Existence of an $L$, such that $\big|\lvert y_1\rvert^\alpha-\lvert y_2\rvert^\alpha\big|\le L\lvert y_1-y_2\rvert$, when $|y|\le b$ and $\alpha>1$

Question

I'm trying to prove the uniqueness of solution in following IVP problem

$$\frac{dy}{dx}=\lvert y\rvert^\alpha,\,\,\,y(0)=0 ~~~~~(\alpha>1)$$

One possible way here is to apply the Picard's Theorem. So I have to prove that

$$||y_1|^\alpha-|y_2|^\alpha|\le L|y_1-y_2|$$

where $L$ is a constant for a certain area, say $|y|<b$. When $\alpha=2$, this can be easily done.

$$||y_1|^2-|y_2|^2|=(|y_1|+|y_2|)(|y_1|-|y_2|)\leq L|y_1-y_2|$$

But for an arbitrary real $\alpha>1$ I have no idea how to do it.

Any idea or discussion is welcome. Thanks in advance.

Answer 1

Let $=t_2=\lvert y_2\rvert>\lvert y_1\rvert=t_1$.

Then, for $t=\lvert y\rvert\le b$: $$ \lvert y_2\rvert^a-\lvert y_1\rvert^a=t_2^a-t_1^a=a\int_{t_1}^{t_2}s^{a-1}\,ds\le a\int_{t_1}^{t_2}b^{a-1}\,ds=ab^{a-1}(t_2-t_1)=ab^{a-1}(\lvert y_2\rvert-\lvert y_1\rvert)\le ab^{a-1}\lvert y_2-y_1\rvert. $$ So $L=ab^{a-1}$.

Answer 2

HINT.

Prove that $$ \left\lvert \lvert y_1\rvert^\alpha -\lvert y_2\rvert^\alpha \right\rvert \le C \lvert y_1 - y_2\rvert\max\left( \lvert y_1\rvert^{\alpha-1}, \lvert y_2\rvert^{\alpha-1}\right), $$ by setting $f(y)=\lvert y\rvert^\alpha$ and using the formula $$ f(y_2)-f(y_1)=\int_{y_1}^{y_2} \frac{df}{dy}\,dy.$$ (Here $C>0$ is some absolute constant. I think that here one can take $C=\alpha$. The formula holds for all $\alpha>0$ but of course it is significant only for $\alpha>1$, because otherwise the terms $|y_1|^{\alpha-1}, \lvert y_2\rvert^{\alpha-1}$ are unbounded around $0$).