Laplace operator

The Laplace operator is defined by the expression

and in the Cartesian coordinate system is described by the formula

Let us find an expression for the Laplace operator in a curvilinear orthogonal coordinate system. To do this, we write the gradient and divergence in a curvilinear coordinate system

Substituting these expressions into the Laplace operator, we get

Example 1. Find an expression for the Laplace operator in a cylindrical coordinate system.

Remark 1. The Laplace operator in a polar coordinate system is defined by the formula

Example 2. Find an expression for the Laplace operator in a spherical coordinate system.

Solution. Substituting the values ​​of the Lame coefficients, we get

Laplace's equation

Laplace's equation is an equation of the form.

This equation is called an elliptic type equation. It is often encountered in problems related to determining the potential of various stationary fields. In particular, the problem of determining the temperature field, electric potential, elastic stresses and strains is associated with solving the Laplace equation. Note that in mathematical physics equations of hyperbolic and parabolic types are also studied.

There are many different methods for solving elliptic type equations. Among them are the method of separation of variables, the source function method, potential theory, the method of analytical functions and many others. Let's consider several simple tasks that do not involve the use of special methods.

Cylindrical symmetry. Let us find a solution to the Laplace equation for a function that has cylindrical symmetry, i.e. independent of the polar angle and the variable z. In this case, Laplace's equation, written in a cylindrical coordinate system, has the form

Partial derivatives are replaced here by complete derivatives. From this equation it follows

where and are arbitrary constants that can be found from the boundary conditions.

Spherical symmetry. Let us find a solution to the Laplace equation for a function that has spherical symmetry, i.e. independent of angles and. In this case, Laplace's equation, written in a spherical coordinate system, has the form

It is not difficult to find a solution to this equation

Let's consider the solution of the Poisson equation using specific examples.

Example 1. Find a solution to the Poisson equation inside a circle of radius if

Solution. The desired function has cylindrical symmetry, so we write Poisson’s equation in a cylindrical coordinate system in the form

Let's solve this equation

gradient curvilinear lamé differential

We find the constants and from the boundary condition and the condition of boundedness of the function. Considering that, we get. From the condition we get

Therefore, we have the final answer

It is a special case of the Helmholtz equation. Can be considered in three-dimensional (1), two-dimensional (2), one-dimensional and n-dimensional spaces:

The operator is called the Laplace operator (The Laplace operator is equivalent to taking the gradient and divergence operations sequentially.).

Solution of Laplace's equation

The solutions to Laplace's equation are harmonic functions.

Laplace's equation belongs to elliptic equations. Laplace's inhomogeneous equation becomes Poisson's equation.

Each solution of the Laplace equation in a bounded domain G is uniquely identified by boundary conditions imposed on the behavior of the solution (or its derivatives) on the boundary of the domain G. If the solution is sought in the entire space , the boundary conditions are reduced to the prescription of some asymptotic behavior for f at . The problem of finding such solutions is called a boundary value problem. The most common are the Dirichlet problem, when the value of the function f itself is given on the boundary, and the Neman problem, when the value of f is given along the normal to the boundary.

Laplace's equation in spherical, polar and cylindrical coordinates

Laplace's equation can be written not only in Cartesian coordinates.

In spherical coordinates (Laplace's equation has the following form:

In polar coordinates (coordinate system), the equation is:

In cylindrical coordinates (the equation is:

Many problems in physics and mechanics lead to the Laplace equation, in which a physical quantity is a function only of the coordinates of a point. Thus, Laplace’s equation describes the potential in a region that does not contain gravitating masses, the potential of an electrostatic field in a region that does not contain charges, temperature during stationary processes, etc. A large number of engineering problems associated, in particular, with slow stationary flow around a ship’s hull , stationary filtration of groundwater, the emergence of a field around an electromagnet, as well as a stationary electric field in the vicinity of a porcelain insulator or an electric cable of variable cross-section buried in the ground, comes down to solving the three-dimensional Laplace or Poisson equations. The Laplace operator plays a great role in quantum mechanics.

