On a Method for Solution of the Ordinary Differential Equations Connected with Huygens Equations

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1 On a Method for Solution of the Ordinary Differential Equations Connected with Huygens Equations A. H. Hovhannisyan, B.-W. Schulze The problem of describing the linear second order hyperbolic equations, satisfying the Huygens principle, is known as the Hadamards problem 1], ], 3]. Ch. Huygens 4] formulated this principle as a geometrical method for constructing the wave front, and it was used for explaining the main properties of light propagation. Its mathematical formulation within the theory of Cauchy problem for hyperbolic partial differential equations has been given by well-known French mathematician J. Hadamard in his 193 Yale lectures 1]. It has received a good deal of attention and nowadays is a classical problem in mathematical physics, but is has turned out to be very difficult and, at present it is still far from its complete solution. Let us consider the Cauchy problem for hyperbolic second order differential equations n n g ij (x)u x i x j + b i (x)u x i + c(x)u = 0, (1) i,j=1 u S = f, u ν S = g. () Here S is a space-like manifold of the dimension n 1 and u ν is a derivative with respect to the normal ν to S. DEFINITION 1. Equation (1) satisfies the Huygens principle when the value of the solution of the Cauchy problem (1), () for any point P = (x 1 0,...,xn 0 ) is determined by the Cauchy data given in the intersection of the initial manifold S and the characteristic conoid with vertex at P. It should be noted that the Huygens principle is invariant relative to the following, so-called elemtary transformations: (a) non-degenerate coordinant transformation; (b) transformation of the unknown function u λu, λ(x) 0; (c) multiplication of the equation by the function λ(x) 0. Two equations are called equivalent when one of them is derived from the other by means of elementary transformations. In our further considerations, these elementary transformations are necessarily taken into account. Papers 5] and 6] contain descriptions of all the Huygens equations of the type n L (k) n u] u tt u xx a i (x t)u y i y i + ak (x t)b(y k )u = 0, (3) k = 1,...,n, where n 4 is an even number, and a i > 0, i = 1,...,n. We briefly describe these results assuming k = 1 for definiteness. We denote where µ(y 1 ) is an arbitrary solution of the equation 1 l µ = y 1 µ (y 1 ) µ(y 1 ), l µ = y 1 µ (y 1 ) µ(y 1 ), µ b(y 1 )µ = 0. (4) 1 A.H. Hovhannisyan acknoweldges DAAD for financial support during his stay in Germany. 1

2 Then the L (1) n may be written in the form L (1) n = P + a 1 (x t)l µl µ, where We define the operator P = t n x a i (x t) (y i ). i= and call it l µ transform of the operator L (1) n. It is clear that the equality takes place. THEOREM 1. If L (1) n satisfies the Huygens principle, then We denote the operator L (1) n L (1) n = P + a 1 (x t)l µ l µ (5) l µ L (1) (1) n = L n l µ (6) (1) L n+ also satisfies the Huygens principle. with b(y 1 ) 0 as L 0 n. From ] it is known that the equation L 0 n u] = 0 satisfies the Huygens princple for any even n 4. Making use of the operator L 04, theorem 1 allows to construct Huygens operators for n 6 variables. There arises a natural question: whether all the Huygens operators of the type (3) may be obtained through the described construction basing on L 0 4. A positive answer to this question is given in 6]. THEOREM. Let the operator L (1) n satisfy the Huygens princple. Then it may be derived from the operator L 0n through l µ transforms applied no more than n 4 times. The aim of the present paper is to show how helpful the described algorithm for constructing the Huygens equations is and to investigate the equation (4) in detals. Taking account of the equality (6) and equation (4), it follows from theorem that there are operators l µk = y 1 µk, k = 0, 1,...,m 1, m n 4 µ k such that where l µm 1 l µ0 L 0n = L (1) n l µ m 1 l µ0, (7) µ 0 = αy 1 + β, (α, β = const), (l µk l µ k )µ k+1 = 0, k = 0,...,m. (8) If m is the minimum number of transformations which transform the operator L 0n into L (1) n, then the functions µ k (k = 0,...,m 1) is uniquely determined with the accuracy up to a constant factor 6]. The following reccurent relation is easily obtained from (8): µ k+1 = a k + b k µ k µ k µ k (y1 )dy 1. However, it is not always convenient to use this formula because of the necessity to find the indefinite integral of µ k. Therefor, further we study properties of the function b(y 1 ) which allow as to determine conditions, easily checked

