An extension of Gronwall's inequality
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Інститут прикладної математики і механіки НАН України
1999
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| Цитувати: | An extension of Gronwall's inequality / J. Webb // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 196-204. — Бібліогр.: 5 назв. — англ. |
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irk-123456789-1692932020-06-10T01:26:31Z An extension of Gronwall's inequality Webb, J. 1999 Article An extension of Gronwall's inequality / J. Webb // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 196-204. — Бібліогр.: 5 назв. — англ. 0236-0497 http://dspace.nbuv.gov.ua/handle/123456789/169293 en Нелинейные граничные задачи Інститут прикладної математики і механіки НАН України |
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Webb, J. An extension of Gronwall's inequality Нелинейные граничные задачи |
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Webb, J. |
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Webb, J. |
| title |
An extension of Gronwall's inequality |
| title_short |
An extension of Gronwall's inequality |
| title_full |
An extension of Gronwall's inequality |
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An extension of Gronwall's inequality |
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An extension of Gronwall's inequality |
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extension of gronwall's inequality |
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Інститут прикладної математики і механіки НАН України |
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1999 |
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An extension of Gronwall's inequality / J. Webb // Нелинейные граничные задачи: сб. науч. тр. — 1999. — Т. 9. — С. 196-204. — Бібліогр.: 5 назв. — англ. |
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Нелинейные граничные задачи |
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AT webbj anextensionofgronwallsinequality AT webbj extensionofgronwallsinequality |
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2025-07-15T04:02:57Z |
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2025-07-15T04:02:57Z |
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1837684152076337152 |
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An extension of Gronwall’s inequality
c© Jeffrey Webb
1 Introduction
The Gronwall inequality is a well-known tool in the study of differential equations,
Volterra integral equations, and evolution equations [2]. It is often used to establish
a priori bounds which are used in proving global existence, uniqueness, and stability
results. There are a number of versions of the result, a well-known version was first
proved by Bellman according to the book [3]. One classical version reads as follows.
Theorem 1.1. Suppose that φ(t) ≤ c0(t) +
∫ t
0
c1(s)φ(s) ds for a.e. t ∈ [0, T ], where
c0 is non-negative and non-decreasing, c1 ∈ L1
+, and φ ∈ L∞+ . [Subscript + denotes
functions that are ≥ 0 a.e.] Then
φ(t) ≤ c0(t) exp
(∫ t
0
c1(s) ds
)
a.e. . (1.1)
Another somewhat different inequality is the following, apparently first given by
Liang Ou-Iang [5], and also given in [2],(both authors consider continuous functions).
Theorem 1.2. Suppose that φ2(t) ≤ c2
0 + 2
∫ t
0
h(s)φ(s) ds for a.e. t ∈ [0, T ], where c0
is a constant, h ∈ L1
+, and φ ∈ L∞+ . Then, for a.e. t ∈ [0, T ],
φ(t) ≤ c0 +
∫ t
0
h(s) ds.
If h ∈ L2, by using the inequality hφ ≤ h2/2 + φ2/2 this case could be reduced to
the classical case but would give a different type of conclusion.
We have discovered a ‘nonlinear’ version of Gronwall’s inequality that contains both
of these results and many other inequalities of Gronwall type given in the literature,
see for example [3].
To describe the type of results we prove consider the following simple case. Suppose
that a non-negative L∞ function φ satisfies a.e. the inequality
φ2(t) ≤ c2
0 +
∫ t
0
2c1(s)φ(s) + 2c2(s)φ
2(s) ds,
where ci are non-negative L1 functions. Then we prove that
φ(t) ≤ [
c0 +
∫ t
0
c1(s) ds
]
exp
(∫ t
0
c2
)
a.e..
This result contains both of the results just cited and also a special case of a result
of Dafermos [1], see below. We are interested in more general inequalities and in
obtaining explicit L∞ bounds for φ which depend only on c0 and integrals of the given
functions ci. Also we consider the case when the inequalities hold a.e. since this is a
commonly occurring situation.
2 Extended Gronwall inequality
We give a general result which contains Theorems 1.1 and 1.2 and many other results
as well as giving new ones. To make the formula a little easier we insert some constants.
