10.3 Basic Theory of Homogeneous Linear System

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 considered in Sections 2.1, 5.1, and 9.1.

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.

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

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 1). We say that \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is a fundamental set of solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\) on if 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 (10.3.1). In this case we say that (10.3.1) is the general solution of \({\bf y}'=A(t){\bf y}\) on \((a,b)\).

It can be shown that if \(A\) is continuous on \((a,b)\) then \({\bf y}'=A(t){\bf y}\) has infinitely many fundamental sets of solutions on \((a,b)\) (Exercises 15 and 16). The next definition will help to characterize fundamental sets of solutions of \({\bf y}'=A(t){\bf y}\).

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

are \(c_1=c_2=\cdots=c_n=0\). If (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 theorem is analogous to Theorems 5.1.3 and 9.1.2.

Solution Suppose

We must show that \(c_1=c_2=c_3=0\). Rewriting this equation in matrix form yields

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

Since this determinant is never zero, \(c_1=c_2=c_3=0\). ∎

We can use the method in Example 10.3.1 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 linear independence on an interval \((a,b)\) on which \(A\) is continuous. To explain this (and for other purposes later), it’s 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

If

then

This shows that

where

and

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

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

The determinant of \(Y\),

is called the Wronskian of \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\). It can be shown (Exercises 2 and 3) that this definition is analogous to definitions of the Wronskian of scalar functions given in Sections 5.1 and 9.1. The next theorem is analogous to Theorems 5.1.4 and 9.1.3. The proof is sketched in Exercise 4 for \(n=2\) and in Exercise 5 for general \(n\).

The next theorem is analogous to Theorems 5.1.6 and 9.1.4.

We say that \(Y\) in (10.3.4) is a fundamental matrix for \({\bf y}'=A(t){\bf y}\) if any (and therefore all) of the statements (a)-(e) of Theorem 10.3.2 are true for the columns of \(Y\). In this case, (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.

Solution (a) From (10.3.5)

Solution (b) Here

so tr\((A)=-4+5=1\). If \(t_0\) is an arbitrary real number then (10.3.6) implies that

which is consistent with (10.3.9).

Solution (c) Since \(W(t)\ne0\), Theorem 10.3.3 implies that \(\{{\bf y}_1,{\bf y}_2\}\) is a fundamental set of solutions of (10.3.7) and

is a fundamental matrix for (10.3.7). Therefore the general solution of (10.3.7) is

Solution (d) Setting \(t=0\) in (10.3.10) and imposing the initial condition in (10.3.8) yields

Thus,

The solution of this system is \(c_1=-1\), \(c_2=-3\). Substituting these values into (10.3.10) yields

as the solution of (10.3.8).

10.3 Exercises

1. Prove: If \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\) are solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\), then any linear combination of \({\bf y}_1\), \({\bf y}_2\), …, \({\bf y}_n\) is also a solution of \({\bf y}'=A(t){\bf y}\) on \((a,b)\).

2. In Section 5.1 the Wronskian of two solutions \(y_1\) and \(y_2\) of the scalar second order equation

   \[ P_0(x)y''+P_1(x)y'+P_2(x)y=0 \tag*{\rm (A)} \]

   was defined to be

   \[ W=\left|\begin{array}{cc} y_1&y_2 \\ y'_1&y'_2\end{array}\right|. \]

   1. Rewrite (A) as a system of first order equations and show that \(W\) is the Wronskian (as defined in this section) of two solutions of this system.

   2. Apply Eqn. (10.3.6) to the system derived in (a), and show that

      \[ W(x)=W(x_0)\exp\left\{-\int^x_{x_0}{P_1(s)\over P_0(s)}\, ds\right\}, \]

      which is the form of Abel’s formula given in Theorem 9.1.3.

