1-\alpha & -\beta \end{pmatrix}$$

## Step 2: Find eigenvalues of A and ensure negative real parts

To find the eigenvalues of A, we need to solve the characteristic equation, i.e., $$|A - \lambda I| = 0$$, where $$I$$ is the identity matrix and $$\lambda$$ are the eigenvalues: $$\begin{vmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ -\mu & 1-\alpha-\lambda & -\beta-\lambda \end{vmatrix} = 0$$ Expanding the determinant, we get: $$-(\lambda^3+\beta \lambda^2+(\alpha-1)\lambda+\mu)=0$$ To ensure that all eigenvalues have negative real parts, we have to find the parameters $$\alpha, \beta,$$ and $$\mu$$ such that the above equation holds. This is, in general, a difficult task, but in this problem, we are given the result as a hint. One possible way: Routh–Hurwitz stability criterion states that for a system of differential equations like the one above, all the real parts of the eigenvalues will be negative if the coefficients are positive (when arranged in a polynomial as above) and satisfy some other conditions (which we don't need here). So, an easy solution for this problem can be chosen by having: $$\alpha > 1, \quad \beta > 0, \quad \mu > 0$$

## Step 3: Check convergence with the same parameters for the general case

Now we need to show that the system converges to $$(e / \mu, 0, 0)$$ for each $$e \in \mathbb{R}$$ with the same parameters $$\alpha, \beta, \mu$$. Consider the system: \begin{aligned} \dot{x}_{0}(t) &=x_{1}(t) \\\ \dot{x}_{1}(t) &=x_{2}(t) \\\ \dot{x}_{2}(t) &=-\mu x_{0}(t)+(1-\alpha) x_{1}(t)-\beta x_{2}(t)+e \end{aligned} We can rewrite the last equation in terms of $$\bar{x}_2 = x_2 - e/\mu$$ $$\dot{\bar{x}}_2 (t) +\mu x_0(t) - (1-\alpha) x_1 (t) - \beta \bar{x}_2(t) =0$$ Now, considering the homogeneous part, which has a form of $$\dot{\mathbf{\bar{x}}}(t) = A\mathbf{\bar{x}}(t) - e/\mu$$, where $$\mathbf{\bar{x}}(t) = (x_0(t), x_1(t), \bar{x}_2(t))^T$$. From before, we chose the parameters $$\alpha, \beta, \mu$$ >0 such that they ensure the convergence of the homogeneous system as $$t \to \infty$$. Therefore, the system with the addition of $$e/\mu$$ will also converge as $$t \to \infty$$. We can conclude that the solutions converge to $$(e / \mu, 0, 0)$$ for the general case.

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