Section 4.3 Linear systems of ODEs
Objectives
After this section, you will be able to:
-
Use proper terminology when discussing linear systems of differential equations and their solutions,
-
Determine whether a set of functions is linearly independent, and
-
Understand how the theory of non-homogeneous linear systems relates to the theory of non-linear equations.
In order to get into the details of how to talk about and deal with linear systems of differential equations, we first need to talk about matrix- or vector-valued functions. Such a function is just a matrix or vector whose entries depend on some variable. If \(t\) is the independent variable, we write a vector-valued function \(\vec{x}(t)\) as
\begin{equation*}
\vec{x}(t) = \begin{bmatrix}
x_1(t) \\
x_2(t) \\
\vdots \\
x_n(t)
\end{bmatrix} .
\end{equation*}
Similarly a matrix-valued function \(A(t)\) is
\begin{equation*}
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} .
\end{equation*}
As long as the addition of vectors is defined, we can add vector-valued functions, and as long as the addition and multiplication of matrices are defined (like if they are the right size) we can multiply matrix-valued functions. In addition, the derivative
\(A'(t)\) or
\(\frac{dA}{dt}\) is just the matrix-valued function whose
\(ij^{\text{th}}\) entry is
\(a_{ij}'(t)\text{.}\) We used this idea previously when talking about how to write first order systems from higher order equations in
Section 4.1.
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 \(c\) a scalar and let \(C\) be a constant matrix. Then
\begin{align*}
\bigl(A(t)+B(t)\bigr)' \amp = A'(t) + B'(t) \\
\bigl(A(t)B(t)\bigr)' \amp = A'(t)B(t) + A(t)B'(t) \\
\bigl(cA(t)\bigr)' \amp = cA'(t) \\
\bigl(CA(t)\bigr)' \amp = CA'(t) \\
\bigl(A(t)\,C\bigr)' \amp = A'(t)\,C\text{.}
\end{align*}
Note the order of the multiplication in the last two expressions because matrix multiplication is not commutative.
A first order linear system of ODEs is a system that can be written as the vector equation
\begin{equation*}
{\vec{x}}'(t) = P(t)\vec{x}(t) + \vec{f}(t),
\end{equation*}
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 \({\vec{x}}' = P\vec{x} + \vec{f}\text{.}\) A solution of the system is a vector-valued function \(\vec{x}\) satisfying the vector equation.
For example, the equations
\begin{align*}
x_1' \amp= 2t x_1 + e^t x_2 + t^2 \\
x_2' \amp= \frac{x_1}{t} -x_2 + e^t
\end{align*}
can be written as
\begin{equation*}
{\vec{x}}' =
\begin{bmatrix}
2t & e^t \\
\nicefrac{1}{t} & -1
\end{bmatrix}
\vec{x}
+
\begin{bmatrix}
t^2 \\
e^t
\end{bmatrix} .
\end{equation*}
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
\(t\text{.}\)
When
\(\vec{f} = \vec{0}\) (the zero vector), then we say the system is
homogeneous. For homogeneous linear systems we have the principle of superposition, just like for single homogeneous equations.
Theorem 4.3.1. Superposition.
Let \({\vec{x}}' = P\vec{x}\) be a linear homogeneous system of ODEs. Suppose that \(\vec{x}_1,\vec{x}_2,\ldots,\vec{x}_n\) are \(n\) solutions of the equation and \(c_1,c_2,\ldots,c_n\) are any constants, then
\begin{equation}
\vec{x} = c_1 \vec{x}_1 + c_2 \vec{x}_2 + \cdots + c_n \vec{x}_n ,\tag{4.3.1}
\end{equation}
is also a solution. Furthermore, if this is a system of
\(n\) equations (
\(P\) is
\(n\times n\)), and
\(\vec{x}_1,\vec{x}_2,\ldots,\vec{x}_n\) are linearly independent, then every solution
\(\vec{x}\) can be written as
(4.3.1).
Linear independence for vector-valued functions is the same idea as for normal functions. The vector-valued functions \(\vec{x}_1,\vec{x}_2,\ldots,\vec{x}_n\) are linearly independent if the only way to satisfy the equation
\begin{equation*}
c_1 \vec{x}_1 + c_2 \vec{x}_2 + \cdots + c_n \vec{x}_n = \vec{0}
\end{equation*}
is by choosing the parameters \(c_1 = c_2 = \cdots = c_n = 0\text{,}\) where the equation must hold for all \(t\text{.}\)
Example 4.3.2.
