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In this reference the author states what he calls "the theory of local solutions" for separable ordinary differential equations of the form $\frac{dy}{dx} = \frac{f(x)}{g(y)}$. He asserts that it suffices for $f$ and $g$ to be continuous and not to vanish simultaneously in a rectangular area $R$ of the plane in order for a unique solution to exist given an initial condition $(x_0,y_0)\in R$, but I have difficulty in interpreting his claim.

He does not specify what the domain of the solution will be, but since he talks about "local" solutions I believe that he is claiming that for each $(x_0,y_0)\in R$ there exist an open set $I$ of $\mathbb{R}$, contained in the projection of $R$ on the $x$-axis (and containing $x_0$), and a differentiable function $\phi : I \rightarrow \mathbb{R}$ such that

  1. $\phi(x_0) = y_0$
  2. for each $x \in I$ $\phi'(x) = f(x)/g(\phi(x))$
  3. $\phi : I \rightarrow \mathbb{R}$ is unique

I do not understand his requirement that $f(x)$ and $g(y)$ should not vanish simultaneously in $R$. I think that if $g(y_0)=0$, even if $f(x_0) \neq 0$, there should be no solution passing through $(x_0,y_0)$ because there would be an undefined value for the derivative of an eventual solution. Should the text be emended to exclude the possibility of $g(y_0)=0$?

edit: I should add that I also don't understand well what is meant by uniqueness given we are talking of a local solution and the domain of the solution is somewhat arbitrary

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up vote 1 down vote accepted

Since $f(x)$ depends only on $x$ and $g(y)$ only on $y$, I think that what the author wants to say is that $f(x)\ne0$ and $g(y)\ne0$ for all $(x,y)\in R$. Otherwise, as you note, the conclusions would not hold.

Think of the simplest example $y'=1/(2\,y)$; $f(x)=1$ never vanishes, and $g(y)=2\,y$ vanishes only at $y=0$. The solution with $y(x_0)=y_0>0$ is $y=\sqrt{x+y_0^2-x_0}$, defined on $[x_0-y_0^2,\infty)$. The case $y_0<0$ is similar. If $y_0=0$, then there ae two solutions, $y=\pm\sqrt{x-x_0\strut}$, defined on $[x_0,\infty)$, which are not differentiable at $x=x_0$.

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So if we assumed that $f(x) \neq 0$ and $g(y) \neq 0$ for all $(x,y)\in R$ does the result (local existence and uniqueness of the solution) hold? And if we assumed just that $y(x_0)=y_0 \neq 0$? – Cauchy Apr 24 '12 at 23:10
Let $G(y)=\int_{y_0}^yg(s)\,ds$ and $F(x)=\int_{x_0}^xf(s)\,ds$. Then the solution of $y'=f(x)/g(y)$, $y(x_0)=y_0$ is, in implicit form, $G(y)=F(x)$. If $y_0\ne0$, then $G'(y_0)=g(y_0)\ne0$. The implicit function theorem implies that $G$ is invertible on an open interval around $y_0$, and we obtain the solution in the form $y(x)=G^{-1}(F(x))$. So the answer to your question is yes, it is enough to have $g(y_0)\ne0$ to guarantee the existence and uniqueness of solution. – Julián Aguirre Apr 25 '12 at 10:39

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