Grönwall\'s inequality

In mathematics, Grönwall's lemma 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. Mathematics is the body of Knowledge and Academic discipline that studies such concepts as Quantity, Structure, Space and There are two forms of the lemma, a differential form and an integral form, for the latter there are several variants.

Grönwall's lemma is an important tool used for obtaining various estimates in ordinary differential equations. In Mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one Independent variable, and one or more of its In particular, it is used to prove uniqueness of a solution to the initial value problem, see the Picard-Lindelöf theorem. In Mathematics and Logic, the phrase "there is one and only one " is used to indicate that exactly one object with a certain property exists In Mathematics, in the field of Differential equations an initial value problem is an Ordinary differential equation together with specified value called

It is named for Thomas Hakon Grönwall (1877–1932). Thomas Hakon Grönwall ( January 16, 1877, Dylta bruk, Sweden - May 9, 1932, New York, USA was a Swedish mathematician Year 1877 ( MDCCCLXXVII) was a Common year starting on Monday (link will display the full calendar of the Gregorian calendar (or a Common Year 1932 ( MCMXXXII) was a Leap year starting on Friday of the Gregorian calendar. It is also commonly known as Gronwall's lemma (or inequality), which is not wrong, since Grönwall spelled his name as Gronwall in his scientific publications after emigrating to the United States.

The differential form was proven by Grönwall in 1919. Year 1919 ( MCMXIX) was a Common year starting on Wednesday (link will display the full calendar of the Gregorian calendar (or a Common ^[1] The integral form was proven by Richard Bellman in 1943. Richard Ernest Bellman ( August 26, 1920 – March 19, 1984) was an applied mathematician, celebrated for his invention of Year 1943 ( MCMXLIII) was a Common year starting on Friday (the link will display full 1943 calendar of the Gregorian calendar. ^[2]

1 Differential form
- 1.1 Proof
2 Integral form for continuous functions
- 2.1 Proof
3 Integral form with locally finite measures
4 References

Differential form

Let I denote an interval of the real line of the form [a,∞) or [a,b] or [a,b) with a < b. In Mathematics, an interval is a set of Real numbers with the property that any number that lies between two numbers in the set is also included in the set In Mathematics, the real line is simply the set R of singleton Real numbers However this term is usually used when R is to be treated as a Let β and u be real-valued continuous functions defined on I. In Mathematics, a continuous function is a function for which intuitively small changes in the input result in small changes in the output If u is differentiable in the interior I^o 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:

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

for all t in I. In Mathematics, the interior of a set S consists of all points of S that are intuitively "not on the edge of S "

Remark: There are no assumptions on the signs of the functions β and u.

Proof

$v(t) = \exp\biggl(\int_a^t \beta(s)\, \mathrm{d} s\biggr)$

is the solution of

$v'(t) = \beta(t)\,v(t),$

with v(a) = 1, then v(t) > 0 for all t, so

$\frac{d}{dt}\frac{u}{v} = \frac{u'v-v'u}{v^2} \le \frac{\beta u v - \beta v u}{v^2} = 0$

for t>a, so

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

which is Gronwall's inequality.

Integral form for continuous functions

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

$u(t) \le \alpha(t) + \int_a^t \beta(s) u(s)\,\mathrm{d}s,\qquad t\in I,$

then

$u(t) \le \alpha(t) + \int_a^t\alpha(s)\beta(s)\exp\biggl(\int_s^t\beta(r)\,\mathrm{d}r\biggr)\mathrm{d}s,\qquad t\in I.$

(b) If, in addition, the function α is constant, then

Failed to parse (Cannot write to or create math output directory): u(t) \le \alpha\exp\biggl(\int_a^t\beta(s)\,\mathrm{d}s\biggr),\qquad t\in I.

Remarks:

There are no assumptions on the signs of the function 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.

Proof

(a) Define

$v(s) = \exp\biggl({-}\int_a^s\beta(r)\,\mathrm{d}r\biggr)\int_a^s\beta(r)u(r)\,\mathrm{d}r,\qquad s\in I.$

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

$v'(s) = \biggl(\underbrace{u(s)-\int_a^s\beta(r)u(r)\,\mathrm{d}r}_{\le\,\alpha(s)}\biggr)\beta(s)\exp\biggl({-}\int_a^s\beta(r)\mathrm{d}r\biggr), \qquad s\in I,$

where we used the assumed integral inequality for the upper estimate. In Calculus, the product rule also called Leibniz's law (see derivation) governs the differentiation of products of differentiable In Calculus, the chain rule is a Formula for the Derivative of the composite of two functions. The exponential function is a function in Mathematics. The application of this function to a value x is written as exp( x) The fundamental theorem of calculus specifies the relationship between the two central operations of Calculus, differentiation and integration. 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

