3602 General Linear ODE Systems and Independent Solutions

We have studied the homogeneous system of ODE's with constant coefficients, $$\boldsymbol{x'}=A\boldsymbol{x}\tag{1}$$ where $A$ is an $n \times n$ matrix of constants ($n = 2, 3$). We described how to calculate the eigenvalues and corresponding eigenvectors for the matrix $A$, and how to use them to find $n$ independent solutions to the system $(1)$.
With this concrete experience in solving low-order systems with constant coefficients, what can be said when the coefficients are functions of the independent variable $t$? We can still write the linear system in the matrix form $(1)$, but now the matrix entries will be functions of $t$: $$\begin{aligned} x'=a(t)x+b(t)y\\ y'=c(t)x+d(t)y \end{aligned},\begin{pmatrix} x'\\y' \end{pmatrix}=\begin{pmatrix} a(t)&b(t)\\c(t)&d(t) \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}\tag{2}$$ or in more abridged notation, valid for $n \times n$ linear homogeneous systems, $$\boldsymbol{x'}=A(t)\boldsymbol{x}\tag{3}$$ Note how the matrix becomes a function of $t$ - we call it a matrix-valued function of $t$, since to each value of $t$ the function rule assigns a matrix: $$t_0 \rarr A(t_0)=\begin{pmatrix} a(t_0)&b(t_0)\\c(t_0)&d(t_0) \end{pmatrix}$$ In the rest of this chapter we will often not write the variable $t$ explicitly, but it is always understood that the matrix entries are functions of $t$.
We will sometimes use $n = 2$ or $3$ in the statements and examples in order to simplify the exposition, but the definitions, results, and the arguments which prove them are essentially the same for higher values of $n$.
Definition 1 Solutions $\boldsymbol{x}_1(t), \ldots, \boldsymbol{x}_n(t)$ to $(3)$ are called linearly dependent if there are constants $c_i$, not all of which are 0, such that $$c_1\boldsymbol{x}_1(t)+\ldots+c_n\boldsymbol{x}_n(t)=0, \text{ for all } t\tag{4}$$ If there is no such relation, i.e., if $$c_1\boldsymbol{x}_1(t)+\ldots+c_n\boldsymbol{x}_n(t)=0, \text{ for all } t \rArr \text{ all } c_i=0 \tag{5}$$ the solutions are called linearly independent, or simply independent.
The phrase for all $t$ is often in practice omitted, as being understood. This can lead to ambiguity. To avoid it, we will use the symbol $\equiv 0$ for identically 0, meaning zero for all $t$; the symbol $\not\equiv 0$ means not identically 0, i.e., there is some $t$-value for which it is not zero. For example, $(4)$ would be written $$c_1\boldsymbol{x}_1(t)+\ldots+c_n\boldsymbol{x}_n(t)\equiv 0$$ Theorem 1 If $\boldsymbol{x}_1, \ldots, \boldsymbol{x}_n$ is a linearly independent set of solutions to the $n \times n$ system $\boldsymbol{x'} = A(t)\boldsymbol{x}$, then the general solution to the system is $$\boldsymbol{x}=c_1\boldsymbol{x}_1+\ldots+c_n\boldsymbol{x}_n\tag{6}$$ Such a linearly independent set is called a fundamental set of solutions.
This theorem is the reason for expending so much effort to find two independent solutions, when $n = 2$ and $A$ is a constant matrix. In this chapter, the matrix $A$ is not constant; nevertheless, $(6)$ is still true.
Proof. There are two things to prove:
(a) All vector functions of the form $(6)$ really are solutions to $\boldsymbol{x'} = A\boldsymbol{x}$.
This is the superposition principle for solutions of the system; it's true because the system is linear. The matrix notation makes it really easy to prove. We have $$\begin{aligned} &(c_1\boldsymbol{x}_1+\ldots+c_n\boldsymbol{x}_n)'\\ =&c_1\boldsymbol{x}_1'+\ldots+c_n\boldsymbol{x}_n'\\ =&c_1A_1\boldsymbol{x}_1+\ldots+c_nA_n\boldsymbol{x}_n\\ =&A(c_1\boldsymbol{x}_1+\ldots+c_n\boldsymbol{x}_n) \end{aligned}$$ (b) All solutions to the system are of the form $(6)$.
This is harder to prove and will be the main result of the next note.