Laplacian on Scalar Field is Divergence of Gradient

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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:

\(\ds \operatorname {div} \mathbf V\) \(=\) \(\ds \nabla \cdot \mathbf V\)
\(\ds \grad \mathbf U\) \(=\) \(\ds \nabla U\)

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:

\(\ds \nabla^2 U\) \(:=\) \(\ds \nabla \cdot \paren {\nabla U}\)
\(\ds \) \(=\) \(\ds \paren {\sum_{k \mathop = 1}^n \mathbf e_k \dfrac \partial {\partial x_k} } \cdot \paren {\sum_{k \mathop = 1}^n \dfrac {\partial U} {\partial x_k} \mathbf e_k}\) Definition of Del Operator
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \dfrac {\partial^2 U} {\partial {x_k}^2}\)

$\blacksquare$


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:

\(\ds \nabla^2 U\) \(:=\) \(\ds \nabla \cdot \paren {\nabla U}\)
\(\ds \) \(=\) \(\ds \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } \cdot \paren {\dfrac {\partial U} {\partial x} \mathbf i + \dfrac {\partial U} {\partial y} \mathbf j + \dfrac {\partial U} {\partial z} \mathbf k}\) Definition of Del Operator
\(\ds \) \(=\) \(\ds \dfrac {\partial^2 U} {\partial x^2} + \dfrac {\partial^2 U} {\partial y^2} + \dfrac {\partial^2 U} {\partial z^2}\)

$\blacksquare$


Sources