Preface
This section is devoted to fundamental matrices for linear differential equations.
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Introduction to Linear Algebra with Mathematica
Glossary
Fundamental Matrices for Variable Coefficient Equations
A fundamental matrix of a system of n homogeneous linear ordinary differential equations
\[
\dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) , \qquad {\bf x}(t) =
\begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \quad {\bf
P}(t) = \begin{pmatrix} p_{11} (t) & p_{12} (t) & \cdots & p_{1n} (t)
\\ p_{21} (t) & p_{22} (t) & \cdots & p_{2n} (t) \\ \vdots & \vdots &
\ddots & \vdots \\ p_{1n} (t) & p_{2n} (t) & \cdots & p_{nn} (t)
\end{pmatrix} ,
\]
is any nonsingular (its determinant is not zero for all t)
matrix-function Ψ(t) that satisfies the matrix
differential equation:
\[
\dot{\bf \Psi} (t) = {\bf P}(t)\,{\bf \Psi}(t) \qquad \mbox{or} \qquad \frac{\text d}{{\text d}t}\, {\bf \Psi} (t) = {\bf P}(t)\,{\bf \Psi}(t) .
\]
Here dot stands for the derivative with respect to time
variable t. In other words, a fundamental matrix has n linearly
independent columns, each of them is a solution of the homogeneous
vector equation \( \dot{\bf x} (t) = {\bf
P}(t)\,{\bf x}(t) . \) Once a fundamental matrix is
determined, every solution to the system can be written as
\( {\bf x} (t) = {\bf
\Psi}(t)\,{\bf c} , \) for some constant vector c
(written as a column vector of height n). A product of a
fundamental matrix and a nonsingular constant matrix is again a
fundamental matrix. Therefore, a fundamental matrix is not unique.
Theorem 1:
If X(t) is a solution of the n × n
matrix differential equation \( \dot{\bf X} (t) = {\bf
P}(t)\,{\bf X}(t) , \) then for any constant
column-vector c, the n-vector u
= X(t) c is a solution of the vector equation
\( \dot{\bf x} (t) = {\bf
P}(t)\,{\bf x}(t) . \)
Theorem 2:
If an n × n matrix P(t) has
continuous entries on an open interval, then the vector differential
equation \( \dot{\bf x} (t) = {\bf P}(t)\,{\bf
x}(t) \) has an n × n fundamental
matrix \( {\bf X} (t) = \left\{ {\bf x}_1 (t) ,
{\bf x}_2 (t) , \ldots , {\bf x}_n (t) \right\} \) on the
same interval. Every solution x(t) to this system can
be written as a linear combination of the column vectors of the fundamental
matrix in a unique way:
\[
{\bf x} (t) = c_1 {\bf x}_1 (t) + c_2 {\bf x}_2 (t) + \cdots + c_n
{\bf x}_n (t) \qquad\mbox{or in matrix form} \quad {\bf x} (t) = {\bf
X} (t)\, {\bf c}
\]
for appropriate constants c1, c2,
... , cn, where \( {\bf c} =
\left\langle c_1 , c_2 , \ldots , c_n \right\rangle^{\mathrm T}
\) is a column vector of these constants.
