Why : $\int_{\Omega}\operatorname{div^{2}\vec{u}}dx=\int_{\Omega }(\nabla u)(\nabla u)^{t}dx$

Question

Isee this steps in book PDE :

I'm going to understand it :

Steps :

$$u\in\mathbb{R^{N}}$$

And $\Omega \subset\mathbb{R^{N}}$

$$K=\int_{\Omega}\operatorname{div^{2}\vec{u}}dx=\int_{\Omega }\sum_{i,j=1,n}\frac{\partial u_{i}\partial u_{j}}{\partial x_{i}\partial x_{j}}dx$$

integration by part we get :

$$K=\int_{\Omega }\sum_{i,j=1}^{N}\frac{\partial u_{i}\partial u_{j}}{\partial x_{j}\partial x_{i}}dx=\int_{\Omega }(\nabla u)(\nabla u)^{t}dx$$

$$≤\int_{\Omega }|\nabla u|^{2}dx$$

I don't understand the second line how he use integration by part and what the relation to come those to

$ $\nabla =\operatorname{grad}$$ $.$

Now I know that :

$$\nabla u=(\frac{\partial u}{dx_{1}},\frac{\partial u}{dx_{2}},...,\frac{\partial u}{dx_{N}})^{t}$$

And :

$$\operatorname{div\vec{u}}=\sum_{i=1}^{N}\frac{\partial u_{i}}{\partial x_{i}}$$

I have already to see your explain !

Answer 1

It seems that $u$ is a vector-valued function. If $u$ is regular enough and is supposed that $u$ vanishes on the boundary, the second line is obtained using one of Green's identity (commonly called integral by parts), twice. If it is not known whether the function $u$ is regular enough, the "integral by parts" is taken in the context of distributions (motivated by the regular case).

Acutually $\nabla u$ is a $n\times n$ matrix.

The integrand $(\nabla u)(\nabla u)^t$ is the usual inner produt betwenn matrices given by $A\cdot B=\mathrm{tr}(B^TA)$. The last inequality is obtained by using of Young's inequality.

Answer 2

Note that, as mentioned in Alberto Leandro's answer, $\nabla \mathbf u $ is an $n\times n$ matrix. This matrix is given coordinate-wise by $$(\nabla \mathbf u)_{ij} = \frac{\partial u^i}{\partial x_j} .$$ This means that $((\nabla \mathbf u)(\nabla \mathbf u)^T)$ is also an $n\times n$ matrix so, as stated, the equality with $K$ doesn t really make sense (the left hand side is a scalar while the right hand side gives an $n\times n$ matrix) - maybe this is me not knowing some important notation convention. I assume the equality must be intepreted as $$\int_\Omega (\mathrm{div}\,\mathbf u )^2 \, dx= \int_\Omega \mathrm {tr}\, ((\nabla \mathbf u)(\nabla \mathbf u)^T) \, dx.$$

---

Suppose that $\Omega$ is bounded with $C^1$ boundary and $\mathbf u=(u^1 ,\dots ,u^n)\in C^2(\Omega;\mathbb R^n) \cap C^1(\overline{\Omega}; \mathbb R^n)$. Integrating by parts with respect to the $j$-th coordinate gives \begin{align*} \int_\Omega (\mathrm{div}\,\mathbf u )^2 \, dx &=\sum_{i=1}^n \sum_{j=1}^n \int_\Omega \bigg (\frac{\partial u^i}{\partial x_i} \bigg ) \bigg ( \frac{\partial u^j}{\partial x_i} \bigg ) \, dx \\ &= -\sum_{i=1}^n \sum_{j=1}^n \int_\Omega u^j\frac{\partial^2 u^i}{\partial x_ix_j} \, dx+ \sum_{i=1}^n \sum_{j=1}^n \int_{\partial \Omega }u^j\bigg ( \frac{\partial u^i}{\partial x_i} \bigg ) \nu_j \, d \mathcal H^{n_1}_x \\ &= -\sum_{i=1}^n \sum_{j=1}^n \int_\Omega u^j\frac{\partial^2 u^i}{\partial x_ix_j} \, dx+ \int_{\partial \Omega }(\mathbf u \cdot \nu)(\mathrm{div}\,\mathbf u ) \, d \mathcal H^{n_1}_x \end{align*} Here $\partial \Omega$ denotes the boundary of $\Omega$, $\nu$ is the outward pointing unit normal of $\partial \Omega$, and $d \mathcal H^{n-1}_x$ is the $(n-1)$-surface measure on $\partial \Omega$. I'm assuming you have some kind of boundary condition on $\mathbf u$, most likely $\mathbf u=0$ on $\partial \Omega$, so that $$\int_{\partial \Omega }(\mathbf u \cdot \nu)(\mathrm{div}\,\mathbf u ) \, d \mathcal H^{n_1}_x=0 $$ otherwise the result is not true. Hence, $$ \int_\Omega (\mathrm{div}\,\mathbf u )^2 \, dx=-\sum_{i=1}^n \sum_{j=1}^n \int_\Omega u^j\frac{\partial^2 u^i}{\partial x_ix_j} \, dx$$ Then integrating by parts again except now with respect to $x_i$ gives \begin{align*} -\sum_{i=1}^n \sum_{j=1}^n \int_\Omega u^j\frac{\partial^2 u^i}{\partial x_ix_j} \, dx &= \sum_{i=1}^n \sum_{j=1}^n \int_\Omega \frac{\partial u^j}{\partial x_i}\frac{\partial u^i}{\partial x_j} \, dx - \sum_{i=1}^n \sum_{j=1}^n \int_{\partial \Omega} u^j\frac{\partial u^i}{\partial x_j} \nu^i \, dx\\ &= \int_\Omega \mathrm {tr}\, ((\nabla \mathbf u)(\nabla \mathbf u)^T) \, dx \end{align*} again using the boundary condition. Thus, $$\int_\Omega (\mathrm{div}\,\mathbf u )^2 \, dx= \int_\Omega \mathrm {tr}\, ((\nabla \mathbf u)(\nabla \mathbf u)^T) \, dx $$ as required.