Textbook Yu. N

“Solving equations of higher degrees” - What does it mean to solve an equation? First stage tasks. WARM-UP (checking the physical condition). Solving equations higher degrees. What types of equations are written on the board? Physical education minute. Stage II Independent work option 1 option 2. What is the root of an equation? Scheme for solving a linear equation of a quadratic equation of a biquadratic equation.

“Methods for solving equations and inequalities” - Ancient Egypt. Cubic equations. Non-standard methods solving equations and inequalities. The idea of ​​homogeneity. Graphic method solving equations containing a modulus. Inequalities with modulus. Solving equations for coefficients. The original inequality does not contain a single solution. Sum of the square.

“Equations and Inequalities” - Substitution. Find the abscissa of the intersection point of the function graphs. At what value of a is the number of roots of the equation. "Graphical method. It consists of the following: plotting graphs of two functions in one coordinate system. Solving equations and inequalities." Find the smallest natural solution inequalities.

“Fractional equations” - Solve the resulting equation. Quadratic equation has 2 roots if...... Eliminate roots not included in valid values fraction equations. ... Your letter. High soul." Algorithm for solving fractional rational equations. And remember - what is most important in a person? Fractional rational equations. How many roots does it have? given equation? 4. What is the name of this equation?

“Solving logarithmic equations” - If the equation contains logarithms with for different reasons, then first of all you should reduce all logarithms to one base using transition formulas. Calculate the values ​​of the expression. Definition: Summarize the material on the properties of logarithms, logarithmic function; consider the main solution methods logarithmic equations; develop oral skills.

“Methods for solving logarithmic equations” - Find. Solving logarithmic equations. What is called a logarithm. Systematize students' knowledge. Creative work. Find the error. System of equations. Solving logarithmic equations various methods. I option II option. Specified function. Method for introducing a new variable. Compare. Methods for solving logarithmic equations.

There are 49 presentations in total

REPORTS OF THE ACADEMY OF SCIENCES, 2008, volume 420, no. 6, p. 744-745

MATHEMATICAL PHYSICS

DECAYING SOLUTIONS OF THE VESELOV-NOVIKOV EQUATION

© 2008 Corresponding Member of the RAS I. A. Taimanov, S. P. Tsarev

Received 02/14/2008

Veselov-Novikov equation

u, = e3 u + E3 u + z E(vu) + zE(vu) = o, E V = E u,

where E = (Ex - ¿Ey), E = 1 (Ex + ¿Ey), is a two-dimensional generalization of the Korteweg-de Vries (KdV) equation

and, = 4 ikhx + viih,

to which it goes in the one-dimensional limit: V = u = u(x). Equation (1) specifies the deformations of the two-dimensional Schrödinger operator

specifies the transformation of the solution f of the equation Нф = 0 into the solution b of the equation Н b = 0, where

N = EE + and, and = and + 2 EE 1p w.

In the one-dimensional limit, the Moutard transformation reduces to the well-known Darboux transformation.

The Mutara transformation is extended to transform the solutions of the system

Nf = 0, f(= (E3 + E3 + 3 VE + 3 V *E)f, ^^^

where E V = Ei, EV* = E u, which is invariant under the transformation (extended Moutard transformation)

= ~|((f Eyu-yu Eph)dz- (f Eyu-yu Ef)dz +

of the form H1 = HA + 5I, where A, B - differential operators. Such deformations preserve the “spectrum” of the operator H on zero level energy, transforming the solutions of the equation

Nf = (EE + u)f = 0 (3)

according to (Eg + A)f = 0.

There is a method for constructing new solutions (u, φ) of equation (3) from old solutions (u, φ) of this equation, which reduces to quadratures - the Mutard transformation. It is as follows: let the operator H be given with potential u and the solution w of equation (3): Hm = 0. Then the formula

Ш |[(f Esh - w Eph) dz - (f Esh - w Ef) dz ]

Institute of Mathematics named after S.L. Sobolev Siberian Branch Russian Academy Sciences, Novosibirsk

Krasnoyarsk State pedagogical university

+ [f E yu - yu E f + yu E f - f E yu +

2 2 "2 _ ~ _2 + 2 (Ef Esh - Ef Esh) -2 (Ef Esh - Ef Esh) +

3V(f Ash - w Eph) + 3 V*(w Ef - f Ash)]dt),

and ^ and + 2EE 1psh, V ^ V + 2E21psh,

V* ^ u* + 2E21psh,

where w satisfies (4).

The Veselov-Novikov equation (1) is

compatibility condition of system (4) at V* = V.

If the solution is real, the conditions u = u and

V* = V are preserved and the extended Moutard transformation translates the real solutions and

equation (1) into other real solutions of this equation.