3 and necessary for equation (3) to acquire Huygens properties, and which may indicate us some other method to find functions µ k. First, we give the known results 7]. THEOREM 3. We assume that the operator has Huygens properties for even values n N, m = N 4 and {l µk } m 1 k=0 is a unique sequence of operators determined by the formula (8), so that the equality (7) takes place. Then ] (a) b(y 1 ) = d m 1 d(y 1 ) ln µ k (y 1 ), k=0 (b) l µ k (y 1 ) = P l (y 1 ) k=0 is a polynomial of degree 1 (l + 1)(l + ), 0 l m 1. Point (a) is proved by direct computations and point (b) through the following important result. THEOREM 4. If the sequance of functions {µ k (y 1 )} m 1 k=0 satisfies the conditions of (8), then each of µ k(y 1 ) is a rational function. Using these results, we shall prove some properties of the function b(y 1 ). Let the polynomial P m 1 (y 1 ) has the form P m 1 (y 1 ) = (y 1 a 1 ) k1 (y 1 a s ) ks, (k k s = 1 m(m + 1)). Then from point (a) of theorem 3 we obtain: b(y 1 ) = q=1 i q (y 1 a i ) k q s. (y 1 a i ) Hence, singularities of the function b(y 1 ) are poles of the second order and (4) is a Funchian equation. We separate one of the poles, e.g. a j, 1 j s, and write the function b(y 1 ) in the form b(y 1 ) = b j(y 1 ) (y 1 a j ), where b j (y 1 ) = k j + q=1 i q (y 1 a i ) k q s. (9) (y 1 a i ) i j It immediately follows from this formula that b (1) (a j ) = 0. Let L n be the l µm transform of the operator L (1) n. Then µ m is the solution of equation (4). According to theorem 4, µ m is a rational fuction. We study its behavior in the neighborhood singular point a j 1 j s. With this purpose we write equation (4) as µ m b j(y 1 ) (y 1 a j ) = 0. 3

4 Solution µ m is found in the form of ] µ m = (y 1 a j ) 1 ρ + µ k m(y 1 a j ) k where ρ is the root of the defining equation ρ ρ b j (a j ) = 0. (10) Hence ρ 1, = 1 ± b j(a j ). As µ s is a rational function, then ρ 1 and ρ must be integers ± 4bj (a j ) = q j + 1, q j = 0, ±1,.... Hence b j (a j ) = q j (q j + 1). (11) As the negative values q j do not suggest new values for b j (a j ), it may be assumed that q j = 0, 1,.... We obtain from formula (9) that b j (a j ) = k j. Therefore, multiplicity of the root a j of the polynomial P m 1 may assume the following values: k j = q j(q j + 1), j = 1,,..., s. As the order of the polynomial P m 1 is 1 n 4 m(m + 1) and m we obtain If s = 1, then j=1 q j (q j + 1) n 4 q n 4 n. which is in agreement with the known result ]. Roots of equation(10) are ρ 1 = q j + 1, ρ = q j. Therefore, signularity of the rational function µ m in the point a j may have only the form (y 1 a j ) qj. Since µ m is the solution of a linear equation, its singularities may appear only in the points where singularities of the equation coefficients are. Therefore the function µ m may be found as µ m = T r (y 1 ) (y 1 a 1 ) q1 (y 1, (1) qs a s ) where T r (y 1 ) is a polynomial of the order r. Acording to theorem 3 the polynomial P m = m µ k = P m 1 µ m k=0 4

5 has the order 1 (m 1)(m + ), and the order of P m 1 is 1 m(m + 1), accordingly. Therefore, r = m q k. Writing T r (y 1 ) with indefinite coefficients and substituting µ m into the equation, it is possible to determine these coefficients, and hence, to find µ m. The facts, proved above, may be presented as the following result. THEOREM 5. Let equation (3) satisfy the Huygens principle and let b(y 1 ) 0. Then (a) singularities if the function b(y 1 ) a 1,..., a s are poles of the second order, being s 1 (n )(n 4) m(m+1) 8. (b) if the function b(y 1 ) is presented in the neighbourhood of a singular point a j 1 j s as then b j (a j ) may assume only one of the following values b(y 1 ) = b j(y 1 ) (y 1 a j ), b j (a j ) = q j (q j + 1), q j = 0, 1,..., where q j (q j + 1) j=1 and b j (a j) = 0. (c) solution of equation (4) has the form of (1). Let us consider the Huygens equation 5], 6] (n 4)(n ) 4 L 8 u] u tt u xx a i (x t)u y i y i + a1 (x t) 6y1 (y 1 ) 3 + ] (y 1 ) 3 1] u = 0. The function b(y 1 ) = 6y1 (y 1 ) 3 + ] (y 1 ) 3 1] has the singularities in the points a 1 = 1, a = 1 + i 3, a 3 = 1 i 3. For the b j(y 1 ), i = 1,, 3 we have Operator L 8 is obtained from the operator L 0 8 b 1 (1) = 1, q 1 = 1, ( b 1 1 ) 3 + i ( b 1 1 ) 3 i = 1, q = 1, = 1, q = 1. L 0 8 = t x a i (x t) y i (13) with the operators l µ0, l µ, where µ 0 = y 1 5

