What Determines the Uniqueness of Solutions in Differential Equations?

[evinda]

Hello! (Wave)

The local Lipschitz condition is the following:Let $c>0$ and $f \in C([a,b] \times [y_0-c, y_0+c])$.

If $f$ satisfies in $[a,b] \times [y_0-c,y_0+c]$ the Lipschitz criterion as for $y$, uniformly as for $t$,

$$\exists L \geq 0: \forall t \in [a,b] \ \forall y_1, y_2 \in [y_0-c,y_0+c]:$$

$$|f(t,y_1)-f(t,y_2)| \leq L|y_1-y_2|$$

then the ODE $(1) \left\{\begin{matrix}
y'(t)=f(t,y(t))\\
y(a)=y_0
\end{matrix}\right.$ is solved uniquely, at least at the interval $[a,b']$where $A=\max_{a \leq t \leq b , y_0-c \leq y \leq y_0+c} |f(t,y)| \ $ and

$b'=\min \{ b, a+ \frac{c}{A}\}$.Remark: The continuity of $f, f \in C([a,b] \times \mathbb{R})$ suffices to ensure the existence of a solution of the ODE $(1)$ at an interval $[a,c], c>a$. But, it doesn't ensure us the uniqueness.For example, $f(y)=\sqrt{|y|}$ doesn't satify the local criterion of Lipschitz as for $y$ at none interval that contains $0$.
I tried to show the latter as follows:$$\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}=\frac{|\sqrt{|y_1|}-\sqrt{|y_2|}|}{|y_1-y_2|}=\frac{|\sqrt{|y_1|}-\sqrt{|y_2|}|}{|\sqrt{|y_1|}-\sqrt{|y_2|}||\sqrt{|y_1|}+\sqrt{|y_2|}|}=\frac{1}{|\sqrt{|y_1|}+\sqrt{|y_2|}|}=\frac{1}{\sqrt{|y_1|}+\sqrt{|y_2|}}$$Is it right so far? And how do we justify that $f$ doesn't satisfy the local condition of Lipschitz as for $y$ at none of the intervals that contains $0$.Does it hold because of the fact that it can be that $y_1=y_2=0$?Could we say that $f'(y)=\frac{1}{2 \sqrt{y}} \to +\infty$ as $y \to 0$ and so the local condition of Lipschitz isn't satified for an interval, if it contains $0$?

Also how can we find the intervals at which the local Lipschitz condition is satified?

Do we have to find the $y$ for which there is a $M \in \mathbb{R}$ such that $\frac{1}{2 \sqrt{y}} \leq M$ ? (Thinking)

 

[jvicens]

Hello! To show that $f$ does not satisfy the local Lipschitz condition as for $y$ at any interval containing $0$, we can consider the limit as $y_1$ and $y_2$ approach $0$. As you correctly pointed out, the derivative of $f$ with respect to $y$ is unbounded at $y=0$, meaning that as $y_1$ and $y_2$ get closer to $0$, the ratio $\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}$ will approach infinity. This violates the Lipschitz condition, which requires the ratio to be bounded by a constant $L$. Therefore, $f$ does not satisfy the local Lipschitz condition at any interval containing $0$.

To find the intervals at which the local Lipschitz condition is satisfied, we can use the definition of the Lipschitz condition. We need to find a constant $L$ such that for any $t \in [a,b]$ and any $y_1, y_2 \in [y_0-c,y_0+c]$, the ratio $\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}$ is bounded by $L$. This can be done by finding the maximum value of $f$ in the interval $[a,b] \times [y_0-c,y_0+c]$ and using that as the constant $L$. The interval at which the Lipschitz condition is satisfied will then be determined by the maximum value of $f$ and the constant $c$.

I hope this helps! Let me know if you have any further questions.