$v(t) \le\int_a^t\alpha(s)\beta(s)\exp\biggl({-}\int_a^s\beta(r)\,\mathrm{d}r\biggr)\mathrm{d}s.$

Using the definition of v(t) for the first step, and then this inequality and the functional equation of the exponential function, we obtain

$\begin{align}\int_a^t\beta(s)u(s)\,\mathrm{d}s &=\exp\biggl(\int_a^t\beta(r)\,\mathrm{d}r\biggr)v(t)\\ &\le\int_a^t\alpha(s)\beta(s)\exp\biggl(\underbrace{\int_a^t\beta(r)\,\mathrm{d}r-\int_a^s\beta(r)\,\mathrm{d}r}_{=\,\int_s^t\beta(r)\,\mathrm{d}r}\biggr)\mathrm{d}s. \end{align}$

Substituting this result into the assumed integral inequality gives Grönwall's inequality. In Mathematics or its applications a functional equation is an Equation expressing a relation between the value of a function (or functions at a point with its values

(b) If the function α is constant, then part (a) and the fundamental theorem on calculus imply that

$\begin{align}u(t)&\le\alpha+\biggl({-}\alpha\exp\biggl(\int_s^t\beta(r)\,\mathrm{d}r\biggr)\biggr)\biggr|^{s=t}_{s=a}\\ &=\alpha\exp\biggl(\int_a^t\beta(r)\,\mathrm{d}r\biggr),\qquad t\in I.\end{align}$

Integral form with locally finite measures

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 locally finite measure on the Borel σ-algebra of I (we need μ([a,t]) < ∞ for all t in I). In Mathematics, measurable functions are Well-behaved functions between measurable spaces. In Mathematics, a locally finite measure is a measure for which every point of the Measure space has a neighbourhood of Finite measure In Mathematics, the Borel algebra (or Borel &sigma-algebra) on a Topological space X is a &sigma-algebra of Subsets of Assume that u is integrable with respect to μ in the sense that

$\int_a^t|u(s)|\,\mu(\mathrm{d}s)<\infty,\qquad t\in I,$

and that u satisfies the integral inequality

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

If, in addition,

the function α is non-negative or
the function t → μ([a,t]) is continuous for t in I and the function α is integrable with respect to μ in the sense that

$\int_a^t|\alpha(s)|\,\mu(\mathrm{d}s)<\infty,\qquad t\in I,$

then u satisfies Grönwall's inequality

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

for all t in I, where I_s,t denotes to open interval (s,t).

Remarks

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 in (0,1], and let α be the zero function. In Logic, and especially in its applications to Mathematics and Philosophy, a counterexample is an exception to a proposed general rule i In Mathematics, the Lebesgue measure, named after Henri Lebesgue, is the standard way of assigning a Length, Area or Volume to In Mathematics, the unit interval is the interval, that is the set of all Real numbers x such that zero is less than or equal to x

Special cases

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

$u(t) \le \alpha(t) + \int_a^t \alpha(s)\beta(s)\exp\biggl(\int_s^t\beta(r)\,\mathrm{d}r\biggr)\,\mathrm{d}s,\qquad t\in I.$

If the function α is non-negative and the density β of μ is bounded by a constant c, then

$u(t) \le \alpha(t) + c\int_a^t \alpha(s)\exp\bigl(c(t-s)\bigr)\,\mathrm{d}s,\qquad t\in I.$

If, in addition, the non-negative function α is a constant, then

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

Outline of proof

The proof is divided into three steps. In 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.

Detailed proof

Claim 1: Iterating the inequality

For every natural number n including zero,

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

with remainder

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

where

$A_n(s,t)=\{(s_1,\ldots,s_n)\in I_{s,t}^n\mid s_1<s_2<\cdots<s_n\},\qquad n\ge1,$

is an n-dimensional simplex and

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

Proof of Claim 1

We use mathematical induction. In Geometry, a simplex (plural simplexes or simplices) or n -simplex is an n -dimensional analogue of a triangle Mathematical induction is a method of Mathematical proof typically used to establish that a given statement is true of all Natural numbers It is done by proving that For n = 0 this is just the assumed integral inequality, because the empty sum is defined as zero. In Mathematics, the empty sum, or nullary sum, is the result of adding no numbers in Summation for example

