Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question

1 Answer 1

up vote 5 down vote accepted

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.

share|cite|improve this answer
I assume you mean $|N(0,0)| < 1$. You can of course get a stronger existence result if you have $\| N \|_{\infty} < 1$. –  Christopher A. Wong Oct 10 '11 at 5:45
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. –  Jeff Oct 10 '11 at 13:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.