Grönwall's inequality explained

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).^[1]

Differential form

Let

I denote an interval of the real line of the form [a, inf ty) 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 inequalityu'(t) \le \beta(t)u(t), 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 l (\int ^{t} _{a} \beta(s) d s r )

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 l (\int ^{t} _{a} \beta(s) d s r ), t\in I.

Note that

v satisfiesv'(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(d)/(dt) (u(t))/(v(t)) = (u'(t)v(t) - v'(t)u(t))/(v ^{2(t)}) = (u'(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 :(u(t))/(v(t)) \le (u(a))/(v(a)) = u(a), 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 ^{t} _{a} \beta(s) u(s) d s, \forall t\in I,

then

u(t) \le \alpha(t) + \int ^{t\beta(r) d r r ) d s,} _{s} t\in I.

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

u(t) \le \alpha(t)\exp l (\int ^{t\beta(s) d s r ),} _{a} 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 l ({ - }\int ^{s\beta(r)u(r) d r,} _{a} 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) = l (\underbrace{u(s) - \int ^{s\beta(r)u(r) d r}} _{\le\alpha(s)} r )\beta(s)\exp l ({ - }\int ^{s\beta(r) d r r ),} _{a} 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 givesv(t) \le\int ^{s\beta(r) d r r ) d s.} _{a}

Using the definition of

v(t) from the first step, and then this inequality and the property e ^{ae} ^{b=e} ^{a+b} , we obtain\begin{align}\int ^{s\beta(r) d r}} _{= \int ^{t\beta(r) d r} _{s}} r ) d s. \end{align}

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

\begin{align}u(t)& \le\alpha(t) + \alpha(t)\int ^{t\beta(r) d r r ) r ) r |} _{s} ^{s = t} ^{s = a} \\ & = \alpha(t)\exp l (\int ^{t\beta(r) d r r ),} _{a} t\in I.\end{align}

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

Notes and References

1. Book: Pachpatte. B.G.. Inequalities for differential and integral equations. 1998. Academic Press. San Diego. 9780080534640.