Induction step from n to n + 1: Inserting the assumed integral inequality for the function u into the remainder gives

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

with

$\tilde R_n(t):=\int_{[a,t)} \biggl(\int_{[a,q)} u(s)\,\mu(\mathrm{d}s)\biggr)\mu^{\otimes n}(A_n(q,t))\,\mu(\mathrm{d}q),\qquad t\in I.$

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

$\tilde R_n(t) =\int_{[a,t)} u(s)\underbrace{\int_{(s,t)} \mu^{\otimes n}(A_n(q,t))\,\mu(\mathrm{d}q)}_{=\,\mu^{\otimes n+1}(A_{n+1}(s,t))}\,\mu(\mathrm{d}s) =R_{n+1}(t),\qquad t\in I.$

Hence Claim 1 is proved for n + 1. In Mathematical analysis, Fubini's theorem, named after Guido Fubini, states that if \int_{A\times B} |f(xy|\d(xy

Claim 2: Measure of the simplex

For every natural number n including zero and all s < t in I

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

with equality in case t → μ([a,t]) is continuous for t in I.

Proof of Claim 2

For n = 0, the claim is true by our definitions. Therefore, consider n ≥ 1 in the following.

Let S_n denote the set of all permutations of the indices in {1,2,. In several fields of Mathematics the term permutation is used with different but closely related meanings . . ,n}. For every permutation σ in S_n define

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

These sets are disjoint for different permutations and

$\bigcup_{\sigma\in S_n}A_{n,\sigma}(s,t)\subset I_{s,t}^n.$

Therefore,

$\sum_{\sigma\in S_n} \mu^{\otimes n}(A_{n,\sigma}(s,t)) \le\mu^{\otimes n}\bigl(I_{s,t}^n\bigr)=\bigl(\mu(I_{s,t})\bigr)^n.$

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

Assume now that t → μ([a,t]) is continuous for t in I. Then, for different indices i and j in {1,2,. . . ,n}, the set

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

is contained in a hyperplane, hence by an application of Fubini's theorem its measure with respect to the n-fold product of μ is zero. A hyperplane is a concept in Geometry. It is a higher-dimensional generalization of the concepts of a line in Euclidean plane geometry and a In Mathematical analysis, Fubini's theorem, named after Guido Fubini, states that if \int_{A\times B} |f(xy|\d(xy Since

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

the claimed equality follows.

Proof of Grönwall's inequality

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

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

Since μ is locally finite on I, we have μ(I_a,t) < ∞. Hence, the integrability assumption on u implies that

$\lim_{n\to\infty}R_n(t)=0,\qquad t\in I.$

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

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

for all s < t in I. In Mathematics, the Exponential function can be characterized in many ways 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 in I, Claim 2 gives

$\sum_{k=0}^{n-1} \mu^{\otimes k}(A_k(s,t)) =\sum_{k=0}^{n-1} \frac{\bigl(\mu(I_{s,t})\bigr)^k}{k!} \to\exp\bigl(\mu(I_{s,t})\bigr)\qquad\text{as }n\to\infty$

and the integrability of the function α permits to use the dominated convergence theorem to derive Grönwall's inequality. In Measure theory, a branch of Mathematical analysis, Lebesgue 's dominated convergence theorem provides Sufficient conditions under which two

References

^ T. H. Gronwall: Note on the derivative with respect to a parameter of the solutions of a system of differential equations, Ann. of Math 20 (1919), 292–296. The Annals of Mathematics (ISSN 0003-486X abbreviated as Ann of Math Year 1919 ( MCMXIX) was a Common year starting on Wednesday (link will display the full calendar of the Gregorian calendar (or a Common
^ Richard Bellman, The stability of solutions of linear differential equations, Duke Math. J. 10 (1943), 643–647. Richard Ernest Bellman ( August 26, 1920 – March 19, 1984) was an applied mathematician, celebrated for his invention of Duke Mathematical Journal is an American mathematical research journal published by Duke University Press. Year 1943 ( MCMXLIII) was a Common year starting on Friday (the link will display full 1943 calendar of the Gregorian calendar.

This article incorporates material from Gronwall's lemma on PlanetMath, which is licensed under the GFDL. PlanetMath is a free, collaborative online Mathematics Encyclopedia.

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Contents

Differential form

Proof

Integral form for continuous functions

Proof

Integral form with locally finite measures

Remarks

Special cases

Outline of proof

Detailed proof

Claim 1: Iterating the inequality

Proof of Claim 1

Claim 2: Measure of the simplex

Proof of Claim 2

Proof of Grönwall's inequality

References