THE GRONWALL’S INEQUALITY
MINGFENG ZHAO
May 26, 2011
Theorem 1 (Gronwall’s Inequality: Differential Form). Let u ∈ C 1 ([a, ∞)), α ∈ C([a, ∞)) be such
that
u0 (t) ≤ α(t)u(t),
for all t > a.
Then
Rt
u(t) ≤ u(a)e
Proof. Look at v(t) = u(t)e−
Rt
a
α(s)ds
a
α(s)ds
.
, then v satisfies v(a) = u(a), and
v 0 (t) = e−
Rt
a
α(s)ds
[u0 (t) − α(t)u(t)] ≤ 0.
So v(t) is decreasing, by integrating from a to t, we get
v(t) − v(a) ≤ 0,
which implies v(t) ≤ v(a) = u(a), i.e,
u(t) ≤ u(a)e
Rt
a
α(s)ds
.
Theorem 2 (Bellman’s Inequality, or Gronwall’s Inequality: Integral Form). Let β, u ∈ C([a, ∞)), α
is a function on [a, ∞), and let
α− (t) = max{−α(t), 0} ∈ L∞
loc ([a, ∞)).
Then
1
2
MINGFENG ZHAO
Rt
(1) If β ≥ 0, and u(t) ≤ α(t) +
a
β(s)u(s)ds, then
t
Z
u(t) ≤ α(t) +
α(s)β(s)e
Rt
β(r)dr
s
for all t ≥ a.
ds,
a
(2) If β ≥ 0, and α is increasing, and u(t) ≤ α(t) +
u(t) ≤ α(t)e
Rt
a
β(s)ds
Rt
a
β(s)u(s)ds, then
for all t ≥ a.
,
In particular, if β ≥ 0, and α = 0, then u(t) ≤ 0 for all t ≥ a.
Proof. 1. Look at v(t) = e−
Rt
v 0 (t)
a
Rt
β(r)dr
a
= e−
Rt
a
β(s)u(s)ds, since β, u are continuous, then
β(r)dr
Z t
β(s)u(s)ds + β(t)u(t)
−β(t)
a
=
β(t)e
−
Rt
a
β(r)dr
t
Z
[u(t) −
β(s)u(s)ds]
a
≤
α(t)β(t)e−
Rt
a
β(r)dr
.
Integrating form a to t, then
t
Z
α(s)β(s)e−
v(t) − v(a) = v(t) − v(a) = v(t) =≤
Rs
a
β(r)dr
a
which implies
e−
Rt
a
β(r)dr
t
Z
Z
t
α(s)β(s)e−
β(s)u(s)ds ≤
a
Rs
a
β(r)dr
ds.
a
Hence
Z
t
β(s)u(s)ds ≤ e
Rt
a
β(r)dr
a
Z
t
α(s)β(s)e−
Rs
a
β(r)dr
a
Z
t
=
Rt
α(s)β(s)e
s
β(r)dr
ds.
a
Therefore, we get
Z
t
u(t) ≤ α(t) +
β(s)u(s)ds
a
Z
≤ α(t) +
t
α(s)β(s)e
a
Rt
s
β(r)dr
ds.
ds
ds,
THE GRONWALL’S INEQUALITY
3
2. By part (1), we have known that
Z
u(t) ≤ α(t) +
t
α(s)β(s)e
Rt
s
β(r)dr
ds,
a
Since α is increasing, then α(s) ≤ α(t) when a ≤ s ≤ t. Therefore,
Z t
Rt
β(r)dr
β(s)e s
ds
u(t) ≤ α(t) 1 +
a
"
Z
t
= α(t) 1 −
Rt
de
s
β(r)dr
#
ds
a
t Rt
= α(t) 1 − e s β(r)dr a
h
i
Rt
= α(t) 1 − 1 + e a β(r)dr
= α(t)e
Rt
a
β(r)dr
.
Theorem 3 (Bihari’s Inequality). Let u and f be two non-negative continuous functions on [0, ∞).
And w is a continuous increasing function on [0, ∞), and w(t) > 0 in (0, ∞). If there exist some
positive constant α > 0 such that u satisfies
f
Z
u(t) ≤ α +
(s)w(u(s))ds
for all t ≥ 0.
0
Then
−1
u(t) ≤ G
Z t
G(α) +
f (s)ds ,
for all 0 ≤ t ≤ T .
0
Where G is defined by
Z
x
G(x) =
x0
1
dy, x > 0, x0 > 0.
w(y)
And G−1 is the inverse function of G, and T is chosen such that
Z
G(α) +
0
t
f (s)ds ∈ Domain of G−1 ,
for all 0 ≤ t ≤ T .
4
MINGFENG ZHAO
Proof. Define
t
Z
f (s)w(u(s))ds,
y(t) = α +
for all t 6= 0.
0
Since f, u, w are continuous functions, then we have
y 0 (t) = f (t)w(u(t)).
Since
t
Z
0 ≤ u(t) ≤ α +
f (s)w(u(s))ds = y(t),
0
and w is increasing function, then
w(u(t)) ≤ w(y(t)).
Since f (t) ≥ 0, so we get
y 0 (t) = f (t)w(u(t)) ≤ f (t)w(y(t)).
That is, we get
y 0 (t)
≤ f (t).
w(y(t))
By f (0) = α, then
Z
y(t)
α
1
dr
w(r)
t
Z
=
0
y 0 (s)
ds Let r = y(s)
w(y(s))
t
Z
≤
f (s)ds.
0
Therefore, we have
Z
G(u(t)) − G(α)
u(t)
=
x0
Z
u(t)
=
α
Z
y(t)
≤
α
Z
≤
1
dr −
w(r)
α
x0
1
dr
w(r)
1
dr
w(r)
1
dr
w(r)
t
f (s)ds.
0
Z
Since u(t) ≤ y(t)
THE GRONWALL’S INEQUALITY
5
That is,
Z
t
G(u(t)) ≤ G(α) +
f (s)ds.
0
Since G is increasing function, so G−1 is also increasing function. Therefore,
Z t
u(t) ≤ G−1 G(α) +
f (s)ds ,
for all 0 ≤ t ≤ T .
0
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu