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Multi-dimensional hyperbolic partial differential equationsFirst-order systems and applications$

Sylvie Benzoni-Gavage and Denis Serre

Print publication date: 2006

Print ISBN-13: 9780199211234

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780199211234.001.0001

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(p.443) A BASIC CALCULUS RESULTS

(p.443) A BASIC CALCULUS RESULTS

Source:
Multi-dimensional hyperbolic partial differential equations
Publisher:
Oxford University Press

The celebrated Gronwall Lemma is used repeatedly in this book. We state our most useful versions of it for convenience.

Lemma A.1 (Basic Gronwall Lemma) If u and f are smooth functions of t Є [0, T] such that

u ( t ) C 0 + C 1 0 t ( u ( τ ) + f ( τ ) ) d τ t [ 0 , T ] ,
with C0 ∈ R and C1 〉 0 then
u ( t ) e C 1 t ( C 0 + C 1 0 t f ( τ ) d τ ) t [ 0 , T ] .

Proof The only trick in the proof is to show the final estimate for the righthand side

v ( t ) : = C 0 + C 1 0 t ( u ( τ ) + f ( τ ) ) d τ

of the original one. Since

v ( t ) = C 1 ( u ( t ) + f ( t ) ) C 1 ( v ( t ) + f ( t ) )

we easily get the inequality

e C 1 t v ( t ) v ( 0 ) + C 1 0 t e C 1 τ f ( τ ) d τ ,

of which the claimed estimate is only a rougher version. □

A slightly more elaborate version that we often use is the following.

Lemma A.2 (Gronwall Lemma) If u and f are smooth functions of t Є [0, T] such that

u ( t ) C 0 e γ t + C 1 0 t e γ ( t - τ ) ( u ( τ ) + f ( τ ) ) d τ t [ 0 , T ]

with C0 Є R and C1 〉 0 then

u ( t ) C 0 e ( C 1 + γ ) t + C 1 0 t e ( C 1 + γ ) ( t - τ ) f ( τ ) d τ t [ 0 , T ] .

(p.444) Lemma A.3 (‘Multidimensional’ Gronwall Lemma) Assume ℒ ⊂ R d+1 is a lens foliated by hypersurfaces Hθ and denote

θ = ε [ 0 , θ ] ε

for θ Є [0, 1]. If u is a smooth function in the neighbourhood of L such that

θ | u | C ( 0 | u | + θ | u | ) θ [ 0 , 1 ]

then there exists C depending only on C and L such that

1 | u | C 0 | u | .

Proof The proof relies on the same trick as before but requires a little multidimensional calculus. Introducing parametric equations x = X(y, θ), t = T(y, θ) (y Є Ω ⊂ R d) for Hθ we have

θ | u | = Ω | u ( X ( y , θ ) , T ( y , θ ) ) | | d y X | 2 + | d y T | 2 d y , θ u = 0 θ Ω | u ( X ( y , ε ) , T ( y , ε ) ) | | J ( y , ε ) | d y d ε , J = | d y X θ X d y T θ T | .

Hence

d d θ θ | u | = Ω | u ( X ( y , θ ) , T ( y , ε ) ) | | J ( y , θ ) | d y max Ω ¯ × [ 0 , 1 ] | J | | d y X | 2 + | d y T | 2 θ | u | .

Then we easily get the wanted estimate with

C = max Ω ¯ × [ 0 , 1 ] exp ( C | J | | d y X | 2 + | d y T | 2 ) .

Lemma A.4 (Discrete Gronwall Lemma) if a is a non-negative continuous function of s Є [0,t] and b is a non-decreasing function of s Є [0,t] such that

a ( s + ε ) a ( s ) ε C ( ε + b ( s ) + a ( s + ε ) ) .
for all (ε, s) with 0 〈 ε ≤ ε0 ∊ (0 t), s ∊ [0 t −ε], then
a ( t ) e C t ( a ( 0 ) + b ( t ) ) .

(p.445) Proof The proof is fully elementary. Take D 〉 C and consider n Є N such that ε n : = t n 1 satisfies

e D ε n 1 C ε n .

For all s Є [0,t − εn], we have

a ( s + ε n ) 1 1 C ε n a ( s ) + C ε n 1 C ε n ( ε n + b ( s ) ) .

Therefore,

e D t a ( t ) a ( 0 ) = k = 0 n e D ( k + 1 ) ε n a ( ( k + 1 ) ε n ) e D k ε n a ( k ε n ) k = 0 n e D k ε n ( e D ε n 1 C ε n 1 ) a ( k ε n ) + k = 0 n e D ( k + 1 ) ε n C ε n 1 C ε n ( ε n + b ( k ε n ) ) C 1 C ε n ( ε n + b ( t ) ) 0 x e D s d s C D ( 1 C ε n ) ( ε n + b ( t ) ) .

We get the final estimate by letting n go to ∞ and then D go to C. □