How to prove $\mathcal{K}:\nabla \nabla T$ = $ k \cdot \nabla^2 T$

Question

I'm trying to demonstrate that: $\mathcal{K}:\nabla \nabla T$ == $ k \cdot \nabla^2 T$ where $\mathcal{K}$ is a second-order tensor.

Considering that:

\begin{equation} \mathcal{K} = kI \tag{1} \end{equation}

$k$ is constant, and $I$ is the identity tensor (second-order).

I know that $\nabla\nabla T = \nabla^2 T$ which means that if:

$\mathcal{K}:\nabla \nabla T = k \cdot \nabla^2 T$

Then the double dot product "$\mathcal{K}:$" has to be equal to the dot product "$kI \cdot$" with $\nabla^2T$ but I don't know how to demonstrate that.

Answer 1

I will assume that by $$\mathbf K:\nabla\nabla \mathbf T$$ Taking $T$ to be a $(p,q)$ tensor, that you mean, in indices, $$K^{lm}\nabla_l\nabla_mT^{i_1\dots i_p}{}_{j_1\dots j_q}$$ I will stick to the case of $T$ being a $(2,0)$ tensor, but this generalizes with no problems.

The equation $\mathbf K=k\mathbf I$ means $$K^i{}_j=k\delta^i_j$$ Equivalently we can raise an index $$K^{ij}=kg^{ij}$$ So $$(K:\nabla\nabla T)^{ij}=K^{lm}\nabla_l\nabla_mT^{ij} \\ =kg^{lm}\nabla_l\nabla_m T^{ij}$$ But of course $g^{lm}\nabla_l=\nabla^m$ hence $$(K:\nabla\nabla T)^{ij}=k \nabla^m\nabla_m T^{ij}=k\Delta T^{ij}$$ Since $\nabla^m\nabla_m=\Delta$ by definition.