用中文回答问题,阅读文章,下述公式$$\int_{\mathbb{R}^N}|\nabla\omeg...

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답변에 사용된 모델 O3 by Chat01

질문

SomeExistenceResults on aClass ofGeneralized.pdf

用中文回答问题,阅读文章,下述公式RNωn2=RN[1+G1(vn)g(G1(vn))g(G1(vn))]2vn24RNvn2,\int_{\mathbb{R}^N}|\nabla\omega_n|^2=\int_{\mathbb{R}^N}\left[1+\frac{G^{-1}(v_n)g^{\prime}(G^{-1}(v_n))}{g(G^{-1}(v_n))}\right]^2|\nabla v_n|^2\leq4\int_{\mathbb{R}^N}|\nabla v_n|^2,怎么得到的,请详细说明步骤

사고

Explaining the inequality

The user asks how to derive the inequality involving the gradient terms. To break this down:

  1. We define the function ωn=G1(vn)g(G1(vn))\omega_n = G^{-1}(v_n) \cdot g(G^{-1}(v_n)).
  2. Next, we compute ωn\nabla \omega_n using the chain rule, considering the composite functions in ωn\omega_n.
  3. Then we apply the inequality related to the function gg, which ensures that the multiplier of the squared gradient terms remains between 1 and 2, making the integral bounded by 4 times the integral of the gradient of vnv_n, as shown in the article's lemma 2.6.

Explaining the inequality steps

To derive the inequality, I first need to use the derivative of G1G^{-1}. Since GG is the antiderivative of gg, I know that G(t)=g(t)G'(t) = g(t), and using this, I can compute (G1)(s)(G^{-1})'(s).

Next, applying the chain and product rule for the gradient of ( \omega_n = G^{-1}(v

Thought for 19 seconds

답변

推导思路概览

  1. 引入 ωn\omega_n
    文中先定义

    ωn(x)=G1 ⁣(vn(x))  g ⁣(G1 ⁣(vn(x))), \omega_n(x)=G^{-1}\!\bigl(v_n(x)\bigr)\;g\!\bigl(G^{-1}\!(v_n(x))\bigr),

    目的是把原来的序列 vnv_n 通过一个光滑可逆变换投射到新的序列 ωn\omega_n,以便后面估计梯度范数。

  2. ωn\nabla\omega_n(链式法则与乘积法则)

    t=G1(vn(x)),vn=G(t),t=G1 ⁣(vn)=1g(t). t=G^{-1}(v_n(x)),\qquad v_n=G(t),\qquad t'=G^{-1}\!{}'(v_n)=\frac{1}{g(t)}.

    ωn(t)=tg(t)\omega_n(t)=t\,g(t) 看作 tt 的函数,再把 tt 视作 xx 的函数,应用链式法则+乘积法则:

    ωn=[g(t)+tg(t)]  t=[g(t)+tg(t)]  vng(t)=[1+tg(t)g(t)]vn. \nabla\omega_n =\bigl[g(t)+t\,g'(t)\bigr]\;\nabla t =\bigl[g(t)+t\,g'(t)\bigr]\;\frac{\nabla v_n}{g(t)} =\Bigl[1+\frac{t\,g'(t)}{g(t)}\Bigr]\nabla v_n .
  3. 平方并在 RN\mathbb R^{N} 上积分

    RNωn2=RN[1+G1(vn)g(G1(vn))g(G1(vn))]2vn2. \int_{\mathbb R^{N}}|\nabla\omega_n|^{2} =\int_{\mathbb R^{N}} \Bigl[1+\frac{G^{-1}(v_n)\,g'(G^{-1}(v_n))}{g(G^{-1}(v_n))}\Bigr]^{2} |\nabla v_n|^{2}.
  4. 用函数 gg 的结构性质控制系数

    文章对 gg 施加了

    β:=suptRtg(t)g(t)10tg(t)g(t)1. \beta:=\sup_{t\in\mathbb R}\frac{t\,g'(t)}{g(t)}\le 1 \qquad\Longrightarrow\qquad 0\le\frac{t\,g'(t)}{g(t)}\le 1 .

    因此

    11+tg(t)g(t)2, 1\le 1+\frac{t\,g'(t)}{g(t)}\le 2 ,

    把上式平方后得到

    [1+tg(t)g(t)]24. \Bigl[1+\frac{t\,g'(t)}{g(t)}\Bigr]^{2}\le 4 .

    这里还用到了 Lemma 2.1 的性质 (g6):
      1G1(t)g(G1(t))t2\;1\le\displaystyle\frac{G^{-1}(t)\,g(G^{-1}(t))}{t}\le2, 与上界 β1\beta\le1 的假设是一致的。

  5. 得到最终不等式

    将系数的估计放回积分式:

    RNωn24RNvn2, \int_{\mathbb R^{N}}|\nabla\omega_n|^{2} \le 4\int_{\mathbb R^{N}}|\nabla v_n|^{2},

    即得所需公式。


关键点小结

  • 链式法则 给出了 ωn\nabla\omega_nvn\nabla v_n 的精确关系;
  • 结构假设 β1\beta\le1 与 Lemma 2.1 中的 (g6) 性质 保证了前置系数始终不超过 2,从而平方后不超过 4;
  • 因此可把 ωn2|\nabla\omega_n|^{2} 的积分压到 vn2|\nabla v_n|^{2} 的 4 倍以内,满足在后续引理 2.6 中对 ωn\omega_n 有界性的需求。

