A version of existence and uniqueness of solution for an initival value problem of ODE

Question

From Wikipedia

Local existence and uniqueness theorem simplified

The theorem can be stated simply as follows.[19] For the equation and initial value problem: $$ y' = F(x,y)\,,\quad y_0 = y(x_0) $$ if $F$ and $∂F/∂y$ are continuous in a closed rectangle $$ R=[x_0-a,x_0+a]\times [y_0-b,x_0+b] $$ in the $x-y$ plane, where $a$ and $b$ are real (symbolically: $a, b ∈ ℝ$) and $×$ denotes the cartesian product, square brackets denote closed intervals, then there is an interval $$ I = [x_0-h,x_0+h] \subset [x_0-a,x_0+a] $$ for some $h ∈ ℝ$ where the solution to the above equation and initial value problem can be found. That is, there is a solution and it is unique.

It seems to say this version of local existence and uniqueness of solution is a simplification of some other version.

So I wonder if it is a special case of Picard–Lindelöf theorem?

Thanks!

Answer 1

Yes, it is a special case of Picard–Lindelöf Theorem. Those assumption imply that $F$ is Lipschitz continuous in $y$. This follows from Mean Value Theorem as follows.

Since $\frac{\partial F}{\partial y}\in C(R)$, it is bounded i.e. there exists $M>0$ such that $$ |\frac{\partial F}{\partial y}(x,y)|\leqslant M,\text{ for all }(x,y)\in R. $$ Hence it follows from Mean Value Theorem that, for some $\theta\in (0,1)$, $$ |f(x,y_1)-f(x,y_2)|=\left|\frac{\partial F}{\partial y}\left(x,\theta y_1+(1-\theta)y_2\right)\right||y_1-y_2|\leqslant M|y_1-y_2|,\text{ for }(x,y_1),(x,y_2)\in R. $$