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# Laplace operator

In mathematics, the Laplace operator or Laplacian is a differential operator given by the divergence of the gradient of a function on Euclidean space. It is usually denoted by the symbols ∇·∇, 2. The Laplacian ∇·∇f(p) of a function f at a point p, is (up to a factor) the rate at which the average value of f over spheres centered at p deviates from f(p) as the radius of the sphere shrinks towards 0. In a Cartesian coordinate system, the Laplacian is given by the sum of second partial derivatives of the function with respect to each independent variable. In other coordinate systems such as cylindrical and spherical coordinates, the Laplacian also has a useful form.

The Laplace operator is named after the French mathematician Pierre-Simon de Laplace (1749–1827), who first applied the operator to the study of celestial mechanics, where the operator gives a constant multiple of the mass density when it is applied to the gravitational potential due to the mass distribution with that given density. Solutions of the equation ∇·∇f = 0, now called Laplace's equation, are the so-called harmonic functions, and represent the possible gravitational fields in regions of vacuum.

The Laplacian occurs in differential equations that describe many physical phenomena, such as electric and gravitational potentials, the diffusion equation for heat and fluid flow, wave propagation, and quantum mechanics. The Laplacian represents the flux density of the gradient flow of a function. For instance, the net rate at which a chemical dissolved in a fluid moves toward or away from some point is proportional to the Laplacian of the chemical concentration at that point; expressed symbolically, the resulting equation is the diffusion equation. For these reasons, it is extensively used in the sciences for modelling all kinds of physical phenomena. The Laplacian is the simplest elliptic operator, and is at the core of Hodge theory as well as the results of de Rham cohomology. In image processing and computer vision, the Laplacian operator has been used for various tasks such as blob and edge detection.

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• ✪ Laplace transform 1 | Laplace transform | Differential Equations | Khan Academy
• ✪ 3.1.2 Laplace's Equation in 1D
• ✪ Laplacian operator | Vector Calculus | LetThereBeMath |
• ✪ Derive the Laplacian for a Spherical Coordinate System in 4 Steps
• ✪ Self-adjointness of the Laplacian