   Show answer

   \(\dst{{\bf y}'=\left[\begin{array}{cc}0&1\\[3pt]-\dst{P_2(x)\over P_0(x) }&-\dst{P_1(x)\over P_0(x)}\end{array}\right]{\bf y}}\)  

3. In Section 9.1 the Wronskian of \(n\) solutions \(y_1\), \(y_2\), …, \(y_n\) of the \(n-\)th order equation

   \[ P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=0 \tag*{\rm (A)} \]

   was defined to be

   \[ W=\left|\begin{array}{cccc} y_1&y_2&\cdots&y_n \\[6pt] y'_1&y'_2&\cdots&y_n'\\[6pt] \vdots&\vdots&\ddots&\vdots\\[6pt] y_1^{(n-1)}&y_2^{(n-1)}&\cdots&y_n^{(n-1)} \end{array}\right|. \]

   1. Rewrite (A) as a system of first order equations and show that \(W\) is the Wronskian (as defined in this section) of \(n\) solutions of this system.

   2. Apply Eqn. (10.3.6) to the system derived in (a), and show that

      \[ W(x)=W(x_0)\exp\left\{-\int^x_{x_0}{P_1(s)\over P_0(s)}\, ds\right\}, \]

      which is the form of Abel’s formula given in Theorem 9.1.3.

   Show answer

   \({\bf y}'=\left[\begin{array}{cccc} 0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&1\\[3pt] -\dst{P_n(x)\over P_0(x)}&-\dst{P_{n-1}(x)\over P_0(x)}&\cdots& -\dst{P_1(x)\over P_0(x)}\end{array}\right]{\bf y}\)

4. Suppose

   \[ {\bf y}_1=\twocol{y_{11}}{y_{21}}\mbox{\quad and \quad} {\bf y}_2=\twocol{y_{12}}{y_{22}} \]

   are solutions of the \(2\times 2\) system \({\bf y}'=A{\bf y}\) on \((a,b)\), and let

   \[ Y=\twobytwo {y_{11}} {y_{12}} {y_{21}} {y_{22}}\mbox{\quad and \quad} W=\left|\begin{array}{cc} y_{11}&y_{12}\\y_{21}&y_{22}\end{array}\right|; \]

   thus, \(W\) is the Wronskian of \(\{{\bf y}_1,{\bf y}_2\}\).

   1. Deduce from the definition of determinant that

      \[ W'=\left|\begin{array}{cc} {y'_{11}}&{y'_{12}}\\ {y_{21}}& {y_{22}}\end{array}\right| +\left|\begin{array}{cc} {y_{11}}&{y_{12}}\\ {y'_{21}}&{y'_{22}}\end{array}\right|. \]

   2. Use the equation \(Y'=A(t)Y\) and the definition of matrix multiplication to show that

      \[ [y'_{11}\quad y'_{12}]=a_{11} [y_{11}\quad y_{12}]+a_{12} [y_{21} \quad y_{22}] \]

      and

      \[ [y'_{21}\quad y'_{22}]=a_{21} [y_{11}\quad y_{12}]+a_{22} [y_{21}\quad y_{22}]. \]

   3. Use properties of determinants to deduce from (a) and (a) that

      \[ \left|\begin{array}{cc} {y'_{11}}&{y'_{12}}\\ {y_{21}}& {y_{22}}\end{array}\right|=a_{11}W\mbox{\quad and \quad} \left|\begin{array}{cc} {y_{11}}&{y_{12}}\\ {y'_{21}}&{y'_{22}}\end{array}\right|=a_{22}W. \]

   4. Conclude from (c) that

      \[ W'=(a_{11}+a_{22})W, \]

      and use this to show that if \(a<t_0<b\) then

      \[ W(t)=W(t_0)\exp\left(\int^t_{t_0} \left[a_{11}(s)+a_{22} (s) \right]\, ds\right)\quad a<t<b. \]

5. Suppose the \(n\times n\) matrix \(A=A(t)\) is continuous on \((a,b)\). Let

   \[ Y= \left[\begin{array}{cccc} y_{11}&y_{12}&\cdots&y_{1n} \\ y_{21}&y_{22}&\cdots&y_{2n} \\ \vdots&\vdots&\ddots&\vdots \\ y_{n1}&y_{n2}&\cdots&y_{nn} \end{array}\right], \]

   where the columns of \(Y\) are solutions of \({\bf y}'=A(t){\bf y}\). Let

   \[ r_i=[y_{i1}\, y_{i2}\, \dots\, y_{in}] \]

   be the \(i\)th row of \(Y\), and let \(W\) be the determinant of \(Y\).

