ИСПОЛЬЗОВАНИЕ ЗАКОНОВ СОХРАНЕНИЯ ДЛЯ РЕШЕНИЯ ЗАДАЧИ О ВОЛНЕ НАГРУЗКИ В УПРУГО-ПЛАСТИЧЕСКОМ СТЕРЖНЕ

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Introduction. Conservation laws, in relation to the differential equations, were published in Emma Neter's article [1] more than 100 years ago. She established the general principle connecting symmetry groups and con- servation laws for the differential equations deduced from the variation principle. For more than 70 years all results in this field were based on this article. More general con- cepts allowing to calculate conservation laws for any sys- tems of the differential equations appeared in A.M. Vino- gradov's works [2; 3]. For a rather long time conservation laws occurred in literature as purely mathematical result, far from applications. In the works [4-6] it was shown how conservation laws can be used for the solution of Cauchy and Riemann problems and also accurate solu- tions of these tasks were made. Later the method of conservation laws was applied to solution of free-boundary problem: elastic plasticity tasks [7-10]. For the first time, a special case of the task of the wave distribution, which is solved by means of conserva- tion laws, is constructed in work [11; 12]. In this work more general case is considered and also condition under which there is a solution of Cauchy problem is formu- lated. Derivations of the main equations 1. We will consider the process of plastic deformations Supposing s(e) is a steadily increasing along e func- tion (fig. 1) and for all e derivative ds / de is a steadily decreasing function (i. е. d 2s / d 2e < 0 ). Fig, 1. The process of propagation of plastic deformations in a semi-infinite elastoplastic rod Рис. 1. Процесс распространения пластических деформаций в полубесконечном упруго-пластическом стержне propagation in a semi-infinite elastic plastic rod caused by For tensions s £ ss ( ss is tensile yield) dependence dynamic loading time, p (t ) applied to the end of the rod, s(e) , according to Hook’s law, is linear: which is not decreasing in (i. e. dp / dt ³ 0 ). We shall find a solution in the Langrangian coordinate system: we will take a rod axis for the axis x , we will choose the origin of coordinates x = 0 on the left end of s = Ee , (3) where Е is elasticity modulus (Young’s modulus). Wherein the values of Young’s modulus E have been sorted out as to under s = s dependence (3) is continuthe rod. Suppose that during deformation there is no lat- eral bulging of the rod and that the influence of transverse deformations of the rod on the process of propagation of longitudinal waves is negligible. Let us consider small s ous. From the equation of through flow in case of minor deformations we obtain the following formula deformations of the rod and assume that the rod density in the course of deformation does not change. The only component of tension tensor, other than zero, will be sxx = s other than zero components of a tensor of defor- mations will be exx = e and e yy = ue . In this case, motion equation exclusive of massive ex- ternal forces is as follows [13]: de = dv . dt dx Taking into account the dependence load and introducing notation a2 (s) = ¶s , ¶e s = s(e) (4) under (5) r ¶v = ¶s , (1) where d s / d e is the rate of change to the curve s(e) ; ¶t ¶x where s = sxx - component of the stress tensor; v - para is a constant, 0 < a < 1 ; a2 (s) = s2b velocity of longiticles velocity along the axis Оx , r - density. tudinal waves propagation in the rod, we have Since density is constant, without generality loss we de = de ¶s = 1 ¶s . further assume that r = 1 . Accepting the defining relation of the deformation dt ds ¶t a2 (s) ¶t (6) plasticity theory (for the uniaxial stress) as follows s = s(e). (2) Placing the relation (4) in (6), we obtain the system of two equations of partial derivatives of the first order [13]: dv = ds , dt dx ¶v = 1 ¶s , ¶x a2 (s) ¶t (7) In this case the plane xOt splits into two domains: elastic bounded by axes х and direct ts P and plastic do- For two functions v(x, t), s(x, t). main, placed above the line ts P . It should be noted that In this equation a(e) is the velocity of longitudinal the equation of this line is as follows: x = a0 (t - ts ) , waves propagation in the rod. Since the velocity of longitudinal waves propagation generally is the tension function, then, equation system (7) is the system of quasilinear equations with partial de- rivatives of the first order of hyperbolic type. For it we will determine characteristics and relations under charac- teristics. Characteristics of equation system (7) are determined by integrating of differential equations’ characteristics: where ts time