The above representation of a solutions as a linear combination of linearly independent function-vectors is referred to as the general solution to the homogeneous vector differential equation \( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) . \)
Theorem 3:
The general solution of a nonhomogeneous linear vector equation
\( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) + {\bf
f} (t) \) is the sum of the general solution of the complement
homogeneous equation
\( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t)
\) and a particular solution of the inhomogeneous
equation. That is, every solution to
\( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) + {\bf
f} (t) \) is of the form
\[
{\bf x} (t) = c_1 {\bf x}_1 (t) + c_2 {\bf x}_2 (t) + \cdots + c_n
{\bf x}_n (t) + {\bf x}_p (t)
\]
for some constants c1, c2,
... , cn, where
\[
{\bf x}_h (t) = c_1 {\bf x}_1 (t) + c_2 {\bf x}_2 (t) + \cdots + c_n
{\bf x}_n (t)
\]
is the general solution of the homogeneous linear equation
\( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t)
\) and xp (t) is a particular
solution of the nonhomogeneosu equation
\( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) + {\bf
f} (t) . \)
Theorem:
Superposition Principle for inhomogeneous equations
Let P(t) be an n × n matrix function
that is continuous on an interval [a,b], and
let x1(t) and x2(t)
be two vector solutions of the nonhomogeneous equations
\[
\dot{\bf x}_1 (t) = {\bf P}(t)\,{\bf x}_1 (t) + {\bf f}_1 (t) , \qquad
\dot{\bf x}_2 (t) = {\bf P}(t)\,{\bf x}_2 (t) + {\bf f}_2 (t) , \quad
t \in [a,b] ,
\]
respectively. Then their sum \( {\bf x} (t) = {\bf
x}_1 (t) + {\bf x}_2 (t) \)
Corollary:
The difference between any two solutions of the nonhomogeneous vector
equation \( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) + {\bf
f} (t) \) is a solution of the complementary homogeneous
equation
\( \dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) . \)
Example 1:
It is not hard to verify that the vector functions
\[
{\bf x}_1 (t) = \begin{bmatrix} 1 \\ t \end{bmatrix} , \qquad {\bf x}_2 (t) = \begin{bmatrix} t^2 \\ t \end{bmatrix}
\]
are two linearly independent solutions to the following homogeneous
vector differential equation
\[
\dot{\bf x} (t) = {\bf P} (t)\, {\bf x} (t) , \qquad {\bf P} (t) =
\frac{1}{t \left( 1 - t^2 \right)} \begin{bmatrix} -2t^2 & 2t \\ 0 & 1- t^2
\end{bmatrix} .
\]
Therefore, the corresponding fundamental matrix is
\[
{\bf X} (t) = \begin{bmatrix} 1 & t^2 \\ t & t \end{bmatrix}, \qquad
\det {\bf X} (t) = t - t^3 = t \left( 1 - t^2 \right) .
\]
■
The determinant \( W(t) = \det\,{\bf X}(t)
\) of a square matrix \( {\bf X}(t) =
\left[ {\bf x}_1 (t) , {\bf x}_2 (t) , \ldots , {\bf x}_n (t) \right]
\) formed from the set of n vector
functions x1, x2,
... , xn, is called the Wronskian of these
column vectors x1, x2,
... , xn.
Theorem 5:
[N. Abel]
Let P(t) be an n × n matrix function
with entries pij(t) (i,j = 1,2,
... ,n) that are continuous functions on some
interval. Let xk(t), k = 1,2,
... , n, be n solutions to the homogeneous vector
differential equation \( \dot{\bf x} (t) = {\bf
P}(t)\,{\bf x}(t) . \) Then the Wronskian of the set of
vector solutions is
\[
W(t) = \det {\bf X}(t) = W(t_0 ) \,\exp \left\{ \int_{t_0}^t \mbox{tr}\,{\bf
P}(t)\,{\text d}t \right\} ,
\]
with t0 being a point within an interval where the
trace tr P(t) = p11
+ p22 + ... + pnn is continuous.
■ Corollary 2:
Let x1(t), x2(t),
... , xn(t) be column solutions of the
homogeneous vector equation \( \dot{\bf x} (t) =
{\bf P}(t)\,{\bf x}(t) \) on some interval |a,b|, where
n × n matrix function P(t) is
continuous. Then the corresponding matrix \( {\bf
X}(t) = \left[ {\bf x}_1 (t) , {\bf x}_2 (t) , \ldots , {\bf x}_n
(t) \right] \) of these column vectors is either a singular
matrix for all t ∈ |a,b| or else nonsingular. In other
words, det X(t) is either identically zero or it never
vanishes on the interval |a,b|.
Corollary 3:
Let P(t) be an n × n matrix function
that is continuous on an interval |a,b|. If
\( \left\{ {\bf x}_1 (t) , {\bf x}_2 (t) , \ldots ,
{\bf x}_n (t) \right\} \) is a linearly independent set of
solutions to the homogeneous differential equation \(
\dot{\bf x} (t) = {\bf P}(t)\,{\bf x}(t) \) on |a,b|, then
the Wronskian
\[
W(t) = \det {\bf X}(t) = \det \left[ {\bf x}_1 (t) , {\bf x}_2 (t) , \ldots ,
{\bf x}_n (t) \right]
\]
is not zero at every point t in |a,b|.
