Laplacian on Scalar Field is Divergence of Gradient

Theorem
Let $\R^n$ denote the real Cartesian space of $n$ dimensions.

Let $U$ be a scalar field over $\R^n$.

Let $\nabla^2 U$ denote the laplacian on $U$.

Then:
 * $\nabla^2 U = \operatorname {div} \grad U$

where:
 * $\operatorname {div}$ denotes the divergence operator
 * $\grad$ denotes the gradient operator.

Proof
From Divergence Operator on Vector Space is Dot Product of Del Operator and definition of the gradient operator:

where $\nabla$ denotes the del operator.

Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis on $\R^n$.

Hence:

Also presented as
In Cartesian $3$-space $\R^3$, where:


 * $U$ is defined as $\map U {x, y, z}$


 * $\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis on $\R^3$.

this is usually presented as: