Consider the following integral equation

$\phi(x) = f(x) + \frac{1}{x}\int_0^x N(x,y)\phi(y)\;dy$,

where $f$ and $N$ are continuous and bounded functions. Are solutions $\phi$ of the above equation unique? If so, can one get an estimate of the form

$\sup_{(0,x)} |\phi| \leq C \sup_{(0,x)}|f|$ ?

Additional info: Assume that $\phi(0)=\phi(1)=f(0)=0$ and that $\|N\|_\infty\geq 1$. I am only interested in a solution (or lack thereof) on the interval $[0,1]$.

As a note, without the $\frac{1}{x}$ multiplier, this is a Volterra equation of the second kind and existence/uniqueness of a solution $\phi$ is well-known and the above estimate is indeed satisfied.


1 Answer 1


Not in general, since if $N(x,y)=a>1$ then the equation with $f=0$ has the solution $\phi(x)=x^{a-1}$. If, however, $N(0,0)<1$ (assuming as in the question that $N$ is continuous) then one can use the Banach fixed-point theorem (on short intervals) to get a unique solution.

  • $\begingroup$ I assume you mean $|N(0,0)| < 1$. You can of course get a stronger existence result if you have $\| N \|_{\infty} < 1$. $\endgroup$ Oct 10, 2011 at 5:45
  • $\begingroup$ Thanks for the replies. I had noticed the contraction argument for $\|N\|_\infty < 1$, but the case I am interested in, I have $\|N\|_\infty > 1$, but I also have $\phi(0)=\phi(1)=0$ (and of course then $f(0)=0$). I should have added these to the problem statement (which I will do now). I would be happy with either uniqueness or non-existence. $\endgroup$
    – Jeff
    Oct 10, 2011 at 13:08

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