# Does non-uniqueness of solution to 1st order ODE implies the existence of infinitely many solutions?

Consider the following initial value problem (IVP) to the first order ODE:

$$\tag{1} \dot x = f(t, x), \ \ \ x(t_0) = x_0.$$

The Existence and Uniqueness Theorem describes when one has exactly one solutions. This is true (e.g.) when $$f$$ is continuously differentiable.

There are examples where uniqueness fails and existence fails (here or here):

In all of the examples where I am aware of, whenever one has more than one solutions, one actually has infinitely many. Hence my question:

Can an IVP has more than one solutions, but only finitely many solutions?

• I am new to this topic , so i will add more details of what i try if i got some ideas Dec 5, 2018 at 16:28
• How about $(y')^2=1$ with $y(0)=0$? The two solutions are $y=x$ and $y=-x$. Dec 5, 2018 at 16:31
• @BarryCipra thanks, but what about first order differential equaltion? Dec 5, 2018 at 16:35
• I don't why there is a negative vote! Dec 5, 2018 at 16:36
• @gt6989b, OK, I was going by a restricted definition of "first-order." It does seem clear, from the comments, that the OP is interested in the explicit variety, so the exercise suggested in Artem's answer would seem to be of interest. Dec 5, 2018 at 16:51

The answer is positive: in general (apart of some artificial examples) if an IVP for an ODE $$\dot x=f(t,x),\quad x(t)\in \mathbb R^n$$ have non unique solution it implies that there are uncountably many solutions. In these general setting this is a very nontrivial theorem which can be found in Hartman's book (Kneser's theorem). If, however, you are dealing with an ODE $$\dot x=f(t,x),\quad x(0)=x_0,$$ where $$x(t)$$ is one-dimensional, it is a good (and simple) exercise to prove that if there are two solutions to this problem then there are infinitely (uncountably) many solutions.