how to solve general integral equations with both variable lower and upper bounds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:16:33Z http://mathoverflow.net/feeds/question/42190 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42190/how-to-solve-general-integral-equations-with-both-variable-lower-and-upper-bounds how to solve general integral equations with both variable lower and upper bounds Haining Yu 2010-10-14T18:12:29Z 2010-10-14T18:19:52Z <p>I am interested in solving the following equations:<br> $f(x) + \int_{\alpha(x)}^{\beta(x)}K(x,t)u(t)dt = 0$<br> and<br> $f(x) + \int_{\alpha(x)}^{\beta(x)}K(x,t)u(t)dt = u(x)$<br> 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.</p> <p>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.</p> <p>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?</p> <p>Thanks!</p> http://mathoverflow.net/questions/42190/how-to-solve-general-integral-equations-with-both-variable-lower-and-upper-bounds/42193#42193 Answer by Piero D'Ancona for how to solve general integral equations with both variable lower and upper bounds Piero D'Ancona 2010-10-14T18:19:52Z 2010-10-14T18:19:52Z <p>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 $A$.</p> <p>If $\alpha,\beta$ are unbounded, then we need some thinking...</p>