6 and µ 1 is a solution of the equation and has the form Then The solution of equation according to the (1) can be find in the form Substituding this function in (14) we obtain µ y i µ = 0 µ 1 = (y1 ) 3 1 y 1. P 0 = µ 0 = y 1, P 1 = µ 0 µ 1 = (y 1 ) 3 1. µ 6y1 (y 1 ) 3 + ] (y 1 ) 3 1] µ = 0 (14) µ = T (y 1 ) (y 1 ) 3 1 = 6 α i (y 1 ) i (y 1 ) α 6 36α 6 + 6α 6 = 0, 0α 5 30α 5 + 6α 5 = 0, 1α 4 4α 4 + 6α 4 = 0, 6α 3 30α 6 18α 3 + 6α 3 = 0, α 0α 5 1α + 6α = 0, 1α 4 6α 1 + 6α 1 = 0, From here we have 6α 3 + α 0 = 0, α = 0. α 5 = α 4 = α = 0, α 0 = α 3 = 5α 6 and α 1, α 6 are any numbers. If we take α 6 = 0, α 1 = 1, then If α 6 = 1, α 1 = 0 µ = y 1 (y 1 ) 3 1. µ = 5 + 5(y1 ) 3 (y 1 ) 6 (y 1 ) 3. 1 Therefore, the common solution of the equation (14) has the form y 1 µ = C 1 (y 1 ) C 5 + 5(y 1 ) 3 (y 1 ) 6 (y 1 ) 3. (15) 1 6

7 For C 1 = 1, C = 0 we have and we obtain the known Huygens equation If C 1 = 0, C = 1, then u tt u xx u tt u xx 8 P = y 1, b(y 1 ) = (y 1 ) 8 a i (x t)u yi y i + a1 (x t) (y 1 ) u = 0. In the case C 1 0, C 0, it is follow that and the following Huygens equation u tt u xx 8 P = 5 + 5(y 1 ) 3 (y 1 ) 6 a i (x t)u y i y i + a1 (x t) 6(y1 ) (y 1 ) 4 150y 1 ] (y 1 ) 6 5(y 1 ) 3 5] u = 0. P = y 1 + γ5 + 5(y 1 ) 3 (y 1 ) 6 ] a i (x t)u y i y i + a1 (x t) + 36γ(y1 ) 5 + 6γ y 1 (y 1 ) (y 1 ) 3 50] {y 1 + γ5 + 5(y 1 ) 3 (y 1 ) 6 ]} u = 0. (16) The solution of the equation (14) can be find in the form too, where Substituding this function in the equation () gives µ = (y1 1) T 3 (t 1 ) (y 1 ) 3 + y T 3 (y 1 ) = 3 α i (y 1 ) i. i=0 µ = (y1 1) (y 1 ) 3 + 3(y 1 ) ] (y 1 ) + y 1. (17) + 1 Using this solution it is possible to find the common solution of the equation (14). The function (17) can be find from (15) if we take C 1 = 9, C = 1 ( γ = 1 9). In this case the equation (16) can be written in the form u tt u xx 8 a i (x t)u y i y i + 6 a1 (x t) (y 1 1) + 6 (y 1 ) 4 + 4(y 1 ) 3 6(y 1 ) + y 1 + ] ] (y1 ) 3 + 3(y 1 ) + 6(y 1 ) ] u = 0. To constuct the new Huygens equation, we have to know the factorization of the denominator of the function b(y 1 ). However using the fact, that P 3 = P µ 3 is the polynomial of degree 10, it is possible to assume, that the equation µ γ(y1 ) 5 + 6γ y 1 (y 1 ) (y 1 ) 3 50] {y 1 + γ5 + 5(y 1 ) 3 (y 1 ) 6 ]} µ 3 = 0 (18) has the solution of the form µ 3 = T 10 (y 1 ) y 1 + γ5 + 5(y 1 ) 3 (y 1 ) 6 ]. 7

