7. (Iterated kernel, resolvent kernel) Show that in the Neumann series in Prob. 4 we can write (S")(t)= Koax(1 T)u(r) dt n=2,3,..., where the iterated kernel kin) is given by km,(1,7)= -| ---[k(5,4)(61, 62) K(tu-1, 7) dtg - dios so that the Neumann series can be written X(t)= v(t) + 4 o k(1, T)U(T) dT + kz,(1,7)U(T) dT + ... 326 Banach Fixed Point Theorem or, by the use of the resolvent kernel k defined by k(t, t, k) = { v'ky+v(1,7) (k(1) = k) 1-0 it can be written x(t)= v(t) + u S* Kl, 7, d)u(7) dt. 8. It is of interest that the Neumann series in Prob. 4 can also be obtained by substituting a power series in Ho *(t)= vo(t)+uv (t)+u? uz(t) +... into (1), integrating termwise and comparing coefficients. Show that this gives vo(t)= v(t), -B* k, T), - (T) d, n=1,2,.... Assuming that lu(t) $co and k(1) Sc show that |v.(t) calc(b-a)]", so that (5) implies convergence. 7. (Iterated kernel, resolvent kernel) Show that in the Neumann series in Prob. 4 we can write (S")(t)= Koax(1 T)u(r) dt n=2,3,..., where the iterated kernel kin) is given by km,(1,7)= -| ---[k(5,4)(61, 62) K(tu-1, 7) dtg - dios so that the Neumann series can be written X(t)= v(t) + 4 o k(1, T)U(T) dT + kz,(1,7)U(T) dT + ... 326 Banach Fixed Point Theorem or, by the use of the resolvent kernel k defined by k(t, t, k) = { v'ky+v(1,7) (k(1) = k) 1-0 it can be written x(t)= v(t) + u S* Kl, 7, d)u(7) dt. 8. It is of interest that the Neumann series in Prob. 4 can also be obtained by substituting a power series in Ho *(t)= vo(t)+uv (t)+u? uz(t) +... into (1), integrating termwise and comparing coefficients. Show that this gives vo(t)= v(t), -B* k, T), - (T) d, n=1,2,.... Assuming that lu(t) $co and k(1) Sc show that |v.(t) calc(b-a)]", so that (5) implies convergence