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)
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.