Examples of problem solving

EXAMPLE 1

Exercise	Find the field between two coaxial cylinders with radii and , the potential difference between which is equal to
Solution	Let us write Laplace's equation in cylindrical coordinates, taking into account axial symmetry: It has a solution +B. Let's choose the zero potential on the outer cylinder, find it, and get: Hence We get: As a result we have:
Answer	The field between two coaxial cylinders is given by the function

EXAMPLE 2

Exercise	Investigate the stability of the equilibrium of a positively charged particle in an electric field (Earnshaw’s theorem).
Solution	Let us place the origin of coordinates at the equilibrium position of the particle. In this case, we can assume that the potential is represented in the form:

Any part of a control system, be it a controller, an object or a sensor, has an input and an output. Using inputs and outputs, they interact with other elements of the system and with the external environment. When an input signal acts on a system element, some internal state changes occur in this element, which lead to a change in the output signal. That is, an element of the system represents a certain function of the dependence of y on x. This can be depicted in Figure 1.

Figure 1 – control system element with input and output

Determining the function F(x) is, in fact, the main problem solved within the framework of the theory of automatic control. Knowing the F(x) of an object will help you create the correct algorithm for controlling it, the F(x) of the sensor will determine the nature of the feedback, and the synthesis of F(x) will make the system truly operational. F itself is also sometimes called an operator because it operates on an input signal.

The basic operations in TAU are integration and differentiation. Suppose the signal increases over some time, which is often very typical for signals in control systems, then to describe this process it should be “assembled” by an integral over the entire time interval:

Differentiation is also extremely useful in the theory of automatic control. The differentiation operator, as opposed to the integration operator, takes the derivative of the input signal, that is:

Here a very important concept in TAU arises - the Laplace operator p, which is intended to replace the notation d/dt, in other words

Also in some sources this operator is represented as the product of the imaginary unit and the angular frequency, that is, p=jω. But we won’t touch the frequency range for now, because this is a broad topic, and we’ll just remember two simple rules:

What does signal integration and differentiation look like? The integration of the step waveform is shown in Figure 2a. Everything is simple here, the signal will be incremented at each integration step until it reaches the initially specified value in time t1. What if we differentiate such a signal? No way! This is a threat to the safety of the Universe; such a signal will pierce the vault of heaven and rush to infinity towards the stars (Figure 2b)! In short, mathematics says that the derivative of an instantaneously changed signal is equal to infinity, and since infinity is an ideal and unattainable value, such an operation makes no sense in the real world. Otherwise, they say that such an operation is not physically feasible. In general, p is not used in its pure form, but is used only as part of more complex expressions, where this p will be compensated in some way.

Figure 2 – signal integration and differentiation

Now that we know about the ratio of the output signal to the input signal and about the Laplace operator, we can move on to the concept of a transfer function. Essentially, the transfer function, written as W(p), represents the output/input ratio. A system written through transfer functions is more visual, and more or less simple methods of analysis and synthesis can be applied to it. But we’ll talk about them later, and now let’s look at a simple example of how such functions are obtained.

Suppose we have a link, the processes occurring in which are described by the following equation:

On the left is the output quantity (and its derivative), on the right is the input quantity (in complex expressions there may also be derivatives there). T is some time constant, K is some coefficient. Now we replace it with the Laplace operator:

As noted above, the transfer function is equal to the output/input ratio:

This is how we obtained the transfer function of the first order inertial link. In TAU there are several standard links (including this one), from which any system, any link of any complexity can be made. Now we just note that transfer functions, depending on the orders of the numerator and denominator, can be correct or incorrect. The above function is correct, also said to be strictly correct, because the order of the denominator is greater than the order of the numerator. And that’s good, it’s feasible. Below are a couple more features.

Function type 1 is also correct, but not strictly. The degree of the numerator is equal to the degree of the denominator, but that’s okay, it’s also realizable. But a function like 2 is not realizable due to the presence of a square in the numerator and the absence of a square or a higher power in the denominator, that is, in this case there will be some kind of uncompensated derivative. Thus, the order in the transfer functions must be strictly monitored!

