Linear System of Differential Equations - 4

This lecture by MIT professor Gilbert Strang is the most relevant for this problem. The lecture videos by Strang are very helpful! System of linear differential equations

Let the system of equations be represented as: $\displaystyle \frac{d \vec{\bold{v}}}{dt} = \bold{A}\vec{\bold{v}}$ ,

Where $\bold{A}$ is the coefficient matrix of the system and $\vec{\bold{v}}$ is the vector $\displaystyle \begin{bmatrix} x_{1} \\ x_{2} \\ \end{bmatrix}$ .

The system is:

$\displaystyle 5x_1 - 1x_2 = \frac{dx_1}{dt}$

$\displaystyle (3+k_1)x_1 + (k_2 - 1)x_2 = \frac{dx_2}{dt}$

The coefficient matrix $\bold{A}$ is:

$\begin{bmatrix} 5 & -1 \\ 3+k_1 & k_2 -1 \end{bmatrix}$

Therefore the system as a linear transformation is:

$\begin{bmatrix} 5 & -1 \\ 3+k_1 & k_2 -1 \end{bmatrix}$ $\displaystyle \begin{bmatrix} x_{1} \\ x_{2} \\ \end{bmatrix}$ = $\displaystyle \frac{d \vec{\bold{v}}}{dt}$

The general solution of $\bold{\vec{v}}$ is:

$\displaystyle \bold{\vec{v}} = c_1 e^{\lambda _{1} t} \vec{\bold{e}_1} + c_2 e^{\lambda _{2} t } \vec{\bold{e}_2}$ ,

where $\lambda_1$ and $\lambda_2$ are the eigenvalues, and $\vec{\bold{e}}_1$ and $\vec{\bold{e}}_2$ are the eigenvectors of the system coefficient matrix.

Therefore, when solving $\det(\bold{A} - \lambda \bold{I}) = 0$ , we get:

$\displaystyle \lambda _{1} = -4 = \frac{2+k_1 - \sqrt{(2+k_1)^2 - 8 +4k_1 +20k_2}}{2}$

$\displaystyle \lambda _{2} = -3 = \frac{2+k_1 + \sqrt{(2+k_1)^2 - 8 +4k_1 +20k_2}}{2}$

Solving the simultaneous equation for $k_1$ and $k_2$ , we get $\boxed{k_1 = -9}$ and $\displaystyle \boxed{k_2 = - \frac{1}{5}}$

With the forcing function, the system matrix becomes:

$\begin{pmatrix} 3 + k_1 & -1 + k_2 \\ 5 & -1 \end{pmatrix}$

We need the matrix to have eigenvalues of $-4$ and $-3$ . To solve for the eigenvalues, set the determinant of the following matrix to zero:

$\begin{pmatrix} 3 + k_1 - \lambda & -1 + k_2 \\ 5 & -1 - \lambda \end{pmatrix}$

This yields:

$\lambda^2 + \lambda (-2 - k_1) + 2 - k_1 - 5 k_2 = 0$

The system of equations to solve becomes:

$\lambda_1 = -4 = \frac{2 + k_1 - \sqrt{(2 + k_1)^2 - 8 + 4 k_1 + 20 k_2}}{2} \\ \lambda_2 = -3 = \frac{2 + k_1 + \sqrt{(2 + k_1)^2 - 8 + 4 k_1 + 20 k_2}}{2}$

Solving yields:

$k_1 = - 9 \\ k_2 = -\frac{1}{5}$

Because the solutions consist of decaying exponentials, the energy of the system decreases over time.

Given that

$\begin{cases} \dot x_1 = (k_1+3)x_1 + (k_2-1)x_2 & ...(1) \\ \dot x_2 = 5x_1 - x_2 & ...(2) \\ x_1 = c_1 e^{-4t} + c_2 e^{-3t} & ...(3) \\ x_2 = c_3 e^{-4t} + c_4 e^{-3t} & ...(4) \end{cases}$

Substitute $(3)$ and $(4)$ in $(1)$ and $(2)$ and differentiate $(3)$ and $(4)$ with respect to $t$ .

$\begin{cases} \dot x_1 = (k_1c_1+k_2c_3+3c_1-c_3)e^{-4t} + (k_1c_2+k_2c_4+3c_2-c_4)e^{-3t} & ...(1a) \\ \dot x_2 = (5c_1-c_3)e^{-4t} + (5c_2-c_4)e^{-3t} & ...(2a) \\ \dot x_1 = -4c_1 e^{-4t} -3 c_2 e^{-3t} & ...(3a) \\ \dot x_2 = -4 c_3 e^{-4t} -3 c_4 e^{-3t} & ...(4a) \end{cases}$

From $(2a) = (4a)$ , $\implies \begin{cases} 5c_1-c_3 = -4 c_3 & \implies c_1 = - \frac 35 c_3 \\ 5c_2-c_4 = -3 c_3 & \implies c_2 = - \frac 25 c_4 \end{cases}$

From $(1a) = (3a)$ :

$\begin{aligned} k_1c_1+k_2c_3+3c_1-c_3 & = - 4c_1 & \small \color{#3D99F6} \text{Since }c_1 = - \frac 35 c_3 \\ - \frac 35 k_1 + k_2 - \frac 95 - 1 & = \frac {12}5 \\ \implies - 3k_1 + 5k_2 & = 26 \quad ...(5) \end{aligned}$

$\begin{aligned} k_1c_2+k_2c_4+3c_2-c_4 & = - 3c_2 & \small \color{#3D99F6} \text{Since }c_2 = - \frac 25 c_4 \\ - \frac 25 k_1 + k_2 - \frac 65 - 1 & = \frac 65 \\ \implies - 2k_1 + 5k_2 & = 17 \quad ...(6) \end{aligned}$

From $(6)-(5): \ k_1 = - 9$ . From $(6): \ k_2 = - \frac 15$ . Therefore, $5(k_1+k_2) = 5 \left(-9 - \frac 15\right) = \boxed {-46}$ .