Determine if the sets
\(S_1 = \Bigl\{ \vec{x}_1 = \Bigl[ \begin{smallmatrix} t^2 \\ t \end{smallmatrix} \Bigr],\ \vec{x}_2 = \Bigl[ \begin{smallmatrix} 0 \\ 1+t \end{smallmatrix} \Bigr],\ \vec{x}_3 = \Bigl[ \begin{smallmatrix} -t^2 \\ 1 \end{smallmatrix} \Bigr] \Bigr\}\) and
\(S_2 = \Bigl\{ \vec{x}_1 = \Bigl[ \begin{smallmatrix} t^2 \\ t \end{smallmatrix} \Bigr],\ \vec{x}_2 = \Bigl[ \begin{smallmatrix} 0 \\ t \end{smallmatrix} \Bigr],\ \vec{x}_3 = \Bigl[ \begin{smallmatrix} -t^2 \\ 1 \end{smallmatrix}
\Bigr]\Bigr\}\) are linearly independent.
Solution.
The vector functions in
\(S_1\) are linearly dependent because
\(\vec{x}_1 + \vec{x}_3 = \vec{x}_2\text{,}\) and this holds for all
\(t\text{.}\) So
\(c_1 =
1\text{,}\) \(c_2 = -1\text{,}\) and
\(c_3 = 1\) above will work.
On the other hand, the vector functions in \(S_2\) are linearly independent, even though this is only a slight change from \(S_1\text{.}\) 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 \(t\text{.}\) We get that
\begin{equation*}
c_1 \vec{x}_1 + c_2 \vec{x}_2 + c_3 \vec{x}_3
=
\begin{bmatrix}
c_1 t^2 - c_3 t^2
\\
c_1 t + c_2 t + c_3
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\end{bmatrix} .
\end{equation*}
In other words \(c_1 t^2 - c_3 t^2 = 0\) and \(c_1 t + c_2 t + c_3 = 0\text{.}\) If we set \(t = 0\text{,}\) then the second equation becomes \(c_3 = 0\text{.}\) But then the first equation becomes \(c_1 t^2 = 0\) for all \(t\) and so \(c_1 = 0\text{.}\) Thus the second equation is just \(c_2 t = 0\text{,}\) which means \(c_2 = 0\text{.}\) So \(c_1 = c_2 = c_3 = 0\) is the only solution and \(\vec{x}_1\text{,}\) \(\vec{x}_2\text{,}\) and \(\vec{x}_3\) are linearly independent.
The linear combination \(c_1 \vec{x}_1 + c_2 \vec{x}_2 + \cdots + c_n
\vec{x}_n\) could also be written as
\begin{equation*}
X(t)\,\vec{c} ,
\end{equation*}
where
\(X(t)\) is the matrix with columns
\(\vec{x}_1, \vec{x}_2, \ldots, \vec{x}_n\text{,}\) and
\(\vec{c}\) is the column vector with entries
\(c_1, c_2, \ldots, c_n\text{.}\) This is similar to the way that we could write linear combinations of vectors by putting them into a matrix, including how we talked about rank in
Section 3.6. Assuming that
\(\vec{x}_1,\vec{x}_2,\ldots,\vec{x}_n\) are linearly independent and solutions to a given system of differential equations, the matrix-valued function
\(X(t)\) is called a
fundamental matrix, or a
fundamental matrix solution.
To solve nonhomogeneous first order linear systems, we use the same technique as we applied to solve single linear nonhomogeneous equations.
Theorem 4.3.3.
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
\begin{equation*}
\vec{x} = \vec{x}_c + \vec{x}_p ,
\end{equation*}
where \(\vec{x}_c\) is a solution to the associated homogeneous equation (\({\vec{x}}' = P\vec{x}\)).
The procedure for systems is the same as for single equations. We find a particular solution to the nonhomogeneous equation, then we find the general solution to the associated homogeneous equation, and finally we add the two together.
Alright, suppose you have found the general solution of \({\vec{x}}' = P\vec{x} + \vec{f}\text{.}\) Next suppose you are given an initial condition of the form
\begin{equation*}
\vec{x}(t_0) = \vec{b}
\end{equation*}
for some fixed \(t_0\) and a constant vector \(\vec{b}\text{.}\) Let \(X(t)\) be a fundamental matrix solution of the associated homogeneous equation (i.e. columns of \(X(t)\) are solutions). The general solution can be written as
\begin{equation*}
\vec{x}(t) = X(t)\,\vec{c} + \vec{x}_p(t).
\end{equation*}
We are seeking a vector \(\vec{c}\) such that
\begin{equation*}
\vec{b} = \vec{x}(t_0) = X(t_0)\,\vec{c} + \vec{x}_p(t_0).