Material from Wikipedia - the free encyclopedia

The Laplace operator is equivalent to taking the gradient and divergence operations sequentially: $\backslash Delta=\backslash operatorname(div)\backslash ,\backslash operatorname(grad)$, thus, the value of the Laplace operator at a point can be interpreted as the density of sources (sinks) of the potential vector field $\backslash \; \backslash operatorname(grad)F$ at this point. In a Cartesian coordinate system, the Laplace operator is often denoted as follows $\backslash Delta=\backslash nabla\backslash cdot\backslash nabla=\backslash nabla^2$, that is, in the form of a scalar product of the Nabla operator with itself. The Laplace operator is symmetric.

Another definition of the Laplace operator

The Laplace operator is a natural generalization to functions of several variables of the ordinary second derivative of a function of one variable. In fact, if the function $\backslash f(x)$ has a point in the vicinity $\backslash x\_0$ continuous second derivative $\backslash f$(x), then, as follows from the Taylor formula

$\backslash \; f(x\_0+r)=f(x\_0)+rf"(x\_0)+\backslash frac(r^2)(2)f$(x_0)+o(r^2), at $r\backslash to\; 0,$, $\backslash f(x\_0-r)=f(x\_0)-rf"(x\_0)+\backslash frac(r^2)(2)f$(x_0)+o(r^2), at $r\backslash to\; 0,$

the second derivative is the limit

$\backslash f$(x_0)=\lim\limits_(r \to 0) \frac(2)(r^2) \left\( \frac(f(x_0+r)+f(x_0-r))(2)-f (x_0)\right\).

If, going to the function $\backslash F$ from $\backslash k$ variables, proceed in the same way, that is, for a given point $M\_0(x\_1^0,x\_2^0,\; ...\; ,x\_k^0)$ look at it $\backslash k$-dimensional spherical neighborhood $\backslash Q\_r$ radius $\backslash r$ and the difference between the arithmetic mean

$\backslash \; \backslash frac(1)(\backslash sigma(S\_r))\backslash int\backslash limits\_(S\_r)Fd\backslash sigma$

functions $\backslash F$ on the border $\backslash S\_r$ such a neighborhood with boundary area $\backslash \backslash sigma(S\_r)$ and meaning $\backslash F(M\_0)$ in the center of this neighborhood $\backslash \; M\_0$, then in the case of continuity of the second partial derivatives of the function $\backslash F$ in the vicinity of a point $\backslash \; M\_0$ Laplacian value $\backslash \backslash Delta\; F$ there is a limit at this point

$\backslash \; \backslash Delta\; F(M\_0)=\backslash lim\backslash limits\_(r\; \backslash to\; 0)\; \backslash frac(2k)(r^2)\; \backslash left\backslash (\backslash frac(1)(\backslash sigma(S\_r))\backslash int\backslash limits\_(S\_r\; )F(M)d\backslash sigma\; -F(M\_0)\; \backslash right\backslash ).$

Simultaneously with the previous presentation for the Laplace operator of the function $\backslash F$, which has continuous second derivatives, the formula is valid

$\backslash \; \backslash Delta\; F(M\_0)=\backslash lim\backslash limits\_(r\; \backslash to\; 0)\; \backslash frac(2(k+2))(r^2)\; \backslash left\backslash (\backslash frac(1)(\backslash omega(Q\_r))\backslash \; int\backslash limits\_(Q\_r)F(M)d\backslash omega\; -F(M\_0)\; \backslash right\backslash ),$ Where $\backslash \backslash omega(Q\_r)$- volume of the surroundings $\backslash Q\_r.$

This formula expresses the direct connection between the Laplacian of a function and its volumetric average in the neighborhood of a given point.

The proof of these formulas can be found, for example, in.

The above limits, in all cases where they exist, can serve as a definition of the Laplace operator of the function $\backslash F.$ This definition is preferable to the usual definition of the Laplacian, which assumes the existence of second derivatives of the functions under consideration, and coincides with the usual definition in the case of continuity of these derivatives.

Expressions for the Laplace operator in various curvilinear coordinate systems

In arbitrary orthogonal curvilinear coordinates in three-dimensional space $q\_1,\backslash \; q\_2,\backslash \; q\_3$:

$\backslash Delta\; f\; (q\_1,\backslash \; q\_2,\backslash \; q\_3)\; =\; \backslash operatorname(div)\backslash ,\backslash operatorname(grad)\backslash ,f(q\_1,\backslash \; q\_2,\backslash \; q\_3)\; =$ $=\backslash frac(1)(H\_1H\_2H\_3)\backslash left[\; \backslash frac(\backslash partial)(\backslash partial\; q\_1)\backslash left(\backslash frac(H\_2H\_3)(H\_1)\backslash frac(\backslash partial\; f)(\backslash partial\; q\_1)\; \backslash right)\; +\; \backslash frac(\backslash partial)(\backslash partial\; q\_2)\backslash left(\backslash frac(H\_1H\_3)(H\_2)\backslash frac(\backslash partial\; f)(\backslash partial\; q\_2)\; \backslash right)\; +\; \backslash frac(\backslash partial)(\backslash partial\; q\_3)\backslash \; left(\backslash frac(H\_1H\_2)(H\_3)\backslash frac(\backslash partial\; f)(\backslash partial\; q\_3)\; \backslash right)\backslash right],$ Where $H\_i\backslash$- Lamé coefficients.

Cylindrical coordinates

In cylindrical coordinates outside the line $\backslash r=0$:

$\backslash Delta\; f$

= (1 \over r) (\partial \over \partial r)

\left(r (\partial f \over \partial r) \right)

+ (\partial^2f \over \partial z^2) + (1 \over r^2) (\partial^2 f \over \partial \varphi^2)

Spherical coordinates

In spherical coordinates outside the origin (in three-dimensional space):

$\backslash Delta\; f$

= (1 \over r^2) (\partial \over \partial r)

\left(r^2 (\partial f \over \partial r) \right)

+ (1 \over r^2\sin^2 \theta) (\partial^2 f \over \partial \varphi^2)

$\backslash Delta\; f$

= (1 \over r) (\partial^2 \over \partial r^2)

\left(rf \right)

+ (1 \over r^2 \sin \theta) (\partial \over \partial \theta)

\left(\sin \theta (\partial f \over \partial \theta) \right)

+ (1 \over r^2 \sin^2 \theta) (\partial^2 f \over \partial \varphi^2).

In case $\backslash f=f(r)$ V n-dimensional space:

$\backslash Delta\; f\; =\; (d^2\; f\backslash over\; dr^2)\; +\; (n-1\; \backslash over\; r\; )\; (df\backslash over\; dr).$

Parabolic coordinates

In parabolic coordinates (in three-dimensional space) outside the origin:

\Delta f= \frac(1)(\sigma^(2) + \tau^(2)) \left[ \frac(1)(\sigma) \frac(\partial )(\partial \sigma) \left (\sigma \frac(\partial f)(\partial \sigma) \right) + \frac(1)(\tau) \frac(\partial )(\partial \tau) \left(\tau \frac(\ partial f)(\partial \tau) \right)\right] + \frac(1)(\sigma^2\tau^2)\frac(\partial^2 f)(\partial \varphi^2)

Cylindrical parabolic coordinates

In the coordinates of a parabolic cylinder outside the origin:

$\backslash Delta\; F(u,v,z)\; =\; \backslash frac(1)(c^2(u^2+v^2))\; \backslash left[\; \backslash frac(\backslash partial^2\; F\; )(\backslash partial\; u^2)+\; \backslash frac(\backslash partial^2\; F\; )(\backslash partial\; v^2)\backslash right]\; +\; \backslash frac(\backslash partial^2\; F\; )(\backslash partial\; z^2).$

General curvilinear coordinates and Riemannian spaces

Let on a smooth manifold $X$ a local coordinate system is specified and $g\_(ij)$- Riemannian metric tensor on $X$, that is, the metric has the form

$ds^2\; =\backslash sum^n\_(i,j=1)g\_(ij)\; dx^idx^j$ .

Let us denote by $g^(ij)$ matrix elements $(g\_(ij))^(-1)$ And

$g\; =\; \backslash operatorname(det)\; g\_(ij)\; =\; (\backslash operatorname(det)\; g^(ij))^(-1)$.

