Grönwall's inequality

Template:Short description 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 Template:Math and Template:Math 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 $[a,\mathrm{\infty })$ or $[a,b]$ or $[a,b)$ with $a<b$. Let $\beta$ and $u$ be real-valued continuous functions defined on $I$. If $u$ is differentiable in the interior $I^{\circ }$ of $I$ (the interval $I$ without the end points $a$ and possibly $b$) and satisfies the differential inequality

$u^{\prime }(t)\le \beta (t)u(t),t\in I^{\circ },$

then $u$ is bounded by the solution of the corresponding differential equation $v^{\prime }(t)=\beta (t)v(t)$:

$u(t)\le u(a)\exp (\underset{a}{\overset{t}{\int }}\beta (s)\mathrm{d}s)$

for all $t\in I$.

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

Proof

Define the function

$v(t)=\exp (\underset{a}{\overset{t}{\int }}\beta (s)\mathrm{d}s),t\in I.$

Note that $v$ satisfies

$v^{\prime }(t)=\beta (t)v(t),t\in I^{\circ },$

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

$\frac{d}{dt}\frac{u(t)}{v(t)}=\frac{u^{\prime }(t)v(t)-v^{\prime }(t)u(t)}{v^2(t)}=\frac{u^{\prime }(t)v(t)-\beta (t)v(t)u(t)}{v^2(t)}\le 0,t\in I^{\circ },$

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 $I$:

$\frac{u(t)}{v(t)}\le \frac{u(a)}{v(a)}=u(a),t\in I,$

which is Grönwall's inequality.

Integral form for continuous functions

Let Template:Math denote an interval of the real line of the form Template:Closed-open or Template:Closed-closed or Template:Closed-open with Template:Math. Let Template:Math, Template:Math and Template:Math be real-valued functions defined on Template:Math. Assume that Template:Math and Template:Math are continuous and that the negative part of Template:Math is integrable on every closed and bounded subinterval of Template:Math.

• (a) If Template:Math is non-negative and if Template:Math satisfies the integral inequality

$u(t)\le \alpha (t)+\underset{a}{\overset{t}{\int }}\beta (s)u(s)\mathrm{d}s,\mathrm{\forall }t\in I,$

then

$u(t)\le \alpha (t)+\underset{a}{\overset{t}{\int }}\alpha (s)\beta (s)\exp (\underset{s}{\overset{t}{\int }}\beta (r)\mathrm{d}r)\mathrm{d}s,t\in I.$

• (b) If, in addition, the function Template:Math is non-decreasing, then

$u(t)\le \alpha (t)\exp (\underset{a}{\overset{t}{\int }}\beta (s)\mathrm{d}s),t\in I.$

Remarks:

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

Proof

(a) Define

$v(s)=\exp (-\underset{a}{\overset{s}{\int }}\beta (r)\mathrm{d}r)\underset{a}{\overset{s}{\int }}\beta (r)u(r)\mathrm{d}r,s\in I.$

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

$v^{\prime }(s)=\left(\underset{\le \alpha (s)}{\underbrace{u(s)-\underset{a}{\overset{s}{\int }}\beta (r)u(r)\mathrm{d}r}}\right)\beta (s)\exp (-\underset{a}{\overset{s}{\int }}\beta (r)\mathrm{d}r),s\in I,$

where we used the assumed integral inequality for the upper estimate. Since Template:Math 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 Template:Math to Template:Math gives

$v(t)\le \underset{a}{\overset{t}{\int }}\alpha (s)\beta (s)\exp (-\underset{a}{\overset{s}{\int }}\beta (r)\mathrm{d}r)\mathrm{d}s.$

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

$\begin{array}{cc}\underset{a}{\overset{t}{\int }}\beta (s)u(s)\mathrm{d}s& =\exp (\underset{a}{\overset{t}{\int }}\beta (r)\mathrm{d}r)v(t)\\ & \le \underset{a}{\overset{t}{\int }}\alpha (s)\beta (s)\exp \left(\underset{=\underset{s}{\overset{t}{\int }}\beta (r)\mathrm{d}r}{\underbrace{\underset{a}{\overset{t}{\int }}\beta (r)\mathrm{d}r-\underset{a}{\overset{s}{\int }}\beta (r)\mathrm{d}r}}\right)\mathrm{d}s.\end{array}$

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

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

$\begin{array}{cc}u(t)& \le \alpha (t)+\alpha (t)\underset{a}{\overset{t}{\int }}\beta (s)\exp (\underset{s}{\overset{t}{\int }}\beta (r)dr)ds\\ & \le \alpha (t)+\alpha (t)(-\exp (\underset{s}{\overset{t}{\int }}\beta (r)\mathrm{d}r\left){|}_{s=a}^{s=t}\right)\\ & =\alpha (t)\exp (\underset{a}{\overset{t}{\int }}\beta (r)\mathrm{d}r),t\in I.\end{array}$

Integral form with locally finite measures

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

$\underset{[a,t)}{\int }|u(s)|\mu (\mathrm{d}s)<\mathrm{\infty },t\in I,$

and that Template:Math satisfies the integral inequality

$u(t)\le \alpha (t)+\underset{[a,t)}{\int }u(s)\mu (\mathrm{d}s),t\in I.$

If, in addition,

• the function Template:Math is non-negative or
• the function Template:Math is continuous for Template:Math and the function Template:Math is integrable with respect to Template:Math in the sense that

$\underset{[a,t)}{\int }|\alpha (s)|\mu (\mathrm{d}s)<\mathrm{\infty },t\in I,$

then Template:Math satisfies Grönwall's inequality

$u(t)\le \alpha (t)+\underset{[a,t)}{\int }\alpha (s)\exp (\mu (I_{s,t}))\mu (\mathrm{d}s)$

for all Template:Math, where Template:Math denotes to open interval Template:Open-open.