■
Example 2:
The matrix
\[
{\bf P} (t) =
\frac{1}{t \left( 1 - t^2 \right)} \begin{bmatrix} -2t^2 & 2t \\ 0 & 1- t^2
\end{bmatrix} \qquad (t \ne 0,1,-1)
\]
has the trace tr \( {\bf P} = \frac{1 - 3\,t^2}{t
\left( 1 - t^2 \right)} . \) Integrating the latter, we get
\[
\int {\bf P} (t) \,{\text d} t = \int \frac{1 - 3\,t^2}{t
\left( 1 - t^2 \right)} \,{\text d} t = \ln \left( t- t^3 \right) .
\]
From Abel's theorem, it follows that the Wronskian is
\[
W(t) = C\, e^{\int \mbox{tr}\,{\bf P}(t)\,{\text d}t} = C\,
\left( t- t^3 \right) .
\]
On the other hand, direct calculations show that the Wronskian of the given
functions x1(t) and x2(t) is
\[
W(t) = \det \begin{bmatrix} 1& t^2 \\ t&t \end{bmatrix} = t - t^3 \ne 0 \quad
\mbox{for} \quad t \ne 0, 1, -1 .
\]
End of Example 2
■ Let us consider the initial value problem
\[
\frac{{\text d}{\bf x}}{{\text d}t} = {\bf P}(t)\, {\bf x} (t) , \qquad
{\bf x} (t_0 ) = {\bf x}_0 .
\]
The general solution of the homogeneous equation is
\[
{\bf x} (t) = {\bf X} (t)\, {\bf c} ,
\]
where \( {\bf c} = \left\langle c_1 , c_2 , \ldots , c_n
\right\rangle^{\mathrm T} \) is the column vector of arbitrary
constants. To satisfy the initial condition, we set
\[
{\bf X} (t_0 )\, {\bf c} = {\bf x}_0 \qquad\mbox{or} \qquad {\bf c} =
{\bf X}^{-1} (t_0 )\, {\bf x}_0 .
\]
Therefore, the solution to the initial value problem becomes
\[
{\bf x}(t) = {\bf \Phi} (t, t_0 )\,{\bf x}_0 = {\bf X} (t)\,{\bf X}^{-1}
(t_0 )\, {\bf x}_0 .
\]
The square matrix \( {\bf \Phi} (t, s) = {\bf X} (t)\,
{\bf X}^{-1} (s) \) is usually referred to as a
propagator matrix.
Theorem 6:
Let X(t) be a fundamental matrix for the homogeneous linear
system \( \dot{\bf x} = {\bf P}(t)\,{\bf x} (t) , \)
meaning that X(t) is a solution of the matrix equation
\( \dot{\bf X} = {\bf P}(t)\,{\bf X} (t) \) and
det X(t) ≠ 0. Then the unique solution of the initial value
problem
\[
\dot{\bf x}(t) = {\bf P}(t)\,{\bf x} (t) , \qquad {\bf x} (t_0 ) =
{\bf x}_0
\]
is given by \( {\bf x}(t) = {\bf \Phi} (t, t_0 )\,{\bf x}_0
. \)
Corollary 4:
For a fundamental matrix X(t), the propagator matrix
Φ(t, t0) is the unique solution of the
following matrix initial value problem
\[
\frac{\text d}{{\text d}t}\,{\bf \Phi} \left( t, t_0 \right) = {\bf P}(t)\,
{\bf \Phi} \left( t, t_0 \right) , \qquad {\bf \Phi} \left( t_0 , t_0 \right)
= {\bf I} ,
\]
where I is the identity matrix. Hence,
Φ(t, t0) is a fundamental matrix of the
homogeneous vector differential equation
\( \dot{\bf x} = {\bf P}(t)\,{\bf x} (t) . \)
Corollary 5:
Let X(t) and Y(t) be two fundamental matrices of
the homogeneous vector equation
\( \dot{\bf x} = {\bf P}(t)\,{\bf x} (t) . \)
Then there exists a nonsingular constant square matrix C such that
\( {\bf X} (t) = {\bf Y} (t)\, {\bf C} , \ \det{\bf C}
\ne 0 . \) This means that the solution space of the matrix equation
\( \dot{\bf X} = {\bf P}(t)\,{\bf X} (t) \) is 1.