Vector field divergence $F$, specified by the coordinates $F^i$(and representing the first order differential operator $\backslash sum\_i\; F^i\backslash frac(\backslash partial)(\backslash partial\; x^i)$) on the manifold X calculated by the formula

$\backslash operatorname(div)\; F\; =\; \backslash frac(1)(\backslash sqrt(g))\backslash sum^n\_(i=1)\backslash frac(\backslash partial)(\backslash partial\; x^i)(\backslash sqrt(g)F^i\; )$,

and the gradient components of the function f- according to the formula

$(\backslash nabla\; f)^j\; =\backslash sum^n\_(i=1)g^(ij)\; \backslash frac(\backslash partial\; f)(\backslash partial\; x^i).$

Laplace operator - Beltrami on $X$:

$\backslash Delta\; f\; =\; \backslash operatorname(div)\; (\backslash nabla\; f)=\; \backslash frac(1)(\backslash sqrt(g))\backslash sum^n\_(i=1)\backslash frac(\backslash partial)(\backslash partial\; x^i)\backslash \; Big(\backslash sqrt(g)\; \backslash sum^n\_(k=1)g^(ik)\; \backslash frac(\backslash partial\; f)(\backslash partial\; x^k)\backslash Big).$

Meaning $\backslash Delta\; f$ is a scalar, that is, it does not change when transforming coordinates.

Application

Using this operator it is convenient to write Laplace's, Poisson's and wave equations. In physics, the Laplace operator is applicable in electrostatics and electrodynamics, quantum mechanics, in many equations of continuum physics, as well as in the study of the equilibrium of membranes, films or interfaces with surface tension (see Laplace pressure), in stationary problems of diffusion and thermal conductivity, which reduce, in the continuous limit, to the usual equations of Laplace or Poisson or to some of their generalizations.

Variations and generalizations

• The D'Alembert operator is a generalization of the Laplace operator for hyperbolic equations. Includes the second derivative with respect to time.
• The vector Laplace operator is a generalization of the Laplace operator to the case of a vector argument.

See also

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Literature

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Differential calculus Basics Private views Differential operators (in different coordinates) First order Second order Higher orders Related Topics