질문

用中文回答问题,阅读文章,下述公式RN(IαG1(vn+)p)G1(vn+)pC(RNvnpr2)2r0,\int_{\mathbb{R}^N}\left(I_\alpha*|G^{-1}(v_n^+)|^p\right)|G^{-1}(v_n^+)|^p\leq C\left(\int_{\mathbb{R}^N}|v_n|^{\frac{pr}{2}}\right)^{\frac{2}{r}}\to0,怎么得到的,请详细说明步骤

사고

Explaining the inequality

The user requests help in understanding the inequality, which relates to Hardy-Littlewood-Sobolev (HLS) inequality and properties of G1(vn+)G^{-1}(v_n^+).

We should start by defining rr as 2NN+α\frac{2N}{N+\alpha}, and then apply the HLS inequality, combining LaL^a and LbL^b spaces to estimate the convolution. After that, we’ll use the properties of G1(vn+)G^{-1}(v_n^+) to relate it to vnv_n, leveraging that G1(t)Ct1/2G^{-1}(t) \leq C|t|^{1/2}. Finally, we note that vn0v_n \to 0 in Lpr/2L^{pr/2}, justifying the inequality. This is all detailed in Lemma 2.7.

Explaining the steps

To explain the inequality, I must follow these steps:

  1. Set r=2NN+αr = \frac{2N}{N + \alpha}, ensuring the Hardy-Littlewood-Sobolev inequality holds appropriately.
  2. Apply the inequality for f=g=G1(vn+)pf = g = |G^{-1}(v_n^+)|^p, then use the LrL^r norm to bound the convolution.
  3. Use property (g7) of G1G^{-1} to relate it to vnv_n, with G1(s)Cs1/2G^{-1}(s) \leq C |s|^{1/2}.
  4. Conclude that since vn0v_n \to 0 in Lpr/2L^{pr/2}, the

Thought for 31 seconds

답변

推导步骤

  1. 选取 Hard-Littlewood-Sobolev (H–L-S) 指数

    r=2NN+α(>1),1r+1rαN=1. r=\frac{2N}{N+\alpha}\quad(>1),\qquad \frac1r+\frac1r-\frac{\alpha}{N}=1 .
  2. 利用 H–L-S 不等式估计卷积项
    对任何 fLr(RN)f\in L^{r}(\mathbb R^{N})gLr(RN)g\in L^{r}(\mathbb R^{N})

    RN(Iαf)g    CfLrgLr. \int_{\mathbb R^{N}} (I_\alpha *f)\,g \;\le\;C\,\lVert f\rVert_{L^{r}}\, \lVert g\rVert_{L^{r}} .

    f=g=G1(vn+)p, f=g=|G^{-1}(v_n^+)|^{p},

    RN(IαG1(vn+)p)G1(vn+)pC(G1(vn+)pLr)2=C(RNG1(vn+)pr) ⁣2/r. \int_{\mathbb R^{N}} \bigl(I_\alpha*|G^{-1}(v_n^+)|^{p}\bigr) |G^{-1}(v_n^+)|^{p} \le C\, \Bigl(\,\|\,|G^{-1}(v_n^+)|^{p}\|_{L^{r}}\Bigr)^2 = C\, \Bigl(\int_{\mathbb R^{N}} |G^{-1}(v_n^+)|^{pr}\Bigr)^{\!2/r}.
  3. G1G^{-1} 的增长性质把被积函数换成 vnv_n
    文章性质 (g7) 给出

    G1(t)    Ct12,tR. |G^{-1}(t)|\;\le\;C\,|t|^{\frac12},\qquad\forall t\in\mathbb R .

    因而

    G1(vn+)pr    Cvnpr2, |G^{-1}(v_n^+)|^{pr} \;\le\;C\,|v_n|^{\frac{pr}{2}},

    代入上式得到

    RN(IαG1(vn+)p)G1(vn+)p    C(RNvnpr2) ⁣ ⁣2/r. \int_{\mathbb R^{N}} \bigl(I_\alpha*|G^{-1}(v_n^+)|^{p}\bigr) |G^{-1}(v_n^+)|^{p} \;\le\; C\left( \int_{\mathbb R^{N}} |v_n|^{\frac{pr}{2}} \right)^{\!\!2/r}.
  4. 极限趋零的理由
    在 Lemma 2.7 的“vanishing”情形中,可得

    vn0于 Ls(RN),  s(2,2), v_n\to0\quad\text{于 }L^{s}(\mathbb R^{N}),\;s\in(2,2^{*}),

    pr2=pNN+α(2,2)(  2N+αN<p<2N+αN2). \frac{pr}{2} =\frac{pN}{N+\alpha}\in(2,2^{*}) \qquad\bigl(因\;2\frac{N+\alpha}{N}<p<2\frac{N+\alpha}{N-2}\bigr).

    因此
    ( \int_{\mathbb R^{N}}|v_n|^{pr/2}\to0 ),
    进而左端整体趋于 0。

综上便得到文中的不等式及其极限性质。

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