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I am referring to this paper, p. 22.

On that page, from the inequality

$$ \begin{align*} &\mathbb{H}(f^{n+1})-\mathbb{H}(f^n)+\Delta t\left(\frac{1}{\varepsilon^2}-\eta\bar{m}_2\right)\Vert f^{n+1}-\rho^{n+1}\mathcal{M}\Vert_{2,\gamma}^2+\Delta t\eta\underline{m}_2\Vert\rho^{n+1}\Vert_2^2\\ &\leq\Delta t\eta\left(\sqrt{\overline{m}_4-\underline{m}_2^2}+\frac{C_P\sqrt{\overline{m}_2}}{\varepsilon}\right)\Vert f^{n+1} - \rho^{n+1}\mathcal{M}\Vert_{2,\gamma}\Vert\rho^{n+1}\Vert_2\tag{1} \end{align*} $$ the authors conclude that for any $\eta\in (0,\eta_2)$ with $$ \eta_2:=\frac{\underline{m}_2}{(\sqrt{\overline{m}_4-\underline{m}_2^2}+C_P\sqrt{\overline{m}_2})^2+\underline{m}_2\overline{m}_2}\tag{2} $$ and $\varepsilon\in (0,1)$ one has $$ \mathbb{H}(f^{n+1})-\mathbb{H}(f^n)+\Delta t K(\eta)(\Vert f^{n+1}-\rho^{n+1}\mathcal{M}\Vert_{2,\gamma}^2+\Vert\rho^{n+1}\Vert_2^2)\leq 0\tag{3} $$ with $$ K(\eta)=\frac{1}{2}\min(1-\eta\overline{m}_2,\eta\underline{m}_2).\tag{4} $$

Could somebody please explain to me how to deduce $(3)$?

I am trying for hours, but I simply do not get it. Is there some trick behind it?

I guess its a pure algebraic deduction and one does not need to know how all the notations appearing here are actually defined (this is why I omit the various definitions here...).


Actually, I have no concrete idea how to start.

To shorten the notation a little bit, I set $$ x:=\Vert f^{n+1}-\rho^{n+1}\mathcal{M}\Vert_{2,\gamma},\qquad y:=\Vert\rho^{n+1}\Vert_2. $$

Since $(1)$ is equivalent to $$\small{ \begin{align*} &\mathbb{H}(f^{n+1})-\mathbb{H}(f^n)+\Delta t\left(\frac{1}{\varepsilon^2}-\eta\bar{m}_2\right)x^2+\Delta t\eta\underline{m}_2y^2-\Delta t\eta\left(\sqrt{\overline{m}_4-\underline{m}_2^2}+\frac{C_P\sqrt{\overline{m}_2}}{\varepsilon}\right)xy\leq 0 \end{align*}} $$

I guess that, in order to show $(3)$, one has to prove that, for $\eta\in (0,\eta_2)$ and $\varepsilon\in (0,1)$, $$ \begin{align*} K(\eta)(x^2+y^2)\leq\left(\frac{1}{\varepsilon^2}-\eta\bar{m}_2\right)x^2+\eta\underline{m}_2y^2-\eta\left(\sqrt{\overline{m}_4-\underline{m}_2^2}+\frac{C_P\sqrt{\overline{m}_2}}{\varepsilon}\right)xy \end{align*} $$

(I did not manage to continue from here, however. It nearly drives me crazy.)

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1 Answer 1

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Remark: Actually, the reasoning is easy.

Let $$X = \Vert f^{n+1}-\rho^{n+1}\mathcal{M}\Vert_{2,\gamma}, \quad Y = \Vert\rho^{n+1}\Vert_2$$ and $$a = 1- \eta\overline{m}_2, \quad b = \eta\underline{m}_2, \quad c = \sqrt{\overline{m}_4-\underline{m}_2^2}, \quad d = C_P\sqrt{\overline{m}_2}, \quad r = \min(a, b)$$ and $$\quad s = \frac{1}{\varepsilon} > 1.$$

(1) is written as $$\mathbb{H}(f^{n+1})-\mathbb{H}(f^n) + \Delta t\cdot (s^2 - 1 + a) X^2 + \Delta t\cdot b Y^2 \le \Delta t\cdot \eta (c + d s) XY$$ or \begin{align*} &\mathbb{H}(f^{n+1})-\mathbb{H}(f^n) + \Delta t\cdot \frac{r}2 (X^2 + Y^2)\\ \le{}& \Delta t\cdot \eta (c + d s) XY - \Delta t\cdot (s^2 - 1 + a) X^2 - \Delta t\cdot b Y^2 + \Delta t\cdot \frac{r}2 (X^2 + Y^2). \end{align*}