Theorem 2.1. Let k ≥ 1 be an integer. Suppose that
φk(t) ≤ ck
0(t) +
∫ t
0
kc1(s)φ(s) + kc2(s)φ
2(s) + · · ·+ kck(s)φ
k(s) ds, for a.e. t ∈ [0, T ],
(2.1)
where c0 ≥ 0 is non-decreasing, φ ∈ L∞+ , and ci ∈ L1
+ for i ≥ 1. Then,
φ(t) ≤ Wk(c0, c1, . . . , ck−1) exp
(∫ t
0
ck
)
a.e. (2.2)
where the function Wk is defined recursively by
W1(c0) =c0
Wk+1(c0, c1, . . . , ck) =Wk
([
ck
0 + k
∫ t
0
c1
]1/k
, c2, . . . , ck
)
, k ≥ 1.
We use in the proof the classical Gronwall inequality quoted above. For complete-
ness, and because we use similar arguments without further explanation, we give a
proof here.
Proof of Theorem 1.1. Fix τ in [0, T ] and let w(t) := c0(τ) +
∫ t
0
c1(s)φ(s) ds. Then w
is absolutely continuous (AC), φ(t) ≤ w(t) for a.e. t ∈ [0, τ ] and
w′(t) = c1(t)φ(t) ≤ c1(t)w(t) for a.e. t ∈ [0, τ ].
Write C1(t) :=
∫ t
0
c1(s) ds. The function w exp(−C1) is AC because C1 is AC, exp is
Lipschitz on bounded intervals, so the composition exp(−C1) is AC, and a product of
AC functions is also AC. Therefore, for a.e. t ∈ [0, τ ],
[w(t) exp(−C1(t))]
′ ≤ 0, (2.3)
and hence, for all t ∈ [0, τ ],
w(t) exp(−C1(t)) ≤ w(0).
Since w(0) = c0(τ) this gives w(t) ≤ c0(τ) exp(C1(t)) for all t ∈ [0, τ ]. In particular,
w(τ) ≤ c0(τ) exp(C1(τ)) and, as τ is arbitrary, (1.1) holds.
Proof of Theorem 2.1. The proof is by induction. As before we replace c0(t) by c0(τ)
and reduce the proof to the case where c0 is a constant which we suppose to be positive
(otherwise add ε > 0). The case k = 1 is the classical result proved above. Suppose
the inequality holds for an integer k and consider the case for k + 1. Thus suppose
that for a.e. t ∈ [0, T ],
φk+1(t) ≤ ck+1
0 +
∫ t
0
(k+1)c1(s)φ(s)+(k+1)c2(s)φ
2(s)+ · · ·+(k+1)ck+1(s)φ
k+1(s) ds.
Write, for t ∈ [0, T ],
uk+1(t) := ck+1
0 +
∫ t
0
(k+1)c1(s)φ(s)+(k+1)c2(s)φ
2(s)+ · · ·+(k+1)ck+1(s)φ
k+1(s) ds.
Since uk+1 is AC and u ≥ c0 > 0, it follows that u is AC. Therefore u′ exists a.e. and
for a.e. t ∈ [0, T ],
uku′ =c1φ + c2φ
2 + · · ·+ ck+1φ
k+1
≤c1u + c2u
2 + · · ·+ ck+1u
k+1.
Hence
uk−1u′ ≤ c1 + c2u + · · ·+ ck+1u
k, a.e. .
So we have for all t ∈ [0, T ],
uk(t)− uk(0) ≤
∫ t
0
kc1 + kc2u + · · ·+ kck+1u
k,
that is,
uk(t) ≤ [
ck
0 +
∫ t
0
kc1
]
+
∫ t
0
kc2u + · · ·+ kck+1u
k.
This shows, by the result for the case k, that for a.e. t ∈ [0, T ],
u(t) ≤ Wk([c
k
0 +
∫ t
0
kc1]
1/k, c2, . . . , ck) exp
(∫ t
0
ck+1
)
.
As u is continuous this holds for all t ∈ [0, T ] and therefore holds almost everywhere
for φ. This completes the inductive step.
The special case k = 2 includes the classical case and the result quoted in Theo-
rem 1.2 since we may take c2 = 0. It includes a special case of a result of Dafermos
[1] who shows that if
φ2(t) ≤ M2φ2(0) +
∫ t
0
2Ng(s)φ(s) + 2γφ2(s) ds
where N,M, γ are non-negative constants and φ ∈ L∞+ , g ∈ L1
+ then
φ(t) ≤ (
Mφ(0) + N
∫ t
0
g(s) ds
)
exp(γt).
