mhill
Aug19-08, 10:50 AM
let be y_a (x) the solution to
\delta (x-a) = \int_{0}^{1} K(x,s)y(s,a)
then i think that the solution to
f(x) = \int_{0}^{1} K(x,s)g(s) is given by g(s)= \int_{-\infty}^{\infty} y(s,t)f(s) dt
since the operator is linear, i think that if this is deduced for a differential operator then it should work also for integral (linear) operators too.
\delta (x-a) = \int_{0}^{1} K(x,s)y(s,a)
then i think that the solution to
f(x) = \int_{0}^{1} K(x,s)g(s) is given by g(s)= \int_{-\infty}^{\infty} y(s,t)f(s) dt
since the operator is linear, i think that if this is deduced for a differential operator then it should work also for integral (linear) operators too.