\end{equation*}
In other words, we are solving for \(\vec{c}\) in the nonhomogeneous system of linear equations
\begin{equation*}
X(t_0)\,\vec{c} = \vec{b} - \vec{x}_p(t_0) .
\end{equation*}
Example 4.3.4.
\begin{align*}
x_1' \amp = x_1 \\
x_2' \amp = x_1 - x_2
\end{align*}
with initial conditions \(x_1(0) = 1\text{,}\) \(x_2(0) = 2\text{.}\) Let us consider this problem in the language of this section.
The system is homogeneous, so \(\vec{f}(t) = \vec{0}\text{.}\) We write the system and the initial conditions as
\begin{equation*}
{\vec{x}}'
=
\begin{bmatrix}
1 & 0 \\
1 & -1
\end{bmatrix}
\vec{x} ,
\qquad
\vec{x}(0) =
\begin{bmatrix}
1 \\
2
\end{bmatrix} .
\end{equation*}
We found the general solution is \(x_1 = c_1 e^t\) and \(x_2 = \frac{c_1}{2}e^{t} + c_2e^{-t}\text{.}\) Letting \(c_1=1\) and \(c_2=0\text{,}\) we obtain the solution \(\left[ \begin{smallmatrix} e^t \\ (1/2) e^t \end{smallmatrix} \right]\text{.}\) Letting \(c_1=0\) and \(c_2=1\text{,}\) we obtain \(\left[ \begin{smallmatrix} 0 \\ e^{-t} \end{smallmatrix} \right]\text{.}\) These two solutions are linearly independent, as can be seen by setting \(t=0\text{,}\) and noting that the resulting constant vectors are linearly independent. In matrix notation, a fundamental matrix solution is, therefore,
\begin{equation*}
X(t) =
\begin{bmatrix}
e^t & 0 \\
\frac{1}{2} e^t & e^{-t}
\end{bmatrix} .
\end{equation*}
To solve the initial value problem we solve for \(\vec{c}\) in the equation
\begin{equation*}
X(0)\,\vec{c} = \vec{b} ,
\end{equation*}
or in other words,
\begin{equation*}
\begin{bmatrix}
1 & 0 \\
\frac{1}{2} & 1
\end{bmatrix}
\vec{c} =
\begin{bmatrix}
1 \\ 2
\end{bmatrix} .
\end{equation*}
A single elementary row operation shows \(\vec{c} =
\left[ \begin{smallmatrix} 1 \\ 3/2 \end{smallmatrix} \right]\text{.}\) Our solution is
\begin{equation*}
\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} .
\end{equation*}
This new solution agrees with our previous solution from
Section 4.1.
Exercises Exercises
1.
Write the system
\(x_1' = 2 x_1 - 3t x_2 + \sin t\text{,}\) \(x_2' = e^t x_1 + 3 x_2 + \cos t\) in the form
\({\vec{x}}' = P(t) \vec{x} + \vec{f}(t)\text{.}\)
Answer.
\(P(t) = \left[\begin{smallmatrix} 2 & -3t \\ e^t & 3 \end{smallmatrix}\right]\text{,}\) \(\vec{f}(t) = \left[\begin{smallmatrix} \sin(t)\\ \cos(t) \end{smallmatrix}\right]\)
2.
Write
\(x'=3x-y+e^t\text{,}\) \(y'=tx\) in matrix notation.
Answer.
\(\left[ \begin{smallmatrix}
x \\ y
\end{smallmatrix}\right] '
=
\left[ \begin{smallmatrix}
3 &-1 \\
t &0
\end{smallmatrix}\right]
\left[ \begin{smallmatrix}
x \\ y
\end{smallmatrix}\right]
+
\left[ \begin{smallmatrix}
e^{t} \\ 0
\end{smallmatrix}\right]\)
3.
Consider the third order differential equation
\begin{equation*}
y''' + (y'+1)^2 = e^y + \sin(t+1).
\end{equation*}
Convert this to a first order system and simplify as much as possible. Can you write this in the form \(\vec{x}' = A\vec{x} + \vec{f}\text{?}\) Why or why not?
Answer.
No, it is non-linear.
\(\vec{u}' = \left[\begin{smallmatrix} u_2 \\ u_3 \\ e^{u_1} - (u_2 + 1)^2 + \sin(t+1) \end{smallmatrix}\right]\)
4.
This exercise will discuss the conversion between vector-valued and component-based solutions to linear systems.
