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How to begin to solve ODE problems where no equations are known, only the evaluations of equations at certain points? I tried googling this and searching the exchange here, but I'm not even sure what the name for this type of problem is, so I am having difficulty finding relevant results. Seems like all solutions for ODE require having some equation already and starting from there.

Here is an example on the simpler side that I am currently working on where I am trying to find f that satisfies all of the following: $$ f'(0) = \infty $$ $$ f'(1) = \infty $$ $$ f'(x) > 0 $$ $$ f(0) = 0 $$ $$ f(1) = 1 $$ $$ f(x) = 1 - f(1 - x) $$ and smooth and continuous for all $x \in [0, 1]$

Simply, a continuously increasing function from 0 to 1 with vertical slope at the edges and symmetry about the middle point. I know the requirements, but I have no equations to start from.

And this just an example that I happen to have handy, so I am not necessarily looking for just an answer to this set; I would like an understanding of how to approach this type of problem.

(not sure what all tags this should have, feel free to let me know in comments)

BitsAreNumbersToo
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  • Why do you even think this or similar constrains could lead to a ODE ? – trula Jul 21 '23 at 16:49
  • @Gonçalo This may be part of the difficulty I have in finding information on this topic. Even the first few results I find on the exchange don't offer me much insight into solving this problem. Here is a great example of what may be the right type of question without much insight: https://math.stackexchange.com/questions/3730475/ or this one which only solves the one in the question, but offers little insight to the general case: https://math.stackexchange.com/questions/440762 – BitsAreNumbersToo Jul 21 '23 at 17:00
  • @trula This one perhaps does not require any ODE, others have constraints similar to those where f is also a function of its derivative. If there is a way to separate these into ones that require ODE and ones that don't, and that helps to solve these, I would be interested in learning more. – BitsAreNumbersToo Jul 21 '23 at 17:02

1 Answers1

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As others have pointed out, what you have is really more of a functional equation than an ODE. (Your equations don't define a vector field.) However, it's still an interesting question. Here's one possible approach. A well known function that has a cusp, i.e. achieves an infinite slope at a finite value of $x$, is $y=x^{2/3}$. We can use this to piece together a function that satisfies all of your conditions. The idea is to take a copy of $y=x^{2/3}$ and "glue to it" a copy that is flipped the other way so that you get an infinite slope at $x=1$ as well. To keep the symmetry condition, we need to glue them together at $x=1/2$. It takes a bit of figuring to get the scaling right, but it can be achieved if you define $$f(x)=\begin{cases} \frac{1}{2}x^{2/3} & \text{if } x < 1/2 \\ -\frac{1}{2}(2-2x)^{2/3}+1 & \text{if } x\ge 1/2 \end{cases} $$

The result is $C^1$ but not smooth. I should mention that the result is far from unique - imagine any small perturbation that you do to both halves of the function in a symmetric way that preserves the property of monotonically increasing- any such perturbed function will still be a solution.

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mwalth
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  • Thank you for your answer. This doesn't provide me alot of insight into how to solve these problems, I am curious how to approach them. In case you are just curious about the solution, here is a solution that also meets the smoothness requirements: x^0.5(1-x)+(1-(1-x)^0.5)x – BitsAreNumbersToo Jul 21 '23 at 21:42
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    I think that the point is, unlike ODEs, the types of equations you're considering have very non-unique solutions. For this reason, you can't expect that there is any one path to find a solution, because there are so many solutions. My feeling is that the best you can do is clever construction. – mwalth Jul 21 '23 at 21:48
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    However, there is a theory of functional equations, you may want to consult the literature. For example: https://www.msri.org/people/staff/levy/files/MCL/Efthimiou/100914book.pdf – mwalth Jul 21 '23 at 21:50
  • @BitsAreNumbersToo What kind of functional equations are you interested in ? Because it is a broad and tortuous subject, as mentioned by mwalth. – Abezhiko Jul 22 '23 at 16:00