Linear Algebra, Theory And Applications, 2012a

422 APPLICATIONS TO DIFFERENTIAL EQUATIONS
Proof:
(
(M (t) N (t))
′ ) ij
( )
(
) ′
′ ∑
≡ (M (t) N (t)) ij = M (t) ik
N (t) kj
k
= ∑ ( ) ′
(M (t) ik
) ′ N (t) kj
+ M (t) ik
N (t) kj
k
≡
∑ k
(
′ ) M (t) N (t) ik kj + M (t) (
′ )
ik N (t)
kj
≡ (M ′ (t) N (t)+M (t) N ′ (t)) ij
In the study of differential equations, one of the most important theorems is Gronwall’s
inequality which is next.
Theorem C.4.3 Suppose u (t) ≥ 0 and for all t ∈ [0,T] ,
u (t) ≤ u 0 +
where K is some nonnegative constant. Then
∫ t
0
Ku(s) ds. (3.10)
u (t) ≤ u 0 e Kt . (3.11)
Proof: Let w (t) = ∫ t
u (s) ds. Then using the fundamental theorem of calculus, (3.10)
0
w (t) satisfies the following.
u (t) − Kw(t) =w ′ (t) − Kw(t) ≤ u 0 ,w(0) = 0. (3.12)
Multiply both sides of this inequality by e −Kt and using the product rule and the chain
rule,
e −Kt (w ′ (t) − Kw(t)) = d (
e −Kt w (t) ) ≤ u 0 e −Kt .
dt
Integrating this from 0 to t,
∫ t
( )
e −Kt w (t) ≤ u 0 e −Ks ds = u 0 − e−tK − 1
.
K
Now multiply through by e Kt to obtain
( )
w (t) ≤ u 0 − e−tK − 1
e Kt = − u 0
K
K + u 0
K etK .
0
Therefore, (3.12) implies
(
u (t) ≤ u 0 + K − u 0
K + u 0
K etK) = u 0 e Kt .
With Gronwall’s inequality, here is a theorem on uniqueness of solutions to the initial
value problem,
x ′ = Ax + f (t) , x (a) =x a , (3.13)
in which A is an n × n matrix and f is a continuous function having values in C n .
Theorem C.4.4 Suppose x and y satisfy (3.13). Then x (t) =y (t) for all t.