Английская Википедия:Grönwall's inequality

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Шаблон: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 Шаблон:Math and Шаблон:Math being constants).[1] Richard Bellman proved a slightly more general integral form in 1943.[2]

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).[3]

Differential form

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

<math>u'(t) \le \beta(t)\,u(t),\qquad t\in I^\circ,</math>

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

<math>u(t) \le u(a) \exp\biggl(\int_a^t \beta(s)\, \mathrm{d} s\biggr)</math>

for all <math>t \in I</math>.

Remark: There are no assumptions on the signs of the functions <math>\beta</math> and <math>u</math>.

Proof

Define the function

<math>v(t) = \exp\biggl(\int_a^t \beta(s)\, \mathrm{d} s\biggr),\qquad t\in I.</math>

Note that <math>v</math> satisfies

<math>v'(t) = \beta(t)\,v(t),\qquad t\in I^\circ,</math>

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

<math>\frac{d}{dt}\frac{u(t)}{v(t)} = \frac{u'(t)\,v(t)-v'(t)\,u(t)}{v^2(t)} = \frac{u'(t)\,v(t) - \beta(t)\,v(t)\,u(t)}{v^2(t)} \le 0,\qquad t\in I^\circ,</math>

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

<math>\frac{u(t)}{v(t)}\le \frac{u(a)}{v(a)}=u(a),\qquad t\in I,</math>

which is Grönwall's inequality.

Integral form for continuous functions

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

<math>u(t) \le \alpha(t) + \int_a^t \beta(s) u(s)\,\mathrm{d}s,\qquad \forall t\in I,</math>
then
<math> 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.</math>
<math>u(t) \le \alpha(t)\exp\biggl(\int_a^t\beta(s)\,\mathrm{d}s\biggr),\qquad t\in I.</math>

Remarks:

Proof

(a) Define

<math>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.</math>

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

<math>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,</math>

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

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

Using the definition of <math>v(t)</math> from the first step, and then this inequality and the functional equation of the exponential function, we obtain

<math>\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}</math>

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

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

<math>\begin{align}u(t)&\le\alpha(t)+\biggl({-}\alpha(t)\exp\biggl(\int_s^t\beta(r)\,\mathrm{d}r\biggr)\biggr)\biggr|^{s=t}_{s=a}\\

&=\alpha(t)\exp\biggl(\int_a^t\beta(r)\,\mathrm{d}r\biggr),\qquad t\in I.\end{align}</math>

Integral form with locally finite measures

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

<math>\int_{[a,t)}|u(s)|\,\mu(\mathrm{d}s)<\infty,\qquad t\in I,</math>

and that Шаблон:Math satisfies the integral inequality

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

If, in addition,

<math>\int_{[a,t)}|\alpha(s)|\,\mu(\mathrm{d}s)<\infty,\qquad t\in I,</math>

then Шаблон:Math satisfies Grönwall's inequality

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

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

Remarks

Special cases

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

=\alpha(t)\exp(c(t-a)),\qquad t\in I.</math>

Outline of proof

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

Detailed proof

Claim 1: Iterating the inequality

For every natural number Шаблон:Math including zero,

<math>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)</math>

with remainder

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

where

<math>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,</math>

is an Шаблон:Math-dimensional simplex and

<math>\mu^{\otimes 0}(A_0(s,t)):=1.</math>

Proof of Claim 1

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

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

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

with

<math>\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.</math>

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

<math>\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.</math>

Hence Claim 1 is proved for Шаблон:Math.

Claim 2: Measure of the simplex

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

<math>\mu^{\otimes n}(A_n(s,t))\le\frac{\bigl(\mu(I_{s,t})\bigr)^n}{n!}</math>

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

Proof of Claim 2

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

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

<math>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)}\}.</math>

These sets are disjoint for different permutations and

<math>\bigcup_{\sigma\in S_n}A_{n,\sigma}(s,t)\subset I_{s,t}^n.</math>

Therefore,

<math>\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.</math>

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

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

<math>\{(s_1,\ldots,s_n)\in I_{s,t}^n\mid s_i=s_j\}</math>

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

<math>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\},</math>

the claimed equality follows.

Proof of Grönwall's inequality

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

<math>|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.</math>

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

<math>\lim_{n\to\infty}R_n(t)=0,\qquad t\in I.</math>

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

<math>\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)</math>

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

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

<math>\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</math>

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

See also

References

  1. ↑ Шаблон:Citation
  2. ↑ Шаблон:Citation
  3. ↑ Шаблон:Cite book
  4. ↑ Шаблон:Citation

Шаблон:PlanetMath attribution

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