#### Transcription

I'll now introduce you to the concept of the Laplace Transform. And this is truly one of the most useful concepts that you'll learn, not just in differential equations, but really in mathematics. And especially if you're going to go into engineering, you'll find that the Laplace Transform, besides helping you solve differential equations, also helps you transform functions or waveforms from the time domain to this frequency domain, and study and understand a whole set of phenomena. But I won't get into all of that yet. Now I'll just teach you what it is. Laplace Transform. I'll teach you what it is, make you comfortable with the mathematics of it and then in a couple of videos from now, I'll actually show you how it is useful to use it to solve differential equations. We'll actually solve some of the differential equations we did before, using the previous methods. But we'll keep doing it, and we'll solve more and more difficult problems. So what is the Laplace Transform? Well, the Laplace Transform, the notation is the L like Laverne from Laverne and Shirley. That might be before many of your times, but I grew up on that. Actually, I think it was even reruns when I was a kid. So Laplace Transform of some function. And here, the convention, instead of saying f of x, people say f of t. And the reason is because in a lot of the differential equations or a lot of engineering you actually are converting from a function of time to a function of frequency. And don't worry about that right now. If it confuses you. But the Laplace Transform of a function of t. It transforms that function into some other function of s. and And does it do that? Well actually, let me just do some mathematical notation that probably won't mean much to you. So what does it transform? Well, the way I think of it is it's kind of a function of functions. A function will take you from one set of-- well, in what we've been dealing with-- one set of numbers to another set of numbers. A transform will take you from one set of functions to another set of functions. So let me just define this. The Laplace Transform for our purposes is defined as the improper integral. I know I haven't actually done improper integrals just yet, but I'll explain them in a few seconds. The improper integral from 0 to infinity of e to the minus st times f of t-- so whatever's between the Laplace Transform brackets-- dt. Now that might seem very daunting to you and very confusing, but I'll now do a couple of examples. So what is the Laplace Transform? Well let's say that f of t is equal to 1. So what is the Laplace Transform of 1? So if f of t is equal to 1-- it's just a constant function of time-- well actually, let me just substitute exactly the way I wrote it here. So that's the improper integral from 0 to infinity of e to the minus st times 1 here. I don't have to rewrite it here, but there's a times 1dt. And I know that infinity is probably bugging you right now, but we'll deal with that shortly. Actually, let's deal with that right now. This is the same thing as the limit. And let's say as A approaches infinity of the integral from 0 to Ae to the minus st. dt. Just so you feel a little bit more comfortable with it, you might have guessed that this is the same thing. Because obviously you can't evaluate infinity, but you could take the limit as something approaches infinity. So anyway, let's take the anti-derivative and evaluate this improper definite integral, or this improper integral. So what's anti-derivative of e to the minus st with respect to dt? Well it's equal to minus 1/s e to the minus st, right? If you don't believe me, take the derivative of this. You'd take minus s times that. That would all cancel out, and you'd just be left with e to the minus st. Fair enough. Let me delete this here, this equal sign. Because I could actually use some of that real estate. We are going to take the limit as A approaches infinity. You don't always have to do this, but this is the first time we're dealing with improper intergrals. So I figured I might as well remind you that we're taking a limit. Now we took the anti-derivative. Now we have to value it at A minus the anti-derivative valued at 0. And then take the limit of whatever that ends up being as A approaches infinity. So this is equal to the limit as A approaches infinity. OK. If we substitute A in here first, we get minus 1/s. Remember we're, dealing with t. We took the integral with respect to t. e to the minus sA, right? That's what happens when I put A in here. Now what happens when I put t equals 0 in here? So when t equals 0, it becomes e to the minus s times 0. This whole thing becomes 1. And I'm just left with minus 1/s. Fair enough. And then let me scroll down a little bit. I wrote a little bit bigger than I wanted to, but that's OK. So this is going to be the limit as A approaches infinity of minus 1/s e to the minus sA minus 1/s. So plus 1/s. So what's the limit as A approaches infinity? Well what's this term going to do? As A approaches infinity, if we assume that s is greater than 0-- and we'll make that assumption for now. Actually, let me write that down explicitly. Let's assume that s is greater than 0. So if we assume that s is greater than 0, then as A approaches infinity, what's going to happen? Well this term is going to go to 0, right? e to the minus-- a googol is a very, very small number. And an e to the minus googol is an even smaller number. So then this e to the minus infinity approaches 0, so this term approaches 0. This term isn't affected because it has no A in it, so we're just left with 1/s. So there you go. This is a significant to moment in your life. You have just been exposed to your first Laplace Transform. I'll show you in a few videos, there are whole tables of Laplace Transforms, and eventually we'll prove all of them. But for now, we'll just work through some of the more basic ones. But this can be our first entry in our Laplace Transform table. The Laplace Transform of f of t is equal to 1 is equal to 1/s. Notice we went from a function of t-- although obviously this one wasn't really dependent on t-- to a function of s. I have about 3 minutes left, but I don't think that's enough time to do another Laplace Transform. So I will save that for the next video. See you soon.

## Definition

The Laplace operator is a second order differential operator in the n-dimensional Euclidean space, defined as the divergence (∇·) of the gradient (f ). Thus if f is a twice-differentiable real-valued function, then the Laplacian of f is defined by

${\displaystyle \Delta f=\nabla ^{2}f=\nabla \cdot \nabla f}$

(1)

where the latter notations derive from formally writing

${\displaystyle \nabla =\left({\frac {\partial }{\partial x_{1}}},\ldots ,{\frac {\partial }{\partial x_{n}}}\right).}$

Equivalently, the Laplacian of f is the sum of all the unmixed second partial derivatives in the Cartesian coordinates xi:

${\displaystyle \Delta f=\sum _{i=1}^{n}{\frac {\partial ^{2}f}{\partial x_{i}^{2}}}}$

(2)

As a second-order differential operator, the Laplace operator maps Ck functions to Ck−2 functions for k ≥ 2. The expression (1) (or equivalently (2)) defines an operator Δ : Ck(ℝn) → Ck−2(ℝn), or more generally an operator Δ : Ck(Ω) → Ck−2(Ω) for any open set Ω.

