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Slm2004
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I am reading papers about yamabe flow. I have a problem about how people derive it as a gradient flow.

Suppose we have $(M,g_0)$, $g(t)=u^{\frac{4}{n-2}}(t)g_0$ is another conformal metric. Let $R=R(t)$ be the scalar curvature and $s=\frac{\int_M Rd\mu}{\int_M d\mu}$ be the average scalar curvature at time $t$. I know many people use $$u_t=(s-R)u$$ as yamabe flow. I want to view it as the gradient flow the following functional $$E(u)=\frac{\int_M R\,d\mu}{V(t)^{\frac{n-2}{n}}}=\frac{\int_M \frac{4(n-1)}{n-2}|\nabla u|^2+R_0u^2\,d\mu_0}{\left(\int_M u^{\frac{2n}{n-1}}d\mu_0\right)^{\frac{n-2}{n}}}$$$$E(u)=\frac{\int_M R\,d\mu}{V(t)^{\frac{n-2}{n}}}=\frac{\int_M \frac{4(n-1)}{n-2}|\nabla u|^2+R_0u^2\,d\mu_0}{\left(\int_M u^{\frac{2n}{n-2}}d\mu_0\right)^{\frac{n-2}{n}}}$$ where $V(t)$ is the volume.

By some lengthy calculation, the frechet derivative of $E$ is $$\langle E'(u),v\rangle=\frac{2}{V(t)^{\frac{n-2}{n}}}\int_{M}(R-s)u^{-1}v\,d\mu$$ It seems that $E'(u)=\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$ in the $L^2$ sense. So the $L^2$ gradient flow of this functional is $$u_t=-\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$$ This is totally different from the yamabe flow as I mentioned. I expect we should have $u$ instead of $u^{-1}$ on the right hand side.

So what is the problem? how should I accommodate them?

I am reading papers about yamabe flow. I have a problem about how people derive it as a gradient flow.

Suppose we have $(M,g_0)$, $g(t)=u^{\frac{4}{n-2}}(t)g_0$ is another conformal metric. Let $R=R(t)$ be the scalar curvature and $s=\frac{\int_M Rd\mu}{\int_M d\mu}$ be the average scalar curvature at time $t$. I know many people use $$u_t=(s-R)u$$ as yamabe flow. I want to view it as the gradient flow the following functional $$E(u)=\frac{\int_M R\,d\mu}{V(t)^{\frac{n-2}{n}}}=\frac{\int_M \frac{4(n-1)}{n-2}|\nabla u|^2+R_0u^2\,d\mu_0}{\left(\int_M u^{\frac{2n}{n-1}}d\mu_0\right)^{\frac{n-2}{n}}}$$ where $V(t)$ is the volume.

By some lengthy calculation, the frechet derivative of $E$ is $$\langle E'(u),v\rangle=\frac{2}{V(t)^{\frac{n-2}{n}}}\int_{M}(R-s)u^{-1}v\,d\mu$$ It seems that $E'(u)=\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$ in the $L^2$ sense. So the $L^2$ gradient flow of this functional is $$u_t=-\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$$ This is totally different from the yamabe flow as I mentioned. I expect we should have $u$ instead of $u^{-1}$ on the right hand side.

So what is the problem? how should I accommodate them?

I am reading papers about yamabe flow. I have a problem about how people derive it as a gradient flow.

Suppose we have $(M,g_0)$, $g(t)=u^{\frac{4}{n-2}}(t)g_0$ is another conformal metric. Let $R=R(t)$ be the scalar curvature and $s=\frac{\int_M Rd\mu}{\int_M d\mu}$ be the average scalar curvature at time $t$. I know many people use $$u_t=(s-R)u$$ as yamabe flow. I want to view it as the gradient flow the following functional $$E(u)=\frac{\int_M R\,d\mu}{V(t)^{\frac{n-2}{n}}}=\frac{\int_M \frac{4(n-1)}{n-2}|\nabla u|^2+R_0u^2\,d\mu_0}{\left(\int_M u^{\frac{2n}{n-2}}d\mu_0\right)^{\frac{n-2}{n}}}$$ where $V(t)$ is the volume.

By some lengthy calculation, the frechet derivative of $E$ is $$\langle E'(u),v\rangle=\frac{2}{V(t)^{\frac{n-2}{n}}}\int_{M}(R-s)u^{-1}v\,d\mu$$ It seems that $E'(u)=\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$ in the $L^2$ sense. So the $L^2$ gradient flow of this functional is $$u_t=-\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$$ This is totally different from the yamabe flow as I mentioned. I expect we should have $u$ instead of $u^{-1}$ on the right hand side.

So what is the problem? how should I accommodate them?

Source Link
Slm2004
Slm2004
  • 633
  • 5
  • 9

Derivation of yamabe flow

I am reading papers about yamabe flow. I have a problem about how people derive it as a gradient flow.

Suppose we have $(M,g_0)$, $g(t)=u^{\frac{4}{n-2}}(t)g_0$ is another conformal metric. Let $R=R(t)$ be the scalar curvature and $s=\frac{\int_M Rd\mu}{\int_M d\mu}$ be the average scalar curvature at time $t$. I know many people use $$u_t=(s-R)u$$ as yamabe flow. I want to view it as the gradient flow the following functional $$E(u)=\frac{\int_M R\,d\mu}{V(t)^{\frac{n-2}{n}}}=\frac{\int_M \frac{4(n-1)}{n-2}|\nabla u|^2+R_0u^2\,d\mu_0}{\left(\int_M u^{\frac{2n}{n-1}}d\mu_0\right)^{\frac{n-2}{n}}}$$ where $V(t)$ is the volume.

By some lengthy calculation, the frechet derivative of $E$ is $$\langle E'(u),v\rangle=\frac{2}{V(t)^{\frac{n-2}{n}}}\int_{M}(R-s)u^{-1}v\,d\mu$$ It seems that $E'(u)=\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$ in the $L^2$ sense. So the $L^2$ gradient flow of this functional is $$u_t=-\frac{2}{V(t)^{\frac{n-2}{n}}}(R-s)u^{-1}$$ This is totally different from the yamabe flow as I mentioned. I expect we should have $u$ instead of $u^{-1}$ on the right hand side.

So what is the problem? how should I accommodate them?