Textbook Yu. N. Makarychev, etc. Lesson topic: Whole equations. Decaying equations Decaying equations method for their solution

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

Definition 47.16.The second-order surface is called decaying if it consists of two surfaces of the first order.

As an example, consider the surface given by the equation

The left-hand side of equality (35.21) can be decomposed into factors

(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 row of the table), pass through the OZ axis. Hence , equation (35.21) defines a disintegrating surface, or rather, two intersecting planes.

Task: Prove that if a surface is both cylindrical and conical at the same time, and also consists of more than one straight line, then it disintegrates, i.e. contains a certain plane.

Consider now the equation (35.30)

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

Note that any pair of planes and can be specified by the following second-order equation. 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.

The equation the same (35.31)

in general, it is equivalent to one linear equation y \u003d 0 and represents one plane (according to paragraph 36 of item 36.2, 12th row of the table, this equation defines a plane).

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

By analogy with equation (35.30) (at), it is sometimes said that equality (35.20) defines two merged parallel planes.

We now turn to degenerate cases.

1.Equation (35.20)

Note that a point M (x, y, z) belongs to the set given by equation (35.20) if and only if its first two coordinates x \u003d y \u003d 0 (and its third coordinate z can be anything). This means that equation (35.20) defines one straight line - the axis of the application OZ.

Note that the equation of any straight line (see paragraph 40, item 40.1, as well as paragraph 37, system (37.3)) can be defined by the following second-order equation. Equality (35.20) is canonicalsecond 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 one.

2. The equation (47.7)

Equation (47.7) can be satisfied by only one triple of numbers x \u003d y \u003d z \u003d 0. Thus, equality (47.7) in space sets only one point О (0; 0; 0) - origin of coordinates; the coordinates of any other point in space cannot satisfy equality (47.7). Note also that a set consisting of one point can be specified by the following second-order equation:

3. Equation (35.23)

And this equation cannot be satisfied by the coordinates of any point in space, i.e. it defines an 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)

The coordinates of any point in space also cannot satisfy this equation, therefore it defines an empty set. By analogy with the similar equation (35.30), this "surface" is also called imaginary parallel planes.

5. The equation (47.22)

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

All cases are considered.

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 February 14, 2008

Veselov-Novikov equation

u, \u003d e3 u + E3 u + s E (vu) + zE (vu) \u003d o, E V \u003d E u,

where E \u003d (Ex - ¿Ey), E \u003d 1 (Ex + ¿Ey), is a two-dimensional generalization of the Korteweg-de Vries (KdV) equation

and, \u003d 4 uhxx \u200b\u200b+ viih,

into which it goes in the one-dimensional limit: V \u003d u \u003d u (x). Equation (1) defines deformations of the two-dimensional Schrödinger operator

specifies the transformation of the solution φ of the equation Hf \u003d 0 into the solution b of the equation H b \u003d 0, where

H \u003d EE + u, u \u003d u + 2 EE 1n w.

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

Moutard transformation expands to system solutions transformation

Hf \u003d 0, f (\u003d (E3 + E3 + 3 VE + 3 V * E) f, ^^^

where Э V \u003d Эи, ЭV * \u003d Э и, which is invariant under transformation (extended Moutard transformation)

\u003d ~ | ((f Eyyu-Eph) dz- (f Eyu-Ef) dz

of the form H1 \u003d HA + 5H, where A, B are differential operators. Such deformations preserve the "spectrum" of the operator H at the zero energy level, transforming the solutions of the equation

Hf \u003d (EE + u) f \u003d 0 (3)

according to (Eg + A) φ \u003d 0.

There is a method for constructing new solutions (u, φ) of equation (3) from old solutions (u, φ) of this equation, which is reduced to quadratures - the Moutard transformation. It consists in the following: let an operator H with a potential u be given and a solution w of equation (3): Hw \u003d 0. Then the formula

W | [(f Esh - w Ef) dz - (f Esh - w Ef) dz]

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

Krasnoyarsk State Pedagogical University

+ [f E u - u E f + u E f - f E u +

2 2 "2 _ ~ _2 + 2 (E f Esh - Ef E w) -2 (E f Esh - Ef E w) +

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

u ^ u + 2EE lnm, V ^ V + 2E21nm

V * ^ u * + 2E21psh,

where w satisfies (4).

Veselov-Novikov equation (1) is

the compatibility condition for system (4) at V * \u003d V.

When the solution w is real, the conditions u \u003d u u

V * \u003d V are preserved and the extended Moutard transform translates real solutions and

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

All rational solitons of the KdV equation are obtained by iterating the Darboux transformation from the potential u \u003d 0. Moreover, all the resulting potentials are singular.

In the two-dimensional case, a similar construction can lead to nonsingular and even rapidly decreasing potentials already after two iterations.