■
Example 3:
Consider the initial value problem
\[
\dot{\bf x} (t) = {\bf P}(t)\,{\bf x} (t) , \quad {\bf x} (2) = {\bf x}_0
\qquad\mbox{where} \quad
{\bf P}(t) = \frac{1}{t \left( 1 - t^2 \right)} \begin{bmatrix} -2t^2 & 2t \\
0 & 1- t^2 \end{bmatrix} \quad {\bf x}_0 = \begin{bmatrix} 2 \\ 1
\end{bmatrix} .
\]
We know from the previous example, that a fundamental matrix for corresponding
homogeneous vector equation is
\[
{\bf X} (t) = \begin{bmatrix} 1 & t^2 \\ t & t \end{bmatrix} .
\]
Since
\[
{\bf X}^{-1} (t) = \frac{1}{t \left( 1 - t^2 \right)} \begin{bmatrix}
t & -t^2 \\ -t & 1 \end{bmatrix} \qquad \Longrightarrow \qquad
{\bf X}^{-1} (2) = \frac{1}{6} \begin{bmatrix} -2&4 \\ 2& -1 \end{bmatrix} ,
\]
we get the propagator matrix
\[
{\bf \Phi} (t,2) = {\bf X} (t) {\bf X}^{-1} (2) = \frac{1}{6} \begin{bmatrix}
2t^2 -2 & 4- t^2 \\ 0 & 3t \end{bmatrix} .
\]
Then the solution of the given initial value problem becomes
\[
{\bf x} (t) = {\bf \Phi} (t,2) \, {\bf x}_0 = \frac{1}{6} \begin{bmatrix}
2t^2 -2 & 4- t^2 \\ 0 & 3t \end{bmatrix} \begin{bmatrix} 2 \\ 1
\end{bmatrix} = \frac{1}{2} \begin{bmatrix} t^2 \\ t \end{bmatrix} .
\]
X[t_] = {{1, t^2}, {t, t}}
P[t_] = {{-((2 t^2)/(t - t^3)), (2 t)/(t - t^3)}, {0, 1/(t - t^3) - t^2/(t - t^3)}}
Simplify[D[X[t], t] - P[t].X[t]]
Inverse[{{1, t^2}, {t, t}}] /. t -> 2
X[t].{{-(1/3), 2/3}, {1/3, -(1/6)}}
Simplify[%]
P[t_] = {{-((2 t^2)/(t - t^3)), (2 t)/(t - t^3)}, {0, 1/(t - t^3) - t^2/(t - t^3)}}
Simplify[D[X[t], t] - P[t].X[t]]
Inverse[{{1, t^2}, {t, t}}] /. t -> 2
X[t].{{-(1/3), 2/3}, {1/3, -(1/6)}}
Simplify[%]
{{t^2/2}, {t/2}}
■
Exponential Matrices
Consider autonomous vector linear differential equation of the form
\[
\dot{\bf y} (t) = {\bf A}\, {\bf y} (t) ,
\]
where A is a square n × n matrix and
y(t) is an (n × 1)-column vector of n
unknown functions. Here we use dot to represent the derivative with respect
to t. A solution of the above equation is a curve in n-dimensional space; it is called an integral curve, a trajectory, a streamline, or an orbit. When the independent variable t is associated with time (which is usually the case), we can call a solution
y(t) the state of the system at time t. Since a
constant matrix A is continuous on any interval, all solutions of the
system \( \dot{\bf y} (t) = {\bf A} \, {\bf y} (t)
\) are determined on ( -∞ , ∞ ). Therefore, when
we speak of solutions to the vector equation
\( \dot{\bf y} (t) = {\bf A} \, {\bf y} (t) , \)
we consider solutions on the real axis.
