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?



1 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 $\theta\in (0,1)$, $$ |f(x,y_1)-f(x,y_2)|=\left|\frac{\partial F}{\partial y}\left(x,\theta y_1+(1-\theta)y_2\right)\right||y_1-y_2|\leqslant M|y_1-y_2|,\text{ for }(x,y_1),(x,y_2)\in R. $$


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