Given $$\begin{cases} y'=f(x,y) \\ y(x_0)=y_0 \end{cases}$$ Is the condition f continuous in $(x_0,y_0)$ necessary for existence? I mean, for example the IVP

$$\begin{cases} yy'=-x \\ y(a)=0 \end{cases}$$

$y(x)^2=a^2-x^2$ satisfies the equation except than $y'$ is not continuous there. Are $y(x)=\sqrt{a^2-x^2}$ and $y(x)=-\sqrt{a^2-x^2}$ solutions or that IVP has no solution?

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    $\begingroup$ I'm sure you could say it doesn't exist if you were to be really pedantic with your definitions of such an IVP ($f$ does not even exists at the initial point). But these kinds of systems are commonly studied and an IVP like this is often called a singular IVP. The important thing to understand is that existence theorems gives conditions only for when a solution has to exist. There might be situations where solutions exists even if the theorem is violated. $\endgroup$
    – Winther
    Sep 13, 2019 at 3:03

1 Answer 1


This is straight out of Ordinary Differential Equations by Richard K. Miller and Anthony N. Michel, page 47:

Theorem 2.3: If $f \in C(D)$ and $(\tau,\xi)\in D$, then the IVP has a solution defined on $|t-\tau| \leq c.$

The IVP they're referring to is:

\begin{cases} x'=f(t,x) \\ x(\tau)=\xi \end{cases}

Note that $D \subset \mathbb{R^2}$ is a domain, that is $D$ is an open, connected (and nonempty) set the plane $(t,x)$.

This is usually the first theorem in existence of a solution given in a beginning graduate (or advanced undergraduate) level ODE class.

Notice that $(\tau,\xi) \in D$, otherwise the theorem really doesn't apply and we can't say much about existence.

Going to your example, we have:

\begin{cases} yy'=-x \\ y(a)=0 \end{cases}

Note that in order to have a solution, we need that $f(x,y)= -\frac{x}{y}$ be continuous on a domain $D \subset \mathbb{R^2}$ and have $(a,0) \in D$. However, notice that $f$ is defined and continuous on the set $$U = \{(x,y) \in \mathbb{R^2}:\hspace{1mm} x\in \mathbb{R} ,y \neq0 \}$$

Note also that $U = D_1 \cup D_2 $, so a domain $D$ for which $f$ is continuous (and the theorem applies) is either the upper (or lower) half plane. Here $D_1$ and $D_2$ are in fact the lower and upper half planes.

The initial condition for this example is not in either of these domains (and hence not in $U$).

However, as the comment above by Winther states, this is just a theorem which gives conditions for when a solution has to exist. More precisely, the condition is sufficient but not necessary for existence of a solution.


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