A blowup proof of Aubin’s theorem in the Yamabe problem

There is a classic blowup analysis proof for Aubin’s theorem, due to Uhlenbeck’s renormalization method described below to give another proof that the Yamabe problem. See the chapter5 at Schoen & Yau              ‘ Lectures on differential geometry’.

Yamabe’s approach was to consider first the perturbed functional

\displaystyle  Q_s(u)\doteqdot\frac{\int_M\Big(|\nabla u|^2+\frac{n-2}{4(n-1)}R_gu^2\Big)d\mu_g}{\big(\int_M|u|^sd\mu_g\big)^{2/s}}

where s\in(0,\frac{2n}{n-2}] and u\in H^1(M)\setminus\{0\}. Set

\displaystyle  \lambda_s\doteqdot\inf\{Q_s(u):u\in H^{1}(M)\setminus\{0\}\}\quad\text{and}\; Y(M)=\lambda_{2^*}

By using a direct minimizing procedure, it can be shown that for 2 < s < 2^*, there exists a smooth positive function us such that its L^s-norm is equal to one, Q_s(u_s) = \lambda_s and us satisfies the equation

\displaystyle  \Delta_gu_s-\frac{n-2}{4(n-1)}R_gu_s+\lambda_su^{s-1}_s=0,\quad \text{in}\;M

The direct method does not work when s=2^* because the Sobolev embedding H^1(M)\to L^{2^*}(M)  is continuous but not compact.

However, if one can show that u_s is uniformly bounded, i.e.there exists a positive constant c such that u_s \le c in M for 2 < s < 2^*, then there exists a sequence \{s_i\} \in 2 < s < 2^* such that and u_{s_i} converges to a smooth positive function u which satisfies the Yamabe equation .

We discuss a blow-up argument. Suppose that no such upper bound c exists. It follows that there exist sequences \{s_k\} \subset (2, 2^*) and \{y_k\} \subset M such that

\displaystyle  s_k\to 2^*\quad\text{and}\quad m_k\doteqdot u_{s_k}(y_k)=\max u_{s_k}\to\infty,\quad as\;k\to\infty

As M is compact, we may assume that y_k \to y_0 as k \to\infty. For a normal coordinate system centered at y_0 and with radius \rho, let the coordinates of y_k be x_k, k = 1, 2, .....

In the local coordinates,

\displaystyle  g_{ij}(x)=\delta_{ij}+O(\rho^2),\quad \det g=1+O(\rho^2)

u_{k}=u_{s_k} satisfies the equation

\displaystyle  \frac1{\sqrt{\det g}}\partial_j\Big(\sqrt{\det g}g^{ij}\partial_iu_{k}\Big)-\frac{n-2}{4(n-1)}R_gu_{k}+\lambda{k}u^{{s_k}-1}_{k}=0,\quad \text{in}\;B_0(\rho)

The idea here is to consider the normalized function

\displaystyle  v_k\doteqdot\frac{u(\delta_kx+x_k)}{m_k}

where \delta_k=m_k^{(2-s_k)/2}. We have x_k \to 0 and \delta_k \to 0 as k \to\infty. Here v_k is defined on a ball in \mathbb{R}^n of radius \rho_k = (\rho-|x_k|)/\delta_k and \rho_k\to\infty as k\to\infty.

By the argument of diagonal subsequence and the property of normal coordinates , one observes that a subsequence of \{v_k\} converges to a smooth positive function v which is a nonnegative solution of the equation

\displaystyle  \Delta_0 v+\lambda v^{\frac{n+2}{n-2}}=0,\quad\text{in}\;\mathbb{R}^n\hfill (1)

where \lambda=\lim\limits_{k\to\infty}\lambda_k,and \Delta_0 is the standard Laplacian on \mathbb{R}^n.

