10.2: Basic Theory of Homogeneous Linear Systems

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In this section we consider homogeneous linear systems \({\bf y}'= A(t){\bf y}\), where \(A=A(t)\) is a continuous \(n\times n\) matrix function on an interval \((a,b)\). The theory of linear homogeneous systems has much in common with the theory of linear homogeneous scalar equations, which we've previously considered in earlier sections. The theory requires knowledge of linear algebra and will therefore be omitted; however, one should note the similarity to the theory of linear homogeneous scalar equations.

Whenever we refer to solutions of \({\bf y}'=A(t){\bf y}\) we’ll mean solutions on \((a,b)\). Since \({\bf y}\equiv{\bf 0}\) is obviously a solution of \({\bf y}'=A(t){\bf y}\), we call it the trivial solution. Any other solution is nontrivial.

Definition 10.2.1

If \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\) are vector functions defined on an interval \((a,b)\) and \(c_1\), \(c_2\), …, \(c_n\) are constants, then

\[\label{eq:10.3.1} {\bf y}=c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n\]

is a linear combination of \({\bf y}_1\), \({\bf y}_2\), …,\({\bf y}_n\).

It’s easy show that if \({\bf y}_1\), \({\bf y}_2\), …,\({\bf y}_n\) are solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\), then so is any linear combination of \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\) (Exercise 10.2.1).

Definition 10.2.2

We say that a set \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) of \(n\)-vector functions is linearly independent on \((a,b)\) if the only constants \(c_1\), \(c_2\), …, \(c_n\) such that

\[\label{eq:10.3.2} c_{1}{\bf y}_{1}(t)+c_{2}{\bf y}_{2}(t)+\cdots +c_{n}{\bf y}_{n}(t)=0,\quad a<t<b,\]

are \(c_1=c_2=\cdots=c_n=0\). If Equation \ref{eq:10.3.2} holds for some set of constants \(c_1\), \(c_2\), …, \(c_n\) that are not all zero, then \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is linearly dependent on \((a,b)\)

The next definition will help to characterize fundamental sets of solutions of \({\bf y}'=A(t){\bf y}\).

Definition 10.2.3

We say that \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is a fundamental set of solutions of an \(n\times n\) system \({\bf y}'=A(t){\bf y}\) on \((a,b)\) if \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is linear independent and every solution of \({\bf y}'=A(t){\bf y}\) on \((a,b)\) can be written as a linear combination of \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\), as in Equation \ref{eq:10.3.1}. In this case we say that Equation \ref{eq:10.3.1} is the general solution of \({\bf y}'=A(t){\bf y}\) on \((a,b)\).

Theorem 10.2.1

Suppose the \(n\times n\) matrix \(A(t)\) is continuous on \((a,b)\). Then a set \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) of \(n\) solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\) is a fundamental set if and only if the determinant

\[\begin{align*} \left|\begin{array}{ccc}{\bf y}_1&{\bf y}_2&{\bf y}_3&\cdots&{\bf y}_n\end{array}\right|&\ne 0\end{align*}\]

on \((a,b)\).

Example 10.2.1

Show that if

\[{\bf y}_1=\left[\begin{array}{c}e^t\\0\\e^{-t}\end{array}\right],\quad {\bf y}_2=\left[\begin{array}{c}0\\e^{3t}\\1\end{array}\right], \quad \text{and} \quad {\bf y}_3=\left[\begin{array}{c}e^{2t}\\e^{3t}\\0\end{array}\right]\nonumber \]

are solutions to some \(3\times 3\) system \({\bf y}'=A(t){\bf y}\), then they form a fundamental set of solutions on every interval \((a,b)\).

Solution

Expanding the determinant of this system in cofactors of the entries of the first row yields

\[\begin{align*} \left|\begin{array}{ccc}e^t&0&e^{2t}\\0&e^{3t}&e^{3t}\\e^{-t}&1&0 \end{array}\right|&=e^t \left|\begin{array}{cc}e^{3t}&e^{3t}\\1&0\end{array}\right|-0 \left|\begin{array}{cc}0&e^{3t}\\e^{-t}&0\end{array}\right| +e^{2t}\left|\begin{array}{cc}0&e^{3t}\\e^{-t}&1\end{array}\right| \\ &=e^t(-e^{3t})+e^{2t}(-e^{2t})=-2e^{4t}.\end{align*}\]

Since this determinant is never zero, \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) must form a fundamental set of solutions.