## Motivation

### Diffusion

In the physical theory of diffusion, the Laplace operator (via Laplace's equation) arises naturally in the mathematical description of equilibrium.[1] Specifically, if u is the density at equilibrium of some quantity such as a chemical concentration, then the net flux of u through the boundary of any smooth region V is zero, provided there is no source or sink within V:

${\displaystyle \int _{\partial V}\nabla u\cdot \mathbf {n} \,dS=0,}$

where n is the outward unit normal to the boundary of V. By the divergence theorem,

${\displaystyle \int _{V}\operatorname {div} \nabla u\,dV=\int _{\partial V}\nabla u\cdot \mathbf {n} \,dS=0.}$

Since this holds for all smooth regions V, it can be shown that this implies

${\displaystyle \operatorname {div} \nabla u=\Delta u=0.}$

The left-hand side of this equation is the Laplace operator. The Laplace operator itself has a physical interpretation for non-equilibrium diffusion as the extent to which a point represents a source or sink of chemical concentration, in a sense made precise by the diffusion equation.

### Density associated with a potential

If φ denotes the electrostatic potential associated to a charge distribution q, then the charge distribution itself is given by the negative of the Laplacian of φ:

${\displaystyle q=-\varepsilon _{0}\Delta \varphi ,}$

where ε0 is the electric constant.

This is a consequence of Gauss's law. Indeed, if V is any smooth region, then by Gauss's law the flux of the electrostatic field E is proportional to the charge enclosed:

${\displaystyle \int _{\partial V}\mathbf {E} \cdot \mathbf {n} \,dS=\int _{V}\operatorname {div} \mathbf {E} \,dV={\frac {1}{\varepsilon _{0}}}\int _{V}q\,dV.}$

where the first equality is due to the divergence theorem. Since the electrostatic field is the (negative) gradient of the potential, this now gives

${\displaystyle -\int _{V}\operatorname {div} (\operatorname {grad} \varphi )\,dV={\frac {1}{\varepsilon _{0}}}\int _{V}q\,dV.}$

So, since this holds for all regions V, we must have

${\displaystyle \operatorname {div} (\operatorname {grad} \varphi )=-{\frac {1}{\varepsilon _{0}}}q}$

The same approach implies that the negative of the Laplacian of the gravitational potential is the mass distribution. Often the charge (or mass) distribution are given, and the associated potential is unknown. Finding the potential function subject to suitable boundary conditions is equivalent to solving Poisson's equation.

### Energy minimization

Another motivation for the Laplacian appearing in physics is that solutions to Δf = 0 in a region U are functions that make the Dirichlet energy functional stationary:

${\displaystyle E(f)={\frac {1}{2}}\int _{U}\Vert \nabla f\Vert ^{2}\,dx.}$

To see this, suppose f : U → ℝ is a function, and u : U → ℝ is a function that vanishes on the boundary of U. Then

${\displaystyle \left.{\frac {d}{d\varepsilon }}\right|_{\varepsilon =0}E(f+\varepsilon u)=\int _{U}\nabla f\cdot \nabla u\,dx=-\int _{U}u\,\Delta f\,dx}$

where the last equality follows using Green's first identity. This calculation shows that if Δf = 0, then E is stationary around f. Conversely, if E is stationary around f, then Δf = 0 by the fundamental lemma of calculus of variations.