DECAYING SOLUTIONS TO THE EQUATION

walkie-talkies. Namely, let u0 \u003d 0 and ω1 ω2 be real solutions of system (4):

u, \u003d r (z, z) + f (z, z), \u003d i (z, z) + i (z, z), (5)

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

fg \u003d Yyyy "yag \u003d yyyy"

Each of the functions uj u2 defines the (extended) Moutard transformation of the potential u \u003d 0 and the corresponding solutions of system (4). Let's designate them as Mu and Ma. The resulting potentials we

we denote by u1 \u003d Myu (u0), u2 \u003d Myu (u0).

Let δ1 e My (ω2), i.e. b1 is obtained from ω2 by transforming M ω. Note that the Moutard transformation for φ depends on the constant of integration. We choose a constant such that b1 is a real function. The choice of a constant allows us to frequently control the nonsingularity of the iterated potential (we will use this in specific examples).

A simple check shows that b2 \u003d - b1 f

e mu (yuh). The well-known lemma holds, which is true for an arbitrary potential u0.

Lemma 1. Let u12 \u003d M01 (u2) and u21 \u003d M02 (u2). Then u12 \u003d u21.

For the case u0 \u003d 0, we have Lemma 2. Let ω1 and ω2 have the form (5). Then the potential u \u003d Mb (My (u0)), where u0 \u003d 0 and b1 e My (u2), is given by the formula

u \u003d 2EE 1nI ((/ I - fya) +) ((f "I - fя") dr + + (GY - G I) - GZ) + + GY "" - G "" z + 2 (zi - zi ")) dz).

Note that even for stationary initial solutions ω1, ω2 of system (4), we can obtain a solution to the Veselov-Novikov equation with nontrivial dynamics in r.

Theorem 1. Let U (z, z) be the rational potential obtained by the double Moutard transformation from ω1 \u003d iz2 - i ~ z, ffl2 \u003d z2 + (1 +

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

U \\ t \u003d 0 \u003d 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) \u003d U (-z, - z) is regular for all t\u003e 0.

Rational potential (1), given in the work, decreases as r-6 and gives a stationary nonsingular solution of the Veselov-Novikov equation. Choosing f (z) \u003d a3z3 + a2z2 + a1z2 + a0 + 6a3t, g (z) \u003d b3z3 + b2z2 + b1z2 + b0 + 6b3t, it is easy to obtain solutions of the Veselov-Novikov equation, decreasing at infinity, non-singular at t \u003d 0 and having singularities at finite times t\u003e t0.

Note that solutions of the Korteweg-de Vries equation with smooth, rapidly decreasing initial data remain nonsingular for t\u003e 0 (see, for example,).

This work was carried out with partial financial support from the Russian Foundation for Basic Research (project codes 06-01-00094 for I.A.T. and 06-01-00814 for S.P.Ts.).

LIST OF REFERENCES

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

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

The teacher greets the students and announces:

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

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

On the back of the board, two students have already prepared solution # 273 and are ready to answer students' questions

Guys, I propose to recall a little the theoretical information that we learned in the previous lesson. I ask you to answer the questions

Which one-variable equation is called an integer? Give examples

How do you find the degree of an entire equation?

What form can the equation of the first degree be reduced to

What will be the solution to such an equation

What form can an equation of the second degree be reduced to?

How to solve such an equation?

How many roots will it have?

What form can the equation of the third degree be reduced to?

Equation of the fourth degree?

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, which are reduced to the form A (x) * B (x) \u003d 0, where A (x) and B (x) are polynomials with respect to X.

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

3) Equations solved by changing the variable.

INSTRUCTIONS

Now each group will receive cards in which the method of solution is described in detail, you need to jointly analyze these equations and complete the tasks on this topic. In your group, check the answers with those of your comrades, find mistakes and come to a common answer.

After each group has worked out their equations, they will need to be explained to the other groups at the blackboard. Think about who you are delegating from the group.

WORK IN GROUPS

During group work, the teacher observes how the children reason, whether the teams have formed, whether the children have leaders.

Provides assistance if necessary. If a group coped with the task earlier than others, then the teacher still has the equations from this card of increased complexity in stock.

CARD PROTECTION

The teacher offers to decide, if the guys have not done this yet, who will defend the card at the blackboard.

The teacher can, during the work of leaders, correct their speech if they make mistakes.

So, guys, you listened to each other, the equations for your own solution are written on the board. Get down to business

Ur. Igr.

IIgr.

IIIgr.

You need to solve those equations that you do not have.

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

So guys, today we studied the solution of new equations in groups. Do you think our work went well?

Have we reached our goal?

What was stopping you in your work?

The teacher evaluates the most active children.

THANK YOU FOR THE LESSON!!!

In the near future it is advisable to carry out an independent work containing the equations analyzed in this lesson.

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"Fractional Equations" - Solve the resulting equation. Quadratic equation has 2 roots if …… Exclude the roots that are not included in the admissible values \u200b\u200bof the fractions of the equation. … Your letter. High soul ". Algorithm for solving fractional rational equations. And remember - what is the main thing in a person? Fractional rational equations. How many roots does this equation have? 4. What is the name of this equation?

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