An excerpt characterizing the Laplace Operator

Princess Marya, sitting in the living room and listening to these talk and gossip of the old people, did not understand anything of what she heard; she only thought about whether all the guests noticed her father’s hostile attitude towards her. She did not even notice the special attention and courtesies that Drubetskoy, who had been in their house for the third time, showed her throughout this dinner.
Princess Marya, with an absent-minded, questioning look, turned to Pierre, who, the last of the guests, with a hat in his hand and a smile on his face, approached her after the prince had left, and they alone remained in the living room.
-Can we sit still? - he said, throwing his fat body into a chair next to Princess Marya.
“Oh yes,” she said. “Didn’t you notice anything?” said her look.
Pierre was in a pleasant, post-dinner state of mind. He looked ahead and smiled quietly.
“How long have you known this young man, princess?” - he said.
- Which one?
- Drubetsky?
- No, recently...
- What do you like about him?
- Yes, he is a nice young man... Why are you asking me this? - said Princess Marya, continuing to think about her morning conversation with her father.
“Because I made an observation, a young man usually comes from St. Petersburg to Moscow on vacation only for the purpose of marrying a rich bride.
– You made this observation! - said Princess Marya.
“Yes,” Pierre continued with a smile, “and this young man now behaves in such a way that where there are rich brides, there he is.” It’s like I’m reading it from a book. He is now undecided who to attack: you or mademoiselle Julie Karagin. Il est tres assidu aupres d'elle. [He is very attentive to her.]
– Does he go to them?
- Yes, very often. And do you know a new style of grooming? - Pierre said with a cheerful smile, apparently in that cheerful spirit of good-natured ridicule, for which he so often reproached himself in his diary.
“No,” said Princess Marya.
- Now, in order to please Moscow girls - il faut etre melancolique. Et il est tres melancolique aupres de m lle Karagin, [one must be melancholic. And he is very melancholy with m elle Karagin,” said Pierre.
– Vraiment? [Really?] - said Princess Marya, looking into Pierre’s kind face and never ceasing to think about her grief. “It would be easier for me,” she thought, if I decided to trust someone with everything I feel. And I would like to tell Pierre everything. He is so kind and noble. It would make me feel better. He would give me advice!”
– Would you marry him? asked Pierre.
“Oh, my God, Count, there are moments when I would marry anyone,” Princess Marya suddenly said to herself, with tears in her voice. “Oh, how hard it can be to love a loved one and feel that... nothing (she continued in a trembling voice) you can’t do for him except grief, when you know that you can’t change it.” Then one thing is to leave, but where should I go?...
- What are you, what’s wrong with you, princess?
But the princess, without finishing, began to cry.
- I don’t know what’s wrong with me today. Don't listen to me, forget what I told you.
All Pierre's gaiety disappeared. He anxiously questioned the princess, asked her to express everything, to confide in him her grief; but she only repeated that she asked him to forget what she said, that she did not remember what she said, and that she had no grief other than the one he knew - the grief that Prince Andrei’s marriage threatens to quarrel with his father son.
– Have you heard about the Rostovs? – she asked to change the conversation. - I was told that they would be here soon. I also wait for Andre every day. I would like them to see each other here.
– How does he look at this matter now? - Pierre asked, by which he meant the old prince. Princess Marya shook her head.
- But what to do? There are only a few months left until the year is over. And this cannot be. I would only like to spare my brother the first minutes. I wish they would come sooner. I hope to get along with her. “You have known them for a long time,” said Princess Marya, “tell me, hand on heart, the whole true truth, what kind of girl is this and how do you find her?” But the whole truth; because, you understand, Andrei is risking so much by doing this against his father’s will that I would like to know...
A vague instinct told Pierre that these reservations and repeated requests to tell the whole truth expressed Princess Marya’s ill will towards her future daughter-in-law, that she wanted Pierre not to approve of Prince Andrei’s choice; but Pierre said what he felt rather than thought.
“I don’t know how to answer your question,” he said, blushing, without knowing why. “I absolutely don’t know what kind of girl this is; I can't analyze it at all. She's charming. Why, I don’t know: that’s all that can be said about her. “Princess Marya sighed and the expression on her face said: “Yes, I expected and was afraid of this.”
– Is she smart? - asked Princess Marya. Pierre thought about it.
“I think not,” he said, “but yes.” She doesn't deserve to be smart... No, she's charming, and nothing more. – Princess Marya again shook her head disapprovingly.
- Oh, I so want to love her! You will tell her this if you see her before me.
“I heard that they will be there one of these days,” said Pierre.
Princess Marya told Pierre her plan about how, as soon as the Rostovs arrived, she would become close to her future daughter-in-law and try to accustom the old prince to her.

Boris did not succeed in marrying a rich bride in St. Petersburg and he came to Moscow for the same purpose. In Moscow, Boris was indecisive between the two richest brides - Julie and Princess Marya. Although Princess Marya, despite her ugliness, seemed more attractive to him than Julie, for some reason he felt awkward courting Bolkonskaya. On her last meeting with her, on the old prince’s name day, to all his attempts to talk to her about feelings, she answered him inappropriately and obviously did not listen to him.
Julie, on the contrary, although in a special way peculiar to her, willingly accepted his courtship.
Julie was 27 years old. After the death of her brothers, she became very rich. She was now completely ugly; but I thought that she was not only just as good, but even much more attractive than she was before. She was supported in this delusion by the fact that, firstly, she became a very rich bride, and secondly, that the older she became, the safer she was for men, the freer it was for men to treat her and, without taking on any obligations, take advantage of her dinners, evenings and the lively company that gathered at her place. A man who ten years ago would have been afraid to go every day to the house where there was a 17-year-old young lady, so as not to compromise her and tie himself down, now went to her boldly every day and treated her not as a young bride, but as a acquaintance who has no gender.
The Karagins' house was the most pleasant and hospitable house in Moscow that winter. In addition to parties and dinners, every day a large company gathered at the Karagins, especially men, who dined at 12 o'clock in the morning and stayed until 3 o'clock. There was no ball, party, or theater that Julie missed. Her toilets were always the most fashionable. But, despite this, Julie seemed disappointed in everything, telling everyone that she did not believe in friendship, nor in love, nor in any joys of life, and expected peace only there. She adopted the tone of a girl who had suffered great disappointment, a girl as if she had lost a loved one or had been cruelly deceived by him. Although nothing of the kind happened to her, they looked at her as if she were one, and she herself even believed that she had suffered a lot in life. This melancholy, which did not prevent her from having fun, did not prevent the young people who visited her from having a pleasant time. Each guest, coming to them, paid his debt to the melancholic mood of the hostess and then engaged in small talk, and dancing, and mental games, and burime tournaments, which were in fashion with the Karagins. Only some young people, including Boris, delved deeper into Julie’s melancholy mood, and with these young people she had longer and more private conversations about the vanity of everything worldly, and to them she opened her albums covered with sad images, sayings and poems.