   1. Deduce from the definition of determinant that

      \[ W'=W_1+W_2+\cdots+W_n, \]

      where, for \(1 \le m \le n\), the \(i\)th row of \(W_m\) is \(r_i\) if \(i \ne m\), and \(r'_m\) if \(i=m\).

   2. Use the equation \(Y'=A Y\) and the definition of matrix multiplication to show that

      \[ r'_m=a_{m1}r_1+a_{m2} r_2+\cdots+a_{mn}r_n. \]

   3. Use properties of determinants to deduce from (b) that

      \[ \det (W_m)=a_{mm}W. \]

   4. Conclude from (a) and (c) that

      \[ W'=(a_{11}+a_{22}+\cdots+a_{nn})W, \]

      and use this to show that if \(a<t_0<b\) then

      \[ W(t)=W(t_0)\exp\left( \int^t_{t_0}\big[a_{11}(s)+a_{22}(s)+\cdots+a_{nn}(s)]\, ds\right), \quad a < t < b. \]

6. Suppose the \(n\times n\) matrix \(A\) is continuous on \((a,b)\) and \(t_0\) is a point in \((a,b)\). Let \(Y\) be a fundamental matrix for \({\bf y}'=A(t){\bf y}\) on \((a,b)\).

   1. Show that \(Y(t_0)\) is invertible.

   2. Show that if \({\bf k}\) is an arbitrary \(n\)-vector then the solution of the initial value problem

      \[ {\bf y}'=A(t){\bf y},\quad {\bf y}(t_0)={\bf k} \]

      is

      \[ {\bf y}=Y(t)Y^{-1}(t_0){\bf k}. \]

7. Let

   \[ A=\twobytwo2442, \quad {\bf y}_1=\left[\begin{array}{c} e^{6t} \\ e^{6t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r} e^{-2t} \\ -e^{-2t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r}-3 \\ 9\end{array}\right]. \]

   1. Verify that \(\{{\bf y}_1,{\bf y}_2\}\) is a fundamental set of solutions for \({\bf y}'=A{\bf y}\).

   2. Solve the initial value problem

      \[ {\bf y}'=A{\bf y},\quad {\bf y}(0)={\bf k}. \tag*{\rm(A)} \]

   3. Use the result of Exercise 6(b) to find a formula for the solution of (A) for an arbitrary initial vector \({\bf k}\).

   Show answer

   (b)  \({\bf y}=\dst{\left[\begin{array}{c}3e^{6t}-6e^{-2t}\\3e^{6t}+6e^{-2t} \end{array}\right]}\)   (c)  \({\bf y}= \dst{{1\over2}\left[\begin{array}{cc}e^{6t}+e^{-2t}&e^{6t}-e^{-2t} \\e^{6t}-e^{-2t}&e^{6t}+e^{-2t}\end{array}\right]}{\bf k}\)

8. Repeat Exercise 7 with

   \[ A=\twobytwo {-2} {-2} {-5}1, \quad {\bf y}_1=\left[\begin{array}{r} e^{-4t} \\ e^{-4t}\end{array}\right], \quad {\bf y}_2=\left[ \begin{array}{r}-2e^{3t} \\ 5e^{3t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 10 \\-4\end{array}\right]. \]
   Show answer

     (b) \({\bf y}=\dst{\left[\begin{array}{c}6e^{-4t}+4e^{3t}\\6e^{-4t}-10e^{3t} \end{array}\right]}\)   (c)  \({\bf y}= \dst{{1\over7}\left[\begin{array}{cc}5e^{-4t}+2e^{3t}&2e^{-4t}-2e^{3t} \\5e^{-4t}-5e^{3t}&2e^{-4t}+5e^{3t}\end{array}\right]}{\bf k}\)

9. Repeat Exercise 7 with

   \[ A=\twobytwo{-4} {-10} 3 7, \quad {\bf y}_1=\left[\begin{array}{r}-5e^{2t} \\ 3e^{2t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r} 2e^t \\-e^t \end{array}\right], \quad {\bf k}=\left[\begin{array}{r}-19 \\ 11\end{array} \right ]. \]
   Show answer

   (b) \({\bf y}=\dst{\left[\begin{array}{c}-15e^{2t}-4e^t\\9e^{2t}+2e^t \end{array}\right]}\)    (c) \({\bf y}=\dst{\left[\begin{array}{cc}-5e^{2t}+6e^t&-10e^{2t}+10e^t \\3e^{2t}-3e^t&6e^{2t}-5e^t\end{array}\right]}{\bf k}\)