All rational solitons of the KdV equation are obtained by iterations of the Darboux transformation from the potential u = 0. In this case, all the resulting potentials are singular.

IN two-dimensional case a similar construction can lead to nonsingular and even rapidly decreasing potentials after two iterations

DISCONTINUING SOLUTIONS OF THE EQUATION

walkie-talkies Namely, let u0 = 0 and ωωω2 be real solutions of system (4):

ω, = ​​Γ(z, z) + /(z, z), = π(z, z) + π(z, z), (5)

where / and i are holomorphic in r and satisfy the equations

fg = G yyy" yg = yyyy"

Each of the functions ωωω2 specifies the (extended) Moutard transformation of the potential u = 0 and the corresponding solutions of system (4). Let us denote them by Mu and Ma. The resulting potentials we

let us denote by u1 = Mu (u0), u2 = Mu (u0).

Let b1 e Mu(ω2), i.e. b1 is obtained from ω2 by transforming Mω. Note that the Moutard transformation for φ depends on the integration constant. We choose a constant such that b1 is a real function. The choice of a constant allows you to often control the nonsingularity of the iterated potential (we will use this in specific examples).

A simple check shows that b2 = - b1 e

e Mu (yuh). There is a well-known lemma that is true for arbitrary potential u0.

Lemma 1. Let u12 = M01 (their) and u21 = M02 (u2). Then u12 = u21.

For the case u0 = 0, Lemma 2 holds. Let ω1 and ω2 have the form (5). Then the potential u = Mb (Mu (u0)), where u0 = 0 and b1 e Mu (u2), is given by the formula

u = 2EE 1pI((/I - yG) + )((f "I - fya")yg + + (GY - GY) yg) +1(G" i - fya"" + 2 (f "I" - G I) + + GY " " - G " "i + 2 (g i - g i")) yg).

Note that even for stationary initial solutionsω1, ω2 of system (4), we can obtain a solution to the Veselov-Novikov equation with nontrivial dynamics in g.

Theorem 1. Let U (z, z) be the rational potential obtained by double Moutard transformation from ω1 = iz2 - i~z , ffl2 = z2 + (1 +

I)z + ~z + (1 - i) z. The potential U is nonsingular and decreases as r-3 as r ^ Solution of the Veselov-Novikov equation (1) with initial data

U\t = 0 = U becomes singular in a finite time and has a singularity of the form

(3 x4 + 4 x3 + 6 x2 y2 + 3 y4 + 4 y3 + 30 - 12 t)

Comment. The Veselov-Novikov equation is invariant under the transformation t^-t, z^-z. It is easy to see that the solution to this

equation with initial data U(z, z, 0) = U (-z, - z) is regular for all t > 0.

Rational potential (1), given in the work, decreases as r-6 and gives a stationary non-singular solution to the Veselov-Novikov equation. When choosing f (z) = a3z3 + a2z2 + a1z2 + a0 + 6a3t, g(z) = b3z3 + b2z2 + b1z2 + b0 + 6b3t, it is easy to obtain solutions of the Veselov-Novikov equation, decreasing at infinity, nonsingular at t = 0 and having features when finite times t > t0.

Note that solutions of the Korteweg-de Vries equation with smooth, rapidly decreasing initial data remain nonsingular for t > 0 (see, for example, ).

The work was carried out with partial financial support from the Russian Foundation basic research(project codes 06-01-00094 for I.A.T. and 06-01-00814 for S.P.C.).

REFERENCES

1. Veseloe AP, Novikov S P. // DAN. 1984. T. 279. No. 1. P. 20-24.

2. Dubrovin B. A., Krichever I.M., Novikov SP. // DAN. 1976. T. 229. No. 1. P. 15-19.


The teacher welcomes the students and announces:

Today we will continue to work with you on the topic: entire equations

We have to strengthen the skills of solving equations with a degree higher than the second; learn about the three main classes of entire equations, master methods for solving them

On back side board, two students have already prepared solution No. 273 and are ready to answer students’ questions

Guys, I suggest you remember a little about those theoretical information which we learned in the previous lesson. Please answer the questions

What equation with one variable is called an integer? Give examples

How to find the degree of an entire equation?

What form can a first degree equation be reduced to?

What will be the solution to such an equation

What form can a second degree equation be reduced to?

How to solve such an equation?

How many roots will it have?

To what form can a third degree equation be reduced?

Fourth degree equation?

How many roots can they have?

Today, guys, we will learn more about entire equations: we will study ways to solve 3 main classes of equations:

1) Biquadratic equations

These are equations of the form
, where x is a variable, a, b, c are some numbers and a≠0.

