# 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 '18 at 16:28
• How about $(y')^2=1$ with $y(0)=0$? The two solutions are $y=x$ and $y=-x$. Dec 5 '18 at 16:31
• @BarryCipra thanks, but what about first order differential equaltion? Dec 5 '18 at 16:35
• I don't why there is a negative vote! Dec 5 '18 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 '18 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.