(3) is written as $$\mathbb{H}(f^{n+1})-\mathbb{H}(f^n) + \Delta t\cdot \frac{r}2 (X^2 + Y^2) \le 0.$$

Thus, it suffices to prove that $$\eta (c + d s) XY - (s^2 - 1 + a) X^2 - b Y^2 + \frac{r}2 (X^2 + Y^2) \le 0.$$

Since $\frac{a}{2}X^2 + \frac{b}{2}Y^2 \ge \frac{r}{2}(X^2 + Y^2)$ (using $a \ge r$ and $b \ge r$), it suffices to prove that $$\eta (c + d s) XY - (s^2 - 1 + a) X^2 - b Y^2 + \frac{a}{2}X^2 + \frac{b}{2}Y^2 \le 0$$ or $$\eta (c+ds)XY \le \frac{b}{2}Y^2 + (s^2 + a/2 - 1)X^2.$$

By AM-GM, it suffices to prove that $$4 \cdot \frac{b}{2} \cdot (s^2 + a/2 - 1) - \eta^2 (c + ds)^2 \ge 0$$ or $$4 \cdot \frac{\eta\underline{m}_2}{2} \cdot (s^2 + (1- \eta\overline{m}_2)/2 - 1) - \eta^2 (c + ds)^2 \ge 0$$ or $$4 \cdot \frac{\underline{m}_2}{2} \cdot (s^2 + (1- \eta\overline{m}_2)/2 - 1) - \eta (c + ds)^2 \ge 0$$ or $$(2s^2 - 1)\underline{m}_2 \ge (c^2 + d^2s^2 + 2cds + \overline{m}_2\underline{m}_2)\eta$$ or $$\eta \le \frac{(2s^2 - 1)\underline{m}_2}{c^2 + d^2s^2 + 2cds + \overline{m}_2\underline{m}_2}.$$

Thus, it suffices to prove that $$\eta_2 = \frac{\underline{m}_2}{(c + d)^2 + \overline{m}_2\underline{m}_2} \le \frac{(2s^2 - 1)\underline{m}_2}{c^2 + d^2s^2 + 2cds + \overline{m}_2\underline{m}_2}$$ which is easy.

We are done.

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  • $\begingroup$ Could you please comment on how you apply the AM-GM inequality? We have that $4XY\leq 2(X^2+Y^2)$. $\endgroup$
    – selector
    Commented Jun 14, 2023 at 8:36
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    $\begingroup$ @selector By AM-GM, we have $u + v \ge 2\sqrt{uv}$. So, $\frac{b}{2}Y^2 + (s^2 + a/2 - 1)X^2 \ge 2\sqrt{\frac{b}{2}Y^2 \cdot (s^2 + a/2 - 1)X^2}$. Then square both sides. $\endgroup$
    – River Li
    Commented Jun 14, 2023 at 8:40
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    $\begingroup$ Let $u := \frac{b}{2}Y^2$ and $v := (s^2 + a/2 - 1)X^2$. Using AM-GM, we have $u + v \ge 2\sqrt{uv} = 2\sqrt{\frac{b}{2}Y^2 \cdot (s^2 + a/2 - 1)X^2}$. It suffices to prove that $2\sqrt{\frac{b}{2}Y^2 \cdot (s^2 + a/2 - 1)X^2} \ge \eta (c+ds)XY$. Squaring both sides, it suffices to prove that $4 \cdot \frac{b}{2} \cdot (s^2 + a/2 - 1) - \eta^2 (c + ds)^2 \ge 0$. $\endgroup$
    – River Li
    Commented Jun 14, 2023 at 8:54
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    $\begingroup$ @selector Because $\eta\in (0,\eta_2)$, $\eta \overline{m}_2 \le \eta_2 \overline{m}_2 =\frac{\underline{m}_2 \overline{m}_2}{(\sqrt{\overline{m}_4-\underline{m}_2^2}+C_P\sqrt{\overline{m}_2})^2+\underline{m}_2\overline{m}_2} \le 1$. $\endgroup$
    – River Li
    Commented Jun 14, 2023 at 9:03
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    $\begingroup$ @selector You are welcome. $\endgroup$
    – River Li
    Commented Jun 14, 2023 at 9:05

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