We can use the method in Example 10.2.1 to test \(n\) solutions \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) of any \(n\times n\) system \({\bf y}'=A(t){\bf y}\) for a fundammental set on an interval \((a,b)\) on which \(A\) is continuous. To explain this (and for other purposes later), it is useful to write a linear combination of \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\) in a different way. We first write the vector functions in terms of their components as

\[{\bf y}_1=\left[\begin{array}{c} y_{11}\\y_{21}\\ \vdots\\ y_{n1}\end{array}\right],\quad {\bf y}_2=\left[\begin{array}{c} y_{12}\\y_{22}\\ \vdots\\ y_{n2}\end{array}\right],\dots,\quad {\bf y}_n=\left[\begin{array}{c} y_{1n}\\y_{2n}\\ \vdots\\ y_{nn}\end{array}\right].\nonumber \]

If

\[{\bf y}=c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n\nonumber \]

then

\[\begin{align*} {\bf y}&= c_1\left[\begin{array}{c} y_{11}\\y_{21}\\ \vdots\\ y_{n1}\end{array}\right]+ c_2\left[\begin{array}{c} y_{12}\\y_{22}\\ \vdots\\ y_{n2}\end{array}\right]+\cdots +c_n\left[\begin{array}{c} y_{1n}\\y_{2n}\\ \vdots\\ y_{nn}\end{array}\right]\\[4pt] &=\left[\begin{array}{cccc} y_{11}&y_{12}&\cdots&y_{1n} \\ y_{21}&y_{22}&\cdots&y_{2n}\\ \vdots&\vdots&\ddots&\vdots \\ y_{n1}&y_{n2}&\cdots&y_{nn} \\ \end{array}\right]\col cn.\end{align*}\]

This shows that

\[\label{eq:10.3.3} c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n=Y{\bf c},\]

where

\[{\bf c}=\col cn\nonumber \]

and

\[\label{eq:10.3.4} Y=[{\bf y}_1\; {\bf y}_2\; \cdots\; {\bf y}_n]= \left[\begin{array}{cccc} y_{11}&y_{12}&\cdots&y_{1n} \\ y_{21}&y_{22}&\cdots&y_{2n}\\ \vdots&\vdots&\ddots&\vdots \\ y_{n1}&y_{n2}&\cdots&y_{nn} \\ \end{array}\right];\]

that is, the columns of \(Y\) are the vector functions \({\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\).

For reference below, note that

\[\begin{aligned} Y'&=[{\bf y}_1'\; {\bf y}_2'\; \cdots\; {\bf y}_n']\\ &=[A{\bf y}_1\; A{\bf y}_2\; \cdots\; A{\bf y}_n]\\ &=A[{\bf y}_1\; {\bf y}_2\; \cdots\; {\bf y}_n]=AY;\end{aligned}\]

that is, \(Y\) satisfies the matrix differential equation

\[Y'=AY.\nonumber \]

The determinant of \(Y\),

\[\label{eq:10.3.5} W=\left|\begin{array}{cccc} y_{11}&y_{12}&\cdots&y_{1n} \\ y_{21}&y_{22}&\cdots&y_{2n}\\ \vdots&\vdots&\ddots&\vdots \\ y_{n1}&y_{n2}&\cdots&y_{nn} \\ \end{array}\right|\]

is called the Wronskian of \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\). It can be shown that this definition is analogous to definitions of the Wronskian of scalar functions given earlier.

The next theorem is analogous to earlier theorems as well.

Theorem 10.2.2

Suppose the \(n\times n\) matrix \(A=A(t)\) is continuous on \((a,b)\) and let \({\bf y}_1\), \({\bf y}_2\), …,\({\bf y}_n\) be solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\). Then the following statements are equivalent; that is, they are either all true or all false:

1. The general solution of \({\bf y}'=A(t){\bf y}\) on \((a,b)\) is \({\bf y}=c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n\), where \(c_1\), \(c_2\), …, \(c_n\) are arbitrary constants.
2. \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is a fundamental set of solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\).
3. \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is linearly independent on \((a,b)\).
4. The Wronskian of \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is nonzero at all points in \((a,b)\).

