Existence and uniqueness of solution of : $y'(x) = (x-y(x))^{4/5}, \space y(3)=3$

Question

Study the Existence and Uniqueness of the IVP : $$y'(x) = (x-y(x))^{4/5}, \space y(3)=3$$

Regarding existence, I'd first assume $f$ the function such :

$$f(x,y)=(x-y)^{4/5}$$

We observe that $f(x,y)$ is not continuous in $ \mathbb R^2$ since we need $x-y\geq0$.

Letting $D$ be a domain such :

$$D=\{(x,y)\in \mathbb R^2: |x-3| \leq ε_1, \space |y-3|\leq ε_2 \}$$

we can say that there exists $ε_1,ε_2 \in \mathbb R$ such that $f$ is continuous in $D$, which means that the IVP has a solution in $D$.

Question : Is the above statement correct ? Shall I go into more details explaining why $f$ can be continuous in such a domain ?

Finally, regarding uniqueness, I know that showing that $f(x,y)=(x-y)^{4/5}$ is a Lipschitz function is enough. :

$$|f(x,y_2) -f(x,y_1)|=|(x-y_2)^{4/5}-(x-y_1)^{4/5}|$$

The function $g(y)=(x-y)^{4/5}$ is continuous in the interval $[y_1,y_2]$ with $y_1,y_2 \in D_f$ and $g$ is also differentiable in $(y_1,y_2)$ , which means that we can apply the Mean Value Theorem and yield :

$$g'(ξ)=\frac{g(y_2)-g(y_1)}{y_2-y_1}\Rightarrow \bigg|\frac{4}{5(x-ξ)^{1/5}}\bigg||y_2-y_1|=|(x-y_2)^{4/5}-(x-y_1)^{4/5}|$$

Here, we can see that there isn't a bound for the expression $\bigg|\frac{4}{5(x-ξ)^{1/5}}\bigg|$, thus meaning that the solution is not unique.

Answer 1

The function $f(x,y) := (x-y)^{4/5}$ is continuous is $\mathbb{R}^2$, being the composition of the continuous functions $\phi(t) := t^{4/5} = \sqrt[5]{t^4}$, $t\in\mathbb{R}$ and $g(x,y) := x-y$, $(x,y)\in\mathbb{R}^2$.

By Peano's existence theorem, the Cauchy problems associated with $f$ do have solutions.

The function $f$ is not (locally) Lipschitz continuous, but this is only a sufficient condition in order to get uniqueness of solution of Cauchy problems.

It is easy to see that $y(x)$ is a solution of $y'=f(x,y)$ if and only if $z(x) := y(x) - x$ is a solution of $z' = h(x,z)$, with $h(x,z) := z^{4/5} - 1$. In particular, $y(x)$ is a solution of the given Cauchy problem if and only if $z(x) := y(x) - x$ is a solution of the Cauchy problem $$ \begin{cases} z' = z^{4/5} - 1,\\ z(3) = 0. \end{cases} $$ This Cauchy problem admits, locally, a unique solution, implicitly defined by $$ \int_0^z \frac{1}{s^{4/5}-1}\, ds = x-3. $$ It can be proved (but this requires a simple qualitative study) that this unique local solution can be prolonged to a unique global solution.

Answer 2

The function $f$ is not defined in your set $D$ since $D$ contains points where $x<y$. So your statement is not correct.

Also for uniqueness, the theorem says that if $f$ is Lipschitz in $y$, then the solution is unique, but the converse is not true. So you cannot say that since $f$ is not Lipschitz, there is no uniqueness.

It is true that one could define $t^{4/5}=(t^4)^{1/5}$ but then you lose important properties like $t^{ab}=(t^a)^b$. In general it is better not to define $t^a$ if $a$ is a rational and $t<0$. A way out in this case would be to extend $f$ to be zero in the region $x<y$. So now the problem $y'(x)=f(x,y)$, $y(3)=3$ has a solution since $f$ is continuous in $\mathbb R^2$. Then you can follow Riegel's proof to get existence and show that the solution that you get stays in the region $x-y\ge 0$ so it is a solution of your original equation.