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I need the following result (which I believe to be true though I was too lazy to write down a complete proof).

Let $f$ be a function of two complex variables analytic at the origin and $a\not\in\mathbb{N}$. Then the differential equation $$z'=\frac{az}{x}+f(x,z)$$ has an unique solution $z(x)$ which is analytic at $x=0$.

Can anyone tell me a reference for this?

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    $\begingroup$ What is $z'$? $dz/dx?$. $\endgroup$
    – Alexandre Eremenko
    Commented Feb 26, 2018 at 14:01
  • $\begingroup$ Of course. $___$ $\endgroup$
    – Alex Gavrilov
    Commented Feb 26, 2018 at 14:05
  • $\begingroup$ Take $a=1/2$, $f=0$, so that the equation reads $2x z'=z$. Multiplying both sides by $z$, calling $y=z^2$, you get $xy'=y$ and thus $y=\alpha x$. If $z$ is analytic, $\alpha\not=0$ and $\zeta= \alpha^{-1/2}z$, you found an analytic function $\zeta$ such that $\zeta^2=x$, which is not possible. Thus $z=0$. $\endgroup$
    – Bazin
    Commented Feb 27, 2018 at 14:27

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These questions were studied for the first time by Briot and Bouquet in 19 century. For a modern reference see for example the book of E. Hille, Ordinary differential equations in the complex domain. The chapter is called Some equations of Briot and Bouguet. Your case is actually simple: plug a formal power series for $z(x)$ and see that all coefficients can be uniquely defined. Then prove convergence by Cauchy majorant method. Your statement is Theorem 11.1.1 on p. 403 in Hille.

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  • $\begingroup$ Take $a=1/2$, $f=0$, so that the equation reads $2x z'=z$. Multiplying both sides by $z$, calling $y=z^2$, you get $xy'=y$ and thus $y=\alpha x$. If $z$ is analytic, $:alpha\not=0$,and $\zeta= \alpha^{-1/2}z$, you found an analytic function $\zeta$ such that $\zeta^2=x$, which is not possible. Thus $z=0$. $\endgroup$
    – Bazin
    Commented Feb 27, 2018 at 14:27
  • $\begingroup$ @Bazin: Yes, in this example zero is the unique analytic function which satisfies your equation. The solution is always unique.. $\endgroup$
    – Alexandre Eremenko
    Commented Feb 27, 2018 at 15:39
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Your equation falls in the category of regular singular differential equation. Writing your equation as $$ x z'=a z+g(x,z), \tag{$\ast$}$$ the singularity is called regular because the exponent of the singularity at zero is the same as the order of derivatives. Examples are Bessel equations $ x^2z''+xz'+(x^2-\nu^2) z=0, $ which can be written as a $2\times 2$ system of your type. Going back to $(\ast)$ for $g=0$, you get on the real line all the homogeneous distributions of degree $a$, that is $$ z_{\alpha, \beta}(x)=\alpha x_+^a+\beta x_-^a,\quad x_+^a=H(x) x^a, \quad x_-^a=H(-x)(-x)^a, $$ where $H=\mathbf 1_{\mathbb R_+}$. On the complex plane, you need a priori to withdraw a half-line starting at 0 to define a logarithm and you get $$ z=\alpha e^{a \text{Log x}}, $$ which has no holomorphic extension except if $a\in \mathbb N$ or $\alpha =0$. In general setting $t=\text{Log x}$, $z(e^t)=\zeta (t)$, you find the equation $$ \dot \zeta=a\zeta+ g(e^t,\zeta), $$ which has no singular point.

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