Grönwall's inequality

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.

Grönwall's inequality is an important tool to obtain various estimates in the theory of ordinary and stochastic differential equations. In particular, it provides a comparison theorem that can be used to prove uniqueness of a solution to the initial value problem; see the Picard–Lindelöf theorem.

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 inequality was first proven by Grönwall in 1919 (the integral form below with and being constants).
Richard Bellman proved a slightly more general integral form 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).

Differential form

Let $I$ denote an interval of the real line of the form , \infty) or [a, b/math> or $[a, b)$ with $a < b$. Let $\beta$ and $u$ be real-valued continuous functions defined on $I$. If $u$ is derivative, differentiable in the Interior (topology), interior $I^\circ$ of $I$ (the interval $I$ without the end points $a$ and possibly $b$) and satisfies the differential inequality

:$u'(t) \le \beta(t)\,u(t),\qquad t\in I^\circ,$

then $u$ is bounded by the solution of the corresponding differential ''equation'' $v'(t) = \beta(t) \, v(t)$:

:$u(t) \le u(a) \exp\biggl(\int_a^t \beta(s)\, \mathrm s\biggr)$

for all $t \in I$.

Remark: There are no assumptions on the signs of the functions $\beta$ and $u$.

Proof

Define the function

:$v(t) = \exp\biggl(\int_a^t \beta(s)\, \mathrm s\biggr),\qquad t\in I.$

Note that $v$ satisfies

:$v'(t) = \beta(t)\,v(t),\qquad t\in I^\circ,$

with $v(a) = 1$ and $v(t) > 0$ for all $t \in I$. By the quotient rule

:$\frac\frac = \frac = \frac \le 0,\qquad t\in I^\circ,$

Thus the derivative of the function $u(t)/v(t)$ is non-positive and the function is bounded above by its value at the initial point $a$ of the interval $I$:

:$\frac\le \frac=u(a),\qquad t\in I,$

which is Grönwall's inequality.

Integral form for continuous functions

Let denote an interval of the real line of the form or or with . Let , and be real-valued functions defined on . Assume that and are continuous and that the negative part of is integrable on every closed and bounded subinterval of .
* (a) If  is non-negative and if satisfies the integral inequality
::$u(t) \le \alpha(t) + \int_a^t \beta(s) u(s)\,\mathrms,\qquad \forall t\in I,$
:then
::$u(t) \le \alpha(t) + \int_a^t\alpha(s)\beta(s)\exp\biggl(\int_s^t\beta(r)\,\mathrmr\biggr)\mathrms,\qquad t\in I.$
* (b) If, in addition, the function is non-decreasing, then

::$u(t) \le \alpha(t)\exp\biggl(\int_a^t\beta(s)\,\mathrms\biggr),\qquad t\in I.$

Remarks:
* There are no assumptions on the signs of the functions and .
* Compared to the differential form, differentiability of is not needed for the integral form.
* For a version of Grönwall's inequality which doesn't need continuity of and , see the version in the next section.

Proof

(a) Define

:$v(s) = \exp\biggl(\int_a^s\beta(r)\,\mathrmr\biggr)\int_a^s\beta(r)u(r)\,\mathrmr,\qquad s\in I.$

Using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain for the derivative

:$v'(s) = \biggl(\underbrace_\biggr)\beta(s)\exp\biggl(\int_a^s\beta(r)\mathrmr\biggr), \qquad s\in I,$

where we used the assumed integral inequality for the upper estimate. Since and the exponential are non-negative, this gives an upper estimate for the derivative of $v(s)$. Since $v(a)=0$, integration of this inequality from to gives

:$v(t) \le\int_a^t\alpha(s)\beta(s)\exp\biggl(\int_a^s\beta(r)\,\mathrmr\biggr)\mathrms.$

Using the definition of $v(t)$ from the first step, and then this inequality and the property $e^ae^b=e^$, we obtain

:$\begin\int_a^t\beta(s)u(s)\,\mathrms &=\exp\biggl(\int_a^t\beta(r)\,\mathrmr\biggr)v(t)\\ &\le\int_a^t\alpha(s)\beta(s)\exp\biggl(\underbrace_\biggr)\mathrms. \end$

Substituting this result into the assumed integral inequality gives Grönwall's inequality.

(b) If the function is non-decreasing, then part (a), the fact , and the fundamental theorem of calculus imply that

:$\beginu(t)& \le\alpha(t)+\alpha(t)\int_a^t\beta(s)\exp\left(\int_s^t\beta(r)dr\right)ds\\ &\le\alpha(t)+\alpha(t)\biggl(\exp\biggl(\int_s^t\beta(r)\,\mathrmr\biggr)\biggr, ^_\biggr)\\ &=\alpha(t)\exp\biggl(\int_a^t\beta(r)\,\mathrmr\biggr),\qquad t\in I.\end$

Integral form with locally finite measures

Let denote an interval of the real line of the form or or with . Let and be measurable functions defined on  and let be a continuous non-negative measure on the Borel σ-algebra of satisfying for all (this is certainly satisfied when is a locally finite measure). Assume that is integrable with respect to in the sense that

:$\int_, u(s), \,\mu(\mathrms)<\infty,\qquad t\in I,$

and that satisfies the integral inequality

:$u(t) \le \alpha(t) + \int_ u(s)\,\mu(\mathrms),\qquad t\in I.$

If, in addition,
* the function is non-negative or
* the function is continuous for and the function is integrable with respect to in the sense that

:: $\int_, \alpha(s), \,\mu(\mathrms)<\infty,\qquad t\in I,$

then satisfies Grönwall's inequality

:$u(t) \le \alpha(t) + \int_\alpha(s)\exp\bigl(\mu(I_)\bigr)\,\mu(\mathrms)$

for all , where denotes to open interval .