Any fundamental matrix is a constant multiple of the exponential matrix:
\[
{\bf \Phi}(t) = e^{{\bf A}\,t} {\bf C}, \qquad \det{\bf C} \ne 0.
\]
The exponential matrix function is a unique solution of the following matrix initial value problem:
\begin{equation} \label{EqExp.1}
\frac{\text d}{{\text d}t} {\bf X}(t) = {\bf A}\,{\bf X}(t) , \qquad {\bf X}(0) = {\bf I} ,
\end{equation}
where I is the identity matrix.
With this in hand, the propagator is expressed as
\[
{\bf \Phi}(t,s) = e^{{\bf A}\,(t-s)} .
\]
Then the solution of the initial value problem
\[
\frac{\text d}{{\text d}t} {\bf y}(t) = {\bf A}\,{\bf y}(t) , \qquad {\bf y}(t_0) = {\bf c} ,
\]
is expressed through the propagator:
\[
{\bf y}(t) = {\bf \Phi}(t,t_0 ) {\bf c} = e^{{\bf A}\,(t-t_0 )} {\bf c} .
\]
Mathematica has a couple of options to determine a fundamental matrix. It has a build-in command MatrixExp[A t] that determined a fundamental matrix for any square matrix A. For a diagonalizable matrix A, another way to find the fundamental matrix is to use two lines approach:
{roots,vectors} = Eigensystem[A]
Phi[t_] = Transpose[Exp[roots t]*vectors]
Phi[t_] = Transpose[Exp[roots t]*vectors]
Example 4:
Consider a linear system of differential equations
\[
\dot{\bf y}(t) = {\bf A}\,{\bf y}(t), \qquad\mbox{where}\quad {\bf y}(t) =
\begin{bmatrix} y_1 (t) \\ y_2 (t) \end{bmatrix}, \quad {\bf A} =
\begin{bmatrix} 2&1 \\ 6&3 \end{bmatrix} .
\]
First, we find eigenvalues and eigenvectors:
Eigenvalues[{{2, 1}, {6, 3}}]
{5, 0}
Eigenvectors[{{2, 1}, {6, 3}}]
{{1, 3}, {-1, 2}}
We check with Mathematica that vectors v1 = [1, 3]
and v1 = [-1, 2] are eigenvectors corresponding eigenvalues
λ1 = 5 and λ1 = 0, respectively.
A = {{2, 1}, {6, 3}};
v1 = {1, 3};
v2 = {-1, 2};
A.v1 - 5*v1
v1 = {1, 3};
v2 = {-1, 2};
A.v1 - 5*v1
{0, 0}
A.v2
{0, 0}
Now we use Mathematica to determine the fundamental matrix. First, we define two linearly independent solutions:
lambda1=5;
lambda2=0;
y1[t_] = Exp[lambda1*t]*v1
Out[13]= {E^(5 t), 3 E^(5 t)}
y2[t_] = Exp[lambda2*t]*v2
Out[14]= {-1, 2}
The general solution:
y[t_] = c1*y1[t]+c2*y2[t]
Out[15]= {-c2 + c1 E^(5 t), 2 c2 + 3 c1 E^(5 t)}
( * check *)
Simplify[y'[t]-A.y[t]=={0,0}]
Simplify[y'[t]-A.y[t]=={0,0}]
Out[16]= True
To find the fundamental matrix:
W[t_]=Transpose[{y1[t],y2[t]}]
Out[17]= {{E^(5 t), -1}, {3 E^(5 t), 2}}
Det[W[t]]
Out[18]= 5 E^(5 t)
Simplify[W'[t]-A.W[t]==0,Trig->False]
Out[19]= {{0, 0}, {0, 0}} == 0
Simplify[W'[t] - A.W[t] == {{0, 0}, {0, 0}}]
Out[20]= True
End of Example 4
■
Example 5:
Let us consider a differential equation with diagonazable matrix:
\[
\frac{{\text d}{\bf y}}{{\text d}t} = {\bf A}\,{\bf y}, \qquad \mbox{with} \qquad {\bf A} = \begin{bmatrix} 3&2&4 \\ 2&0&2 \\ 4&2&3
\end{bmatrix} .
