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?



A formula in harmonic heat flow
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.
$$
R^N(X,Y,X,Y)=\langle A(X,X),A(Y,Y)\rangle-\langle A(X,Y),A(X,Y)\rangle.
$$


