Integration by Parts with multi-variable functions I want to prove that 
$$ - \int\limits_{\mathbb R^n} y_t \Delta y =  \int\limits_{\mathbb R^n} \sum_{j=1}^n \left(\frac{\partial y}{\partial x_j}\right)\left(\frac{\partial y_t}{\partial x_j}\right) $$ 
Here $ \Delta y  =  \sum\limits_{j=1}^n \frac{\partial^2 y}{\partial x_j^2} $, and $y = y(t,x_1, \cdots x_n ) \in  C^\infty ([0,\infty) \times \mathbb R^n )$,  $y_t(t, \cdot) \in L^2$.
 A: We begin writing $$\int_{\mathbb R^n}y_t\Delta y=\sum_{j=1}^n\int_{\mathbb R^n}y_t\partial_{jj}y=\sum_{j=1}^n\int_{\mathbb R^{n-1}}\left(\int_{\mathbb R}y_t\partial_{jj}ydx_j\right)\widehat{dxj},$$
where $\widehat{dxj}=dx_1\ldots dx_{j-1}dx_{j+1}\ldots dx_n$. Now, on each integral 
$I_j:=\int_{\mathbb R}y_t\partial_{jj}ydx_j$, we do an integration by parts to get $I_j=-\int_{\mathbb R}\partial_jy_t\partial_jydx_j$, assuming that $\lim_{|x_j|\to\infty}y_t(x)\partial_jy(x)=0$. We get 
$$-\int_{\mathbb R^n}y_t\Delta ydx=\sum_{j=1}^n\int_{\mathbb R^{n-1}}\left(\int_{\mathbb R}\partial_jy_t\partial_jydx_j\right)\widehat{dx_j}=\sum_{j=1}^n\int_{\mathbb R^n}\partial_j y_t\partial_jydx,$$
which is the wanted result.
A: You could use the formula from vector calculus $$\textrm{div}(aF)=(\textrm{grad}\,a)\cdot F+a\,\textrm{div}F,$$ for an arbitrary scalar function $a$ and a vector function $F.$
Taking $F=\textrm{grad}\,y,$ $a=y_t$ you get $$\textrm{div}(y_t\,\textrm{grad}\,y)=\textrm(\textrm{grad}\,y_t)\cdot(\textrm{grad}\,y)+y_t\Delta y.$$
Integrating termwise on $\mathbb{R}^n$ you get $$\int_{\mathbb{R}^n}\textrm(\textrm{grad}\,y_t)\cdot(\textrm{grad}\,y)+\int_{\mathbb{R}^n}y_t\Delta y=\lim_{R\to\infty}\int_{|x|=R}y_t\,\textrm{grad}\,y\cdot\hat{\bf{n}}.$$
Now you need adeguate conditions to make the rhs vanishing
