Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 12 / 1

Choose $0<R<R_1$, $0<S<S_1$ and $0<T<T_1$. On this smaller closed polydisc, the holomorphic function $A$ and its $w$-derivative are bounded; write $|A|\le M$ and $|{\partial }_{w}A|\le L$. Choose $0<r<R$ with $rM<S/2$ and $rL<1$, replacing a zero bound by any positive upper bound if necessary, and choose $0<\epsilon <T$.

Consider functions $u(z,\alpha )$ continuous on $\overline{D(0,r)}\times \overline{D(0,\epsilon )}$, jointly holomorphic in the interior, with $u(0,\alpha )=0$ and $\Vert u{\Vert }_{\infty }\le S/2$. 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 $z{\int }_0^{1}A(tz,u(tz,\alpha ),\alpha )dt$, an integral of jointly holomorphic functions locally uniformly bounded in both variables. Convexity of the $w$-disc and the derivative bound giveThus the Banach fixed-point theorem gives a fixed point $f(z,\alpha )$. 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 $(z,\alpha )$. Uniqueness among arbitrary local solutions follows first on a small disc where their values stay within $|w|<S$, 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 $w_{\ast }$ and put $u=f-w_0$. The equation becomesIts right-hand side is jointly holomorphic in $(z,u,w_0)$ near $(0,0,w_{\ast })$. The preceding construction supplies a common disc and a jointly holomorphic solution for $w_0$ near $w_{\ast }$. Consequently $f=u+w_0$ is jointly holomorphic in $z$ and its initial value $w_0$.

The assertion is true. The assumed finite multiplicity means that $f_0-w_1$ is not identically zero and has an isolated zero of order $k$ at $z_1$. Choose $\rho >0$ so that $\overline{D(z_1,\rho )}\subset D(0,r)$ and $f_0-w_1$ has no other zeros in this closed disc. On its boundary,The joint holomorphy established above gives $f_{\alpha }\to f_0$ uniformly on that circle as $\alpha \to 0$. For sufficiently small $|\alpha |$, therefore, $|f_{\alpha }-f_0|<\eta$. By Rouché's theorem, $f_{\alpha }-w_1$ has exactly $k$ zeros in $D(z_1,\rho )$, counted with multiplicity. In particular it has at least one zero in $D(0,r)$. This preserves the total local zero count, without asserting that the zeros remain together.

The assertion is false for $k\ge 2$, but true for $k=1$. For $k\ge 2$, take the jointly holomorphic right-hand side $A_{\alpha }(z,w)=k{z}^{k-1}+\alpha$. Its solution with zero initial value isAt $\alpha =0$, the value zero has multiplicity $k$ at the origin. For $\alpha \ne 0$, however, every zero is a simple zero: $f_{\alpha }^{\prime }(0)=\alpha$, and at a nonzero zero $z^{k-1}=-\alpha$,Thus there is no zero of multiplicity $k$ anywhere, even though the $k$ simple zeros stay close to zero for small $\alpha$. When $k=1$, part (a) gives exactly one zero counted with multiplicity in a small disc, so that zero is simple. Alternatively, the holomorphic implicit function theorem gives its holomorphic location as a function of $\alpha$.