# Ngô Quốc Anh

## September 15, 2013

### Some integral identities on manifolds with boundary

In this note, I summary several useful integral identities on Riemannian manifolds with boundary.

1. Suppose that $f$ is a function and $X$ is a $1$-form, then

$\displaystyle\boxed{\int_M {f\text{div}Xd{v_g}} = - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}.}$

To prove this, we write everything in local coordinates as follows

$\begin{array}{lcl} \displaystyle\int_M {f \text{div} Xd{v_g}} &=& \displaystyle\int_M {f{\nabla _i}{X^i}d{v_g}} \hfill \\ &=& \displaystyle - \int_M {{\nabla _i}f{X^i}d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}} \hfill \\ &=& \displaystyle - \int_M {\left\langle {\nabla f,X} \right\rangle_g d{v_g}} + \int_{\partial M} {f\left\langle {X,\nu } \right\rangle_g d{\sigma _g}}\end{array}$

as claimed.

2. Using the previous identity, we can prove the following

$\displaystyle \boxed{\int_M {{{\left\langle {X,\nabla (\text{div} X)} \right\rangle }_g}d{v_g}} = - \int_M {|\text{div} X|_g^2d{v_g}} + \int_{\partial M} {\text{div} X{{\left\langle {X,\nu } \right\rangle }_g}d{\sigma _g}}}$

where $X$ is again a vector field on $M$. To prove this, we simply apply the previous identity with $f$ replaced by $\text{div}(X)$ to get the desired result.

3. Now we have a look-like integration by parts for scalar functions, actually, for any vector field $X$ over $M$, we have

$\displaystyle \boxed{\int_M {{{\left\langle {X,\Delta X} \right\rangle }_g}d{v_g}} = - \int_M {|\nabla X|_g^2d{v_g}} + \int_{\partial M} {{X^k}{{\left\langle {\nabla {X_k},\nu } \right\rangle }_g}d{\sigma _g}} .}$

To prove this, we write

$\begin{array}{lcl}\displaystyle\int_M {{{\left\langle {X,\Delta X} \right\rangle }_g}d{v_g}} &=& \displaystyle\int_M {{X^k}\underbrace {{\nabla ^j}{\nabla _j}{X_k}}_{\Delta {X_k}}d{v_g}} \hfill \\ &=& \displaystyle - \int_M {{\nabla ^j}{X^k}{\nabla _j}{X_k}d{v_g}} + \int_{\partial M} {{X^k}{\nabla _j}{X_k}{\nu ^j}d{\sigma _g}} \hfill \\ &=& \displaystyle - \int_M {|\nabla X|_g^2d{v_g}} + \int_{\partial M} {{X^k}{{\left\langle {\nabla {X_k},\nu } \right\rangle }_g}d{\sigma _g}} \end{array}$

as claimed.

4. We end this note by talking the following identity which can be thought of a generalization of the third identity. We prove

$\displaystyle \boxed{\int_M {\left\langle {\alpha ,\nabla \beta } \right\rangle_g d{v_g}} = - \int_M {\left\langle {\text{div} \alpha ,\beta } \right\rangle_g d{v_g}} + \int_M {\text{div} Td{v_g}}}$

where $\alpha$ is an $(r,s)$-tensor, $\beta$ is an $(r-1,s)$-tensor, and $T$ is a vector field given by

$\displaystyle {T_j} = \alpha _{j{i_2} \cdots {i_r}}^{{k_1} \cdots {k_s}}\beta _{{i_2} \cdots {i_r}}^{{k_1} \cdots {k_s}}.$

The proof is straightforward as one can easily see that

$\displaystyle \text{div} T = {\left\langle {\alpha ,\nabla \beta } \right\rangle _g} + {\left\langle {\text{div}\alpha ,\beta } \right\rangle _g}.$

In view of the divergence theorem (or the Stokes theorem), we can further write

$\displaystyle\int_M {{{\left\langle {\alpha ,\nabla \beta } \right\rangle }_g}d{v_g}} = - \int_M {{{\left\langle {\text{div} \alpha ,\beta } \right\rangle }_g}d{v_g}} + \int_{\partial M} {{{\left\langle {T,\nu } \right\rangle }_g}d{\sigma _g}} .$

Let us go back to the third identity, we shall use $\alpha = \nabla X$ and $\beta=X$ to get

$\displaystyle\int_M {|\nabla X|_g^2d{v_g}} = - \int_M {{{\left\langle {\Delta X,X} \right\rangle }_g}d{v_g}} + \int_{\partial M} {{{\left\langle {T,\nu } \right\rangle }_g}d{\sigma _g}}$

where $T$ is given by

$\displaystyle {T_j} = ({\nabla _j}{X_k}){X^k}.$

From this, we obtain the desired result through

$\displaystyle {\left\langle {T,\nu } \right\rangle _g} = {T_j}{\nu ^j} = ({\nabla _j}{X_k}){X^k}{\nu ^j} = {X^k}{\left\langle {\nabla {X_k},\nu } \right\rangle _g}.$