Systems of linear differential equation

Consider

$$\left\{\begin{array}{ccc} x_{1}^{\prime} &= a_{11}x_{1} + a_{12}x... ...{n1}x_{1} + a_{n2}x_{2} +& \cdots + a_{nn}x_{n} + f_{n}(t) \end{array}\right .$$

Using the coefficient matrix $A$ and the column vector $X$ anf $F$, we can express

$${\bf X}^{\prime} = A{\bf X} + {\bf F}$$

where

$${\bf X} = \left(\begin{array}{c} x_{1}(t)\\ x_{2}(t)\\ \vdots\\... ...\begin{array}{c} f_{1}(t)\\ f_{2}(t)\\ \vdots\\ f_{n}(t) \end{array}\right)$$

The differentiable functions $x_{1},x_{2},\ldots,x_{n}$ satisfying the above equation is called the solution. When ${\bf F} = {\bf0}$, we say that the differential equation is homogeneous equation.

We will explain how to solve ${\bf X}^{\prime} = A{\bf X}$. Let ${\bf X} = {\bf C}e^{\lambda x}$. Then

$$\lambda {\bf C}e^{\lambda x} = A {\bf C}e^{\lambda x}$$

Since $e^{\lambda x} \neq 0$, we have

$$A{\bf C} - {\bf C}\lambda = {\bf0}$$

or

$$(A - \lambda I){\bf C} = {\bf0}$$

Here $I$ denotes the unit matrix of the degree $n$. Now if ${\bf C} = {\bf0}$, then the equation is obviously satisfied. So, we need to find the ${\bf C} \neq {\bf0}$ satisfying the above equation. Note that if ${\rm det}(A - \lambda I) \neq 0$, then ${\bf C} = 0$. Thus to find the ${\bf C} \neq {\bf0}$, we have to solve $\det(A - \lambda I) = 0$. The $\lambda$ is called the eigenvalue of the matrix $A$ and the nonzero C satisfying $(A - \lambda I){\bf C} = {\bf0}$ is called the eigenvector for $A$.

Example 3..1   Solve the following differential equation

$${\bf X}^{\prime} = \left(\begin{array}{rrr} 1&0&1\\ 0&1&-1\\ 1&0&1 \end{array}\right){\bf X}$$

SOLUTION Let ${\bf X} = {\bf C}e^{\lambda t}$. Then we have

$$(A - \lambda I){\bf C} = {\bf0}\ (*)$$

Now

$$\det(A - \lambda I) = \left(\begin{array}{rrr} 1-\lambda&0&1\\ 0&1-\lambda&-1\\ 1&0&1-\lambda \end{array}\right) = \lambda(1-\lambda)(\lambda -2)$$

Thus the eigenvalues are $\lambda = 0,1,2$. Now we find the eigenvector for $\lambda = 0$. Substitute $\lambda = 0$ into the equation (*). Then

$$A{\bf C} = \left(\begin{array}{rrr} 1&0&1\\ 0&1&-1\\ 1&0&1 \end... ...ght)\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3} \end{array}\right) = {\bf0}$$

To solve this system, we use Gaussian elimination.

$$A = \left(\begin{array}{rrr} 1&0&1\\ 0&1&-1\\ 1&0&1 \end{array}\rig... ...ightarrow} \left(\begin{array}{rrr} 1&0&1\\ 0&1&-1\\ 0&0&0 \end{array}\right)$$

Now we can let $c_{3} = 1$. Then $c_{1} = -1$, $c_{2} = 1$ Thus, the eigenvector C is $\left(\begin{array}{r} -1\\ 1\\ 1 \end{array}\right)$ and ${\bf X}_{1} = \left(\begin{array}{r} -1\\ 1\\ 1 \end{array}\right)$ is a solution.
For the eigenvector for $\lambda = 1$.

$$A - I = \left(\begin{array}{rrr} 0&0&1\\ 0&0&-1\\ 1&0&0 \end{array}\rig... ...ightarrow} \left(\begin{array}{rrr} 1&0&0\\ 0&0&1\\ 0&0&-1 \end{array}\right)$$

$$\stackrel{\begin{array}{c} {}_{R_{2}+R_{3}} \end{array}}{\longrightarrow} \left(\begin{array}{rrr} 1&0&0\\ 0&0&1\\ 0&0&0 \end{array}\right)$$

