Integral Equations

with the initial conditiony(x 0 ) = y 0 , (32)where f(x, y) is defined and continuous in a two-dimensional domain G which contains thepoint (x 0 , y 0 ).Integrating (31) subject to (32) we obtainφ(x) = y 0 +∫ xx 0f(t, φ(t))dt, (33)which is called the Volterra integral equation of the second kind with respect to the unknownfunction φ(t). This equation is equivalent to the initial value problem (31) and (32). Note thatthis is generally a nonlinear integral equation with respect to φ(t).Consider now a linear ODE of the second order with variable coefficientswith the initial conditiony ′′ + A(x)y ′ + B(x)y = g(x) (34)y(x 0 ) = y 0 , y ′ (x 0 ) = y 1 , (35)where A(x) and B(x) are given functions continuous in an interval G which contains the pointx 0 . Integrating y ′′ in (34) we obtainy ′ (x) = −∫ xx 0A(t)y ′ (x)dx −∫ xx 0B(x)y(x)dx +Integrating the first integral on the right-hand side in (36) by parts yieldsy ′ (x) = −A(x)y(x) −∫ xIntegrating a second time we obtainy(x) = −∫ xx 0A(x)y(x)dx−Using the relationship∫ xwe transform (38) to obtainy(x) = −∫ xx 0(B(x) − A ′ (x))y(x)dx +x 0∫ xx 0(B(t)−A ′ (t))y(t)dtdx+∫ xx 0∫ xx 0f(t)dtdx =∫ xx 0{A(t)+(x−t)[(B(t)−A ′ (t))]}y(t)dt+∫ x∫ x∫ xx 0g(x)dx + y 1 . (36)x 0g(x)dx + A(x 0 )y 0 + y 1 . (37)x 0∫ xx 0g(t)dtdx+[A(x 0 )y 0 +y 1 ](x−x 0 )+y 0 .(38)x 0(x − t)f(t)dt, (39)∫ xx 0(x−t)g(t)dt+[A(x 0 )y 0 +y 1 ](x−x 0 )+y 0 .Separate the known functions in (40) and introduce the notation for the kernel functionThen (40) becomes(40)K(x, t) = −A(t) + (t − x)[(B(t) − A ′ (t))], (41)f(x) =∫ xx 0(x − t)g(t)dt + [A(x 0 )y 0 + y 1 ](x − x 0 ) + y 0 . (42)∫ xy(x) = f(x) + K(x, t)y(t))dt, (43)x 0which is the Volterra IE of the second kind with respect to the unknown function φ(t). Thisequation is equivalent to the initial value problem (34) and (35). Note that here we obtain alinear integral equation with respect to y(x).9