Prove that the following IVP has a solution in a range of $x=0$ : $$y'(x) = (x-y(x))^{8/9}, \space y(0) = 0$$
Discussion :
I found such exercises be really tight from a theoritical aspect, since what I am going to elaborate really causes a mess. After asking this highly discussed and upvoted question regarding a similar issue, I really did not seem to find an end point to how such an issue should be worked around.
Let's discuss it over though.
First of all, it's important to notice that if we let $f$ be the function :
$$f(x,y)=(x-y)^{8/9}$$
then it's really important to notice that the Domain that the function is defined at, is :
$$D_f=\{ (x,y) \in \mathbb R^2 : x \geq y \}$$
This is a really small detail which makes a big difference on how to handle exercise, since :
$$(x-y)^{8/9} \neq \sqrt[9]{(x-y)^8} \quad \text{for} \quad x<y$$
This means that the function $f(x,y)$ is not continuous in a domain as :
$$D=\{(x,y)\in \mathbb R^2 : |x|\leq ε_1, \space |y|\leq ε_2\}$$
which means that you cannot make a conclusion regarding the existence of a solution around $x=0$ or in an other way, around the initial value $y(0)=0$.
Now, on the other hand, if you decide to define $f$ in a way such that :
$$f(x,y)= \sqrt[9]{(x-y)^8}$$
it's easy to see that it's continuous in $\mathbb R^2$ which means that there is a solution around $x=0$, but for me this is literally nonsense and really bad mathematically regarding strict definitions, since it may work and it may be defined by a matter of taste, but strict mathematical properties regarding powers will not hold.
An other way, mentioned as well in my post above, is that you could extend $f$ to be the following :
$$f(x,y) = \begin{cases} (x-y)^{8/9} \space \space, \space x \geq y \\ 0 \quad \quad \quad \quad, \space x<y \end{cases}$$ then $f$ is continuous and thus has a solution. But going this way, we have never handled such a problem that way in university and it also seems like changing the given exercise, which I do not really like in the matter of providing a perfectly acceptable answer.
I would really appreciate any thorough help that will lead me in understanding how such an issue could be handled, as from what I can see on our textbook, problems like that often pop up (with different powers) and are not even discussed in it. I'm looking forward to seeing a way of proving the existence without manipulating stuff.
It is true that the general definition of $z^{8/9}$ for negative $z$ is multivalued (and the principal branch is not real), but in a real-variables context the standard way of handling it is to take the unique real $9$'th root of $z^8$, or (what is the same thing) the $8$'th power of the real $9$'th root of $z$.