Pachpatte [4] has an extension of this special case of Dafermos’s result for continuous
functions when the integral is replaced by a suitable multiple integral, which has
applications to certain higher order ODE’s. In Dafermos, [1], an extra term is included
with γ replaced by γ + βt in the first inequality. We shall use the same idea to give
another result below which will extend Dafermos’s result and allow some t dependence
on the functions ci. We first consider some special cases of the inequality proved in
Theorem 2.1.
Example 2.2. If u ≤ a0 +
∫ t
0
a1(s)u
m/k(s) ds where m < k are positive integers, and
a1 ∈ L1
+, u ∈ L∞+ , then
u ≤ [
a
1−m/k
0 +
∫ t
0
(1−m/k)a1(s) ds
] 1
1−m/k .
As usual we may suppose that a0 > 0. Write v = u1/k so that the inequality reads
vk ≤ a0 +
∫ t
0
a1(s)v
m(s) ds.
This is now a special case of Theorem 2.1 so we can write
v ≤ Wk(a
1/k
0 , 0, . . . , a1/k, 0, . . . , 0),
where a1/k occurs in the (m + 1)-st place. Thus,
v ≤Wk−m+1(a
1/k
0 , a1/k, 0, . . . , 0)
=Wk−m
([
a
k−m
k
0 +
∫ t
0
(k −m)a1(s)/k ds
] 1
k−m , 0, . . . , 0
)
.
Since Wr(c0, 0, . . . , 0) = c0 for every r, this establishes the result.
By an approximation argument this result shows the following: if
u(t) ≤ a0 +
∫ t
0
a1(s)u
α(s) ds, for a.e. t ∈ [0, T ],
where 0 < α < 1, then
u(t) ≤ [
a1−α
0 +
∫ t
0
(1− α)a1(s) ds
] 1
1−α , a.e..
This last inequality is well-known and is usually shown via a comparison theorem: If
h is continuous and u(t) ≤ u0 +
∫ t
0
h(s, u(s)) ds then u(t) ≤ v(t) where v is the solution
of the differential equation v′ = h(t, v), v(0) = u0; [“the faster runner goes further”].
Example 2.3. Given the inequality
u ≤ a0 +
∫ t
0
a1u
m/k + a2u
j/k
we can immediately write down an inequality for u, namely
u1/k ≤ Wk(a
1/k
0 , 0, . . . , 0, a1/k, 0, . . . , 0, a2/k, 0, . . . , 0),
where a1/k (respectively a2/k) occurs in the (m + 1) (resp. (j + 1)) place. Hence, as
above, we obtain
u1/k ≤ Wk−m−(j−m)
({[
a
k−m
k
0 +
∫ t
0
(k −m)a1(s)/k ds
] k−j
k−m + k−j
k
∫ t
0
a2(s) ds
}
, 0, . . . , 0
)
,
that is,
u ≤ {[
a
1−m
k
0 +
∫ t
0
(1−m/k)a1(s) ds
] 1−j/k
1−m/k + (1− j/k)
∫ t
0
a2(s) ds
}1/(1−j/k)
.
By an approximation argument this gives the following result. The inequality
u ≤ a0 +
∫ t
0
a1u
α + a2u
β,
where 0 < α < β < 1, implies
u ≤ {[
a1−α
0 +
∫ t
0
(1− α)a1(s) ds
] 1−β
1−α + (1− β)
∫ t
0
a2(s) ds
}1/(1−β)
.
It is not clear whether the comparison result applies to this situation, because
possibly noncontinuous functions are involved. Also one encounters the problem of
solving the differential equation v′ = a1v
α + a2v
β where ai need not be constants.
For example, in the simple case when we have
u(t) ≤ a0 +
∫ t
0
a1(s)u
1/3 + a2(s)u
2/3 ds
we obtain the estimate u ≤ w := W3(a
1/3
0 , a1/3, a2/3)3 which can be computed directly
to be
w(t) =
([
a
2/3
0 + 2A1(t)/3
]1/2
+ A2(t)/3
)3
,
where Ai(t) =
∫ t
0
ai(s) ds. To apply the comparison result we have to take a0 > 0
or else find a maximal (or minimal) solution because of lack of uniqueness. A small
calculation shows that w′ ≥ a1w
1/3 +a2w
2/3 so that w(t) ≥ v(t) where v is the solution
of v′ = a1v
1/3 + a2v
2/3, v(0) = a0. In other words the estimate we obtain is not as
good as could (in theory) be obtained from the comparison result. However, even with
a1 and a2 constant, the solution v in this case is found only in implicit form, so the
estimate obtained from the comparison result is of less practical use than our explicit
estimate.