\]
First, we check its eigenvalues and corresponding eigenvectors:
A = {{3, 2, 4}, {2, 0, 2}, {4, 2, 3}};
Eigenvalues[A]
Eigenvalues[A]
Out[2]= {8, -1, -1}
Eigenvectors[A]
{{2, 1, 2}, {-1, 0, 1}, {-1, 2, 0}}
Therefore, the given matrix A has three linearly independent eigenvectors; therefore, it is diagonalizable and its minimal polynomial is
\[
\psi (\lambda ) = \left( \lambda -8 \right)\left( \lambda +1 \right)
\]
We can build the coresponding exponential matrix using Sylvester's formula:
\[
e^{{\bf A}\,t} = e^{8t} {\bf Z}_8 + e^{-t} {\bf Z}_{-1} ,
\]
where
\[
{\bf Z}_8 = \frac{1}{9} \left( {\bf A} + {\bf I} \right) = \frac{1}{9} \begin{bmatrix} 4&2&4 \\ 2&1&2 \\ 4&2&4
\end{bmatrix} , \qquad {\bf Z}_{-1} = -\frac{1}{9} \left( {\bf A} -8 {\bf I} \right) = \frac{1}{9}\begin{bmatrix} \phantom{-}5&-2&-4 \\ -2&\phantom{-}8&-2 \\ -4&-2&\phantom{-}4
\end{bmatrix} .
\]
A = {{3, 2, 4}, {2, 0, 2}, {4, 2, 3}};
Z8 = (A + IdentityMatrix[3])/9
Z1 = -(A - 8* IdentityMatrix[3])/9
We can also check our answer with a standard Mathematica command:
Z8 = (A + IdentityMatrix[3])/9
Z1 = -(A - 8* IdentityMatrix[3])/9
MatrixExp[A t]
{{1/9 E^-t (5 + 4 E^(9 t)), 2/9 E^-t (-1 + E^(9 t)),
4/9 E^-t (-1 + E^(9 t))}, {2/9 E^-t (-1 + E^(9 t)),
1/9 E^-t (8 + E^(9 t)),
2/9 E^-t (-1 + E^(9 t))}, {4/9 E^-t (-1 + E^(9 t)),
2/9 E^-t (-1 + E^(9 t)), 1/9 E^-t (5 + 4 E^(9 t))}}
Now we check that the exponential matrix is a solution of the matrix diferential equation \eqref{EqExp.1}. We check it in two ways. First we consider Sylvester's formula that leads to
\[
\frac{\text d}{{\text d}t} e^{{\bf A}\,t} = \frac{\text d}{{\text d}t} \left( e^{8t} {\bf Z}_8 + e^{-t} {\bf Z}_{-1} \right) = 8\, e^{8t} {\bf Z}_8 - e^{-t} {\bf Z}_{-1} .
\]
Therefore, it is sufficient to show the validity of the following equations:
\begin{align*}
8\,{\bf Z}_8 &= {\bf A}\, {\bf Z}_8 ,
\\
- {\bf Z}_{-1} &= {\bf A}\, {\bf Z}_{-1} .
\end{align*}
Indeed, Mathematica confirms
8*Z8 - A.Z8
A.Z1 + Z1
A.Z1 + Z1
{{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}
If you would like to use the standard Mathematica command, we need to differentiate the exponential function.
