# Existence and uniqueness ODE

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?

• 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. Sep 13, 2019 at 3:03

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.