Example 2.4. The inequality
u(t) ≤ a0 +
∫ t
0
a1(s)u
m/k(s) + a2(s)u(s) ds
has been studied by Perov when m/k is replaced by α, 0 ≤ α < 1. [In fact he has a
result also for α > 1.] We can immediately write down an inequality for u, namely
u1/k ≤ Wk(a
1/k
0 , 0, . . . , 0, a1/k, 0, . . . , 0) exp
(∫ t
0
a2(s)/k ds
)
.
where a1/k is in the (m + 1)-place. As in example 2.2 this gives
u1/k ≤ [
a
1−m/k
0 +
∫ t
0
(1−m/k)a1(s) ds
] 1
k−m exp
(∫ t
0
a2(s)/k ds
)
.
Hence we obtain
u ≤ [
a
1−m/k
0 +
∫ t
0
(1−m/k)a1(s) ds
] 1
1−m/k exp
(∫ t
0
a2(s) ds
)
.
By a simple approximation argument we can then obtain: the inequality
u(t) ≤ a0 +
∫ t
0
a1(s)u
α(s) + a2(s)u(s) ds
implies that
u ≤ [
a1−α
0 +
∫ t
0
(1− α)a1(s) ds
] 1
1−α exp
(∫ t
0
a2(s) ds
)
.
This is a minor change from Perov’s result.
Remark 2.5. We could, of course, write down other consequences of the theorem but
such inequalities await application. Another ‘nonlinear’ Gronwall inequality is that
attributed to Bihari but apparently proved earlier by Lasalle [3].
If u(t) ≤ a + b
∫ t
0
k(s)g(u(s)) ds where g is nondecreasing, then
u(t) ≤ G−1[G(a) + b
∫ t
0
k(s) ds]
where G(s) =
∫ s
0
dξ
g(ξ)
. This can, in theory, be applied to some of the situations we
include, but the problems of calculating G and G−1 do not appeal in general. Also
such inequalities do not fit our aim of giving explicit bounds in terms of the given
data.
We now give a further extension which allows for noninteger powers.
Theorem 2.6. Let m ≥ 1 be an integer and p ≥ m be a real number. Suppose that
φp(t) ≤ cp
0(t) +
∫ t
0
pc1(s)φ
p−m(s) + pc2(s)φ
p−m+1(s) + · · ·+ pcm+1(s)φ
p(s) ds, (2.4)
for a.e. t ∈ [0, T ], where c0 ≥ 0 is non-decreasing, φ ∈ L∞+ , and ci ∈ L1
+, i ≥ 1. Then,
φ(t) ≤ Wm+1(c0, c1, c2, . . . , cm) exp
(∫ t
0
cm+1
)
, a.e. . (2.5)
Proof. As usual we may and do suppose that c0 is a positive constant and we omit
‘almost everywhere’ in the following. Write
up(t) = cp
0 +
∫ t
0
pc1(s)φ
p−m(s) + pc2(s)φ
p−m+1(s) + · · ·+ pcm+1(s)φ
p(s) ds.
Then we obtain
up−1u′ ≤ c1u
p−m + · · ·+ cm+1u
p
so that
um−1u′ ≤ c1 + c2u + · · ·+ cm+1u
m.
Hence um ≤ cm
0 +
∫ t
0
mc1 +
∫ t
0
mc2u + · · ·+ mcm+1u
m, which gives
u(t) ≤ Wm([cm
0 +
∫ t
0
mc1]
1/m, c2, . . . , cm) exp
(∫ t
0
cm+1
)
,
and yields the conclusion.
Remark 2.7. When p−m = j/k it is possible to reduce the inequality in the hypoth-
esis of Theorem 2.6 to a special case of the one in Theorem 2.1. For example, given
the inequality
u5/2(t) ≤ c
5/2
0 +
∫ t
0
5
2
c1(s)u
1/2(s) + 5
2
c2(s)u
3/2(s) + 5
2
c3(s)u
5/2(s) ds
we can say at once from Theorem 2.6 that
u ≤ W3(c0, c1, c2) exp
(∫ t
0
c3(s) ds
)
.
However writing v = u1/2 the given inequality may be written
v5(t) ≤ (c
1/2
0 )5 +
∫ t
0
5(c1/2)v + 5(c2/2)v3 + 5(c3/2)v5.