Remarks

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

Special cases

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

$u(t)\le \alpha (t)+\underset{a}{\overset{t}{\int }}\alpha (s)\beta (s)\exp (\underset{s}{\overset{t}{\int }}\beta (r)\mathrm{d}r)\mathrm{d}s,t\in I.$

• If the function Template:Math is non-negative and the density Template:Math of Template:Math is bounded by a constant Template:Math, then

$u(t)\le \alpha (t)+c\underset{a}{\overset{t}{\int }}\alpha (s)\exp (c(t-s))\mathrm{d}s,t\in I.$

• If, in addition, the non-negative function Template:Math is non-decreasing, then

$u(t)\le \alpha (t)+c\alpha (t)\underset{a}{\overset{t}{\int }}\exp (c(t-s))\mathrm{d}s=\alpha (t)\exp (c(t-a)),t\in I.$

Outline of proof

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

Detailed proof

Claim 1: Iterating the inequality

For every natural number Template:Math including zero,

$u(t)\le \alpha (t)+\underset{[a,t)}{\int }\alpha (s)\sum _{k=0}^{n-1}{\mu }^{\otimes k}(A_k(s,t))\mu (\mathrm{d}s)+R_n(t)$

with remainder

$R_n(t):=\underset{[a,t)}{\int }u(s){\mu }^{\otimes n}(A_n(s,t))\mu (\mathrm{d}s),t\in I,$

where

$A_n(s,t)=\{(s_1,\dots ,s_n)\in I_{s,t}^n\mid s_1<s_2<\cdots <s_n\},n\ge 1,$

is an Template:Math-dimensional simplex and

${\mu }^{\otimes 0}(A_0(s,t)):=1.$

Proof of Claim 1

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

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

$R_n(t)\le \underset{[a,t)}{\int }\alpha (s){\mu }^{\otimes n}(A_n(s,t))\mu (\mathrm{d}s)+{\tilde{R}}_n(t)$

with

${\tilde{R}}_n(t):=\underset{[a,t)}{\int }(\underset{[a,q)}{\int }u(s)\mu (\mathrm{d}s)){\mu }^{\otimes n}(A_n(q,t))\mu (\mathrm{d}q),t\in I.$

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

${\tilde{R}}_n(t)=\underset{[a,t)}{\int }u(s)\underset{={\mu }^{\otimes n+1}(A_{n+1}(s,t))}{\underbrace{\underset{(s,t)}{\int }{\mu }^{\otimes n}(A_n(q,t))\mu (\mathrm{d}q)}}\mu (\mathrm{d}s)=R_{n+1}(t),t\in I.$

Hence Claim 1 is proved for Template:Math.

Claim 2: Measure of the simplex

For every natural number Template:Math including zero and all Template:Math in Template:Math

${\mu }^{\otimes n}(A_n(s,t))\le \frac{(\mu (I_{s,t}){)}^n}{n!}$

with equality in case Template:Math is continuous for Template:Math.

Proof of Claim 2

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

Let Template:Math denote the set of all permutations of the indices in Template:Math}. For every permutation Template:Math define

$A_{n,\sigma }(s,t)=\{(s_1,\dots ,s_n)\in I_{s,t}^n\mid s_{\sigma (1)}<s_{\sigma (2)}<\cdots <s_{\sigma (n)}\}.$

These sets are disjoint for different permutations and

$\bigcup _{\sigma \in S_n}{A}_{n,\sigma }(s,t)\subset I_{s,t}^n.$

Therefore,

$\sum _{\sigma \in S_n}{\mu }^{\otimes n}(A_{n,\sigma }(s,t))\le {\mu }^{\otimes n}\left(I_{s,t}^n\right)=(\mu (I_{s,t}){)}^n.$

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

Assume now that Template:Math is continuous for Template:Math. Then, for different indices Template:Math}, the set

$\{(s_1,\dots ,s_n)\in I_{s,t}^n\mid s_i=s_j\}$

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

$I_{s,t}^n\subset \bigcup _{\sigma \in S_n}{A}_{n,\sigma }(s,t)\cup \bigcup _{1\le i<j\le n}\{(s_1,\dots ,s_n)\in I_{s,t}^n\mid s_i=s_j\},$

the claimed equality follows.

Proof of Grönwall's inequality

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

$|R_n(t)|\le \frac{(\mu (I_{a,t}){)}^n}{n!}\underset{[a,t)}{\int }|u(s)|\mu (\mathrm{d}s),t\in I.$

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

$\underset{n\to \mathrm{\infty }}{\lim }{R}_n(t)=0,t\in I.$

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

$\sum _{k=0}^{n-1}{\mu }^{\otimes k}(A_k(s,t))\le \sum _{k=0}^{n-1}\frac{(\mu (I_{s,t}){)}^k}{k!}\le \exp (\mu (I_{s,t}))$

for all Template:Math in Template:Math. If the function Template:Math 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 Template:Math.

In case Template:Math is continuous for Template:Math, Claim 2 gives

$\sum _{k=0}^{n-1}{\mu }^{\otimes k}(A_k(s,t))=\sum _{k=0}^{n-1}\frac{(\mu (I_{s,t}){)}^k}{k!}\to \exp (\mu (I_{s,t}))\text{as }n\to \mathrm{\infty }$

and the integrability of the function Template:Math 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.