I was reading existence and uniquensess theorem , form that I got that if certain condition are satisfied then we can have uniue solution passing through that point
I was thinking if condition not satisfied then what are possible choices for solution to have
Like in linear algebra system of equation either we have unique solution , or no solution or infinitely many solution
There it is not possible to have 2 solution or n solution .
So is this same in case of ODE
Any help will be appreciated
Take the ODE $$\dot{x} = x^{2/3}, \quad x(0) = 0$$ Note that $\dot{x}$ is not Lipschitz at $x=0$, this gives uncountably many solutions for any $c\in\mathbb{R}$ we have $$x_c(t) = \begin{cases} 0 \text{ for } -\infty \leq t \leq c\\ (\frac{t -c}{3})^3 \text{ for } t> c \end{cases}$$ is a solution. There are other cases when there are no solutions as well. Consider $$t \dot{x}(t) = x(t)$$ Then $x(\lambda t) = \lambda x(t)$, for any $\lambda \in \mathbb{R}$. This implies that any solution will have for $t=1$, $\lim_{\epsilon \rightarrow 0} -\epsilon x(-\epsilon) =\lim_{\epsilon \rightarrow 0} \epsilon x(\epsilon)$. If we ask that a "solution" be continuous, then any solution must have that $x(0) =0$. Thus the system $$t \dot{x}(t) = x(t), \quad x(0) = 1$$ has no solution. The difficulty is what we mean by solution. If we want continuity then the vector field must have some regularity around the initial condition.
In general an $n$th order O.D.E. can have at most $n$ linearly independent solutions. The reasons for this are pretty much the same as there being $n$ linearly independent solutions to a system of $n$ equations. For any $n$ solutions we get a system of $n$ solutions to the equation
$$a_ny^{(n)}(0)+....+a_1y^{(1)}(0)+a_0y^{(0)}(0)=c,$$
And since the initial conditions are given we have
$$a_ny^{(n)}(0)+....+a_1y^{(1)}(0)=c-a_0y^{(0)}(0).$$
If we think of $y^{(n)}(0),...,y^{(1)}(0)$ as just coordinates, then this equation can have at most $n$ linearly independent solutions. This should give you some intuition as to why there are $n$ solutions. For a more detailed argument look here.
