Lecture Notes in Differential Equations - Bruce E. Shapiro

232 LESSON 27. HIGHER ORDER EQUATIONS
To prove theorem 27.3 1 we will need a generalization of the fundamental
identity. We first define a the following norm of a function:
n−1
∑
‖f(t)‖ 2 ∣
= ∣f (i) (t) ∣ 2 (27.35)
i=0
= |f(t)| + |f ′ ∣
(t)| + · · · + ∣f (n−1) (t) ∣ 2 (27.36)
That ‖· · · ‖ defines a norm follows from the fact that for any two functions
f(t) and g(t), and for and constant c, the following four properties are true
for all t:
Properties of a Norm of a Function:
1. ‖f(t)‖ ≥ 0
2. ‖f(t)‖ = 0 ⇔ f(t) = 0
3. ‖f(t) + g(t)‖ ≤ ‖f(t)‖ + ‖g(t)‖
4. ‖cf(t)‖ ≤ |c| ‖f(t)‖
For comparison to the norm of a vector space, see definition 15.2 which is
equivalent to this for the vector space of functions.
Lemma 27.5. (Fundamental Identity for nth order equations) Let φ(t) be
a solution of
L n y = 0, y(t 0 ) = y 0 , ..., y (n−1) (t 0 ) = y n (27.37)
then
for all t.
‖φ(t 0 )‖ e −K‖t−t0‖ ≤ ‖φ(t)‖ ≤ ‖φ(t 0 )‖ e K‖t−t0‖ (27.38)
Lemma 27.6. 2 |a| |b| ≤ |a| 2 + |b| 2
Proof. (|a| + |b|) 2 = |a| 2 + |b| 2 − 2 |a| |b| ≥ 0.
Proof. (of Lemma 27.5.) We can assume that a n ≠ 0; otherwise this would
not be an nth-order equation. Further, we will assume that a n = 1; otherwise,
redefine L n by division through by a n .
Let
n−1
∑
u(t) = ‖φ(t)‖ 2 ∣
= ∣φ (i) (t) ∣ 2 n−1
∑
= φ (i) (t)φ (i)∗ (t) (27.39)
i=0
1 This material is somewhat more abstract and the reader may wish to skip ahead to
the examples following the proofs.
i=0