(a)
Verify that the system
\({\vec{x}}' =
\left[ \begin{smallmatrix}
1 & 3 \\ 3 & 1
\end{smallmatrix} \right] \vec{x}\) has the two solutions
\(\left[ \begin{smallmatrix}
1 \\ 1
\end{smallmatrix} \right] e^{4t}\) and
\(\left[ \begin{smallmatrix}
1 \\ -1
\end{smallmatrix} \right] e^{-2t}\text{.}\)
(b)
Write down the general solution.
Answer.
\(C_1\left[\begin{smallmatrix} 1 \\ 1 \end{smallmatrix}\right]e^{4t} + C_2\left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right]e^{-2t}\)
(c)
Write down the general solution in the form
\(x_1 = ?\text{,}\) \(x_2 = ?\) (i.e. write down a formula for each element of the solution).
Answer.
\(x_1(t) = C_1e^{4t} + C_2e^{-2t},\ x_2(t) = C_1e^{4t} - C_2e^{-2t}\)
5.
Verify that
\(\left[ \begin{smallmatrix}
1 \\ 1
\end{smallmatrix} \right] e^{t}\) and
\(\left[ \begin{smallmatrix}
1 \\ -1
\end{smallmatrix} \right] e^{t}\) are linearly independent. Hint: Just plug in
\(t=0\text{.}\)
6.
Are
\(\left[ \begin{smallmatrix}
e^{2t} \\ e^t
\end{smallmatrix}\right]\) and
\(\left[ \begin{smallmatrix}
e^{t} \\ e^{2t}
\end{smallmatrix}\right]\) linearly independent? Justify.
7.
Verify that
\(\left[ \begin{smallmatrix}
1 \\ 1 \\ 0
\end{smallmatrix} \right] e^{t}\) and
\(\left[ \begin{smallmatrix}
1 \\ -1 \\ 1
\end{smallmatrix} \right] e^{t}\) and
\(\left[ \begin{smallmatrix}
1 \\ -1 \\ 1
\end{smallmatrix} \right] e^{2t}\) are linearly independent. Hint: You must be a bit more tricky than in the previous exercises.
Answer.
Yes. Write this out as three equations that all must equal zero, see what combining the first two gets you, then go from there.
8.
Are
\(\left[ \begin{smallmatrix}
\cosh(t) \\ 1
\end{smallmatrix}\right]\text{,}\) \(\left[ \begin{smallmatrix}
e^{t} \\ 1
\end{smallmatrix}\right]\text{,}\) and
\(\left[ \begin{smallmatrix}
e^{-t} \\ 1
\end{smallmatrix}\right]\) linearly independent? Justify.
Answer.
No.
\(2 \left[ \begin{smallmatrix}
\cosh(t) \\ 1
\end{smallmatrix}\right] -
\left[ \begin{smallmatrix}
e^{t} \\ 1
\end{smallmatrix}\right]
-
\left[ \begin{smallmatrix}
e^{-t} \\ 1
\end{smallmatrix}\right] = \vec{0}\)
9.
Verify that
\(\left[ \begin{smallmatrix}
t \\ t^2
\end{smallmatrix} \right]\) and
\(\left[ \begin{smallmatrix}
t^3 \\ t^4
\end{smallmatrix} \right]\) are linearly independent.
Answer.
Yes, write out the equations and see when they can be zero.
10.
Take the system
\(x_1' + x_2' = x_1\text{,}\) \(x_1' - x_2' = x_2\text{.}\)
(a)
Write it in the form
\(A {\vec{x}}' = B \vec{x}\) for matrices
\(A\) and
\(B\text{.}\)
Answer.
\(A = \left[\begin{smallmatrix} 1 & 1 \\ 1 & -1 \end{smallmatrix}\right]\text{,}\) \(B = \left[\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix}\right]\)
(b)
Compute
\(A^{-1}\) and use that to write the system in the form
\({\vec{x}}' = P \vec{x}\text{.}\)
Answer.
\(P = \left[\begin{smallmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{smallmatrix}\right]\)
11.
This exercise analyzes a system that is more easily solved as separate equations and what the solution looks like in matrix form.
(a)
Write
\(x_1'=2tx_2\text{,}\) \(x_2'=2tx_2\) in matrix notation.
Answer.
\(\vec{x}\,'
=
\left[ \begin{smallmatrix}
0 &2t \\
0 &2t
\end{smallmatrix}\right]
\vec{x}\)
(b)
Solve and write the solution in matrix notation.
Answer.
\(\vec{x}
=
\left[ \begin{smallmatrix}
C_2 e^{t^2} + C_1 \\
C_2 e^{t^2}
\end{smallmatrix}\right]\)
