The existence and uniqueness theorem claims that if $F(x,y)$ and $F_y(x,y)$ are continuous around $(x_0,y_0)$, then there exists a unique solution to the following IVP (=Initial Value Problem) in some open interval around $x_0$. $$ \begin{cases} y'=F(x,y)\\ y(x_0)=y_0 \end{cases} $$

I have two basic questions regarding to IVP's and to above theorem:

(a) Suppose that $y(x)$ satisfies the equation $y'=F(x,y)$ in some open interval and suppose that $y(x_0)=y_0$. Can we say that $y(x)$ is a solution of the IVP? To sharpen my question, consider the following IVP $$ \begin{cases} y'=\frac{x-1}{y}\\ y(3)=0 \end{cases} $$ Here $F(x,y)=\frac{x-1}{y}$ is not continuous in $(3,0)$, so we can not apply the Existence Theorem. Nevertheless, the function $$ y(x)=\sqrt{(x+1)(x-3)} $$ satisfies $y(3)=0$ and satisfies the equation in the interval $(3,\infty)$. Can we say that $y(x)$ is a solution to the IVP ?

(b) If the conditions of the Existence and uniqueness Theorem does not apply, then it may happen that the IVP have a unique solution, but unfortunately I have not found any example for that situation to happen. I will be happy if such an example can be given.


1 Answer 1


Note that the domain of the solution is always an open interval, with the initial point an inside point.

So yes, your characterization is the definition of a local solution of the IVP.

And no, your example is not a standard IVP, as the solution is not differentiable in the initial point and the initial point is not inside the domain of the equation. You would have to change that to a generalized IVP concept with $\lim_{x\to 3^+}y(x)=0$.

An example for b) is $y'=1+\sqrt{|y|}$. From the "existence" point-of-view that is not different from the well-known $y'=\sqrt{|y|}$ with its solutions branching off the constant zero solution. But due to the strict positivity the solutions of the former DE are also unique at $y=0$. This leads to the "supplementary" uniqueness theorems for scalar $y'=f(y)$ where the finite or infinite value of $\int \frac{dy}{f(y)}$ around the equilibrium points is the deciding criterion.

  • $\begingroup$ Thanks @Lutz Lehmann. So $y(x)$ is a solution to the "usual" IVP if $y(x)$ satisfies the equation in some open interval containing $x_0$ ? By the way, do you have an example for question (b)? $\endgroup$
    – boaz
    Oct 7, 2021 at 8:24
  • $\begingroup$ Thanks, but what is the unique solution of $y'=1+\sqrt{|y|}$ such that $y(0)=0$? $\endgroup$
    – boaz
    Oct 7, 2021 at 11:47
  • 1
    $\begingroup$ You get $y(x)=x+o(x)$, so the solution crosses this line transversally from one region with local Lipschitz continuity to the other. $\endgroup$ Oct 7, 2021 at 12:04
  • $\begingroup$ Does the IVP $y'=1+|y|$ with y(0)=0 is also a suitable example? why we need the square root? $\endgroup$
    – boaz
    Oct 7, 2021 at 12:15
  • 1
    $\begingroup$ That would be regularly Lipschitz continuous, even with a global Lipschitz constant. The square root gives an unbounded derivative. $\endgroup$ Oct 7, 2021 at 12:18

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