A formula in harmonic heat flow

Question

Assume that $ (M,g) $ and $ (N,h) $ are two smooth closed manifold and $ N $ is embedded isometrically into $ \mathbb{R}^K $ for some $ K\in\mathbb{Z}_+ $. Assume that $ u\in C^{\infty}(M\times\mathbb{R}_+,N) $ satisfies the equation of harmonic heat flow $$ \frac{\partial u}{\partial t}-\Delta_g u=A(u)(\nabla u,\nabla u), $$ where $ u=(u^1,u^2,...,u^K) $ and $ A(u)=(A_{jk}^i(u))_{1\leq i,j,k\leq K} $ is the second fundamental form of $ N $ at $ u(x) $. In the book "The analysis of harmonic maps and their heat flows" by F. Lin and C. Wang, page 115, there is a formula $$ (\partial_t-\Delta_g)|\partial_tu|^2=-|\nabla\partial_tu|^2+R^N(\nabla u,\partial_tu,\nabla u,\partial_t u), $$ where $ R^N $ denote the Riemannian curvature tensor. I have some difficulty in obtaining this formula since the I cannot simply use the arguments in the proof of Bochner formula. Can you give me some hints or references?

Answer 1

Actually there is a typo in the formula, i.e. $$ (\partial_t-\Delta_g)|\partial_tu|^2=-2|\nabla\partial_tu|^2+2R^N(\nabla u,\partial_tu,\nabla u,\partial_t u). $$ Firstly for $ x\in M $, we can choose an orthorgonal basis $ \{e_{\alpha}\} $. Then it follows from simple calculations that $$ \begin{aligned} \partial_t|\partial_tu|^2&=\langle\partial_tu,\partial_{tt}^2u\rangle\\ \nabla_{e_{\alpha}}\nabla_{e_{\alpha}}\langle\partial_tu,\partial_tu\rangle&=2\langle\partial_tu_{,\alpha},\partial_tu_{,\alpha}\rangle+2\langle\partial_tu,\partial_tu_{,\alpha\alpha}\rangle. \end{aligned} $$ Therefore $$ (\partial_t-\Delta_g)|\partial_tu|^2=-2|\partial_t u_{,\alpha}|^2+2\langle\partial_tu,(A(u)(\nabla u,\nabla u))_t\rangle $$ Define $ P(y):\mathbb{R}^K\to T_uN $ as an orthogonal projectopn map. We can obtain that $$ \begin{aligned} (\partial_t-\Delta_g)|\partial_tu|^2&=-2|\nabla \partial_tu|^2-2|A(u)(\partial_tu,\nabla u)|^2+2\langle P(u)\partial_tu,(A(u)(\nabla u,\nabla u))_t\rangle\\ &=-2|\nabla \partial_tu|^2-2|A(u)(\partial_tu,\nabla u)|^2-2\langle \partial_tP(u)\partial_tu,(A(u)(\nabla u,\nabla u))_t\rangle, \end{aligned} $$ where we have used the fact that $$ \langle P(u)\partial_tu,(A(u)(\nabla u,\nabla u))_t\rangle=0,\,\,\langle P(u)\partial_{tt}^2u,(A(u)(\nabla u,\nabla u))_t\rangle=0. $$ Since $ \partial_tP(u)\partial_tu=P(u)_{,\alpha}(\partial_tu,\partial_tu)=-A(u)(\partial_tu,\partial_tu) $, it can be got that $$ (\partial_t-\Delta_g)|\partial_tu|^2=-2|\nabla \partial_tu|^2-2|A(u)(\partial_tu,\nabla u)|^2+2\langle A(u)(\partial_tu,\partial_tu),(A(u)(\nabla u,\nabla u))_t\rangle. $$ Therefore the formula above is true by using Gauss-Codazi equation, i.e. $$ R^N(X,Y,X,Y)=\langle A(X,X),A(Y,Y)\rangle-\langle A(X,Y),A(X,Y)\rangle. $$