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Lecture Notes in Differential Equations - Bruce E. Shapiro

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234 LESSON 27. HIGHER ORDER EQUATIONS<br />

so that<br />

S<strong>in</strong>ce<br />

and<br />

n−1<br />

∑<br />

|u ′ ∣<br />

(t)| ≤ ∣φ (i) (t) ∣ 2 n−2<br />

∑<br />

∣<br />

+ ∣φ (i) (t) ∣ 2 n−1<br />

∣<br />

+ 2 ∣φ (n−1) ∑<br />

∣<br />

(t) ∣ |a i | ∣φ (i) (t) ∣<br />

i=1<br />

i=0<br />

i=0<br />

(27.49)<br />

n−1<br />

∑<br />

n−1<br />

∑<br />

|c i | ≤ |c i | (27.50)<br />

i=1<br />

i=0<br />

n−2<br />

∑<br />

n−1<br />

∑<br />

|c i | ≤ |c i | (27.51)<br />

i=0<br />

for any set of numbers c i , this becomes<br />

From equation (27.39),<br />

From lemma 1,<br />

Therefore,<br />

Hence<br />

i=0<br />

n−1<br />

∑<br />

|u ′ ∣<br />

(t)| ≤ ∣φ (i) (t) ∣ 2 n−1<br />

∑<br />

∣<br />

+ ∣φ (i) ∣<br />

(t)<br />

i=0<br />

i=0<br />

n−1<br />

∣<br />

+2 ∣φ (n−1) ∑<br />

∣<br />

(t) ∣ |a i | ∣φ (i) (t) ∣ (27.52)<br />

i=0<br />

n−1<br />

|u ′ ∣<br />

(t)| ≤ 2u(t) + 2 ∣φ (n−1) ∑<br />

∣<br />

(t) ∣ |a i | ∣φ (i) (t) ∣ (27.53)<br />

i=0<br />

det M ≠ 0 (27.54)<br />

n−1<br />

∑<br />

|u ′ (t)| ≤ 2u(t) + |a i | 2u(t) (27.55)<br />

i=0<br />

[ ]<br />

n−1<br />

∑<br />

= 2u(t) 1 + |a i | = 2Ku(t) (27.56)<br />

i=0<br />

− 2K ≤ u′ (t)<br />

u(t)<br />

Let u(t 0 ) = u 0 and <strong>in</strong>tegrate from t 0 to t<br />

− 2K<br />

∫ t<br />

t o<br />

ds ≤<br />

∫ t<br />

t o<br />

∣ 2<br />

≤ 2K (27.57)<br />

u ′ ∫<br />

(s)<br />

t<br />

u(s) ds ≤ 2K<br />

t o<br />

ds (27.58)

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