The Laplace operator is a differential operator acting in a linear space of smooth functions and is denoted by the symbol. He associates the function F with the function

The Laplace operator is equivalent to taking the gradient and divergence operations sequentially.

Gradient is a vector showing the direction of the fastest increase of a certain quantity, the value of which changes from one point in space to another (scalar field). For example, if we take the height of the Earth's surface above sea level as the height, then its gradient at each point on the surface will show the “direction of the steepest rise.” The magnitude (modulus) of the gradient vector is equal to the growth rate in this direction. For the case of three-dimensional space, the gradient is a vector function with components, where is some scalar function of the coordinates x, y, z.

If is a function of n variables, then its gradient is an n-dimensional vector

The components of which are equal to the partial derivatives of all its arguments. The gradient is denoted by grad, or using the nabla operator,

From the definition of gradient it follows that:

The meaning of the gradient of any scalar function f is that its scalar product with an infinitesimal displacement vector gives the total differential of this function with a corresponding change in coordinates in the space in which f is defined, that is, the linear (in the case of general position it is also the main) part of the change in f when shifted by. Using the same letter to denote a function of a vector and the corresponding function of its coordinates, we can write:

It is worth noting here that since the formula for the total differential does not depend on the type of coordinates x i, that is, on the nature of the parameters x in general, the resulting differential is an invariant, that is, a scalar, under any coordinate transformations, and since dx is a vector, then the gradient calculated in the usual way, turns out to be a covariant vector, that is, a vector represented in a dual basis, which is the only scalar that can be given by simply summing the products of the coordinates of an ordinary (contravariant) one, that is, a vector written in a regular basis.

Thus, the expression (generally speaking for arbitrary curvilinear coordinates) can be quite correctly and invariantly written as:

Or, according to Einstein’s rule, omitting the sum sign,

Divergence is a differential operator that maps a vector field onto a scalar one (that is, the differentiation operation that results in a scalar field when applied to a vector field), which determines (for each point) “how much the field incoming and outgoing from a small neighborhood of a given point diverges” (more precisely, how much the incoming and outgoing flows diverge).

If we take into account that an algebraic sign can be assigned to a flow, then there is no need to take into account the incoming and outgoing flows separately; everything will be automatically taken into account when summing taking into account the sign. Therefore, we can give a shorter definition of divergence:

divergence is a differential operator on a vector field that characterizes the flow of a given field through the surface of a small neighborhood of each internal point of the domain of definition of the field.

The divergence operator applied to the field F is denoted as or

The definition of divergence looks like this:

where ФF is the flow of the vector field F through a spherical surface of area S, limiting the volume V. Even more general, and therefore convenient to use, is the definition when the shape of a region with surface S and volume V is allowed to be any. The only requirement is that it be inside a sphere with a radius tending to zero. This definition, unlike the one given below, is not tied to specific coordinates, for example, Cartesian, which can be an additional convenience in certain cases. (For example, if you choose a neighborhood in the shape of a cube or parallelepiped, the formulas for Cartesian coordinates given in the next paragraph are easily obtained).

thus, the value of the Laplace operator at a point can be interpreted as the density of sources (sinks) of the potential vector field gradF at this point. In the Cartesian coordinate system, the Laplace operator is often denoted as follows, that is, in the form of a scalar product of the Nabla operator with itself.