= Global Picard iteration with a decaying kernel
{c}
{title2=$4|b|<c$}
Consider the <Volterra integral equation>
$$
y(t)=b h(t)+\int_0^tK(t,s)y(s)^2\,ds,\qquad t\geq0,
$$
where $h$ is continuous, $|h|\leq1$, $h(0)=1$, and the continuous kernel satisfies $0\leq K(t,s)\leq e^{-c(t-s)}$ for $0\leq s\leq t$, with $c>0$. If $0<|b|<c/4$, there is exactly one continuous solution on the whole half-line. It is bounded, and <Picard iteration> beginning at $y_0=bh$ converges in <supremum norm> with contraction ratio at most $\gamma=4|b|/c<1$.
For a proof, the integral of the kernel is at most $1/c$. Thus $T$ maps the closed ball of radius $2|b|$ in the <Banach space> of <bounded continuous functions> into itself, since $\|Ty\|_\infty\leq|b|+4b^2/c<2|b|$. Factoring $y^2-v^2$ gives $\|Ty-Tv\|_\infty\leq\gamma\|y-v\|_\infty$ on this ball. The <contraction mapping theorem> gives its unique fixed point. A <first-exit argument> shows that every continuous solution, even one not initially assumed bounded, stays strictly inside the same ball: a first time with $|y|=2|b|$ would contradict $|y|\leq|b|+4b^2/c$. Reapplying the estimate gives the strict global bound $\|y\|_\infty\leq|b|(1+\gamma)<2|b|$, so uniqueness is unrestricted within the continuous solution class. If $b=0$, zero is the only solution by the <Gronwall inequality> on each finite interval.
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