# Ngô Quốc Anh

## January 24, 2015

### Reversed Gronwall-Bellman’s inequality

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 23:01

In mathematics, Gronwall’s inequality (also called Grönwall’s lemma, Gronwall’s lemma or Gronwall–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. The differential form was proven by Grönwall in 1919. The integral form was proven by Richard Bellman in 1943. A nonlinear generalization of the Gronwall–Bellman inequality is known as Bihari’s inequality.

First, we consider the Gronwall inequality.

Type 1. Bounds by integrals based on lower bound $a$.

Let $\beta$ and $u$ be real-valued continuous functions defined on $[a,b]$. If $u$ is differentiable in $(a,b)$ and satisfies the differential inequality $\displaystyle u'(t) \leqslant \beta(t) u(t),$

then $u$ is bounded by the solution of the corresponding differential equation $y'(t) = \beta (t)y(t)$, that is to say $\displaystyle \boxed{u(t) \leqslant u(a) \exp\biggl(\int_a^t \beta(s) ds\biggr)}$

for all $t \in [a,b]$.

Proof. We define $v$, a solution of the equation $y'(t) = \beta (t)y(t)$, i.e. $\displaystyle v(t) = \exp\biggl(\int_a^t \beta(s) ds\biggr),$

for all $t \in [a,b]$. Clearly $v(a)=1$ and $v >0$ for all $t$. By the quotient rule, we know that $\displaystyle\frac{d}{dt}\frac{u(t)}{v(t)} = \frac{u'(t)\,v(t)-v'(t)\,u(t)}{v^2(t)} \leqslant \frac{\beta(t)\,u(t)\,v(t) - \beta(t)\,v(t)\,u(t)}{v^2(t)} = 0.$

Hence the quotient $u/v$ is monotone decreasing in $[a,b]$. In particular, we obtain $\displaystyle \frac{u(t)}{v(t)}\leqslant \frac{u(a)}{v(a)}=u(a),$

for all $t \in [a,b]$, which is Gronwall’s inequality.

Type 2. Bounds by integrals based on upper bound $b$.

Let $\beta$ and $u$ be real-valued continuous functions defined on $[a,b]$. If $u$ is differentiable in $(a,b)$ and satisfies the differential inequality $\displaystyle u'(t) \leqslant \beta(t) u(t),$

then $u$ is bounded by the solution of the corresponding differential equation $y'(t) = \beta (t)y(t)$, that is to say $\displaystyle \boxed{u(t) \leqslant u(a) \exp\biggl(\int_a^b \beta(s) ds\biggr) \exp\biggl(-\int_t^b \beta(s) ds\biggr)}$

for all $t \in [a,b]$.

Proof. We basically use Type 1. Simply writing $\displaystyle u(a) \exp\biggl(\int_a^t \beta(s) ds\biggr) = u(a) \exp\biggl(\int_a^b \beta(s) ds - \int_t^b \beta(s) ds\biggr),$

we obtain $\displaystyle u(t) \leqslant u(a) \exp\biggl(\int_a^b \beta(s) ds\biggr) \exp\biggl(-\int_t^b \beta(s) ds\biggr)$

for all $t \in [a,b]$.

We now consider the Bellman inequality.

Type 1. Bounds by integrals based on lower bound $a$.

Let $\alpha$ $\beta$ and $u$ be real-valued functions defined on $[a,b]$. Assume that $\beta$ and $u$ are continuous and that the negative part of $\alpha$ is integrable on every closed and bounded subinterval of $[a,b]$.

• If $\beta$ is non-negative and if $u$ satisfies the integral inequality $\displaystyle u(t) \leqslant \alpha(t) + \int_a^t \beta(s) u(s) ds,$

for all $t\in [a,b]$ then $\displaystyle \boxed{u(t) \leqslant \alpha(t) + \int_a^t\alpha(s)\beta(s)\exp\biggl(\int_s^t\beta(r)dr\biggr) ds},$

for all $t\in [a,b]$.

• If, in addition, the function $\alpha$ is non-decreasing, then $\displaystyle \boxed{u(t) \leqslant \alpha(t)\exp\biggl(\int_a^t\beta(s) ds\biggr)},$

for all $t\in [a,b]$.

• If, in addition, the function $\alpha \equiv 0$, then $\displaystyle \boxed{u(t) \leqslant \exp\biggl(\int_a^t\beta(s) ds\biggr)},$

for all $t\in [a,b]$.

Proof. If we differentiate the RHS of the integral inequality, we then obtain $\displaystyle \alpha '(t) + \beta(t) u(t).$

Hence, a solution $v$ to $y'(t)=\alpha '(t) + \beta(t) y(t)$ should take the following from $\displaystyle v(t) = \exp \left( - \int_a^t \beta(s) ds\right) \int_a^t \exp \left( \int_a^s \beta(r) dr \right) \alpha ' (s) ds.$

Note that, upon integrating by parts, we further obtain $\displaystyle v(t) = \alpha (t) - \exp \left( - \int_a^t \beta(s) ds\right) \int_a^t \alpha (s) \beta (s) \exp \left( \int_a^s \beta(r) dr \right) ds.$

If we need a plus sign after “= \alpha (t)”, we simply change $\int_a^s \beta(r) dr$ to $\int_s^t \beta(r) dr$, hence we obtain the RHS of the desired inequality.

If the function $\alpha$ is non-decreasing, then $\alpha (s) \leqslant \alpha (t)$, and the fundamental theorem of calculus implies the desired estimate.

Type 2. Bounds by integrals based on upper bound $b$.

Let $\alpha$ $\beta$ and $u$ be real-valued functions defined on $[a,b]$. Assume that $\beta$ and $u$ are continuous and that the negative part of $\alpha$ is integrable on every closed and bounded subinterval of $[a,b]$.

• If $\beta$ is non-negative and if $u$ satisfies the integral inequality $\displaystyle u(t) \leqslant \alpha(t) + \int_t^b \beta(s) u(s) ds,$

for all $t\in [a,b]$ then $\displaystyle \boxed{u(t) \leqslant \alpha(t) - \int_a^t\alpha(s)\beta(s)\exp\biggl(-\int_s^t\beta(r)dr\biggr) ds},$

for all $t\in [a,b]$.

• If, in addition, the function $\alpha$ is non-decreasing, then $\displaystyle \boxed{u(t) \leqslant \alpha(t)\exp\biggl(-\int_a^t\beta(s) ds\biggr)},$

for all $t\in [a,b]$.

• If, in addition, the function $\alpha \equiv 0$, then $\displaystyle \boxed{u(t) \leqslant \exp\biggl(-\int_t^b\beta(s) ds\biggr)},$

for all $t\in [a,b]$.

Proof. We again use Type 1.

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