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.
Grönwall's inequality is an important tool to obtain various estimates in the theory of ordinary and stochastic differential equations.
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 inequality was first proven by Grönwall in 1919 (the integral form below with α and β being constants).
[1] Richard Bellman proved a slightly more general integral form in 1943.
denote an interval of the real line of the form
is bounded by the solution of the corresponding differential equation
is non-positive and the function is bounded above by its value at the initial point
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.
Remarks: (a) Define Using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain for the derivative 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
(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, then u satisfies Grönwall's inequality for all t ∈ I, where Is,t denotes to open interval (s, t).
The idea is to substitute the assumed integral inequality into itself n times.
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.
For every natural number n including zero, with remainder where is an n-dimensional simplex and We use mathematical induction.
For n = 0 this is just the assumed integral inequality, because the empty sum is defined as zero.
Induction step from n to n + 1: Inserting the assumed integral inequality for the function u into the remainder gives with Using the Fubini–Tonelli theorem to interchange the two integrals, we obtain Hence Claim 1 is proved for n + 1.
For every natural number n including zero and all s < t in I with equality in case t ↦ μ([a, t]) is continuous for t ∈ I.
Let Sn denote the set of all permutations of the indices in {1, 2, .
For every permutation σ ∈ Sn define These sets are disjoint for different permutations and Therefore, Since they all have the same measure with respect to the n-fold product of μ, and since there are n!
, n}, the set is contained in a hyperplane, hence by an application of Fubini's theorem its measure with respect to the n-fold product of μ is zero.
Hence, the integrability assumption on u implies that Claim 2 and the series representation of the exponential function imply the estimate for all s < t in I.
If the function α is non-negative, then it suffices to insert these results into Claim 1 to derive the above variant of Grönwall's inequality for the function u.
In case t ↦ μ([a, t]) is continuous for t ∈ I, Claim 2 gives and the integrability of the function α permits to use the dominated convergence theorem to derive Grönwall's inequality.
This article incorporates material from Gronwall's lemma on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.