I want to find the increasing function $y(x): [0,1] \rightarrow [0,1]$ which is defined implicitely as the solution to the following equation:
$f_1(x) = f_2(y(x)) \quad \forall x \in [0,1]$
where $f_1(), f_2()$ are known and $C_1$ on $[0,1]$
$f_1$ and $f_2$ are such that an increasing solution exists (I have existence already proved elsewhere).
I want to show that there exists a unique increasing solution $y(x)$ starting from $y(0) = 0$.
This solution should be unique as long as the set such that $f_1’(x) = 0$ is of null measure. The intuition is quite simple: both function goes through exactly the same points, in the same order (just not at the same speed) (you can see this graphically). So as long as there is no flat part, there should be a unique mapping from $x$ to $y$.
Remark: on the optimal path, given the equation, this equation will also imply that the set such that $f_2’(y(x)) = 0$ is also of null measure, and the null derivative coincides.
Remark 2: Obviously, if $f_2()$ is strictly monotone, we can just invert it to find the solution: $y(x) = f_2^{-1}(f_1(x)) \quad \forall x \in [0,1]$
To solve my problem, since I’m solving for a function, I thought I could resort to properties of differential equations (Picard-Lindelof for e.g.); Differentiating my equation yields:
$f_2’(y(x)) y’(x) = f_1’(x)$.
$\iff y’(x) = f_1’(x) / f_2’(y(x)) := g(x, y(x))$ when $f_2’(y) \neq 0$
By Picard-Lindelof, if $g()$ is $C_1$ on $[0,1]$, then I have a unique solution. But this would correspond to the monotone case in my remark 2 above.
The difficulty I have is with the points with null derivatives. These null derivatives yields problem because there is a indeterminacy in $g()$ on the optimal path ($0$ over $0$, since $y’ > 0$), and at other points the function is not even defined. So I cannot resort to Picard Lindelof.
However, since I’m looking for an increasing solution, the solution should still be unique.
But I cannot find any proof of that, or any variation of Picard-Lindelof doing this kind of stuff (i.e adding increasing constraint into the system).
Hence my first question:
- Question: does a variation of Picard-Lindelof uniqueness result exists for my kind of problem? (i.e. problem with increasing constraint on the solution)
would be great if I could just find a reference for this as I was not able to find any.
Otherwise, I am trying to build the solution myself, resorting to some kind of pasting. But I’m not sure it’s correct to do it this way.
The idea is that, I’m using Picard-Lindelof starting from $y(0)=0$ up to the first point where $f_2’(y) = 0$ that I denote $y_1$. Then I’m doing it again from $y_1$ to $y_2$ (the second point where $f_2’(y) = 0$). And so on and so forth.
There are two ‘problems’ with this solution:
- How do I ‘start’ from the $y_k$ points as generally speaking, $g(x_k, y_k(x_k))$ is not defined (or actually indeterminate $0$ over $0$), so I’m not sure I can write it down.
- I should have trouble with this notion of ‘points’ with null derivatives. What if they are an infinite number of such points? I think I can prove that it is not possible on the compact set [0,1] if $f_1, f_2$ are $C_1$ and if the set such that $f’_1 = 0$ is of null measure. But I’m not sure and there might exist some $x sin(x)$ case I did not think about (even if $x sin(x)$ is not $C_1$ in $0$ so it does not enter my setup).
And more importantly, if I can resort to this concept of points, I can just use the original system directly: because it is piecewise monotone between those points, I can invert it between those points!
All of this to say I’m really not sure about how to build a proof so far.
I have a simple showcase example to illustrate the problem:
Imagine $f_1(x) = 1 - 4 (x - 0.5)^2$, $f_2(y) = 4(\sqrt{y} - y)$.
This example is built such that the unique increasing solution to this problem with initial value $y(0) = 0$ is $y(x) = x^2$.
The question is: how can I recover this using my differential equation $f’_2(y(x)) y’(x) = f’_1(x)$ ?
By my method, I’ll just realize that $y_1 = 0.25$ and $x_1 = 0.5$ (and that there is only one point such that the derivatives are $0$). And I’ll just have to split the Picard-Lindelof in two parts: starting from $(x,y)=(0,0)$ to $(0.5, 0.25)$; and then from $(0.5, 0.25)$ to $(1, 1)$.
But even there, you see that the system of differential equations is ‘indeterminate’ at $(0.5, 0.25)$, so how can I ‘start’ again from there?
That’s why I find my ‘proof’ to be quite clumsy so far, and why I would like something more correct, if possible an already built result that I could cite or from which I could start my proof.
EDIT: since my ODE is separable I should solve it by getting to the initial implicit form, so I’m not sure getting to the differential equation is even necessary; I thought it could be easier to get some standard results about uniqueness though. But maybe I could get them with the initial implicit definition of $y(x)$ by adding a monotonicity constraint: i.e. there exists only one strictly increasing part in the integral curve. But I don’t know how to show this.
Sketch of an answer:
If the set such that $f_1’(x) = 0$ has a finite number of points (instead of imposing measure zero), then I will do the proof as follows:
Denote $x_k$ the ordered set of $K$ points such that $f_1’(x_k) = 0$.
Since we know the functions f_1, f_2, are such that an increasing solution exists, $f_2’(y)$ must also have the same finite number $K$ of points with $f_2’(y_k) = 0$. Using the order of these points and the monotonicity constraint on the solution, we know that $y(x_k) = y_k \forall k$.Now, I can just use the fact that $f_2()$ is piecewise monotone between each $y_k$. By monotonicity, since $y_k = y(x_k)$, then all $y \in [y_k, y_{k+1}]$ are images of $x \in [x_k, x_{k+1}]$.
This way starting from (x, y) = (0,0), we have:
$\forall x \in [0, x_1), \quad f_2(y) = f_1(x)$.
Moreover, on $[0, y(x_1))$, $f_2$ is monotone (and $C_1$). So it is inversible, and we get:
$\forall x \in [0, x_1), \quad y = f_2^{-1}(f_1(x))$.
Then we have $y(x_1) = y_1$. And we repeat the same procedure on $[x_1, x_2)$, and so on and so forth up to $[x_K, 1]$.
This way, I get a unique increasing mapping $y(x)$ exploiting piecewise monotonicity of my function.
Now, I’m not $100\%$ sure of this proof. And more importantly, I’m not sure how to write it down if I have an infinite number of points with null derivative.
I guess the idea would be exactly the same (all that matters is the ordering). It’s just that I’m not sure everything I wrote above would follow through and still be correct. Maybe I did not think about some particular case that would forbid my piecewise inversions.
For example, $f_2(y) = y^3 sin(1/y)$ cannot be considered piecewise monotone near $0$ I suppose... But I’m not sure.
Remark: I thought ODE would have results on this (i.e. proving existence of a unique increasing solution), but I cannot find anything, so in the end using the ODE form of my equation seems quite useless. I just exploit the initial implicit form.