Remarks

* There are no continuity assumptions on the functions and .
* The integral in Grönwall's inequality is allowed to give the value infinity.
* If is the zero function and is non-negative, then Grönwall's inequality implies that is the zero function.
* The integrability of with respect to is essential for the result. For a counterexample, let denote Lebesgue measure on the unit interval , define and for , and let be the zero function.
* The version given in the textbook by S. Ethier and T. Kurtz. makes the stronger assumptions that is a non-negative constant and is bounded on bounded intervals, but doesn't assume that the measure is locally finite. Compared to the one given below, their proof does not discuss the behaviour of the remainder .

Special cases

* If the measure has a density with respect to Lebesgue measure, then Grönwall's inequality can be rewritten as

:: $u(t) \le \alpha(t) + \int_a^t \alpha(s)\beta(s)\exp\biggl(\int_s^t\beta(r)\,\mathrmr\biggr)\,\mathrms,\qquad t\in I.$
* If the function is non-negative and the density of is bounded by a constant , then

:: $u(t) \le \alpha(t) + c\int_a^t \alpha(s)\exp\bigl(c(t-s)\bigr)\,\mathrms,\qquad t\in I.$
* If, in addition, the non-negative function is non-decreasing, then

:: $u(t) \le \alpha(t) + c\alpha(t)\int_a^t \exp\bigl(c(t-s)\bigr)\,\mathrms =\alpha(t)\exp(c(t-a)),\qquad t\in I.$

Outline of proof

The proof is divided into three steps. The idea is to substitute the assumed integral inequality into itself 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 to infinity to derive the desired variant of Grönwall's inequality.

Detailed proof

Claim 1: Iterating the inequality

For every natural number including zero,

:$u(t) \le \alpha(t) + \int_ \alpha(s) \sum_^ \mu^(A_k(s,t))\,\mu(\mathrms) + R_n(t)$

with remainder

:$R_n(t) :=\int_u(s)\mu^(A_n(s,t))\,\mu(\mathrms),\qquad t\in I,$

where

:$A_n(s,t)=\,\qquad n\ge1,$

is an -dimensional simplex and

:$\mu^(A_0(s,t)):=1.$

Proof of Claim 1

We use mathematical induction. For this is just the assumed integral inequality, because the empty sum is defined as zero.

Induction step from to :
Inserting the assumed integral inequality for the function into the remainder gives

:$R_n(t)\le\int_ \alpha(s) \mu^(A_n(s,t))\,\mu(\mathrms) +\tilde R_n(t)$

with

:$\tilde R_n(t):=\int_ \biggl(\int_ u(s)\,\mu(\mathrms)\biggr)\mu^(A_n(q,t))\,\mu(\mathrmq),\qquad t\in I.$

Using the Fubini–Tonelli theorem to interchange the two integrals, we obtain

:$\tilde R_n(t) =\int_ u(s)\underbrace_\,\mu(\mathrms) =R_(t),\qquad t\in I.$

Hence Claim 1 is proved for .

Claim 2: Measure of the simplex

For every natural number including zero and all in

:$\mu^(A_n(s,t))\le\frac$

with equality in case is continuous for .

Proof of Claim 2

For , the claim is true by our definitions. Therefore, consider in the following.

Let denote the set of all permutations of the indices in . For every permutation define

:$A_(s,t)=\.$

These sets are disjoint for different permutations and

:$\bigcup_A_(s,t)\subset I_^n.$

Therefore,

:$\sum_ \mu^(A_(s,t)) \le\mu^\bigl(I_^n\bigr)=\bigl(\mu(I_)\bigr)^n.$

Since they all have the same measure with respect to the -fold product of , and since there are permutations in , the claimed inequality follows.

Assume now that is continuous for . Then, for different indices , the set

:$\$

is contained in a hyperplane, hence by an application of Fubini's theorem its measure with respect to the -fold product of is zero. Since

:$I_^n\subset\bigcup_A_(s,t) \cup \bigcup_\,$

the claimed equality follows.

Proof of Grönwall's inequality

For every natural number , Claim 2 implies for the remainder of Claim 1 that

:$, R_n(t), \le \frac \int_ , u(s), \,\mu(\mathrms),\qquad t\in I.$

By assumption we have . Hence, the integrability assumption on implies that

:$\lim_R_n(t)=0,\qquad t\in I.$

Claim 2 and the series representation of the exponential function imply the estimate

:$\sum_^ \mu^(A_k(s,t)) \le\sum_^ \frac \le\exp\bigl(\mu(I_)\bigr)$

for all in . If the function  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 .

In case is continuous for , Claim 2 gives

:$\sum_^ \mu^(A_k(s,t)) =\sum_^ \frac \to\exp\bigl(\mu(I_)\bigr)\qquad\textn\to\infty$

and the integrability of the function permits to use the dominated convergence theorem to derive Grönwall's inequality.

See also

* Stochastic Gronwall inequality
* Logarithmic norm, for a version of Gronwall's lemma that gives upper and lower bounds to the norm of the state transition matrix.
* Halanay inequality. A similar inequality to Gronwall's lemma that is used for differential equations with delay.
* Chaplygin's theorem. The generalized differential form of Gronwall's inequality, potentially involving any Lipschitz-over-u right part.

References

{{DEFAULTSORT:Gronwall's inequality
Lemmas in mathematical analysis
Ordinary differential equations
Stochastic differential equations
Articles containing proofs
Probabilistic inequalities