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Regarding the duplicate. Yes, I know the other one has a lot of shared text, but those were just definitions/setup and I was being lazy. The core questions are still different unless you believe derivatives are weak derivatives in which case you might need to read up on them. I don't know the exact definition, but I do know they aren't the same...

Now I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me.

Let us define a differential algebra known as implied differentiation. It actually does not have a unique value. Let us denote the implied derivative operator as $I(f)$, where $f$ is any function being implied differentiated. This is of course, nonstandard terminology.

We define the operator I to be $I(f)(x)(g)$ to be:

$$I(f)(x)(g) := g(x) \left(\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} \right) + (1-g(x)) \left(\lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h} \right)$$

Where $g(x)$ is an arbitrary characteristic/indicator function.

By this I mean that the evaluation of $I(f)$ at $x$ is either one of the one-sided limits and the choice of which one to pick comes from your particular choice of $g$.

The general crux of this question is that I wish to determine whether or not the following statement is true. I'm pretty sure it is, though that's just intuition.

Is a function a solution to an ordinary differential equation if and only if it is a continuous solution to the corresponding implied differential equation?

By corresponding equations, I just mean that they are corresponding if they are the same except with all of the derivative operators replaced with the implied derivative operator. So, the equation $D(y) = e^x$ has a corresponding equation of $I(y) = e^x$.

I do not know for sure whether or not anyone will actually be able to prove this statement. I think it is a bit tricky, but even just some advice on how to approach this or how to begin would be greatly appreciated. It isn't for homework or anything like that. It's just a statement and concept I've developed in my head over the years and I want to determine its truthfulness.

Note: if something equivalent to this or very similar that just makes this a special case has been proven in the past feel free to use that as an answer. I'm going under the (possibly mistaken) impression that this hasn't actually been proven before.

UPDATE:

I believe that a possible route to proving this statement might come by proving the following propositions.

The solution sets of sub-differential equations are a super set of the solution sets of the corresponding differential equations

The solution sets of sub-differential equations are a super set of the solution sets of the corresponding implied differential equations

If a solution to a sub-differential equation is continuous then it is a solution to the corresponding differential equation

If a solution to a sub-differential equation is continuous then it is a solution to the corresponding implied differential equation

I believe that the the first two propositions might follow trivially from the definition of the sub-derivative. The third one might have been proven in the past. I am unsure. The fourth one would then be the true meat and bones of the proof.

The subderivative is defined here: https://en.wikipedia.org/wiki/Subderivative

Reasons for wanting this statement proved:

The implied differential operator is defined via the limits; however, the purpose of that is to emulate a differential algebra wherein all step functions have a derivative of 0 and all the other normal rules are preserved. If it wasn't apparent, because of this any implied differential equation involving step functions is trivial to solve (at least in terms of the step functions themselves providing difficulty). If the statement were true, it would provide a new avenue by which to attack differential equations, some of which might be made trivial to solve via this method.

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  • $\begingroup$ But $I(y) = e^x$ doesn't make sense unless $y$ is differentiable, in which case, $I(y) = D(y)$ and it is just the same equation. If $y$ is not differentiable, but does have left and right derivatives, then $I(y)$ is a set of two real numbers for each $x$, whereas $e^x$ is just a single real number. $\endgroup$ – Paul Sinclair Apr 25 '17 at 3:53
  • $\begingroup$ You might want to look at different definitions of sub-differentials, en.wikipedia.org/wiki/Subderivative. Consider also the equivalence of the IVP to the Picard integral equation. $\endgroup$ – Lutz Lehmann Apr 25 '17 at 10:46
  • $\begingroup$ IVP = Initial value problem. If you look for extended frameworks for non-continuous ODE, look for A.F. Filippov, see for instance books.google.de/books?isbn=9401577935. Essentially, you get to look at all mollifications of the right side and then examine if they have limit points when the mollification parameter goes to zero. $\endgroup$ – Lutz Lehmann Apr 25 '17 at 15:47
  • $\begingroup$ I was aware that was an example problem, and I was giving an example of what needs to be clarified about your concept. Whether you consider it a set of possible functions (which might get messy as there is no relation between choices made at different points) or a function returning sets is immaterial. In either case, you have to replace $e^x =$ with $e^x \in$, and deal with the consequences of this change. But fortunately Lutzl is more informed on the subject. $\endgroup$ – Paul Sinclair Apr 25 '17 at 16:27
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One direction is trivial: a solution to any ODE is clearly a continuous solution to the corresponding IDE (implied differential equation).

The opposite direction seems to be false. Consider $I(f)=2\lfloor x\rfloor -1$ on the interval $[-\frac12, \frac12]$. This is solved by $f(x)=|x|$, which is continuous. But $f'=2\lfloor x\rfloor -1$ has no solutions: since RHS has a nonessential discontinuity at 0, it cannot be the derivative of any function.

The solution below doesn't work because Darboux's theorem assumes $f$ is differentiable. I've kept it for archival purposes.


When we restrict our attention to functions defined on a closed interval, we get a simple solution thanks to Darboux's theorem. This theorem implies, in particular, that $f'$ has only essential discontinuities (this fact is mentioned on the Wiki page currently). But the fact that $f$ is a solution to an IDE on the interval implies that it has only nonessential discontinuities.

So in fact, $f'$ is continuous on the whole interval, which means that $f$ is differentiable, which means that $I(f)=D(f)$ (or, more pedantically, $I(f)(a)=\{D(f)(a)\}$ for all $a$ in the interval). This is certainly strong enough to prove the desired result :P

I will spend some time thinking about more exotic scenarios. In particular, I have reason to believe that OP would be particularly interested in the case of half-open intervals (we talked in chat; this isn't true). I am not quite sure how derivatives at endpoints work, so if there anyone can shore up that bit in a line or two, please mention it in the comments :)

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  • $\begingroup$ Just for the record you probably meant the function $ 2\left \lfloor{x}\right \rfloor +1$. $\endgroup$ – lcv Apr 3 at 2:58

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