How to, in general prove, that:
div($T_{\nabla u})=\nabla$div$u$?
$T$:the transpose.
$u$: is a function s.t $(x,t) \rightarrow u(x,t):=x'(t)=(x_1'(t),...,x_n'(t))$
is the velocity field of the fluid.
$\nabla$ :nabla operator.
div: divergence.
In a reference, I have seen that:
1)Tr(Du)=Tr∇u = divu , 2)div(t∇u)=∇(divu) 3)div(2Du) = div(∇u +t∇u)=Δu +∇(divu) 4)div((div u)Id) = ∇(divu). ,Tr:trace,t:transpose.
And point 2 is my question(while divergenve gives a scalar, how might we apply a gradient on it! I am new to Navier Stokes equation lesson). And I need some help pls, or clarification.
If $A = (a_{ij})_{ij}$ then $\operatorname{div}(A)_i$ (that is, the $i$-th component of the vector $\operatorname{div}(A)$) is given by $$\sum_{j}\partial_j a_{ij}.$$
If $A = (\nabla u)^t$, that is, if $A$ is the transpose of $\nabla u$, then $a_{ij} = \partial_i u_j$ and therefore $$\operatorname{div}((\nabla u)^t)_i = \sum_{j}\partial_j\partial_iu_j.$$ On the other hand, since $\operatorname{div}u = \sum_j \partial_ju_j$, the $i$-th component of the vector $\nabla \operatorname{div}(u)$ is precisely $$\partial_i \operatorname{div}(u) = \partial_i \sum_j \partial_j u_j = \sum_j \partial_j\partial_i u_j.$$ Notice that in the very last step we have used Schwartz's theorem to swap the other of the derivatives. Hence these computations are rigorous if $u$ is sufficiently regular. Since you mention Navier-Stokes, chances are that you are working with different classes of functions. However, I hope this clarifies things a little bit!