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Let $u: \mathbb{R}^{n} \rightarrow \mathbb{R}$ and $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$. View $f$ as a column vector, $\nabla u$ as a row vector. Why does $$\nabla \cdot \nabla u(f(x)) = \nabla u \cdot \Delta f + \operatorname{tr}((\nabla f)^{t}D^{2}u\nabla f)$$ where $D^{2}u$ is the Jacobian of $u$?

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    $\begingroup$ Did you try anything? What didn't work for you? If you don't share the results of your efforts, then we can't help you. $\endgroup$ Feb 6, 2014 at 23:43
  • $\begingroup$ I've verified it for $n = 3$, but it was by writing $f = (f_{1}, f_{2}, f_{3})^{t}$ and $x = (x_{1}, x_{2}, x_{3})$. The whole process was fairly tedious, and the process would work for general $n$, however, I was wondering if there was a coordinate free method. $\endgroup$
    – gpy1234
    Feb 7, 2014 at 0:28

2 Answers 2

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Let n=3 and expand the two sides of the equality by the definitions of Laplace and gradient operators. This will help you understand the process and finally you could plug in n.

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We will use Einsteinian notation (it's useful in vector calculus!): we sum with respect the repeating indexes, i.e. instead of writing $\sum_{j=1}^3a_jb_j$ we can write simply $a_jb_j$. Another useful notation: ,j in the subscript means derivative with respect to $j$-th cordinate. In other words, $f_{k,j}$ stands for $\frac{\partial f_k}{\partial x_j}$. As usual, $\vec e_j$ are standard basis vectors. Armed with this, on with the show!

$$\nabla u(f(x)) = u_{,k}f_{k,j}\vec e_j$$ $$\nabla \cdot\nabla u(f(x)) =\nabla\cdot( u_{,k}f_{k,j}\vec e_j) = ( u_{,k}f_{k,j})_{,j} = u_{,ks}f_{s,j}f_{k,j} + u_{,k}f_{k,jj}.$$

The last term is, in fact, $\nabla u\cdot \Delta f$, because

$$\nabla u\cdot \Delta f = (u_{,k}\vec e_k)\cdot(f_{l,jj}\vec e_l) = u_{,k}f_{k,jj}.$$

The first term requires a little bit of tinkering.

$$D^2 u = u_{,ks}\vec e_k \otimes \vec e_s,$$ $$\nabla f = f_{l,m}\vec e_l \otimes \vec e_m,$$ $$\nabla f ^t= f_{p,q}\vec e_q \otimes \vec e_p,$$ therefore

$$(\nabla f)^{t}D^{2}u\nabla f = ( f_{p,q}\vec e_q \otimes \vec e_p)(u_{,ks}\vec e_k \otimes \vec e_s)(f_{l,m}\vec e_l \otimes \vec e_m ) $$$$=( f_{k,q} u_{,ks}\vec e_q \otimes \vec e_s)(f_{l,m}\vec e_l \otimes \vec e_m ) $$ $$= f_{k,q} u_{,ks} f_{s,m}\vec e_q\otimes \vec e_m . $$ Now, the trace of this matrix is equal to $$ tr (f_{k,q} u_{,ks} f_{s,m}\vec e_q\otimes \vec e_m) = u_{,ks} f_{k,m} f_{s,m} , $$which, up to a change of summation variables, coincides with the first term.

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