We now extend our earlier work to inhomogeneous linear systems of the form x0 = Ax + b(t),
where A is an n×n constant matrix and b(t) is a vector of continuous functions. Note that b(t) is allowed to involve the independent variable t but none of the dependent variables. To solve such systems, we will extend the variation of parameters method covered in introductory differential equations courses.
Variation of parameters. Let us briefly review how to solve first-order, linear, homogeneous ODEs with only one dependent variable. Specifically, consider the ODE x0(t) = a(t)x(t) + b(t), where x(t) is scalar-valued, not vector-valued2. The first step is to write the ODE in the form
x0(t) − a(t)x(t) = b(t). (2.14)
We cannot simply integrate both sides with respect to t, because this would lead to an integral equation
x(t) − Z
a(t)x(t) dt = Z
b(t) dt,
which does not help us solve for x(t). Instead, the trick is to multiply Equation (2.14) by an integrating factor
e−Ra(t) dt (2.15)
to obtain
e−Ra(t) dt[x0− a(t)x] = e−Ra(t) dtb(t).
Equivalently,
d dt
n
e−Ra(t) dtx(t) o
= e−Ra(t) dtb(t).
Now we are in a position to integrate both sides with respect to t, after which we may (hopefully) solve for x(t).
Example. Solve the initial value problem dx
dt = −2x
t + ln t x(1) = 1.
2For this scalar ODE, the coefficient function a(t) is actually allowed to depend upon the inde-pendent variable t. Later, when we solve x0 = Ax + b(t), we must insist that A be a constant matrix.
Solution: In our above notation, a(t) = −2/t, from which we calculate Z 2
t dt = 2 ln |t|.
(We need not include the integration constant here. After reading ahead a few lines, make sure you can explain why.) Since our initial condition is given at t = 1, we may drop the absolute value bars. The integrating factor is
e2 ln t = eln t2 = t2.
Multiplying both sides of our ODE by the integrating factor gives t2dx
dt + 2tx = t2ln t.
Equivalently,
d dt
©t2xª
= t2ln t, and integrating both sides yields
t2x = Z
t2ln t dt.
Integrating by parts,
t2x = t3
3 ln t − t3 9 + C,
where C is a constant of integration. The initial condition x(1) = 1 can be used to calculate C = 10/9, which means that the solution of our initial value problem is
x(t) = t
3ln t − t 9+ 10
9t2.
We can extend the variation of parameters technique to inhomogeneous systems of ODEs of the form x0 = Ax + b(t), where A is a constant matrix. To facilitate this process, we first state a lemma which tells us how to write our usual initial value problem as an integral equation. Reformulating ODEs as integral equations will help us frequently in subsequent chapters.
Lemma 2.5.1. Consider the initial value problem x0 = f (x, t) with x(0) = x0. If f is continuous, then this initial value problem can be written as an integral equation
x(t) = x0+ Z t
f (x(s), s) ds. (2.16)
Here, s is a “dummy variable” of integration.
Proof. First, observe that if we set t = 0 in Equation (2.16), then the equation reduces to x(0) = x0, which means the initial condition is satisfied. If we differentiate both sides of (2.16) using the Fundamental Theorem of Calculus, then we obtain x0(t) = f (x(t), t), and we see that the ODE is also satisfied.
Now consider the initial value problem
x0 = Ax + b(t) x(0) = x0, (2.17)
where A is an n×n constant matrix. By analogy with the one-variable problem above, we re-write the equation as x0 − Ax = b(t). If this were a one-variable problem with A constant, we would use e−tA as the integrating factor. We claim that for the initial value problem (2.17), we can use the matrix e−tA as an integrating factor.
Multiplying both sides of our ODE system by e−tA yields e−tA
|{z}
matrix
{x0− Ax}
| {z } vector
= e|{z}−tA matrix
|{z}b(t) vector
.
Equivalently,
d dt
©e−tAxª
= e−tAb(t), and by Lemma 2.5.1 we have
e−tAx(t) = e−tAx(t)¯¯
t=0+ Z t
0
e−sAb(s) ds = Ix(0) + Z t
0
e−sAb(s) ds, where I is the identity matrix. Multiplying through by etA, we have established Theorem 2.5.2. The solution of the inhomogeneous initial value problem (2.17) is given by
x(t) = etAx(0) + etA Z t
0
e−sAb(s) ds. (2.18)
When applying formula (2.18), our main challenge is to actually evaluate the integral of the vector-valued function e−sAb(s).
Example. Solve the initial value problem
x01 = −x1+ x2+ e−t x1(0) = 1 x02 = −x1− x2+ 2e−t x2(0) = 1.
Solution: In matrix form, the system becomes Proposition 2.1.16, we find that
etA = e−t with −s and using these same trigonometric facts, we have
e−sA = es
· cos s − sin s sin s cos s
¸ . Therefore, the integrand in formula (2.18) is
e−sAb(s) = es Integrating each component of this vector separately,
Z t
Putting all of this together, Equation (2.18) tells us that the solution of the initial value problem is Expanding these products, our overall result is
x1(t) = −e−tcos t + 2e−tsin t + 2e−t x2(t) = e−tsin t + 2e−tcos t − e−t.
Readers interested in learning more about the types of linear systems of ODEs for which exact solution is possible are encouraged to explore other resources such as the references listed in the bibliography. The special classes of linear systems we have discussed up to this point will be sufficient for our purposes in the remainder of this text.
EXERCISES
1. If λ is an eigenvalue of an invertible matrix A, show that 1/λ is an eigenvalue of A−1. What are the eigenvectors of A−1 associated with 1/λ?
