In mathematics, Grönwall's inequality (also called Grönwall's lemma or the Grönwall–Bellman inequality) allows one to bound a function that is known to satisfy a certain differential or integral inequality by the solution of the corresponding differential or integral equation. There are two forms of the lemma, a differential form and an integral form. For the latter there are several variants.
It is named for Thomas Hakon Grönwall (1877–1932). Grönwall is the Swedish spelling of his name, but he spelled his name as Gronwall in his scientific publications after emigrating to the United States.
The differential form was proven by Grönwall in 1919.
The integral form was proven by Richard Bellman in 1943.
A nonlinear generalization of the Grönwall–Bellman inequality is known as Bihari–LaSalle inequality. Other variants and generalizations can be found in Pachpatte, B.G. (1998).
Let I denote an interval of the real line of the form [a, ∞) or [a, b] or [a, b) with a < b. Let β and u be real-valued continuous functions defined on I. If u is differentiable in the interiorIo of I (the interval I without the end points a and possibly b) and satisfies the differential inequality
then u is bounded by the solution of the corresponding differential equationv ′(t) = β(t) v(t):
for all t ∈ I.
Remark: There are no assumptions on the signs of the functions β and u.
Let I denote an interval of the real line of the form [a, ∞) or [a, b] or [a, b) with a < b. Let α, β and u be real-valued functions defined on I. Assume that β and u are continuous and that the negative part of α is integrable on every closed and bounded subinterval of I.
(a) If β is non-negative and if u satisfies the integral inequality
(b) If, in addition, the function α is non-decreasing, then
There are no assumptions on the signs of the functions α and u.
Compared to the differential form, differentiability of u is not needed for the integral form.
For a version of Grönwall's inequality which doesn't need continuity of β and u, see the version in the next section.
where we used the assumed integral inequality for the upper estimate. Since β and the exponential are non-negative, this gives an upper estimate for the derivative of v. Since v(a) = 0, integration of this inequality from a to t gives
Using the definition of v(t) for the first step, and then this inequality and the functional equation of the exponential function, we obtain
Substituting this result into the assumed integral inequality gives Grönwall's inequality.
(b) If the function α is non-decreasing, then part (a), the fact α(s) ≤ α(t), and the fundamental theorem of calculus imply that
Let I denote an interval of the real line of the form [a, ∞) or [a, b] or [a, b) with a < b. Let α and u be measurable functions defined on I and let μ be a continuous non-negative measure on the Borel σ-algebra of I satisfying μ([a, t]) < ∞ for all t ∈ I (this is certainly satisfied when μ is a locally finite measure). Assume that u is integrable with respect to μ in the sense that
and that u satisfies the integral inequality
If, in addition,
the function α is non-negative or
the function t ↦ μ([a, t]) is continuous for t ∈ I and the function α is integrable with respect to μ in the sense that
then u satisfies Grönwall's inequality
for all t ∈ I, where Is,t denotes to open interval (s, t).
There are no continuity assumptions on the functions α and u.
The integral in Grönwall's inequality is allowed to give the value infinity.
If α is the zero function and u is non-negative, then Grönwall's inequality implies that u is the zero function.
The integrability of u with respect to μ is essential for the result. For a counterexample, let μ denote Lebesgue measure on the unit interval[0, 1], define u(0) = 0 and u(t) = 1/t for t ∈ (0, 1], and let α be the zero function.
The version given in the textbook by S. Ethier and T. Kurtz. makes the stronger assumptions that α is a non-negative constant and u is bounded on bounded intervals, but doesn't assume that the measure μ is locally finite. Compared to the one given below, their proof does not discuss the behaviour of the remainder Rn(t).
The proof is divided into three steps. The idea is to substitute the assumed integral inequality into itself n times. This is done in Claim 1 using mathematical induction. In Claim 2 we rewrite the measure of a simplex in a convenient form, using the permutation invariance of product measures. In the third step we pass to the limit n to infinity to derive the desired variant of Grönwall's inequality.