Does $y'=|y|^a$ have any global solutions?

Question

Assume the differential equation

$$ y'=|y|^a $$ My intuition tells me that since it involves an absolute value, there might not be any solutions defined everywhere, except for the case $a=0$, where $y(x)=x+c$.

To show this, let $$ f(y)=|y|^a$$

$\bullet\,$For $a<0$:

$f$ is not defined for $y=0$ plus it's not bounded

$\bullet\,$ For $a=0$: $$y'=1 \iff y(x)=x+c, \quad x \in \mathbb{R}$$ $\bullet\,$ For $a>0$:

$f$ is defined $\forall y \in \mathbb{R}$, but it's not bounded

Can we thus conclude that the only global solution of $y'=|y|^a$ is $y(x)=x+c$?

Answer 1

I have summarised some of the information that has already been stated and made an attempt to complete the answer.

I believe that your question may be rephrased as follows: given $a \in \mathbb{R}$ does there exist $y:\mathbb{R}\longrightarrow\mathbb{R}$ such that $(Dy)(t) = |y(t)|^a$ (1) for all $t \in \mathbb{R}$?

Remark: it is assumed that $0^0$ is not defined.

Consider four cases (the first two cases are not necessary to answer your question, but it is always useful to exhibit closed-form analytic solutions when they exist):

Case I: $a = 1$. If $c \in \mathbb{R}$, then $y(t)=ce^{\mathsf{sgn}(c) t}$ is a solution of (1) on $\mathbb{R}$ (this case is not necessary: see Case III).

Case II: $0 < a < 1$. Then $y(t)=sgn(t)(1-a)^{\frac{1}{1-a}}(sgn(t)t)^{\frac{1}{1-a}}$ is a solution of (1) on $\mathbb{R}$ (this case is not necessary: see Case III) - of course, other solutions may be obtained using the expression above.

Case III: $a > 1$. Then $y(t) = 0$ is a solution of (1) on $\mathbb{R}$ (of course, $y(t) = 0$ is also a solution for $0 < a \leq 1$).

Case IV: $a \leq 0$. Suppose that $y(t)$ is a solution of (1) on $\mathbb{R}$. Then, $y(t)\neq 0$ for all $t \in \mathbb{R}$. Given that $(Dy)(t)=|y(t)|^a$, $(Dy)(t) > 0$ for all $t \in \mathbb{R}$. However, then $y$ must be an increasing function. Therefore, by differentiability of $y$, either $y(t) < 0$ for all $t \in \mathbb{R}$ or $y(t) > 0$ for all $t \in \mathbb{R}$.

Suppose $y(t) < 0$ for all $t \in \mathbb{R}$. Then $(Dy)(t)=(-y(t))^a$ for all $t \in \mathbb{R}$. Thus, the second derivative exists and is given by $(D^2 y)(t)=-a(-y(t))^{2a-1}$ for all $t \in \mathbb{R}$. Under assumptions $a \leq 0$ and $y(t) < 0$, $ 0 \leq (D^2 y)(t)$ for all $t \in \mathbb{R}$. Therefore, $y(t)$ is convex on $\mathbb{R}$. Thus, $y(t)$ is constant (link). However, $y(t)$ is also an increasing function. Thus, a contradiction is reached and, therefore, $y(t) > 0$ for all $t \in \mathbb{R}$.

Suppose $y(t) > 0$ for all $t \in \mathbb{R}$. Then, consider a function $q(t)$ such that $q(t)=y(-t)$ for all $t \in \mathbb{R}$. Thus, $q(t)>0$, Also, $(Dq)(t)=-q(t)^a=-(y(-t))^a$ and $(D^2q)(t)=aq(t)^{2a-1}$ for all $t \in \mathbb{R}$. Thus, $(Dq)(t) < 0$ and $(D^2q)(t) \leq 0$ for all $t \in \mathbb{R}$. Hence, $q$ is decreasing and concave on $\mathbb{R}$. However, in this case, it cannot be bounded below (link). Thus, by contradiction, if $a \leq 0$, then (1) does not have any global solutions on $\mathbb{R}$.

In conclusion, if $a > 0$, then there always exists at least one global solution of (1). Otherwise, (1) does not have any global solutions.

Answer 2

There is no requirement that $f$ be bounded. When $a=1/2$ you can use $y(x)=(x^2/4) \mathrm{sgn\,} x$, for example and you can also translate $x\mapsto x+c$ to get other solutions. Also don't forget Cauchy-Lipschitz fails, so $y\equiv0$ is a solution too.