10. Repeat Exercise 7 with

    \[ A=\twobytwo 2 1 1 2, \quad {\bf y}_1=\left[\begin{array}{r} e^{3t} \\ e^{3t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r}e^t \\ -e^t\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 2 \\ 8 \end{array}\right].\]
    Show answer

    (b) \({\bf y}=\dst{\left[\begin{array}{c}5e^{3t}-3e^t\\5e^{3t}+3e^t \end{array}\right]}\)       (c) \(\dst{\bf y}={{1\over2}\left[\begin{array}{cc}e^{3t}+e^t&e^{3t}-e^t \\e^{3t}-e^t&e^{3t}+e^t\end{array}\right]}{\bf k}\)

11. Let

    \begin{eqnarray*} A&=&\threebythree 3 {-1} {-1} {-2} 3 24 {-1} {-2}, \\ {\bf y}_1&=&\left[\begin{array}{c} e^{2t} \\ 0 \\ e^{2t}\end{array} \right], \quad {\bf y}_2=\left[\begin{array}{c} e^{3t} \\-e^{3t} \\ e^{3t}\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{-t} \\-3e^{-t} \\ 7e^{-t} \end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 2 \\-7 \\ 20\end{array}\right]. \end{eqnarray*}

    1. Verify that \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions for \({\bf y}'=A{\bf y}\).

    2. Solve the initial value problem

       \[ {\bf y}'=A{\bf y}, \quad {\bf y}(0)={\bf k}. \tag*{\rm(A)} \]

    3. Use the result of Exercise 6(b) to find a formula for the solution of (A) for an arbitrary initial vector \({\bf k}\).

    Show answer

    (b) \({\bf y }=\dst{\left[\begin{array}{c}e^{2t}-2e^{3t}+3e^{-t}\\2e^{3t}-9e^{-t}\\ e^{2t}-2e^{3t}+21e^{-t}\end{array}\right]}\) (c) \({\bf y}= \dst{{1\over6}\left[\begin{array}{ccc}4e^{2t}+3e^{3t}-e^{-t}&6e^{2t}-6e^{3t} &2e^{2t}-3e^{3t}+e^{-t} \\-3e^{3t}+3e^{-t}&6e^{3t}&3e^{3t}-3e^{-t}\\ 4e^{2t}+3e^{3t}-7e^{-t}&6e^{2t}-6e^{3t}&2e^{2t}-3e^{3t}+7e^{-t} \end{array}\right]}{\bf k}\)

12. Repeat Exercise 11 with

    \begin{eqnarray*} A&=&\threebythree 0 2 2 2 0 2 2 2 0, \\ {\bf y}_1&=&\left[\begin{array}{c}-e^{-2t} \\ 0 \\ e^{-2t} \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{c}-e^{-2t} \\ e^{-2t} \\ 0\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{4t} \\ e^{4t} \\ e^{4t}\end{array} \right], \quad {\bf k}=\left[\begin{array}{r} 0 \\-9 \\ 12\end{array} \right]. \end{eqnarray*}
    Show answer

    (b) \({\bf y }=\dst{{1\over3}\left[\begin{array}{c}-e^{-2t}+e^{4t}\\-10e^{-2t}+e^{4t}\\ 11e^{-2t}+e^{4t}\end{array}\right]}\)   (c) \({\bf y}= \dst{{1\over3}\left[\begin{array}{ccc}2e^{-2t}+e^{4t}&-e^{-2t}+e^{4t} &-e^{-2t}+e^{4t} \\-e^{-2t}+e^{4t}&2e^{-2t}+e^{4t}&-e^{-2t}+e^{4t}\\ -e^{-2t}+e^{4t}&-e^{-2t}+e^{4t}&2e^{-2t}+e^{4t} \end{array}\right]}{\bf k}\)