We say that \(Y\) in Equation \ref{eq:10.3.4} is a fundamental matrix for \({\bf y}'=A(t){\bf y}\) if any (and therefore all) of the statements (a)-(d) of Theorem 10.2.2 are true for the columns of \(Y\). In this case, Equation \ref{eq:10.3.3} implies that the general solution of \({\bf y}'=A(t){\bf y}\) can be written as \({\bf y}=Y{\bf c}\), where \({\bf c}\) is an arbitrary constant \(n\)-vector.

Example 10.2.2

The vector functions

\[{\bf y}_1=\twocol {-e^{2t}}{2e^{2t}}\quad \text{and} \quad {\bf y}_2=\twocol{-e^{-t}}{\phantom{-}e^{-t}}\nonumber \]

are solutions of the constant coefficient system

\[\label{eq:10.3.7} {\bf y}' = \left[\begin{array}{cc}{-4}&{-3}\\{6}&{5}\end{array} \right] {\bf y}\]

on \((-\infty,\infty)\). (Verify.)

1. Compute the Wronskian of \(\{{\bf y}_1,{\bf y}_2\}\) directly from the definition Equation \ref{eq:10.3.5}
2. Find the general solution of Equation \ref{eq:10.3.7}.
3. Solve the initial value problem \[\label{eq:10.3.8} {\bf y}'=\left[\begin{array}{cc}{-4}&{-3}\\{6}&{5}\end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[\begin{array}{r} 4 \\-5\end{array}\right].\]

Solution a

From Equation \ref{eq:10.3.5}

\[\label{eq:10.3.9} W(t)=\left|\begin{array}{cc}-e^{2t}&-e^{-t}\\2e^{2t}&\hfill e^{-t}\end{array}\right|=e^t.\]

Solution b

Since \(W(t)\ne0\), Theorem 10.2.2 implies that \(\{{\bf y}_1,{\bf y}_2\}\) is a fundamental set of solutions of Equation \ref{eq:10.3.7} and

\[Y=\left[\begin{array}{cc}-e^{2t}&-e^{-t}\\2e^{2t}&\hfill e^{-t}\end{array}\right]\nonumber \]

is a fundamental matrix for Equation \ref{eq:10.3.7}. Therefore the general solution of Equation \ref{eq:10.3.7} is

\[\label{eq:10.3.10} {\bf y}=c_1{\bf y}_1+c_2{\bf y}_2= c_1\twocol {-e^{2t}}{2e^{2t}}+c_2\twocol{-e^{-t}}{e^{-t}} =\left[\begin{array}{cc}-e^{2t}&-e^{-t}\\2e^{2t}&\hfill e^{-t}\end{array}\right] \left[\begin{array}{c}c_1\\c_2\end{array}\right].\]

Solution c

Setting \(t=0\) in Equation \ref{eq:10.3.10} and imposing the initial condition in Equation \ref{eq:10.3.8} yields

\[c_1\left[\begin{array}{r}-1 \\2\end{array}\right]+c_2 \left[\begin{array}{r}-1 \\1\end{array}\right]= \left[\begin{array}{r} 4 \\-5\end{array}\right].\nonumber \]

Thus,

\[\begin{aligned} -c_1-c_2&=\phantom{-}4 \\ 2c_1+c_2&=-5.\end{aligned}\nonumber \]

The solution of this system is \(c_1=-1\), \(c_2=-3\). Substituting these values into Equation \ref{eq:10.3.10} yields

\[{\bf y}=-\left[\begin{array}{c}-e^{2t} \\ 2e^{2t}\end{array} \right]-3 \left[\begin{array}{c}-e^{-t} \\ e^{-t}\end{array}\right]= \left[ \begin{array}{c} e^{2t}+3e^{-t} \\-2e^{2t}-3e^{-t} \end{array}\right]\nonumber \]

as the solution of Equation \ref{eq:10.3.8}.