INTRODUCTION

Matrices can be used in an entirely different manner to solve a system of linear first-order differential equations. Recall that the simple linear first-order differential equation x′ = ax, where a is a constant, has the general solution x = ce^at. It seems natural, then, to ask whether we can define a matrix exponential e^At, where A is a matrix of constants, so that e^At is a solution of the system X′ = AX.

Homogeneous Systems

We shall now see that it is possible to define a matrix exponential e^At so that the homogeneous system X′ = AX, where A is an n × n matrix of constants, has a solution

X = e^AtC.(1)

Since C is to be an n × 1 column matrix of arbitrary constants, we want e^At to be an n × n matrix. While the complete development of the meaning and theory of the matrix exponential would require a thorough knowledge of matrix algebra, one way of defining e^At is inspired by the power series representation of the scalar exponential function e^at:

(2)

The series in (2) converges for all t. Using this series, with 1 replaced by the identity matrix I and the constant a replaced with an n × n matrix A of constants, we arrive at a definition for the n × n matrix e^At.

It can be shown that the series given in (3) converges to an n × n matrix for every value of t. Also, in (3), A⁰ = I, A² = AA, A³ = A(A²), and so on.

EXAMPLE 1 Matrix Exponential Using (3)

Compute e^At for the matrix

SOLUTION

From the various powers

we see from (3) that

In view of (2) and the identifications a = 2 and a = 3, the power series in the first and second rows of the last matrix represent, respectively, e^2t and e^3t and so we have

≡

The matrix in Example 1 is an example of a 2 × 2 diagonal matrix. In general, an n × n matrix A is a diagonal matrix if all its entries off the main diagonal are zero; that is,

.

Hence if A is any n × n diagonal matrix it follows from Example 1 that

.

Derivative of e^At

The derivative of the matrix exponential e^At is analogous to that of the scalar exponential; that is, d/dt e^at = ae^at. To justify

(4)

we differentiate (3) term-by-term:

Because of (4), we can now prove that (1) is a solution of X′ = AX for every n × 1 vector C of constants:

X′ = e^AtC = Ae^AtC = A(e^AtC) = AX.(5)

e^At Is a Fundamental Matrix

If we denote the matrix exponential e^At by the symbol Ψ(t), then (4) is equivalent to the matrix differential equation Ψ′(t) = AΨ (see (3) of Section 10.4). In addition, it follows immediately from Definition 10.5.1 that Ψ(0) = e^A0 = I and so det Ψ(0) ≠ 0. It turns out that these two properties are sufficient for us to conclude that Ψ(t) is a fundamental matrix of the system X′ = AX.

Nonhomogeneous Systems

We have seen in (4) of Section 2.3 that the general solution of the single linear first-order differential equation x′ = ax + f(t), where a is a constant, can be expressed as

x = x_c + x_p = ce^at + e^at f(s) ds.

For a nonhomogeneous system of linear first-order differential equations, it can be shown that the general solution of X′ = AX + F(t), where A is an n × n matrix of constants, is

X = X_c + X_p = e^AtC + e^At F(s) ds.(6)

Since the matrix exponential e^At is a fundamental matrix, it is always nonsingular and e^−As = (e^As)⁻¹. Note that e^−As can be obtained from e^At by replacing t by −s.

Computation of e^At

The definition of e^At given in (3) can, of course, always be used to compute e^At. However, the practical utility of (3) is limited by the fact that the entries in e^At are power series in t. With one’s natural desire to work with simple and familiar things, we then try to recognize whether these series define a closed form function. See Problems 1–4 in Exercises 10.5. Fortunately there are many alternative ways of computing e^At. We sketch two of these methods in the discussion that follows.

Using the Laplace Transform

We saw in (5) that X = e^At is a solution of X′ = AX. Indeed, since e^A0 = I, X = e^At is a solution of the initial-value problem

X′ = AX, X(0) = I.(7)

If x(s) = {X(t)} = {e^At}, then the Laplace transform of (7) is

sx(s) − X(0) = Ax(s) or (sI − A)x(s) = I.

Multiplying the last equation by (sI − A)⁻¹ implies x(s) = (sI − A)⁻¹I = (sI − A)⁻¹. In other words, {e^At} = (sI − A)⁻¹ or

e^At = ^–1{(sI – A)^–1}.(8)

EXAMPLE 2 Matrix Exponential

Use the Laplace transform to compute e^At for A = .