By the strong maximum principle, v>0. It is known that \lambda<\lambda(M) if \lambda(M) < 0; and \lambda=\lambda(M) if \lambda(M) \ge 0 . Let d be the diameter of (M, g). By a change of variables we have

\displaystyle  \int_{|x|<\frac d2\delta_k^{-1}}v_k^{s_k}\sqrt{\det g}dx=\delta_k^{\frac{2s_k}{s_k-2}-n}\int_{B_{x_k}(\frac d2)}u_k^{s_k}d\mu_g\le\delta_k^{\frac{2s_k}{s_k-2}-n}\hfill (2)

where B_{x_k}(d/2) denotes the open ball in (M, g) with center at x_k and radius equal to d/2. we note that

 \displaystyle  \frac{2s_k}{s_k-2}-n>0\quad \text{and}\;\to0\quad\text{as}\;k\to\infty.

From (2) the Fatou lemma and \lim\limits_{k\to\infty}\delta_k\to0 , we obtain

\displaystyle  \int_{\mathbb{R}^n}v^{\frac{2n}{n-2}}dx\le1\hfill(3)

A similar argument implies

\displaystyle  \int_{\mathbb{R}^n}|\nabla v|^2dx<\infty.

Let \eta\in C^{\infty}_0(\mathbb{R}^n) be a cutoff function satisfies \eta =1 in B_0(d) and  \eta =0 in \mathbb{R}^n\setminus B_0(2d)

Defined v_R(x)=\eta{\frac xR}v(x), then

\displaystyle  \int_{\mathbb{R}^n}(|\nabla(v-v_R)|^2+|v-v_R|^{2^*})dx\to0,\quad \text{as}\;R\to\infty.\hfill (4)

Multiplies (1) by v_R and integration by parts, we obtain

\displaystyle  \int_{\mathbb{R}^n}\nabla v_R\nabla vdx=\lambda\int_{\mathbb{R}^n}v^{2^*-1}v_Rdx

Taking R\to\infty in above equation and thanks to (4) we get

\displaystyle  \int_{\mathbb{R}^n}|\nabla v|^2dx=\lambda\int_{\mathbb{R}^n}v^{2^*}dx.\hfill(5)

  •  If \lambda\le0, then v=\text{constant}, and (2)  implies v\equiv0, which is a contradiction with v>0.
  •  If \lambda>0, \lambda=\lambda(M). (2) (5) and the best Sobolev imbedding implies

 \displaystyle \Lambda\Big(\int_{\mathbb{R}^n}v^{2^*}dx\Big)^{2/2^*}\le\int_{\mathbb{R}^n}|\nabla v|^2dx=\lambda(M)\int_{\mathbb{R}^n}v^{2^*}dx.

 Thus

\displaystyle  \Lambda\le\lambda(M)\Big(\int_{\mathbb{R}^n}v^{2^*}dx\Big)^{n/2}\le\lambda(M).

  We are led to the contradiction with \lambda(M)<\lambda(\mathbb{S}^n)=\Lambda.
 Therefore, u_s is uniformly bounded.

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2 Responses to A blowup proof of Aubin’s theorem in the Yamabe problem

  1. Ngô Quốc Anh说道:

    It seems there is a typo in your calculation. After introducing the normalized function, I think the constant \delta_k should be m_k^{(2-s_k)}. If I am not wrong, your equation of u_k can be transformed to the following

    \displaystyle \frac{{{m_k}}}{{{\delta _k}}}\frac{1}{{\sqrt {\det g} }}{\partial _j}\left( {\sqrt {\det g} {g^{ij}}{\partial _i}{v_k}} \right) = \frac{{\frac{{n - 2}}{{4(n - 1)}}R({\delta _k}x + {x_k})m_k^{2 - {s_k}}{v_k}(x) - {\lambda _k}{v_k}{{(x)}^{{s_k} - 1}}}}{{m_k^{1 - {s_k}}}}

    which forces \delta_k=m_k^{(2-s_k)}. Please clarify.

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