Laplacian Operator

The Laplacian operator (also called Laplace operator) is a differentiation operator that operates on both vector fields and scalar functions. It is a composite function constructed by taking the divergence of the gradient of a function.

The Laplacian operator (usually shortened to just the Laplacian) has different forms depending on whether the function it is operating on is a vector field or a scalar function. It is also notated by any of:

• $\nabla \cdot \nabla f$
• ${\nabla }^{2}f$
• $\Delta f$

Definition §

Laplacian Operator (Scalar Function) ^definition-scalar

Let $f$ be a scalar function in Euclidean space, $f:D\subseteq {\mathbb{R}}^n\to \mathbb{R}$ where $f(x_1,\dots ,x_n)=y$. If $f$ is a $C^2$ function, then the Laplacian of $f$ is:

$$\begin{array}{rl}{\nabla }^{2}f& =\nabla \cdot \nabla f=\nabla \cdot (\nabla f)\\ & =\nabla \cdot \left(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\dots \frac{\partial f}{\partial x_n}\right)\\ & =\frac{{\partial }^{2}f}{\partial x_1^2}+\frac{{\partial }^{2}f}{\partial x_2^2}+\cdots +\frac{{\partial }^{2}f}{\partial x_n^2}\end{array}\text{\hspace{1em}(Grad Operator)}$$

• ${\nabla }^{2}f:{\mathbb{R}}^n\to \mathbb{R}$
• In the case of the scalar function $f:{\mathbb{R}}^3\to \mathbb{R}$, if we have $f(x,y,z)=w$ we have:

$${\nabla }^{2}f=\nabla \cdot \nabla f=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}$$

Laplacian Operator (Vector Field) ^definition-vector-field

Let $\vec{F}$ be a vector field $\vec{F}:D\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ where $\vec{F}(x_1,\dots ,x_n)=(f_1(x_1,\dots ,x_n),\dots ,f_n(x_1,\dots ,x_n))$. If $\vec{F}$ is $C^2$ (that is, all $f_i$ are $C_2$ scalar functions), then the Laplacian of $\vec{F}$ is:

$${\nabla }^2\vec{F}=\nabla \cdot \nabla \vec{F}=(\nabla \cdot \nabla )\vec{F}=({\nabla }^2{f}_1,\dots ,{\nabla }^2{f}_n)$$

That is, we distribute the Laplacian over the vector to make it act on scalar functions.!

• ${\nabla }^2\vec{F}:{\mathbb{R}}^n\to {\mathbb{R}}^n$
• In the case of a real vector field $\vec{F}:{\mathbb{R}}^3\to {\mathbb{R}}^3$, if we have $\vec{F}(x,y,z)=(u(x,y,z),v(x,y,z),w(x,y,z))$ where $u,v,w:{\mathbb{R}}^3\to \mathbb{R}$ are scalar functions, we have:

$$\begin{array}{rl}{\nabla }^2\vec{F}& =\nabla \cdot \nabla \vec{F}=({\nabla }^{2}u,{\nabla }^{2}v,{\nabla }^{2}w)\\ & =\left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2},\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}+\frac{{\partial }^{2}v}{\partial z^2},\frac{{\partial }^{2}w}{\partial x^2}+\frac{{\partial }^{2}w}{\partial y^2}+\frac{{\partial }^{2}w}{\partial z^2}\right)\end{array}$$

Applications §

Potential Energy §

The Laplacian operator can be used to verify the equations of potential energy:

$${\nabla }^2{U}_g=0$$

Take the equation for gravitational potential energy:

Taking the Cartesian form, the grad returns:

$$\begin{array}{rl}\nabla U_g& =\left(\frac{\partial U_g}{\partial x},\frac{\partial U_g}{\partial y},\frac{\partial U_g}{\partial z}\right)\\ & =(\frac{-x}{(x^2+y^2+z^2{)}^{3/2}},\frac{-y}{(x^2+y^2+z^2{)}^{3/2}},\frac{-z}{(x^2+y^2+z^2{)}^{3/2}})\end{array}$$

And the divergence of that returns:

$$\begin{array}{rl}\nabla \cdot \nabla U_g& =\frac{3x^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2{)}^{5/2}}+\frac{3y^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2{)}^{5/2}}+\frac{3z^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2{)}^{5/2}}\\ & =\frac{3(x^2+y^2+z^2)-3(x^2+y^2+z^2)}{(x^2+y^2+z^2{)}^{5/2}}=0\end{array}$$

Resources §

• https://www.youtube.com/watch?v=gWbfdPUynz0