SOLUTION

First we compute the matrix sI − A and then find its inverse:

Then we decompose the entries of the last matrix into partial fractions:

(9)

Taking the inverse Laplace transform of (9) gives the desired result,

≡

Using Powers A^m

In Section 8.9, we developed a method for computing an arbitrary power A^k, k a nonnegative integer, of an n × n matrix A. Recall from Section 8.9 that we can write

(10)

where the coefficients c_j are the same in each and the last expression is valid for the eigenvalues λ₁, λ₂, … , λ_n of A. We assume here that the eigenvalues of A are distinct. By setting λ = λ₁, λ₂, … , λ_n in the second expression in (10), we were able to find the c_j in the first expression by solving n equations in n unknowns. It will be convenient in the development that follows to emphasize the fact that the coefficients c_j in (10) depend on the power k by replacing c_j by c_j (k). From (3) and (2), we have

(11)

We next use (10) in (11) to replace A^k and λ^k as finite sums followed by an interchange of the order of summations

(12)

(13)

where b_j (t) = (t^k/k!)c_j (k). Analogous to how we used the eigenvalues of A in (10) to determine the c_j, we again use the eigenvalues, but this time in the finite sum (13) to determine a system of equations to determine the b_j; these coefficients, in turn, are used in (12) to determine e^At.

EXAMPLE 3 Matrix Exponential

Use of Computers

For those willing to momentarily trade understanding for speed of solution, e^At can be computed with the aid of computer software; for example, in Mathematica, the function MatrixExp[A t] computes the matrix exponential for a square matrix A; in Maple, the command is exponential(A, t); in MATLAB the function is expm(At). See Problems 27 and 28 in Exercises 10.5.

10.5 Exercises Answers to selected odd-numbered problems begin on page ANS-29.

In Problems 1 and 2, use (3) to compute e^At and e^−At.

1. A =
2. A =

In Problems 3 and 4, use (3) to compute e^At.

3. A =
4. A =

In Problems 5–8, use (1) and the results in Problems 1–4 to find the general solution of the given system.

5. X′ = X
6. X′ = X
7. X′ = X
8. X′ = X

In Problems 9–12, use (6) to find the general solution of the given system.

9. X′ = X +
10. X′ = X +
11. X′ = X +
12. X′ = X +
13. Solve the system in Problem 7 subject to the initial condition

    X(0) = .

14. Solve the system in Problem 9 subject to the initial condition

    X(0) = .

In Problems 15–18, use the method of Example 2 to compute e^At for the coefficient matrix. Use (1) to find the general solution of the given system.

15. X′ = X
16. X′ = X
17. X′ = X
18. X′ = X

In Problems 19–22, use the method of Example 3 to compute e^At for the coefficient matrix. Use (1) to find the general solution of the given system.

19. X′ = X
20. X′ = X
21. X′ = X
22. X′ = X
23. If the matrix A can be diagonalized, then P⁻¹AP = D or A = PDP⁻¹. Use this last result and (3) to show that e^At = Pe^Dt P⁻¹.
24. Use D = and (3) to show that

    

In Problems 25 and 26, use the results of Problems 23 and 24 to solve the given system.

25. X′ = X
26. X′ = X

Computer Lab Assignments

27. 1. Use (1) to find the general solution of X′ = X. Use a CAS to find e^At. Then use the computer to find eigenvalues and eigenvectors of the coefficient matrix A = and form the general solution in the manner of Section 10.2. Finally, reconcile the two forms of the general solution of the system.
    2. Use (1) to find the general solution of X′ = X. Use a CAS to find e^At. In the case of complex output, utilize the software to do the simplification; for example, in Mathematica, if m=MatrixExp[A t] has complex entries, then try the command Simplify [ComplexExpand[m]].
28. Use (1) to find the general solution of

    X′ = X.

    Use a CAS to find e^At.

Discussion Problems

29. Reread the discussion leading to the result given in (8). Does the matrix sI−A always have an inverse? Discuss.
30. In Exercises 8.9 we saw that a nonzero n × n matrix A is nilpotent if m is the smallest positive integer such that A^m = 0. Verify that A = is nilpotent. Discuss why it is relatively easy to compute e^At when A is nilpotent. Compute e^At for the given matrix and then use (2) to solve the system X′ = AX.