7

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?

Arctic Char
  • 16,007
Cloud JR K
  • 2,466
  • I am new to this topic , so i will add more details of what i try if i got some ideas – Cloud JR K Dec 05 '18 at 16:28
  • How about $(y')^2=1$ with $y(0)=0$? The two solutions are $y=x$ and $y=-x$. – Barry Cipra Dec 05 '18 at 16:31
  • @BarryCipra thanks, but what about first order differential equaltion? – Cloud JR K Dec 05 '18 at 16:35
  • I don't why there is a negative vote! – Cloud JR K Dec 05 '18 at 16:36
  • @CloudJR Barry's example IS first order -- it goes by derivatives :) but if your equation is linear, that would be the implication you are looking for – gt6989b Dec 05 '18 at 16:36
  • @gt6989b I'm sorry, i messed up – Cloud JR K Dec 05 '18 at 16:38
  • @gt6989b, actually my example is not first order, which means the equation is of the form $y'=f(t,y)$, which $y'=\pm1$ is not, since $\pm1$ is not a function. (The OP added "first order" after I posted my comment.) – Barry Cipra Dec 05 '18 at 16:38
  • @BarryCipra this is incorrect. You are defining explicit first order equations, and the one you posted is implicit but still first-order. See Wiki ODE page – gt6989b Dec 05 '18 at 16:43
  • 1
    @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. – Barry Cipra Dec 05 '18 at 16:51

1 Answers1

8

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.

John Omielan
  • 47,976
Artem
  • 14,414