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First let us talk about matrix or vector valued functions. Such a function is just a matrix whose entries depend on some variable. If \(t\) is the independent variable, we write a vector valued function \( \vec {x} (t) \) as
\[ \vec {x} (t) = \begin {bmatrix} x_1(t) \\ x_2 (t) \\ \vdots \\ x_n (t) \end {bmatrix} \nonumber \]
Similarly a matrix valued function \( A(t) \) is
\[ A (t) = \begin {bmatrix} a_{11} (t) & a_{12} (t) & \cdots & a_{1n} (t) \\ a_{21} (t) & a_ {22} (t) & \cdots & a_{2n} (t) \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1}(t) & a_{n2}(t) & \cdots & a_{nn}(t) \end {bmatrix} \nonumber \]
We can talk about the derivative \(A'(t)\) or \( \frac {dA}{dt} \). This is just the matrix valued function whose \(ij^{th}\) entry is \(a'_{ij} (t) \).
Rules of differentiation of matrix valued functions are similar to rules for normal functions. Let \(A(t)\) and \(B(t)\) be matrix valued functions. Let \(c\) be a scalar and let \(C\) be a constant matrix. Then
\[\begin{align}\begin{aligned} {(A(t) + B(t))}' &= A' (t) + B' (t) \\ (A(t)B(t))' &= A'(t)B(t) + A(t)B'(t) \\ (cA(t))' &= cA' (t) \\ (CA(t))' &= CA'(t) \\ (A(t)C)' &= A' (t)C \end{aligned}\end{align} \nonumber \]
Note the order of the multiplication in the last two expressions.
A first order linear system of ODEs is a system that can be written as the vector equation
\[ \vec {x} (t) = P(t) \vec {x} (t) + \vec {f} (t) \nonumber \]
where \( P(t) \) is a matrix valued function, and \( \vec {x} (t) \) and \( \vec {f} (t) \) are vector valued functions. We will often suppress the dependence on \(t\) and only write \( \vec {x} = P \vec {x} + \vec {f} \). A solution of the system is a vector valued function \( \vec {x} \) satisfying the vector equation.
For example, the equations
\[\begin{align}\begin{aligned} x'_1 &= 2tx_1 + e^tx_2 + t^2 \\ x'_2 &= \frac {x_1}{t} - x_2 + e^t \end{aligned}\end{align} \nonumber \]
can be written as
\[ \vec {x'} = \begin {bmatrix} 2t & e^t \\ \frac {1}{t} & -1 \end {bmatrix} \vec {x'} + \begin {bmatrix} t^2 \\ e^t \end {bmatrix} \nonumber \]
We will mostly concentrate on equations that are not just linear, but are in fact constant coefficient equations. That is, the matrix \( P\) will be constant; it will not depend on \(t\).
When \( \vec {f} = \vec {0} \) (the zero vector), then we say the system is hom*ogeneous. For hom*ogeneous linear systems we have the principle of superposition, just like for single hom*ogeneous equations.
Theorem \(\PageIndex{1}\)
Superposition
Let \( \vec {x'} = P \vec {x'} \) be a linear hom*ogeneous system of ODEs. Suppose that \( \vec {x}_1, \dots, \vec {x}_n \) are \(n\) solutions of the equation, then
\[ \vec {x} = c_1 \vec {x}_1 + c_2 \vec {x}_2 + \dots + c_n \vec {x}_n \nonumber \]
is also a solution. Furthermore, if this is a system of \(n\) equations \( (P \rm{~is~} n \times n) \), and \( \vec {x}_1, \dots , \vec {x}_n \) are linearly independent, then every solution can be written as \(\eqref{eq:12}\).
Linear independence for vector valued functions is the same idea as for normal functions. The vector valued functions \( \vec {x}_1, \vec {x}_2, \dots, \vec {x}_n \) are linearly independent when
\[ \label{eq:12}c_1 \vec {x}_1 + c_2 \vec {x}_2 + \dots + c_n \vec {x}_n = \vec {0} \]
has only the solution \( c_1 = c_2 = \dots = c_n = 0 \), where the equation must hold for all \(t\).
Example 3.3.1
\( \vec {x}_1 = \begin {bmatrix} t^2 \\ t \end {bmatrix}, \vec {x}_2 = \begin {bmatrix} 0 \\ {1 + t } \end {bmatrix}, \vec {x}_3 = \begin {bmatrix} -t^2 \\ 1 \end {bmatrix} \) are linearly depdendent because \( \vec {x}_1 + \vec {x}_3 = \vec {x}_2\), and this holds for all \(t\). So \(c_1 = 1, c_2 = -1\) and \(c_3 = 1\) above will work.