Dt[MatrixExp[A t], t];
Simplify[%]
Simplify[%]
Out[5]= {{1/9 E^-t (-5 + 32 E^(9 t)), 2/9 E^-t (1 + 8 E^(9 t)),
4/9 E^-t (1 + 8 E^(9 t))}, {2/9 E^-t (1 + 8 E^(9 t)),
8/9 E^-t (-1 + E^(9 t)),
2/9 E^-t (1 + 8 E^(9 t))}, {4/9 E^-t (1 + 8 E^(9 t)),
2/9 E^-t (1 + 8 E^(9 t)), 1/9 E^-t (-5 + 32 E^(9 t))}}
To check that the exponential matrix is the solution of the matrix differential equation \eqref{EqExp.1} directly:
4/9 E^-t (1 + 8 E^(9 t))}, {2/9 E^-t (1 + 8 E^(9 t)),
8/9 E^-t (-1 + E^(9 t)),
2/9 E^-t (1 + 8 E^(9 t))}, {4/9 E^-t (1 + 8 E^(9 t)),
2/9 E^-t (1 + 8 E^(9 t)), 1/9 E^-t (-5 + 32 E^(9 t))}}
Simplify[Dt[MatrixExp[A t], t] - A.MatrixExp[A t]]
Out[6]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}
Finally, we need to show that the exponential matrix satisfies the initial condition
Print[MatrixExp[A 0]]
Out[7]= {{1,0,0},{0,1,0},{0,0,1}}
Note that instead of Dt, we can use the partial derivative operator: D[function,t]The general solution:
CC := {c1, c2, c3} (* vector of arbitrary constants *)
(* note that the upper case letter C is prohibited to use *)
MatrixExp[A t].CC
(* note that the upper case letter C is prohibited to use *)
MatrixExp[A t].CC
Out[9]= {-c1 E^(2 t) (-1 + t) + c2 E^(2 t) t +
c3 E^(2 t) t, -(1/2) c1 E^(2 t) (-4 + t) t +
1/2 c3 E^(2 t) (-2 + t) t + 1/2 c2 E^(2 t) (2 - 2 t + t^2),
1/2 c1 E^(2 t) (-6 + t) t - 1/2 c2 E^(2 t) (-4 + t) t -
1/2 c3 E^(2 t) (-2 - 4 t + t^2)}
c3 E^(2 t) t, -(1/2) c1 E^(2 t) (-4 + t) t +
1/2 c3 E^(2 t) (-2 + t) t + 1/2 c2 E^(2 t) (2 - 2 t + t^2),
1/2 c1 E^(2 t) (-6 + t) t - 1/2 c2 E^(2 t) (-4 + t) t -
1/2 c3 E^(2 t) (-2 - 4 t + t^2)}
End of Example 5
■
Example 6:
We consider a matrix that has pure imaginary eigenvalues:
\[
{\bf A} = \begin{bmatrix} \phantom{-}0 & 1 \\ -1&0
\end{bmatrix} .
\]
A := {{0, 1}, {-1, 0}}
Eigenvalues[A]
Eigenvalues[A]
Out[2]= {I, -I}
Simplify[ComplexExpand[MatrixExp[A t]]]
Out[3]= {{Cos[t], Sin[t]}, {-Sin[t], Cos[t]}}
diag = DiagonalMatrix[{2, -1, 4}]
Out[4]= {{2, 0, 0}, {0, -1, 0}, {0, 0, 4}}
y[t_] = MatrixExp[diag t]
Out[5]= {{E^(2 t), 0, 0}, {0, E^-t, 0}, {0, 0, E^(4 t)}}
y[0]
Out[6]= {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}
y'[t] - diag.y[t]
Out[7]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}
DiagonalMatrix[{2, 3}, -1]
Out[8]= {{0, 0, 0}, {2, 0, 0}, {0, 3, 0}}
DiagonalMatrix[{2, 3}, 1] // MatrixForm
out[9]/MatrixForm=
{"0", "2", "0"},
{"0", "0", "3"},
{"0", "0", "0"}
{"0", "2", "0"},
{"0", "0", "3"},
{"0", "0", "0"}
End of Example 6
■
- Chi-Tsong Chen (1998). Linear System Theory and Design (3rd ed.). New York: Oxford University Press,. ISBN 978-0195117776.
- Vladimir Dobrushkin, Applied Differential Equations. The Primary Course, CRC Press, 2015; http://www.crcpress.com/product/isbn/9781439851043.
- Devi, J.V., Deo, S.G., Khandeparkar, R., Linear Algebra to Differential Equations, 2021, CRC Press, ISBN 9780815361466
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