## Coordinate expressions

### Two dimensions

The Laplace operator in two dimensions is given by:

${\displaystyle \Delta f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}}$

where x and y are the standard Cartesian coordinates of the xy-plane.

{\displaystyle {\begin{aligned}\Delta f&={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}\\&={\frac {\partial ^{2}f}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial f}{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}},\end{aligned}}}

where r represents the radial distance and θ the angle.

### Three dimensions

In three dimensions, it is common to work with the Laplacian in a variety of different coordinate systems.

${\displaystyle \Delta f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}.}$
${\displaystyle \Delta f={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}},}$

where r represents the radial distance, φ the azimuth angle and z the height.

{\displaystyle {\begin{aligned}\Delta f&={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}\\&={\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}(rf)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}\end{aligned}}}

where φ represents the azimuthal angle and θ the zenith angle or co-latitude.

In general curvilinear coordinates (ξ1, ξ2, ξ3):

${\displaystyle \nabla ^{2}=\nabla \xi ^{m}\cdot \nabla \xi ^{n}{\frac {\partial ^{2}}{\partial \xi ^{m}\partial \xi ^{n}}}+\nabla ^{2}\xi ^{m}{\frac {\partial }{\partial \xi ^{m}}}=g^{mn}\left({\frac {\partial ^{2}}{\partial \xi ^{m}\partial \xi ^{n}}}-\Gamma _{mn}^{l}{\frac {\partial }{\partial \xi ^{l}}}\right),}$

where summation over the repeated indices is implied, gmn is the inverse metric tensor and Γl mn are the Christoffel symbols for the selected coordinates.

### N dimensions

In arbitrary curvilinear coordinates in N dimensions (ξ1,..., ξN), we can write the Laplacian in terms of the inverse metric tensor, ${\displaystyle g^{ij}}$:

${\displaystyle \nabla ^{2}={\frac {1}{\sqrt {\det g}}}{\frac {\partial }{\partial \xi ^{i}}}\left({\sqrt {\det g}}g^{ij}{\frac {\partial }{\partial \xi ^{j}}}\right)}$,

from the Voss- Weyl formula[2] for the  divergence.

In spherical coordinates in N dimensions, with the parametrization x = ∈ ℝN with r representing a positive real radius and θ an element of the unit sphere SN − 1,

${\displaystyle \Delta f={\frac {\partial ^{2}f}{\partial r^{2}}}+{\frac {N-1}{r}}{\frac {\partial f}{\partial r}}+{\frac {1}{r^{2}}}\Delta _{S^{N-1}}f}$

where ΔSN−1 is the Laplace–Beltrami operator on the (N − 1)-sphere, known as the spherical Laplacian. The two radial derivative terms can be equivalently rewritten as

${\displaystyle {\frac {1}{r^{N-1}}}{\frac {\partial }{\partial r}}\left(r^{N-1}{\frac {\partial f}{\partial r}}\right).}$

As a consequence, the spherical Laplacian of a function defined on SN−1 ⊂ ℝN can be computed as the ordinary Laplacian of the function extended to N\{0} so that it is constant along rays, i.e., homogeneous of degree zero.

## Euclidean invariance

The Laplacian is invariant under all Euclidean transformations: rotations and translations. In two dimensions, for example, this means that

${\displaystyle \Delta (f(x\cos \theta -y\sin \theta +a,x\sin \theta +y\cos \theta +b))=(\Delta f)(x\cos \theta -y\sin \theta +a,x\sin \theta +y\cos \theta +b)}$

for all θ, a, and b. In arbitrary dimensions,

${\displaystyle \Delta (f\circ \rho )=(\Delta f)\circ \rho }$

whenever ρ is a rotation, and likewise

${\displaystyle \Delta (f\circ \tau )=(\Delta f)\circ \tau }$

whenever τ is a translation. (More generally, this remains true when ρ is an orthogonal transformation such as a reflection.)

