I am self-studying the book Riemann Surfaces by Simon Donaldson. He leaves the proof of the proposition below as an exercise. It is clear to me that one can find local power series solutions at each point along $\gamma$; however, I am not sure how to construct local solutions such that they agree on their common domain of definition. Any help would be appreciated. Thank you!
Proposition
Suppose $\Omega$ is an open subset of $\mathbb{C}$ and $\gamma:[0, 1] \to \Omega$ is a path in $\Omega$. Furthermore, suppose $P$ and $Q$ are holomorphic functions defined on a neighborhood $U$ of $\gamma(0)$ with analytic continuations along $\gamma$. If $u$ is a solution to
\begin{equation} \label{eq:1.1}
u'' + P u' + Q u = 0
\end{equation}
on $U$, then $u$ has an analytic continuation along $\gamma$, through solutions of the equation.
Best Answer
After reading up more on the theory of differential equations in the complex domain using these notes, I was able to construct the following proof, which I'm pretty sure is correct.
Proof
Denote $z_t \equiv \gamma(t)$. By the uniform continuity of $\gamma$, the fact that $P, Q$ have analytic continuations along $\gamma$ by assumption, and the fact that analytic continuations are independent of the specific choice of chain of discs, there exists a chains of discs $D_j := D(z_{t_j}, r_j)$ along $\gamma$ in $\Omega$, where $0 = t_0 < t_1 < \cdots < t_J = 1$, $0 \leq j \leq J$, and $\gamma([t_j, t_{j + 1}]) \subseteq D_j \cap D_{j + 1}$ for $0 \leq j \leq J - 1$, and $P_j, Q_j \in H(D_j)$ such that $P_j = P_{j + 1}$ and $Q_j = Q_{j + 1}$ on $D_j \cap D_{j + 1}$ and $P_0 = P$ and $Q_0 = Q$ on a neighborhood $U_0$ of $z_0$.
Define $u_0$ as follows. Without loss of generality, since we can always increase the fineness of our chain of discs, suppose that $D(z_0, r_0)$ is a subset of $U$ and $U_0$, so $P_0 = P$ and $Q_0 = Q$ on $D(z_0, r_0)$. Since $D(z_0, r_0)$ is simply connected and $P_0 = P$ and $Q_0 = Q$ are holomorphic on $D(z_0, r_0)$, it follows by a theorem from differential equations that there exists a unique holomorphic function $u_0$ on $D(z_0, r_0)$ such that $u_0$ is a solution to the differential equation and satisfies the initial conditions $u_0(z_0) = u(z_0)$ and $u_0'(z_0) = u'(z_0)$. Furthermore, by applying the same theorem to $U$, it follows by uniqueness that the restriction of $u$ to $D(z_0, r_0)$ coincides with $u_0$ on $D(z_0, r_0)$.
Now, construct $u_j(z)$ inductively for $1 \leq j \leq J$ as follows. Suppose we have determined $u_{j - 1}(z)$ on $D(z_{j - 1}, r_{j - 1})$. Let $X_j = D(z_{j - 1}, r_{j - 1}) \cup D(z_j, r_j)$, and define the holomorphic functions $P_{j - 1, j}$ and $Q_{j - 1, j}$ on $X_j$, where $P_{j - 1, j}(z) = P_{j - 1}(z)$ and $Q_{j - 1, j}(z) = Q_{j - 1}(z)$ for $z \in D(z_{j - 1}, r_{j - 1})$ and $P_{j - 1, j}(z) = P_j(z)$ and $Q_{j - 1, j}(z) = Q_j(z)$ for $z \in D(z_j, r_j)$. Since $X_j$ is simply connected and $P_{j - 1, j}$ and $Q_{j - 1, j}$ are holomorphic on $D(z_0, r_0)$, there exists a unique holomorphic function $B_j$ on $X_j$ such that $B_j$ is a solution to the differential equation and satisfies the initial conditions $B_j(z_{j - 1}) = u_{j - 1}(z_{j - 1})$ and $B_j'(z_{j - 1}) = u_{j - 1}'(z_{j - 1})$, so we can define $u_j$ as the restriction of $B_j$ to $D(z_j, r_j)$. Furthermore, by applying the same theorem to $D(z_{j - 1}, r_{j - 1})$, it follows by uniqueness that the restriction of $B_j$ to $D(z_{j - 1}, r_{j - 1})$ coincides with $u_{j - 1}$ on $D(z_{j - 1}, r_{j - 1})$. By induction, it follows that there exists $u_j$ for $0 \leq j \leq J$ such that $u_j = u_{j +1}$ on $D_j \cap D_{j + 1}$ and $u_0 = u$ on a neighborhood of $z_0$, so we have constructed an analytic continuation of $u$ along $\gamma$.