2) Decaying equations that reduce to the form A(x)*B(x)=0, where A(x) and B(x) are polynomials with respect to X.

You have already partially solved decaying equations in the previous lesson.

3) Equations solved using a change of variable.

INSTRUCTIONS

Now each group will receive cards that describe the solution method in detail; you need to work together to analyze these equations and complete tasks on this topic. In your group, check the answers with the answers of your comrades, find errors and come to a single answer.

After each group has worked out their equations, they will need to explain them to the other groups at the board. Consider who you delegate from the group.

WORK IN GROUPS

Teacher during group work watches the guys discuss whether the teams have formed, whether the guys have leaders.

Provides assistance if necessary. If some group completed the task before others, then the teacher has more equations from this card of increased complexity in stock.

CARD PROTECTION

The teacher offers to decide, if the children have not already done so, who will defend the card at the board.

The teacher can correct their speech while the leaders are working if they make mistakes.

So, guys, you listened to each other, the equations for independent decision. Get to work

UR. Igr.

IIgr.

IIIgr.

You need to solve those equations that you don’t have.

No. 276(b,d), 278(b,d), 283(a)

So guys, today we studied solving new equations in groups. Do you think our work was a success?

Have we achieved our goal?

What was stopping you at work?

The teacher evaluates the most active children.

THANKS FOR THE LESSON!!!

In the near future it would be advisable to independent work, containing the equations discussed in this lesson.

It remains to consider the sets defined by equations (35.21), (35.23), (35.30), (35.31), (35.32), (47.7), (47.22) and (35.20)

Definition 47.16.A second-order surface is called disintegrating, if it consists of two first order surfaces.

As an example, consider the surface given by the equation

Left side equalities (35.21) can be factorized

(47.36)

Thus, a point lies on the surface given by equation (35.21) if and only if its coordinates satisfy one of the following equations or . And these are the equations of two planes, which, according to paragraph 36 (see paragraph 36.2, 10th line of the table), pass through the OZ applicate axis. Hence , equation (35.21) specifies a disintegrating surface, or more precisely, two intersecting planes.

Problem: Prove that if a surface is both cylindrical and conical, and also consists of more than one straight line, then it breaks up, i.e. contains some plane.

Consider now the equation (35.30)

It can be decomposed into two linear equations and . Thus, if a point lies on the surface defined by equation (35.30), then its coordinates must satisfy one of the following equations: and . And this, according to paragraph 36 (see paragraph 36.2, 6th row of the table), is the equation of planes parallel to the plane. Thus, equation (35.30) specifies two parallel planes and is also a disintegrating surface.

Note that any pair of planes and can be specified the following equation second order. Equations (35.21) and (35.30) are canonical equations of two planes, that is, their equations in a specially selected coordinate system, where they (these equations) have the simplest form.

Equation same (35.31)

generally equivalent to one linear equation y = 0 and represents one plane (according to paragraph 36 of clause 36.2, 12th line of the table, this equation defines the plane).

Note that any plane can be defined by the following second-order equation.

By analogy with equation (35.30) (for ) it is sometimes said that equality (35.20) specifies two merged parallel planes.

Let's now move on to degenerate cases.

1.Equation (35.20)

Note that the point M(x, y, z) belongs to the set given by the equation(35.20), if and only if its first two coordinates x=y=0 (and its third coordinate z can be anything). And this means that equation (35.20) specifies one straight line – the OZ applicate axis.

Note that any equation of a straight line (see paragraph 40, paragraph 40.1, as well as paragraph 37, system (37.3)) can be specified by the following second-order equation. Equality (35.20) is canonical a second order equation for a straight line, i.e. its second-order equation in a specially selected coordinate system, where it (this equation) has the simplest.



2. Equation (47.7)

Equation (47.7) can only be satisfied by one triple of numbers x=y=z=0. Thus, equality (47.7) in space sets only one point O (0; 0; 0) – origin; the coordinates of no other point in space cannot satisfy equality (47.7). Note also that a set consisting of one point can be defined by the following second-order equation:

3. Equation (35.23)

And this equation cannot be satisfied at all by the coordinates of any point in space, i.e. it defines the empty set. By analogy with equation (33.4)

(see section 47.5, definition 47.8), it is also called an imaginary elliptic cylinder.

4.Equation (35.32)

This equation also cannot be satisfied by the coordinates of any point in space, therefore it defines the empty set. By analogy with similar equation(35.30), this “surface” is also called imaginary parallel planes.

5. Equation (47.22)

And this equation cannot be satisfied by the coordinates of any point in space, and, therefore, it defines the empty set. By analogy with equality (47.17) (see section 47.2), this set is also called an imaginary ellipsoid.

All cases have been considered.



Did you like the article? Share with your friends!