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Setting

Although this detail is not relevant to my question, let me set the problem that my question arise.

We are considering an initial value problem \begin{align*} \begin{cases} u\in L^\infty(I,H^{1}_0)\cap W^{1,\infty}(I,H^{-1})\\ iu_t+\Delta u+\lambda |u|^\alpha u =0 \\ u(0)=\varphi \end{cases} \end{align*} on an interval $I$ containing $0$.

In order to prove the uniqueness of this problem, we use the Duhamel formula. In other words, we need to estimate $$u_1-u_2=i\lambda\int_{0}^{t}\mathcal{T}(t-s)\left[|u_1(s)|^\alpha u_1(s)-|u_2(s)|^\alpha u_2(s) \right]ds$$

Question

The author says that one can easily verify the following inequality.

$$||u_2|^\alpha u_2-|u_1|^\alpha u_1|\leq C (|u_1|^\alpha+|u_2|^\alpha)|u_2-u_1|$$

And this inequality was the core part that allow us to prove the uniqueness of the weak solution. However, I am not the one who can easily prove the inequality. I have tried few days but still cannot figure it out.

I post this question hoping I can have answer from someone. Thanks in advance.

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1 Answer 1

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I assume that $u_1,u_2$ are complex and $\alpha\geqslant 0$ is fixed. Denote $u_1=a$, $u_2=a+b$, $f(z)=|z|^\alpha z$, then $$f(a+b)-f(a)=\int_0^1 \frac{d}{dt}f(a+bt)dt\\=b\int_0^1 |a+bt|^\alpha+\alpha (a+bt)|a+bt|^{\alpha-1}\frac{d}{dt}|ab^{-1}+t|dt.$$ Since $a+bt=:c$ is a point on the segment $[a,b]$ we get $|c|\leqslant \max(|a|,|a+b|)$, using this and $|\frac{d}{dt}|ab^{-1}+t||\leqslant 1$ we get the necessary bound $$|f(a+b)-f(a)|\leqslant C(\alpha)|b|\max(|a|^\alpha,|a+b|^\alpha).$$

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  • $\begingroup$ Thank you for the great answer. You are right! I should have said $u_1$ and $u_2$ are complex valued function and $\alpha\geq 0$ be fixed. If you don't mind, could you elaborate more on the inequality $$\left| \frac{d}{dt}|ab^{-1}+t| \right|\leq 1$$ I don't know how the inequality come from. Even is it true that $|ab^{-1}+t|$ differentiable? $\endgroup$
    – Lev Bahn
    Feb 17, 2021 at 7:19
  • $\begingroup$ Ah... your argument is perfect! Thank you so much! $\endgroup$
    – Lev Bahn
    Feb 17, 2021 at 7:45
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    $\begingroup$ It is differentiable unless $t=-ab^{-1}$ but we may suppose without loss of generality that it does not happen (the fundamental theorem of calculus holds for piecewise smooth continuous functions). The derivative is at most 1, since the map $t\to |ab^{-1}+t|$ is 1-Lip by triangle inequality. $\endgroup$ Feb 17, 2021 at 9:00
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    $\begingroup$ for what it worth, the inequality holds for negative $\alpha$ too, but the argument should be modified: it does not work if there exists a point $p\in [u_1,u_2]$ such that $|p|<\min (|u_1|,|u_2|)/3$, say. But in this case the trivial bound $|u_1|^{\alpha+1}+|u_2|^{\alpha+1}$ works $\endgroup$ Feb 17, 2021 at 9:16

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