point, when s achieves the yield stress ss (fig. 2). In the plastic domain we have a linear problem which can be easily solved applying traditional methods. Hence, we will seek for the Cauchy problem solution for equations (7) only in the plastic domain. dx = ∓a(s)dt. (8) These equations generally cannot be integrated in plain (x, t) before the problem has been solved since а is the tension function s(x, t) . Along characteristics dx = ∓a(s) dt lations are made the following re- dv ∓ 1 a(s) ds = 0. (9) These relations are called differential equations of characteristics in hodograph plane (s, v) . After integrating we obtain ò 1.2 s d s Fig. 2. Characteristics of equations (14) Рис. 2. Характеристики уравнений (14) v = ∓ 1 + C при dx = ∓a(s)dt. 0 a(s1 ) (10) Problem definition. To find the value of function We will now consider the simplest case of load waves v ( x, t ), s( x,t ) at the point M (xm ,tm ) if the values of the propagation in homogeneous half-infinite rod, which at the initial moment was in nonperturbed state. We will consider the equation solution (7) under the required function along points Q(0,tq ), P(xp ,tp ) ts P and tsQ are known. Here are considered as intersection given initial conditions (Cauchy conditions): points of the correspondent characteristics with the axis v(x, 0) = v(x), (11) Оt and the line ts P , drawn from the point M . According to (13) the equations (7)-(10) will be as follows and boundary condition s(0, t) = - p(t), ( p(t) > 0), dx = ∓a(s)dt. (12) dv = ds , ¶s = s2b ¶v , b = a -1 . (14) dt dx ¶t ¶x a where, to ensure the load process there must be p¢(t) > 0 . Characteristics of the present equation system accord- Conditions (11)-(12) mean that at the initial moment the rod is in the deformed and dormant state. Meeting the initial conditions correlates with Cauchy problem solution in the domain (fig. 2), limited by axis х and positive char- acteristics tsQ. 2. For simplicity we will consider the following asser- tions for function (2) s = Ee, under s < sz , (13) ing to (8) are as follows dx = ∓sbdt. Relations on characteristics (9), after integration will be s-b+1 v ∓ -b +1 = C1,2 , where C1, C2 are random constants. We will introduce Riemann’s invariants under the s(e) = 1 ea , under 0 < a < 1 dx = ∓a(s)dt. s-b+1 s-b+1 a formula x = v + -b +1 , h = v - -b +1 , then the system General case is considered similarly. For the continuity of function s(e) suppose E = 1 ea-1. at point es we (14) will be as follows ¶x - sb ¶x = 0, ¶h + sb ¶h = 0. (15) a s ¶t ¶x ¶t ¶x Employment of conservation laws for equations From the first relation we have describing the wave load in the elastic plastic rod. Conservation laws for the equation system (15) is founded as follows [5] ¶t A + ¶x B = 0, bsb-1 æ 1 ç 2 è sb ö A + sb A + B = - b ÷ h h ø 2 along x = x0 . Asb-1 + 2 Ah = 0 ¶ A(x, h) + ¶ B(x, h) = Since sb-1 = 2 , we obtain differential æ b ¶A ¶t ¶B ö ¶x ¶x æ b ¶A ¶B ö ¶h (x - h)(-b +1) equation for A along x = x = çs + ÷ + ç -s + ÷ = 0. 0 è ¶x ¶x ø ¶x è ¶h ¶h ø ¶x From here we obtain the equation to determine A and B - b A + A x - h = 0. ( ) -b +1 h 0 sb ¶A + ¶B = 0, ¶x ¶x -sb ¶A + ¶B = 0. ¶h ¶h (16) By its integrating we obtain b ln (h - x ) = ln A + ln C , Excluding from (16) function B we obtain the equation to determine function A : or 2(b -1) 0 3 8 ¶2 A æ ¶A ¶A ö 1 b è ø 0 b(-b +1) ¶x¶h - ç ¶x - ¶h ÷ x - h = 0. A = C3 h - x0 2(b-1) , B = -sb A -1, along x = x . (19) We will introduce in this equation the notation Similarly along h = h0 we have 8 = w-1. As a result we obtain the Euler- bsb- æ 1 s ö + s - = b(-b +1) 1 ç b ÷ A è 2 ø b Ax Bx Poisson-Darboux equation [14]: = bsb-1 1 b b ¶2 A w æ ¶A ç - ¶A ö = 0. (17) 2 As + 2s Ax = 0. ÷ ¶x¶h x - h è ¶x ¶h ø To determine function B we get a similar equation Therefore along h = h0 b ¶2B w æ ¶B ¶B ö A = C h - x 2(b-1) , B = sb A. (20) + ç - ÷ = 0. 4 0 ¶x¶h x - h è ¶x ¶h ø Matching conditions (19) and (20) at the point x = x0 , Applying (15) we will write the integral about closed path tsQMP h = h0 gives C3 = C4 . Thus, for the final problem solution we have to solve the equation (17) with the restricted conditions (19) and °ò Adx - Bdt =0. (18) (20). To solve this problem Riemann function is used. It We will split this integral into four integrals taken about closed paths ts P, PM , MQ, Qts . looks as follows: w(x0, h0; x, h) = About closed paths ts P and Qts integrals can be calculated after determining A , B inclusive initial and boundary conditions (11), (12). We will determine A and B so that along the charac- teristics PM and MQ integrals transform to zero. We æ x - h öw æ x - h öw = ç 0 0 ÷ ç 0 0 ÷ è x0 - h ø è x - h0 ø ; (x - h)(x - h ) F (w, w;1, t ), (21) have