2
$\begingroup$

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.

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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