Download original notes
PAGE 1

7.2 Matrices and Linear Systems

\( n^{\text{th}} \)-order \( \leftrightarrow \) \( n \) 1st-order eqs. in a system

for example, \( t^2 x'' + t x' + (t^2 - 1)x = e^t \)

\[ x'' + \frac{1}{t} x' + \frac{t^2 - 1}{t^2} x = \frac{e^t}{t^2} \]

let

\[ z_1 = x \]\[ z_2 = x' \]

eq 1: \( z_1' = z_2 \)

eq 2: from the diff. eq

\[ x'' = -\frac{1}{t} x' - \frac{t^2 - 1}{t^2} x + \frac{e^t}{t^2} \]\[ z_2' = -\frac{1}{t} z_2 - \frac{t^2 - 1}{t^2} z_1 + \frac{e^t}{t^2} \]
PAGE 2
\[ z_1' = z_2 \]\[ z_2' = -\frac{t^2 - 1}{t^2} z_1 - \frac{1}{t} z_2 + \frac{e^t}{t^2} \]

as a matrix eq:

\[ \underbrace{\begin{bmatrix} z_1' \\ z_2' \end{bmatrix}}_{\vec{x}'} = \underbrace{\begin{bmatrix} 0 & 1 \\ -\frac{t^2 - 1}{t^2} & -\frac{1}{t} \end{bmatrix}}_{P(t)} \underbrace{\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}}_{\vec{x}} + \underbrace{\begin{bmatrix} 0 \\ \frac{e^t}{t^2} \end{bmatrix}}_{\vec{f}(t)} \]

in the form \( \vec{x}' = P(t) \vec{x} + \vec{f}(t) \)

  • \( \downarrow \)coefficient matrix (refers to \( P(t) \))
  • \( \hookrightarrow \)nonhomogeneous term (refers to \( \vec{f}(t) \))
PAGE 3

Matrix Representation of Systems

Turn into a matrix equation:

\[\begin{aligned} x' &= 8x - 2y + z + t \\ y' &= x - 4z + t^2 \\ z' &= 5y - 2z + t^3 \end{aligned}\]
\[\begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} 8 & -2 & 1 \\ 1 & 0 & -4 \\ 0 & 5 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} + \begin{bmatrix} t \\ t^2 \\ t^3 \end{bmatrix}\]
\[\vec{x}' = P(t)\vec{x} + \vec{f}(t)\]

Basic properties are similar to a first-order scalar equation.

If \(\vec{f}(t) = \vec{0} \rightarrow \vec{x}' = P(t)\vec{x}\) is called a homogeneous system

PAGE 4

The solution \(\vec{x}\) is a vector that satisfies the matrix equation:

\[\vec{x}' = P(t)\vec{x} + \vec{f}(t)\]

If \(P(t)\) is \(n \times n \rightarrow n\) solutions \(\vec{x}_1, \vec{x}_2, \vec{x}_3, \dots, \vec{x}_n\)

(because \(n \times n \rightarrow n\) 1st-order equations in a system \(\leftrightarrow\) one \(n^{\text{th}}\)-order scalar equation.)

Linear combination of the solutions is also a solution.

The general solution to the homogeneous system is:

\[\vec{x} = c_1 \vec{x}_1 + c_2 \vec{x}_2 + \dots + c_n \vec{x}_n\]

These solutions are linearly independent:

\[c_1 \vec{x}_1 + c_2 \vec{x}_2 + \dots + c_n \vec{x}_n = \vec{0} \rightarrow c_1 = c_2 = \dots = c_n = 0\]
PAGE 5

The Wronskian of these solutions is

\[ W(\vec{x}_1, \vec{x}_2, \dots, \vec{x}_n) = \begin{vmatrix} \vec{x}_1 & \vec{x}_2 & \vec{x}_3 & \dots & \vec{x}_n \end{vmatrix} \]

(solutions as columns)

if \( W \neq 0 \) on an interval of \( t \), then the solutions are linearly indp on that interval.

Example

Verify that \( \vec{x}_1 = e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) and \( \vec{x}_2 = e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \) are solutions of \( \vec{x}' = \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} \vec{x} \) and show that they are linearly indp.

PAGE 6

if they are the solutions, then they satisfy the differential eq.

\[ \vec{x}' = \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} \vec{x} \]

\( \vec{x}_1 = e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) sub into system above

\( \vec{x}_1' = 2e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)

\[ 2e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} \left( e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right) \]\[ = e^{2t} \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]\[ 2e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = e^{2t} \begin{bmatrix} 2 \\ 2 \end{bmatrix} \]

same, so \( e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) is a solution

PAGE 7

we can similarly verify that \( \vec{x}_2 = e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \) is also a solution

let's look at the Wronskian of \( \vec{x}_1 = e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) and \( \vec{x}_2 = e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \)

\[ W(\vec{x}_1, \vec{x}_2) = \begin{vmatrix} e^{2t} & e^{-2t} \\ e^{2t} & 5e^{-2t} \end{vmatrix} \]
\( \vec{x}_1 \)\( \vec{x}_2 \)
\[ = e^{2t} \cdot 5e^{-2t} - e^{2t} \cdot e^{-2t} \]\[ = 5 - 1 = 4 \neq 0 \text{ for any } t \rightarrow \vec{x}_1 \text{ and } \vec{x}_2 \text{ indp on } -\infty < t < \infty \]
PAGE 8
\[ \vec{x}' = \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} \vec{x} \quad \vec{x}_1 = e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \quad \vec{x}_2 = e^{-2t} \begin{bmatrix} 1 \\ 5 \end{bmatrix} \]

let's find the eigenvalues and eigenvectors of the coefficient matrix

\[ A = \begin{bmatrix} 3 & -1 \\ 5 & -3 \end{bmatrix} \]\[ \det(A - \lambda I) = \begin{vmatrix} 3 - \lambda & -1 \\ 5 & -3 - \lambda \end{vmatrix} = 0 \]\[ (3 - \lambda)(-3 - \lambda) + 5 = 0 \]\[ -9 - 3\lambda + 3\lambda + \lambda^2 + 5 = 0 \rightarrow \lambda^2 - 4 = 0 \]\[ \lambda = 2, -2 \]
PAGE 9

Eigenvectors

λ = 2

Solve \((A - \lambda I)\vec{v} = \vec{0}\)

\[\begin{bmatrix} 1 & -1 & 0 \\ 5 & -5 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\]

\(x_2 = r\), \(x_1 = x_2 = r\)

\[\vec{v} = r \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Choose \(r = 1\)

\[\vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]
\[\vec{x}_1 = e^{2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]
  • → eigenvalue: 2
  • → corresponding eigenvector: \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\)

\(\vec{x}_2\) is found the same way

PAGE 10
\[\vec{x}' = P(t)\vec{x} + \vec{f}(t)\]

If \(\vec{f}(t) \neq \vec{0}\), then the nonhomogeneous system has solution:

\[\vec{x} = \vec{x}_c + \vec{x}_p\]

→ Complementary solution

solution to

\(\vec{x}' = P(t)\vec{x}\)

(no \(\vec{f}(t)\))

→ Particular solution

(due to presence of \(\vec{f}(t)\))