Given IVP, for $x \in (-5, 5) $ and $ t \in R $
$$ \frac{dx}{dt} = \sqrt{|x|}$$
$$ x(0) = 0 $$
I want to find if it has a unique solution or not.
So I use the Picard theorem:
- Either I check if $\sqrt(|x|) $ is continuous or not and $\partial f/ \partial x$ is continuous or not in our given range (-5,5). If both conditions holds, then I can say that a unique solution exists. And it could be found using Picard iterates.
Here, in 1, we would check for continuity of $\partial f / \partial x$ right? where $f (x, t) = \sqrt(|x|) $ . I mean the variables could be anything like $x,y,t,z,w…$ all we need to check is for the variable given there. (It's because if it's not in $y$, I get confused in visualizing, so maybe I can just put in $y$ in place of $t$ and visualize?)
- It may happen that 1. may fail or it is very difficult to compute the partial derivative, in which case we check for the condition for $f(x, t)$ being lipschitz continuous in the given range.
$$ |f(t,x1) – f(t,x2)| \leq L |x1 – x2| $$
So here , I need to find an L such that this holds:
$$ \frac {|\sqrt|x1| – \sqrt|x2||} {|x1 – x2|} \leq L $$
If the function is differentiable everywhere on the given range, then I can use the Mean Value Theorem.
and for $f(x1) – f(x2) \leq f'(c) (x1 – x2) $ for some $ c$ between $x1$ and $x2$ and for $x1 < x2$ for all $x1, x2 \in (-5,5)$
And then I can take absolute values on both sides and say that we can take any $L$ such that $|f'(c)| \leq L$ we found a bound.
a. How do I find the value of L? I can put $x1, x2$ as the boundary values of the interval, and how do I choose $c$?
b. I can apply MVT only if the function is differentiable, what if the function is not differentiable in the given range?
Also, Picard's theorem is only a sufficient condition, what do I do if the conditions for Picard Theorem fails? Is there any example where Picard's theorem fails and the IVP still has a unique solution? I think such a function would not satisfy the Lipschitz condition. But I need an example.
This is my understanding, Please suggest how do I approach such problems. I don't require a trick or direct observation of this particular problem. I want to learn general construction so I can verify the existence and uniqueness of solution for any IVP.
Best Answer
In this particular problem, if you didn't immediately kind of see what was going on to be able to take a shortcut, you would start with this computation. For $x \neq y$:
$$\frac{|\sqrt{|x|}-\sqrt{|y|}|}{|x-y|}=\frac{||x|-|y||}{|x-y|(\sqrt{|x|}+\sqrt{|y|})} \\ \leq \frac{1}{\sqrt{|x|}+\sqrt{|y|}}.$$
You recognize that this is exact when $x,y$ have the same sign or just one of them is zero (so you haven't done some kind of blunt estimate that completely changes the behavior).
Therefore, how can you make $\sqrt{|x|}+\sqrt{|y|}$ very small? Make $y=0$ and then consider $x \to 0$. You do that and you conclude that your $f(t,x)$ is not Lipschitz in $x$ near zero.
This means that Picard-Lindelof can't be applied if the trajectory would pass through zero at some point. This may or may not imply non-uniqueness.
In this case, it gives a hint that there might be non-uniqueness for solutions that hit zero. And indeed you can find two solutions with say $x(0)=0$ by hand. One of them is identically zero, another is $\operatorname{sign}(t) t^2/2$ which can be seen by separation of variables. More generally there are actually infinitely many solutions to this problem, determined by the interval on which you choose to have $x=0$.
But this phenomenon is not determined solely by the regularity, it is also determined by the detailed behavior of the dynamics. In particular, this behavior does not occur for $x'=\sqrt{|x|}+1$ (which has exactly the same situation with respect to the hypotheses of Picard-Lindelof). In this case separation of variables works without any division by zero anywhere. The problem is that this "singularity" in $x'=\sqrt{|x|}$ requires the dynamics to become very slow as $x$ approaches zero (but not so slow that zero cannot be reached in finite time at all). This doesn't happen with $x'=\sqrt{|x|}+1$.
Generally, there is not really a nice necessary-and-sufficient condition for existence/uniqueness in ODEs. The local version of Picard-Lindelof works in most situations we frequently encounter, but when it is not applicable, we tend to need to do things on an ad hoc basis.