13. Repeat Exercise 11 with

    \begin{eqnarray*} A&=&\threebythree {-1} 2 3 0 1 6 0 0 {-2}, \\ {\bf y}_1&=&\left[\begin{array}{c} e^t \\ e^t \\ 0\end{array}\right], \quad {\bf y}_2=\left[\begin{array}{c} e^{-t} \\ 0 \\ 0\end{array}\right], \quad {\bf y}_3=\left[\begin{array}{c} e^{-2t} \\-2e^{-2t} \\ e^{-2t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 5 \\ 5 \\-1 \end{array}\right]. \end{eqnarray*}
    Show answer

    (b) \({\bf y }=\dst{\left[\begin{array}{c}3e^t+3e^{-t}-e^{-2t}\\3e^t+2e^{-2t}\\ -e^{-2t}\end{array}\right]}\)   (c) \({\bf y}=\dst{\left[\begin{array}{ccc}e^{-t}&e^t-e^{-t} &2e^t-3e^{-t}+e^{-2t}\\0&e^t&2e^t-2e^{-2t}\\ 0&0&e^{-2t}\end{array}\right]}{\bf k}\)

14. Suppose \(Y\) and \(Z\) are fundamental matrices for the \(n\times n\) system \({\bf y}'=A(t){\bf y}\). Then some of the four matrices \(YZ^{-1}\), \(Y^{-1}Z\), \(Z^{-1}Y\), \(Z Y^{-1}\) are necessarily constant. Identify them and prove that they are constant.

    Show answer

    \(YZ^{-1}\) and \(ZY^{-1}\)

15. Suppose the columns of an \(n\times n\) matrix \(Y\) are solutions of the \(n\times n\) system \({\bf y}'=A{\bf y}\) and \(C\) is an \(n \times n\) constant matrix.

    1. Show that the matrix \(Z=YC\) satisfies the differential equation \(Z'=AZ\).

    2. Show that \(Z\) is a fundamental matrix for \({\bf y}'=A(t){\bf y}\) if and only if \(C\) is invertible and \(Y\) is a fundamental matrix for \({\bf y}'=A(t){\bf y}\).

16. Suppose the \(n\times n\) matrix \(A=A(t)\) is continuous on \((a,b)\) and \(t_0\) is in \((a,b)\). For \(i=1\), \(2\), …, \(n\), let \({\bf y}_i\) be the solution of the initial value problem \({\bf y}_i'=A(t){\bf y}_i,\; {\bf y}_i(t_0)={\bf e}_i\), where

    \[ {\bf e}_1=\left[\begin{array}{c} 1\\0\\ \vdots\\0\end{array}\right],\quad {\bf e}_2=\left[\begin{array}{c} 0\\1\\ \vdots\\0\end{array}\right],\quad\cdots\quad {\bf e}_n=\left[\begin{array}{c} 0\\0\\ \vdots\\1\end{array}\right]; \]

    that is, the \(j\)th component of \({\bf e}_i\) is \(1\) if \(j=i\), or \(0\) if \(j\ne i\).

    1. Show that\(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) is a fundamental set of solutions of \({\bf y}'=A(t){\bf y}\) on \((a,b)\).

    2. Conclude from (a) and Exercise 15 that \({\bf y}'= A(t){\bf y}\) has infinitely many fundamental sets of solutions on \((a,b)\).

17. Show that \(Y\) is a fundamental matrix for the system \({\bf y}'=A(t){\bf y}\) if and only if \(Y^{-1}\) is a fundamental matrix for \({\bf y}'=- A^T(t){\bf y}\), where \(A^T\) denotes the transpose of \(A\). Hint: See Exercise 11.

18. Let \(Z\) be the fundamental matrix for the constant coefficient system \({\bf y}'=A{\bf y}\) such that \(Z(0)=I\).

    1. Show that \(Z(t)Z(s)=Z(t+s)\) for all \(s\) and \(t\). Hint: For fixed \(s\) let \(\Gamma_1(t)=Z(t)Z(s)\) and \(\Gamma_2(t)=Z(t+s)\). Show that \(\Gamma_1\) and \(\Gamma_2\) are both solutions of the matrix initial value problem \(\Gamma'=A\Gamma,\quad\Gamma(0)=Z(s)\). Then conclude from Theorem 10.2.1 that \(\Gamma_1=\Gamma_2\).

    2. Show that \((Z(t))^{-1}=Z(-t)\).

    3. The matrix \(Z\) defined above is sometimes denoted by \(e^{tA}\). Discuss the motivation for this notation.