Let $\Omega \subset \mathbb{R}^2$. Then we have \begin{align*} \int_{\Omega} \textbf{v} \cdot \nabla (\nabla\cdot \textbf{u})dX &= \int_{\Omega} \begin{bmatrix} v_1 \\ v_2\end{bmatrix} \cdot \nabla (u_{1,x} + u_{2,y}) dX, \\ &= \int_{\Omega} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \cdot \begin{bmatrix} u_{1,xx} + u_{2,yx} \\ u_{1,xy} + u_{2,yy}\end{bmatrix} dX, \\ &= \int_{\Omega} v_1 (u_{1,xx} + u_{2,xy}) + v_2 (u_{1,xy} + u_{2,yy}) dX. \end{align*} Is there an easy way to "move the derivatives" over to the components of $\textbf{v}$ (by using the Divergence theorem or Integration by parts, perhaps)?

I want to rewrite my starting integral as the sum of an integral over $\Omega$ and an integral over the boundary $\partial \Omega$. The motivation for this is in deriving the weak form a BVP involving linear elasticity.

Are there any vector calculus identities that give me what I want immediately?

See here:

Deriving the weak form for linear elasticity equation


1 Answer 1


Your integrand can be written as follows: $$\vec{v}\cdot\vec{\mathrm{grad}}(\mathrm{div}\vec{u})=\mathrm{div}(\vec{v}\,\mathrm{div}\vec{u})-\mathrm{div}\vec{v}\,\mathrm{div}\vec{u}$$

Plug it into your equation and apply $\textit{Gauss}$ theorem: $$\int_{\Omega}{\mathrm{div}\vec{F}\,dV}=\int_{\partial\Omega}{\vec{F}\cdot\vec{n}\,dS}$$ To arrive to the final formulation you asked for $$\int_{\Omega}{\vec{v}\cdot\vec{\mathrm{grad}}(\mathrm{div}\vec{u})\,dV}=\int_{\Omega}{\mathrm{div}(\vec{v}\,\mathrm{div}\vec{u})\,dV}-\int_{\Omega}{\mathrm{div}\vec{v}\,\mathrm{div}\vec{u}\,dV}=\int_{\partial\Omega}{\vec{v}\cdot\vec{n}\,\mathrm{div}\vec{u}\,dS}-\int_{\Omega}{\mathrm{div}\vec{v}\mathrm{div}\vec{u}\,dV}$$


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