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}.


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