Gronwall’s inequality

hbghlyj

Wikipedia - Grönwall's inequality
people.math.wisc.edu/~robbin/angelic/gronwall.pdf

DE1notes
Theorem 1.2. (Gronwall's inequality): Suppose $A \geq 0$ and $b \geq 0$ are constants and $v$ is a non-negative continuous function satisfying
\[
v(x) \leq b+A\left|\int_a^x v(s) d s\right|\tag{1.20}\label{1.20}
\]
then
\[
v(x) \leq b e^{A|x-a|} .
\]
(The modulus is needed to take care of the case $x \leq a$.)
Proof: We use an integrating factor.
For $x \geq a$ let $V(x)=\int_a^x v(s) d s$, so that $V^{\prime}(x)=v(x)$.
As $x \geq a$ and $v \geq 0$, we have $V(x) \geq 0$, so \eqref{1.20} becomes
\[
V^{\prime}(x) \leq b+A V(x) .
\]
Multiply through by the integrating factor $e^{-A x}$ so
\begin{aligned}
\left(V^{\prime}(x)-A V(x)\right) e^{-A x} &\leq b e^{-A x} \text { that is }\\
\frac{d}{d x}\left(V(x) e^{-A x}\right) & \leq b e^{-A x}, \text { so, integrating and noting that } V(a)=0 \\
V(x) e^{-A x} & \leq \int_a^x b e^{-A s} d s=\frac{b}{A}\left(e^{-A a}-e^{-A x}\right) \text {, so } \\
V(x) & \leq \frac{b}{A}\left(e^{A(x-a)}-1\right)
\end{aligned}Finally, using \eqref{1.20}
\[
v(x) \leq b+A \int_a^x v(s) d s=b+A V(x) \leq b+A \frac{b}{A}\left(e^{A(x-a)}-1\right)=b e^{A(x-a)}
\]
as required. Similarly if $x≤a$.  $\boxed{\,}$

hbghlyj

Remark: Gronwall's inequality says that $v$ is bounded above by the solution of the integral equation one obtains when there is equality in \eqref{1.20}). For, if we differentiate
\[
v(x)=b+A \int_a^x v(s) d s,
\]
we get
\[
v^{\prime}(x)=A v(x), \quad v(a)=b,
\]
which has solution $v(x)=b e^{A(x-a)}$.
Suppose now that $y$ and $z$ are solutions of the ordinary differential equation $y^{\prime}(x)=$ $f(x, y(x))$ with $y(a)=b$ and $z(a)=c$, where $f$ satisfies conditions $(\mathbf{P}(\mathbf{i}))$ and $(\mathbf{P}(\mathbf{i i}))$. Then
\[
y(x)-z(x)=b-c+\int_a^x(f(s, y(s))-f(s, z(s)) d s
\]
so setting $v(x)=|y(x)-z(s)|$ (note Gronwall requires $v$ to be non-negative) we get that
\[
|y(x)-z(x)| \leq|b-c|+\left|\int_a^x L\left| y(s)-z(s)\right|d s\right|
\]
and by Gronwall's inequality
\[\tag{1.21}\label{1.21}
|y(x)-z(x)| \leq|b-c| e^{L|x-a|} \leq|b-c| e^{L h} .
\]
Thus we have related $|y(x)-z(x)|$ to $|b-c|$.
We say a solution is continuously dependent on the initial data if we can make $|y(x)-z(x)|$ as small as we like by taking $|b-c|$ small enough. In other words the error in the solution will be small provided the error in the initial data is small enough. To be precise, in this case, solutions are continuously dependent on the initial data for $x \in[a-h, a+h]$ if for all $\epsilon>0$ there exists $\delta>0$ such that if $y$ and $z$ are as above,
\[
|b-c|<\delta \quad \Rightarrow|y(x)-z(x)| \leq \epsilon, \quad \forall x \in[a-h, a+h]
\]
This is clearly true from \eqref{1.21}, because given $\epsilon>0$, we have $|y(x)-z(x)| \leq \epsilon$ whenever $|b-c|<e^{-L h} \epsilon$, so we can take $\delta=e^{-L h} \epsilon$.

We could also use this to prove uniqueness. If $y$ and $z$ are solutions with $b=c$, then we get $y(x)=z(x)$ for every $x$ so the solution is unique.