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Unformatted text preview: 6 MA 36600 FINAL REVIEW §7.7: Fundamental Matrices. • Consider the initial value problem
d
x = P(t) x,
x(t0 ) = x0 .
dt
Say that x(1) , x(2) , . . . , x(n) is a fundamental set of solutions. Then the general solution is
x(t) = c1 x(1) (t) + c2 x(2) (t) + · · · + cn x(n) (t) = Ψ(t) c in terms of the matrices x11 (t x21 (t) Ψ(t) = .
.
. x12 (t)
x22 (t)
.
.
. x1n (t)
x2n (t) .
.
. ···
···
..
. c1 c2 c = . .
.
. and xn1 (t) xn2 (t) · · · xnn (t)
cn
We call Ψ(t) = xij (t) a fundamental matrix. It satisﬁes the matrix equation
d
Ψ = P(t) Ψ.
dt Since the Wronskian is W = det Ψ, we see that Ψ(t) is a nonsingular matrix. Hence we can choose
the constant c in order to solve the initial value problem:
x0 = Ψ(t0 ) c =⇒ c = Ψ(t0 )−1 x0 . • Say P(t) = A is an n × n constant matrix. The solution to the initial value problem is x(t) = Φ(t) x0
in terms of the matrix exponential
Φ(t) = exp (A t) = ∞
k=0 In general, the initial value problem
d
Ψ = A Ψ,
dt Ψ(0) = Ψ0 Ak tk
t2
= I + A t + A2 + · · · .
k!
2 has the unique solution Ψ(t) = exp (A t) Ψ0 . The order does matter! The matrix Ψ0 exp (A t) is not a solution to the initial value problem.
• We say that an n × n matrix is a diagonal matrix if it is in the form rt r1 0 · · · 0
e1
0
···
0 0 r2 · · · 0 0
er2 t · · ·
0 D=.
=⇒
exp (D t) = .
. ..
.
.
. .
..
.
.
.
.
.
.
..
.
.
.
.
.
.
0 0 · · · rn
0
0
· · · ern t An n × n matrix A is said to be diagonalizable there exists a nonsingular n × n matrix T such that
T−1 A T = D is an n × n diagonal matrix. The exponential of such a matrix is
exp (A t) = T exp (D t) T−1 . • Say A is a 2 × 2 matrix with distinct eigenvalues r1 and r2 having corresponding eigenvectors
ξ
ξ
ξ (1) = 11
and
ξ (2) = 12 .
ξ21
ξ22
Denote the 2 × 2 matrices
r1 0
D=
0 r2 and
ξ
T = 11
ξ21 ξ12
ξ22 =⇒ T−1 A T = D. ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University.
 Spring '09
 EdrayGoins
 Matrices

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