Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 12 1 Solution 2026-10-14
Choose , and . On this smaller closed polydisc, the holomorphic function and its -derivative are bounded; write and . Choose with and , replacing a zero bound by any positive upper bound if necessary, and choose .
Consider functions continuous on , jointly holomorphic in the interior, with and . This is a closed subset of a Banach space in the supremum norm. The integral operatoris well defined on the straight segment and is jointly holomorphic. Indeed, it equals , an integral of jointly holomorphic functions locally uniformly bounded in both variables. Convexity of the -disc and the derivative bound giveThus the Banach fixed-point theorem gives a fixed point . More explicitly, the successive integral iterates starting from zero converge uniformly, and locally uniform convergence of holomorphic functions makes their limit jointly holomorphic. Differentiating the integral equation gives the required complex ordinary differential equation and initial value. This proves local existence, uniqueness and joint holomorphic dependence on . Uniqueness among arbitrary local solutions follows first on a small disc where their values stay within , by the same contraction mapping estimate, and then throughout their common connected domain by the identity theorem.
For holomorphic dependence of ordinary differential equations on parameters in the initial value, fix an admissible initial value and put . The equation becomesIts right-hand side is jointly holomorphic in near . The preceding construction supplies a common disc and a jointly holomorphic solution for near . Consequently is jointly holomorphic in and its initial value .