where 1- t = 0 0 (x0 - h)(x - h0 ) F is hypergeometric ò Adx - Bdt = ò (-sb A - B)dt. PM PM We will calculate the obtained integral in parts polynomial of the second raw. Suppose N is a random point from the ts PMQ do- Q ò (-sb A - B)dt = t (-sb A - B ) M - ò td (-sb A - B ). PM PM Similarly along MQ we obtain main (fig. 3). We will connect the point N with MP characteristics NK - xN , and with QM characteristics NL - hN . As a result the value of function A at the point N will equal ò Adx - Bdt = t (sb A - B) Q - ò td (sb A - B). æ Aw ö M MQ MQ A( N ) = A(M ) w (M ) + ò wç - (x - h) + Ax ÷ d x + Finally obtain KM è ø æ Aw ö d (sb A + B ) = 0, d (sb A - B ) = 0. + ò wç (x - h) + Ah ÷ d h, x=x0 =const h=h0 =const ML è ø where function w is determined by the formula (21). Then, from (22) we obtain identical to the above + formula will define the function B values in point N B ( N ) = B (M ) w (M ) + ò w æ - Bw + B ö dx + tm - tq = °ò tsQ Bdt - °ò Pts Adx - Bdt . (24) è ø KM ç + ò w æ Bw ç (x - h) x ÷ + ö Bh ÷ dh. If the equation is (24) solvable, we can find the intersection point of the characteristic and the initial curve. In this case the Cauchy problem is unsolvable. Equation (24) ML è (x - h) ø can be easily solved applying the method of successive approximations. Fig. 3. The solution of the Cauchy problem Рис. 3. Решение задачи Коши Now from (18), taking into account the obtained rela- tions we have Fig. 4. Finding the intersection point of the characteristic and the curve on which the initial conditions are given Рис. 4. Нахождение точки пересечения характеристики и кривой, на которой заданы начальные условия Conclusion. Knowledge of conservation laws allowed = °ò tsQ °ò ts PMQ Adx - Bdt + Adx - Bdt = °ò Adx - Bdt + t Pts m - tq = 0 . to find coordinates of characteristics’ points of intersec- tion, and therefore to solve the problem discussed in the article. The case when one of characteristics crosses the line where initial conditions are set is considered. In this case, as we know, Cauchy problem cannot be solved. It Hence, we have tm - tq = °ò Bdt - °ò Adx - Bdt . leads to the procedure which, by means of conservation laws, allows to settle the issue of Cauchy problem solvtsQ Pts ability. Similar calculations help to find coordinates of xm References xm - xq = °ò Bdt - °ò Adx - Bdt . tsQ Pts Later, according to values v, s at points Q and P knowing the relations along characteristics PM and QM , we will find values v ( xm ,tm ) and s( xm , tm ) . In conclusion, we will consider the question of Cauchy problem solvability which always arise when solving the systems of non-linear differential equations. As it is known, Cauchy problem is solvable if each char- acteristics crosses the lines Qts, tsP only once [15]. Ap- parently, this question can as well be solved knowing the conservation laws. Suppose that characteristics QM cross the line tsP at the point M (fig. 4). Then we get the con- servation law 1. Neter E. [Invariant variational problems]. Variat- sionnye printsipy mekhaniki. Moscow, Fizmatgiz Publ., 1959, P. 611-630 (In Russ.). 2. Vinogradov A. M., Krasil’shchik I. S., Lychagin V. V. Simmetrii i zakony sokhraneniya [Symmetries and conser- vation laws]. Moscow, Faktorial Publ., 1997, 464 p. 3. Vinogradov A. M. Local symmetries and conser- vation laws. Acta appl. Math. 1984, Vol. 2, No. 1, P. 21-78. 4. Senashov S. I. [On the laws of conservation of the equations of plasticity]. Dokl. AN SSSR. 1991, Vol. 320, No. 3, P. 606-608 (In Russ.). 5. Kiryakov P. P., Senashov S. I.,Yakhno A. N. Prilozhenie simmetriy i zakonov sokhraneniya dlya resheniya differentsial’nykh uravneniy [The application of °ò ts MQ Suppose as before b Adx - Bdt = 0. (22) symmetries and conservation laws for the solution of differential equations]. Novosibirsk, SO RAN Publ., 2001, 192 p. 6. Senashov S. I., Vinogradov A. M. Symmetries and Conservation Laws of 2-Dimensional Ideal Plasticity. 3 0 0 A = C h - x 2(b-1) , B = -sb A -1, along x = x . (23) Proc. of Edinb. Math. Soc. 31, 1988, P. 415-439.

Об авторах

С. И. Сенашов

Сибирский государственный университет науки и технологий имени академика М. Ф. Решетнева

Email: sen@sibsau.ru
Российская Федерация, 660037, г. Красноярск, просп. им. газ. «Красноярский рабочий», 31

И. Л. Савостьянова

Сибирский государственный университет науки и технологий имени академика М. Ф. Решетнева

Российская Федерация, 660037, г. Красноярск, просп. им. газ. «Красноярский рабочий», 31

Е. В. Филюшина

Сибирский государственный университет науки и технологий имени академика М. Ф. Решетнева

Российская Федерация, 660037, г. Красноярск, просп. им. газ. «Красноярский рабочий», 31

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