Linear Algebra, Theory And Applications, 2012a

368 NORMS FOR FINITE DIMENSIONAL VECTOR SPACES
Show this is well defined and does indeed give ∫ b
Ψ(t) dt ∈L(V,W) . Also show the
a
triangle inequality ∣∣ ∣∣∣∣ ∣∣∣∣ ∫ b
∫ b
Ψ(t) dt
a ∣∣ ≤ ||Ψ(t)|| dt
a
where ||·|| is the operator norm and verify the fundamental theorem of calculus holds.
(∫ t
′
Ψ(s) ds)
=Ψ(t) .
a
Also verify the usual properties of integrals continue to hold such as the fact the
integral is linear and
∫ b
a
Ψ(t) dt +
∫ c
b
Ψ(t) dt =
∫ c
a
Ψ(t) dt
and similar things. Hint: On showing the triangle inequality, it will help if you use
the fact that
|w| W
=sup|(w, v)| .
|v|≤1
You should show this also.
22. Prove Gronwall’s inequality. Suppose u (t) ≥ 0 and for all t ∈ [0,T] ,
u (t) ≤ u 0 +
∫ t
where K is some nonnegative constant. Then
0
u (t) ≤ u 0 e Kt .
Ku(s) ds.
Hint: w (t) = ∫ t
u (s) ds. Then using the fundamental theorem of calculus, w (t)
0
satisfies the following.
u (t) − Kw(t) =w ′ (t) − Kw(t) ≤ u 0 ,w(0) = 0.
Now use the usual techniques you saw in an introductory differential equations class.
Multiply both sides of the above inequality by e −Kt and note the resulting left side is
now a total derivative. Integrate both sides from 0 to t and see what you have got. If
you have problems, look ahead in the book. This inequality is proved later in Theorem
C.4.3.
23. With Gronwall’s inequality and the integral defined in Problem 21 with its properties
listed there, prove there is at most one solution to the initial value problem
y ′ = Ay, y (0) = y 0 .
Hint: If there are two solutions, subtract them and call the result z. Then
It follows
and so
z ′ = Az, z (0) = 0.
z (t) =0+
||z (t)|| ≤
∫ t
0
∫ t
0
Az (s) ds
‖A‖||z (s)|| ds
Now consider Gronwall’s inequality of Problem 22.