I would like to proof Grönwall inequality by successive approximation method. Grönwall inequality is as below:
Let $k>0$ and $f(t),g(t)$ are continous and non-negative function in intevral $a\leq t \leq b$.
We also have:
\begin{equation}
f(t)\leq K+ \int_{a}^{t}f(s)g(s)ds \qquad a\leq t \leq b. \qquad(1)
\end{equation}
and would like to prove:
\begin{equation}
f(t)\leq Kexp(\int_{a}^{t}g(s)ds) \qquad a\leq t \leq b. \qquad(2)
\end{equation}
My ideas so far:
since $f(t)\leq K+ \int_{a}^{t}f(s)g(s)ds$, there exists $r(t)<0$ s.t. $f(t)=K+ r(t)+\int_{a}^{t}f(s)g(s)ds$.
Let
\begin{equation}
\begin{aligned}
f_0(t)&=K\\
f_n(t)&=K+r(t)+\int_{a}^{t}f_{n-1}(s)g(s)ds \quad(n=1,2,….)
\end{aligned}
\end{equation}
I would like to proof $|f_{n+1}(t)-f_n(t)|\leq \frac{K}{n!}[\int_{a}^{t}g(s)ds]^n$ by induction. So I can use this inequality to proof the uniformly convergence of $f_n(t)$.
However, $|f_{1}(t)-f_0(t)|=|r(t)+K\int_{a}^{t}g(s)ds|$ which means $n=1$ doesn't satisfy the above inequality. So, I cannot go on. Do I make some mistakes in my proof?
Thanks in advance for any help!
Best Answer
Define a sequence of upper bounds via $$ u_0=f, ~~u_{n+1}(t)=K+\int_0^t g(s)u_n(s)\,ds $$ Then obviously $$ f=u_0\le u_1\le u_2\le... $$ Now employ partial integration, with $G(t)=\int_0^t g(s)\,ds$ and $u_n'(t)=g(t)u_{n-1}(t)$ \begin{align} u_{n+1}(t)&=K+G(t)u_n(t)-\int_0^t G(s)u_n'(s)\,ds \\ &=K+G(t)\left[K+\int_0^t g(s)u_{n-1}(s)\,ds\right]-\int_0^t G(s)g(s)u_{n-1}(s)\,ds \\ &=K(1+G(t))+\int_0^t (G(t)-G(s))g(s)u_{n-1}(s)\,ds \\ &=K(1+G(t))-\frac12\left[(G(t)-G(s))^2u_{n-1}(s)\right]_{s=0}^{s=t} +\frac12\int_0^t (G(t)-G(s))^2u_{n-1}'(s)\,ds \\ &=K\left(1+G(t)+\frac12G(t)^2\right)+\frac12\int_0^t (G(t)-G(s))^2g(s)u_{n-2}(s)\,ds \end{align} etc.