Theorem 2.1 then gives the conclusion
v ≤ W5(c
1/2
0 , c1/2, 0, c2/2, 0) exp
(∫ t
0
c3(s)/2 ds
)
.
A small calculation checks that
W5(c
1/2
0 , c1/2, 0, c2/2, 0) = W3(c0, c1, c2)
1/2
so the apparent two conclusions are but one. However, using Theorem 2.6 is simpler
here.
3 A further extension
We now give a result of the same type where the functions ci are allowed to depend
on t. The extra idea used is that employed by Dafermos [1] in proving the following:
φ2(t) ≤ M2φ2(0) +
∫ t
0
2Ng(s)φ(s) + (2γ + 4βt)φ2(s) ds
where N,M, γ are non-negative constants and φ ∈ L∞+ , g ∈ L1
+ then
φ(t) ≤ (
Mφ(0) + N
∫ t
0
g(s) ds
)
exp(αt + βt2),
where α = γ + β/γ.
We will prove the following result.
Theorem 3.1. Let k ≥ 1 be an integer. Suppose that for i = 1, 2, . . . , k, and for
s, t ∈ [0, T ], we have 0 ≤ gi(s, t) ≤ h(s) where h is an L1 function, and that ∂gi/∂t
exists and satisfies 0 ≤ ∂gi(s, t)/∂t ≤ Agi(s, t) for some non-negative constant A.
Suppose that
φk(t) ≤ ck
0(t) + k
∫ t
0
g1(s, t)φ(s) + · · ·+ gk(s, t)φ
k(s) ds, for a.e. t ∈ [0, T ], (3.1)
where c0 ≥ 0 is non-decreasing, φ ∈ L∞+ . Then,
φ(t) ≤ Wk(c0(t), g1(t, t), . . . , gk−1(t, t)) exp
(∫ t
0
[gk(s, s) + A/k] ds
)
. (3.2)
Proof. Let
uk(t) = ck
0 +
∫ t
0
kg1(s, t)φ(s) + kg2(s, t)φ
2(s) + · · ·+ kgk(s, t)φ
k(s) ds,
where we suppose that c0 is a positive constant. The hypotheses made allow us to
differentiate under the integral sign to obtain
uk−1u′ =
k∑
i=1
gi(t, t)φ
i(t) +
∫ t
0
k∑
i=1
∂gi(s, t)
∂t
φi(s) ds
≤
k∑
i=1
gi(t, t)u
i(t) +
∫ t
0
k∑
i=1
Agi(s, t)φ
i(s) ds
≤
k−1∑
i=1
gi(t, t)u
i(t) + [A/k + gk(t, t)]u
k(t).
This yields the inequality
uk(t) ≤ uk(0) +
∫ t
0
k−1∑
i=1
kgi(s, s)u
i(s) + [A + kgk(s, s)]u
k(s) ds.
Theorem 2.1 now applies to give the conclusion
φ(t) ≤ u(t) ≤ Wk(c0, g1(t, t), . . . , gk−1(t, t)) exp
(∫ t
0
[A/k + gk(s, s)] ds
)
.
This includes Dafermos’s result for we may take g1(s, t) = Ng(s) and g2(s, t) =
γ + 2βt. Then ∂g1/∂t = 0 and ∂g2(s, t)/∂t = 2β ≤ A(γ + 2βt) for A = 2β/γ.
References
[1] Dafermos C.M., The second law of thermodynamics and stability, Arch. Rational
Mech. Anal., 70 (1979), 167–179.
[2] Haraux, A., Nonlinear evolution equations-global behaviour of solutions, Lecture
Notes in Mathematics 841, Springer-Verlag 1981.
[3] Mitrinović, D. S., Pec̆arić, J. E., and Fink, A.M., Inequalities involving functions
and their integrals and derivatives, Kluwer Academic Publishers, Dordrecht /
Boston / London, 1991.
[4] Pachpatte, B.G., On some fundamental integral inequalities arising in the theory
of differential equations, Chinese J. Math. (Taiwan, R.O.C.), 22 (1994), 261-273.
[5] Liang Ou-Iang, The boundedness of solutions of linear differential equations y′′+
A(t)y = 0, Shuxue Jinzhan 3 (1957), 409–415.
Department of Mathematics, University of Glasgow,
Glasgow, G12 8QW, Scotland, U. K.
Phone: +44 141 330 5181
FAX: +44 141 330 4111
E-mail: J.Webb@maths.gla.ac.uk
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