Thus, we can take $c_2 = 1$. Then $c_1 = 0$, $C_3 = 0$ and the eigenvector is $\left(\begin{array}{r} 0\\ 1\\ 0 \end{array}\right)$. ${\bf X}_{2} = \left(\begin{array}{r} 0\\ 1\\ 0 \end{array}\right)e^{t}$ is a solution. Similarly, we find the eigenvector corresponds to $\lambda = 2$

$$A - 2I = \left(\begin{array}{rrr} -1&0&1\\ 0&-1&-1\\ 1&0&-1 \end{array}\... ...ightarrow} \left(\begin{array}{rrr} 1&0&-1\\ 0&1&1\\ 0&0&0 \end{array}\right)$$

Then we can take $c_3 = 1$ and $c_1 = 1, c_2 = -1$. Thus the eigenvector is $\left(\begin{array}{r} 1\\ -1\\ 1 \end{array}\right)$. ${\bf X}_{3} = \left(\begin{array}{r} 1\\ -1\\ 1 \end{array}\right)e^{2t}$ is a solution. Since ${\bf X}_{1},{\bf X}_{2},{\bf X}_{3}$ are linearly independent, we have ${\bf X} = c_{1}{\bf X}_{1} + c_{2}{\bf X}_{2} + c_{3}{\bf X}_{3}$ and the general solution is

$${\bf X} = c_{1}\left(\begin{array}{r} -1\\ 1\\ 1 \end{array}\ri... ...egin{array}{r} 1\\ -1\\ 1 \end{array}\right)e^{2t}\ensuremath{\ \blacksquare}$$

Theorem 3..1   Suppose that ${\bf X}^{\prime} = A{\bf X}$ has the $n$ different eigenvalues $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ and corresponding eigenvectors ${\bf C}_{1},{\bf C}_{2},\ldots,{\bf C}_{n}$. Then the general solution is given by

$${\bf X}(t) = \sum_{i=1}^{n}c_{i}{\bf X}_{i}(t)$$

where ${\bf X}_{i} = {\bf C}_{i}e^{\lambda_{i}t} \ (i = 1,2,\ldots,n)$ are the solutions of ${\bf X}^{\prime} = A{\bf X}$ and linearly independent.

Proof

$$W(e^{\lambda_{1}t},e^{\lambda_{2}t}, \ldots e^{\lambda_{n}t}) = \left\vert\begin{array}{llll} e^{\lambda_{1}t}& e^{\lambda_{2}t}&... ...lambda_{2}t}& \ldots & \lambda_{n}^{n-1}e^{\lambda_{n}t} \end{array}\right\vert$$

$$= e^{\lambda_{1}t}e^{\lambda_{2}t}\cdots e^{\lambda_{n}t}\left\vert... ...{1}^{n-1}& \lambda_{2}^{n-1}& \ldots & \lambda_{n}^{n-1} \end{array}\right\vert$$

$$= \Pi_{1 \leq i < j \leq n-1}(\lambda_j - \lambda_i)$$

Since $\lambda_{1} ,\lambda_{2} , \ldots \lambda_{n}$ are different,

$$W(e^{\lambda_{1}t},e^{\lambda_{2}t},\ldots, e^{\lambda_{n}t}) \neq 0\ensuremath{\ \blacksquare}$$

The $n$ linearly independent solutions of ${\bf X}^{\prime} = A{\bf X}$ is called the fundamental solution.

Example 3..2   Solve the following differential equation

$\left\{\begin{array}{rl} x_{1}^{\prime} &= 8x_{1} + 6x_{2} \\ x_{2}' &= -3x_1 - 2x_2 \end{array}\right .$ .

SOLUTION

$$x_{1}^{\prime} = 8x_{1} + 6x_{1}$$

$$x_{2}^{\prime} = -3x_{1} - 2x_{2}$$

or

$${\bf X}^{\prime} = \left(\begin{array}{rr} 8&6\\ -3&-2 \end{array}\right){\bf X}$$

We find the eigenvalues and eigenvectors.

$$\left\vert\begin{array}{ll} 8-\lambda & 6\\ -3 & -2-\lambda \end{array}\right\vert = -16-6\lambda + \lambda^2 + 18 = \lambda^2 - 6\lambda + 2 = 0$$

Thus we have $\lambda = 3 \pm \sqrt{7}$. For $\lambda_1 = 3 + \sqrt{7}, {\bf C}_{1} = \left(\begin{array}{c} -6\\ 5-\sqrt{7} \end{array}\right)$. For $\lambda_{2} = 3 - \sqrt{7}, {\bf C}_{2} = \left(\begin{array}{c} -6\\ 5 + \sqrt{7} \end{array}\right)$. Thus the general solution is

$${\bf X}(t) = c_{1}\left(\begin{array}{c} -6\\ 5-\sqrt{7} \end{ar... ... 5 + \sqrt{7} \end{array}\right)e^{(3-\sqrt{7})t}\ensuremath{\ \blacksquare}$$