On the other hand if we change the example just slightly \( \vec {x}_1 = \begin {bmatrix} t^2 \\ t \end {bmatrix}, \vec {x}_2 = \begin {bmatrix} 0 \\ t \end {bmatrix}, \vec {x}_3 = \begin {bmatrix} -t^2 \\ 1 \end {bmatrix} \), then the functions are linearly independent. First write \( c_1 \vec {x}_1 + c_2 \vec {x}_2 + c_3 \vec {x}_3 = \vec {0} \) and note that it has to hold for all \(t\). We get that
\( c_1 \vec {x}_1 + c_2 \vec {x}_2 + c_3 \vec {x}_3 = \begin {bmatrix} c_1t^2 - c_3t^3 \\ c_1t + c_2t + c_3 \end {bmatrix} = \begin {bmatrix} 0 \\ 0 \end {bmatrix} \)
In other words \( c_1t^2 - c_3t^3 = 0 \) and \(c_1t + c_2t + c_3 = 0 \). If we set \(t = 0\), then the second equation becomes \(c_3 = 0 \). However, the first equation becomes \(c_1t^2 = 0\) for all \(t\) and so \(c_1 = 0 \). Thus the second equation is just \(c_2t = 0\), which means \(c_2 = 0\). So \(c_1 = c_2 = c_3 = 0 \) is the only solution and \( \vec {x}_1, \vec {x}_2 \) and \(\vec {x}_3\) are linearly independent.
The linear combination \( c_1 \vec {x}_1 + c_2 \vec {x}_2 + \dots + c_n \vec {x}_n \) could always be written as
\[ X (t) \vec {c} \nonumber \]
where \( X (t) \) is the matrix with columns \(\vec {x}_1, \dots , \vec {x}_n \), and \( \vec {c} \) is the column vector with entries \( c_1, \dots , c_n \). The matrix valued function \( X (t) \) is called the fundamental matrix, or the fundamental matrix solution.
To solve nonhom*ogeneous first order linear systems, we use the same technique as we applied to solve single linear nonhom*ogeneous equations.
Theorem \(\PageIndex{2}\)
Let \( \vec {x}' = P \vec {x} + \vec {f} \) be a linear system of ODEs. Suppose \( \vec {x}_p\) is one particular solution. Then every solution can be written as
\[ \vec {x} = \vec {x}_c + \vec {x}_p \nonumber \]
where \( \vec {x}_c \) is a solution to the associated hom*ogeneous equation \( (\vec {x} = P \vec {x}) \).
So the procedure will be the same as for single equations. We find a particular solution to the nonhom*ogeneous equation, then we find the general solution to the associated hom*ogeneous equation, and finally we add the two together.
Alright, suppose you have found the general solution \( \vec {x}' = P \vec {x} + \vec {f} \). Now you are given an initial condition of the form \[ \vec {x} {t_0} = \vec {b} \nonumber \] for some constant vector \( \vec {b} \). Suppose that \( X (t) \) is the fundamental matrix solution of the associated hom*ogeneous equation (i.e. columns of \( X (t) \) are solutions). The general solution can be written as
\[ \vec {x} (t) = X (t) \vec {c} + \vec {x}_p (t) \nonumber \]
We are seeking a vector \(\vec {c} \) such that
\[ \vec {b} = \vec {x} (t_0) = X (t_0) \vec {c} + \vec {x}_p (t_0) \nonumber \]
In other words, we are solving for \( \vec {c} \) the nonhom*ogeneous system of linear equations
\[ X(t_0) \vec {c} = \vec {b} - \vec {x}_p (t_0) \nonumber \]
Example 3.3.2
In Section 3.1 we solved the system
\[\begin{align}\begin{aligned} x_1' &= x_1 \\ x'_2 &= x_1 - x_2 \end{aligned}\end{align} \nonumber \]
with initial conditions \( x_1(0) = 1, x_2 (0) = 2\).
Solution
This is a hom*ogeneous system, so \( \vec {f} (t) = \vec {0} \). We write the system and the initial conditions as
\[ \vec {x} ' = \begin {bmatrix} 1 & 0 \\ 1 & -1 \end {bmatrix} \vec{x}, \quad \vec{x}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \nonumber \]
We found the general solution was \( x_1= C_1 e^t \) and \(x_2=\frac{c_1}{2} e^t + c_2 e^{-t} \). Letting \( C_1=1 \) and \(C_2=0\), we obtain the solution \( \begin{bmatrix} e^t \\ \frac{1}{2}e^t \end{bmatrix} \). Letting \( C_1=0 \) and \(C_2=1\), we obtain \( \begin{bmatrix} 0 \\ e^{-t} \end{bmatrix} \). These two solutions are linearly independent, as can be seen by setting \( t=0 \), and noting that the resulting constant vectors are linearly independent. In matrix notation, the fundamental matrix solution is, therefore,
\[ X(t) = \begin{bmatrix} e^t & 0 \\ \frac{1}{2}e^t & e^{-t} \end{bmatrix} \nonumber \]
Hence to solve the initial problem we solve the equation
\[ X(0)\vec(c) = \vec{b} \nonumber \]
or in other words,
\[ \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix} \vec{c} =\begin{bmatrix} 1 \\ 2 \end{bmatrix} \nonumber \]
\[ \vec{x}(t) = X(t) \vec{c} = \begin{bmatrix} e^t & 0 \\ \frac{1}{2}e^{t} & e^{-t} \end{bmatrix} \begin{bmatrix} 1 \\ \frac{3}{2} \end{bmatrix} = \begin{bmatrix} e^t \\ \frac{1}{2} e^t + \frac{3}{2} e^{-t} \end{bmatrix} \nonumber \]
This agrees with our previous solution from Section 3.1.