2. An n × n matrix A is called idempotent if A2 = A. Show that each eigenvalue of an idempotent matrix is either 0 or 1.
3. Show that if N is a nilpotent matrix (see Definition 2.1.19), then zero is the only eigenvalue of N.
4. Give an example of a 3 × 3 matrix that is diagonalizable but not invertible. Then, give an example of a 3 × 3 matrix that is invertible but not diagonalizable.
5. Square matrices A and B are called simultaneously diagonalizable if they are diag-onalizable by the same invertible matrix P . Show that if A and B are simultaneously diagonalizable, then A and B commute. That is, AB = BA.
6. The cosine of an n × n square matrix A is defined in terms of the Maclaurin series representation of the cosine function:
cos(A) = X∞ k=0
(−1)k A2k
(2k)! = I −A2 2! + A4
4! + · · ·
(a) Suppose that D = diag{λ1, λ2, . . . λn} is a diagonal matrix. What is cos(D)?
(b) Suppose that A is a diagonalizable matrix and A = P DP−1 where D is diagonal.
Show that cos(A) = P [cos(D)]P−1.
(c) Find cos(A) if
A =
· −3 6 4 −1
¸ .
7. Use the Maclaurin series representation of the sine function to define sin(A), where A is an n × n matrix. Use your definition to compute (a) the sine of the 2 × 2 zero matrix, and (b) the sine of the diagonal matrix D = diag{π/2, π/2, π/2}.
8. Consider the two matrices A =
· 1 0 0 0
¸
and B =
· 0 1 0 0
¸ .
(a) Show that A and B do not commute. (b) Show that B is not diagonalizable.
(c) Show that eA+B 6= eAeB.
9. For each of the following matrices A, compute etA: (a) A =
· 2 25/8
2 2
¸
(b) A =
· 0 −1
−1 2
¸
(c) A =
· 2 −1 2 0
¸ .
10. Solve the system
dx1
dt = 3x1+ x2 dx2
dt = x1+ 3x2, with x1(0) = −1 and x2(0) = 4.
11. Find the general solution of the system dx1
dt = x1+ 2x2+ 3x3 dx2
dt = 2x2 + 8x3 dx3
dt = 3x3.
HINT: Do you really need to exponentiate a matrix here?
12. Find the general solution of the system dx1
dt = x1− 5x2 dx2
dt = 5x1+ x2.
13. Solve the initial value problem dx1
dt = 5x1+ 10x2, x1(0) = 1 dx2
dt = −x1+ 3x2, x2(0) = 1.
14. Solve the IVP
dx1
dt = x1− 5x2 dx2
dt = 2x1+ 3x2, with x1(0) = 1 and x2(0) = 1.
15. Find the general solution of the system x0 = Ax where A is the matrix A =
· 0 1
−1 2
¸ .
16. Find the general solution of the system x0 = Ax where A is the matrix
17. Find the general solution of the system x0 = Ax where A is the matrix
A =
18. Solve the initial value problem x0 = Ax where
A =
19. For each of the following matrices A, sketch the phase portrait for the linear system x0 = Ax. In each case, indicate whether the origin is a stable node, unstable node, stable focus, unstable focus, saddle, or center.
A =
20. For the following matrices A, carefully sketch the phase portrait for the linear system x0 = Ax. In each case, identify the stable subspace Es and the unstable subspace Eu. Hint: Try to mimic examples that appear in the text.
21. (Trace and determinant.) Without finding the eigenvalues of the coefficient
matrices of the following systems, determine whether the origin is a stable node, unstable node, stable focus, unstable focus, or center.
(a) x0 =
· −1 1
3 2
¸
x (b) x0 =
· 2 1 1 2
¸ x (c) x0 =
· −2 1 1 −1
¸
x (d) x0 =
· 2 −1 5 −2
¸ x.
22. The trace of any n × n matrix A is equal to the sum of its eigenvalues; the purpose of this exercise is to prove this statement for diagonalizable matrices A.
(a) Suppose A and B are n × n matrices. Show that tr(AB) = tr(BA).
(b) Using the result from Part (a), show that if A and B are similar, then trA = trB.
(c) Suppose that A is diagonalizable and that A = P DP−1 where D is diagonal.
Use the result from Part (b) to explain why the trace of A is equal to the sum of its eigenvalues.
23. Consider the constant-coefficient system x0 = Ax where
A =
4 0 −3
0 −2 0
3 0 4
.
Determine the stable, unstable and center subspaces Es, Eu and Ec associated with the equilibrium at the origin.
24. Consider the system x0 = Ax, where
A =
−3 0 0
0 2 −4
0 4 2
.
Identify Es, Eu and Ec for this system, and provide a rough sketch of the phase portrait.
25. Determine Es, Eu and Ec for the following system, and describe the the flow
starting from initial conditions within each of these subspaces.
26. Each of the following systems contains an unspecified constant α. For each sys-tem, (i) determine the eigenvalues of the coefficient matrix in terms of α; (ii) find the critical values of α at which the qualitative nature of the phase portrait experiences a dramatic change; and (iii) sketch the phase portrait for different choices of α: one just below each critical value, and one just above each critical value.
(a) x0 =
27. Solve the initial value problem dy
28. Solve the initial value problem dy
dx = −y
x + ex2, y(1) = 3.
29. Solve the initial value problem x0 =
30. Solve the initial value problem dx
dt = x + 2t x(0) = 2 dy = −y + et y(0) = 1.
31. Solve the initial value problem dx1
dt = −x1− 2x2+ cos 2t, x1(0) = 3 dx2
dt = 2x1− x2+ sin 2t, x2(0) = 3.