## how to solve general integral equations with both variable lower and upper bounds

I am interested in solving the following equations:
$f(x) + \int_{\alpha(x)}^{\beta(x)}K(x,t)u(t)dt = 0$
and
$f(x) + \int_{\alpha(x)}^{\beta(x)}K(x,t)u(t)dt = u(x)$
when $u(x)$ is the unknown function defined on $[0^+,\infty)$ and all other functions are known and are assumed to have convenient differentiability and other reasonable properties. In particular, both lower and upper bounds $\alpha(x)$ and $\beta(x)$ are not constants.

I know that when both bounds are constant they belong to Fredholm equations, and when only one bound is constant they belong to Volterra equations, where plenty of literature work exists.

However I am not clear about the case when both bounds are variables. Can the equations defined above be conveniently translated into Fredholm or Volterra theory? If they cannot, what is the theory for these type of equations?

Thanks!

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If $\alpha$ and $\beta$ are bounded functions, with $a\le\alpha\le\beta \le b$, then your equations are of standard type since they can be written as $$f(x)+\int_a^b H(x,t) u(t) dt =0$$ where $$H(x,t)=K(x,t)\chi_{[\alpha(x),\beta(x)]}(t).$$ Here $\chi_A$ denotes the characteristic function of the set $A$.
If $\alpha,\beta$ are unbounded, then we need some thinking...
 Thank you for your answer. In my case $\alpha(x)$ and $\beta(x)$ are bounded. So hopefully this saves some discussion on unbounded case. However, using indicator function (or more generally speaking characteristic function) may impose make the kernal function indifferentiable or even discontinuous. Is this considered a big problem, analytically or numerically? Again thank you very much. – Haining Yu Oct 14 2010 at 18:53 Analytically there is no problem. Numerically I do not know for sure, but I think it should be not very different from the case with continuous kernel. – Piero D'Ancona Oct 14 2010 at 19:43