Gronwalls Integral Inequality

Let g(x) and h(x) be continuous functions defined for $x>x_{0}$ such that $g(x),h(x)\ge 0$ Let k>0 be a constant such that $g(x)\le k+\int_{x_{0}}^{x}g(t)h(t)dt$ Then: $$g(x)\le ke^{\int_{x_{0}}^{x}h(t)dt}$$

Proof #

$$ \begin{align*} \frac{g(x)h(x)}{k+\int_{x_{0}}^{x}g(t)h(t)dt}&\le h(x)\\ \text{Integrating both sides}\\ \left. \left(\log \left(k+\int_{x_{0}}^{x} g(t)h(t)dt\right)\right) \right|_{x_{0}}^{x}&\le \int_{x_{0}}^{x}h(t)dt\\ \log\left(k+\int_{x_{0}}^{x}g(t)h(t)dt\right)-\log(k)&\le \int_{x_{0}}^{x}h(t)dt\\ \\ \left(k+\int_{x_{0}}^{x}g(t)h(t)dt\right)k^{-1}&\le e^{\int_{x_{0}}^{x}h(t)dt}\\ \text{Using the given,}\\ g(x)&\le k e^{\int_{x_{0}}^{x}h(t)dt} \end{align*} $$

Corollary #

If $g(x)\ge0$ and k is a constant such that $k>0$, then if $g(x)\le k\int_{x_{0}}^{x}g(t)dt$, Then g(x)=0

We can use Gronwalls Integral Inequality with $g(x)=g(x);h(x)=k$

$$ \begin{align*} \forall \epsilon>0\\ g(x)&\le \epsilon+\int_{x_{0}}^{x}g(t)kdt\\ g(x)&\le \epsilon e^{\int_{x_{0}}^{x}kdt}\\ &\le \epsilon e^{k(x-x_{0})}\\ g(x)&= 0 \end{align*} $$