In fact, the algebra of all scalar linear differential operators, with constant coefficients, that commute with all Euclidean transformations, is the polynomial algebra generated by the Laplace operator.

## Spectral theory

The spectrum of the Laplace operator consists of all eigenvalues λ for which there is a corresponding eigenfunction f with

${\displaystyle -\Delta f=\lambda f.}$

This is known as the Helmholtz equation.

If Ω is a bounded domain in n then the eigenfunctions of the Laplacian are an orthonormal basis for the Hilbert space L2(Ω). This result essentially follows from the spectral theorem on compact self-adjoint operators, applied to the inverse of the Laplacian (which is compact, by the Poincaré inequality and the Rellich–Kondrachov theorem).[3] It can also be shown that the eigenfunctions are infinitely differentiable functions.[4] More generally, these results hold for the Laplace–Beltrami operator on any compact Riemannian manifold with boundary, or indeed for the Dirichlet eigenvalue problem of any elliptic operator with smooth coefficients on a bounded domain. When Ω is the n-sphere, the eigenfunctions of the Laplacian are the well-known spherical harmonics.

## Generalizations

A version of the Laplacian can be defined wherever the Dirichlet energy functional makes sense, which is the theory of Dirichlet forms. For spaces with additional structure, one can give more explicit descriptions of the Laplacian, as follows.

### Laplace–Beltrami operator

The Laplacian also can be generalized to an elliptic operator called the Laplace–Beltrami operator defined on a Riemannian manifold. The d'Alembert operator generalizes to a hyperbolic operator on pseudo-Riemannian manifolds. The Laplace–Beltrami operator, when applied to a function, is the trace (tr) of the function's Hessian:

${\displaystyle \Delta f=\operatorname {tr} {\big (}H(f){\big )}}$

where the trace is taken with respect to the inverse of the metric tensor. The Laplace–Beltrami operator also can be generalized to an operator (also called the Laplace–Beltrami operator) which operates on tensor fields, by a similar formula.

Another generalization of the Laplace operator that is available on pseudo-Riemannian manifolds uses the exterior derivative, in terms of which the “geometer's Laplacian" is expressed as

${\displaystyle \Delta f=\delta df.}$

Here δ is the codifferential, which can also be expressed using the Hodge dual. Note that this operator differs in sign from the "analyst's Laplacian" defined above, a point which must always be kept in mind when reading papers in global analysis. More generally, the "Hodge" Laplacian is defined on differential forms α by

${\displaystyle \Delta \alpha =\delta d\alpha +d\delta \alpha .}$

This is known as the Laplace–de Rham operator, which is related to the Laplace–Beltrami operator by the Weitzenböck identity.

### D'Alembertian

The Laplacian can be generalized in certain ways to non-Euclidean spaces, where it may be elliptic, hyperbolic, or ultrahyperbolic.

In the Minkowski space the Laplace–Beltrami operator becomes the D'Alembert operator or D'Alembertian:

${\displaystyle \square ={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-{\frac {\partial ^{2}}{\partial x^{2}}}-{\frac {\partial ^{2}}{\partial y^{2}}}-{\frac {\partial ^{2}}{\partial z^{2}}}.}$

It is the generalisation of the Laplace operator in the sense that it is the differential operator which is invariant under the isometry group of the underlying space and it reduces to the Laplace operator if restricted to time-independent functions. Note that the overall sign of the metric here is chosen such that the spatial parts of the operator admit a negative sign, which is the usual convention in high energy particle physics. The D'Alembert operator is also known as the wave operator, because it is the differential operator appearing in the wave equations and it is also part of the Klein–Gordon equation, which reduces to the wave equation in the massless case.

The additional factor of c in the metric is needed in physics if space and time are measured in different units; a similar factor would be required if, for example, the x direction were measured in meters while the y direction were measured in centimeters. Indeed, theoretical physicists usually work in units such that c = 1 in order to simplify the equation.