1
$\begingroup$

Let $k(x,t,z)$ be continuous for $0 \leq t \leq x \leq a$, $-\infty < z < \infty$ and satisfy a Lipschitz condition with respect to $z$ and let $h(x)$ be continuous for $0 \leq x \leq a$. Show that the Volterra Integral Equation $u(x) = h(x) + \int^{x}_{0} k(x,t,u(t))dt$ has exactly one continuous solution in $0 \leq x \leq a$.

I know that this is an application of Banach Fixed Point Theorem, just need some guidance on how to apply it directly to this problem.

  • 0
    One way to start is to notice that if there are two such solutions $u$ and $v$ then using the Lipschitz condition we must have $$|u(x) - v(x)| = \left|\int_0^xk(x,t,u(t)) - k(x,t,v(t))\right| \leq \int_0^x M|u(t) - v(t)|{\rm d}t$$ – 2017-02-11
  • 0
    Then we can conclude that $||u(x) - v(x)|| \leq |x| M||u(t) - v(t)|| $, where $|| *||$ is the max norm. – 2017-02-11

0 Answers 0