8 Easy to see that substitution of the function µ = T r(y 1 ) Q s (y 1 ), in the equation µ R Q s = 0 lead to expresion For µ 3 we have Q st r Q s Q st r + (Q s) Q s Q s R]T r = 0. (19) 10 T r = T 10 = α i (y 1 ) i, i=0 Q s = Q 6 = y 1 + γ5 + 5(y 1 ) 3 (y 1 ) 6 ], R = + 36γ(y 1 ) 5 + 6γ y 1 (y 1 ) (y 1 ) 3 50]. Substituding these functions in (19) we can find the common solution of the equation (18) where and µ 3 = C 1 µ (1) 3 + C µ () 3, µ (1) 3 = 3γ (y 1 ) 10 45γ (y 1 ) 7 1γ(y 1 ) 5 105γ(y 1 ) 55γ y 1 7 y 1 + γ5 + 5(y 1 ) 3 (y 1 ) 6 ] µ () 3 = Now let us consider the following Huygens equation L 8 u] u tt u xx (y 1 ) 3 1 y 1 + γ5 + 5(y 1 ) 3 (y 1 ) 6 ]. 6 a i (x t)u y i y i + 6a1 (x t) (y 1 ) u = 0 (0) Operator L 8 can be obtained from L 0 8 (13) by the transformations with operators l µ 0 and l µ0, µ 0 = y 1, µ 1 = (y 1 ). To construct the next Huygens equation we have to solve Then and The coresponding Huygens equation has the form u tt u xx 6 µ 6 (y 1 ) µ = 0. µ = (y1 ) 5 + γ (y 1 ) P = µ 0 µ 1 µ = y 1 (y 1 ) 5 + γ]. a i (x t)u y i y i + a1 (x t) 1(y1 ) 10 36γ(y 1 ) 5 + γ (y 1 ) (y 1 ) 5 + γ] u = 0. (1) 8

9 Note that in spite of the fact that in the equation (0) q =, nevertheless in (1) q i = 1, i = 1,...,6. Therefor the solution of the equation µ 1(y1 ) γ(y 1 ) 5 + γ (y 1 ) (y 1 ) 5 + γ] µ 3 = 0 () can be find in the form Substuding of this function in () gives and the Huygens equation µ 3 = T 10 y 1 (y 1 ) 5 + γ]. µ (1) 3 = 3(y1 ) γ(y 1 ) 5 7γ y 1 (y 1 ) 5 + γ] 10 u tt u xx a i (x t)u yi y i + a1 (x t) 180(y1 ) γ(y 1 ) γ (y 1 ) γ 3 (y 1 ) 3 3(y 1 ) γ(y 1 ) 5 7γ ] u = 0. The second linear independent solution is and the corresponding Huygens equation µ () 3 = (y 1 ) (y 1 ) 5 + γ] 10 u tt u xx a i (x t)u y i y i + 6 a1 (x t) (y 1 ) u = 0. All these examples of the odinary differential equations of the form (4) conectes with Huygens principle can be usefull for the investigating of the following problem. Let us write the equation (4) with the coefficient (a) in theorem 3 in the form µ + k i (y 1 µ = 0, (3) a i ) where a i (i = 1,...,s) are the root of the polynomial P m 1 (y 1 ) in formula (b) and k i are there multiplicities. As it was shown before for an equations, conected with Huygens equations k i = q i (q i + 1) with some positive integers q i. The equations of the form (3) are known as Fuchs equations 9]. The arrising problem is to discribe the a i and k i, i = 1,...,s such that the equation (3) will be conected with Huygens principle, in particular all the solutions of (3) are rational functions. Below we give a necessary and suficient condition on a i and k i, i = 1,..., n for which the solutions of the equation (3) do not consist the logarithmic singularities. In the neighborhood of a singular point a r (r = 1,...,s) an equation (3) can be written in the form (y 1 a r ) µ q r (q r + 1) + k r q k (q k + 1) (y1 a r ) (y 1 a k ) ] µ = 0. If we write ( ) y 1 a r y 1 = (n 1) (y1 a r ) n a k (a k a r ) n n= 9

10 and look for the solution in the form µ(y 1 ) = (y 1 a r ) q m=0 α mr (y 1 a r ) m we obtain m=0 α mr (m + q)(m + q 1)(y 1 a r ) m q r (q r + 1) + q k (q k + 1) k r n= (n 1) (a k a r ) n (y1 a r ) n] + For m = 0 we have + α(y 1 a r ) m = 0. (4) m=0 q 1 = q r, q r = q r + 1 and α 0r an arbitrary number. Let us consider q = q r, when the solution has the singularity. In this case the equation (4) has the form m=0 α mr (m q r )(m q r 1)(y 1 a r ) m q r (q r +1)+ For m = 1 we obtain α 1r = 0 and for m = p α pr p(p q r 1)] = k r q k (q k +1) k r p q k (q k + 1) l=0 n= (n 1)(y 1 a r ) n (a k a r ) n ] m=0 α mr (y 1 a r ) m = 0. (p l 1), r = 1,...,s. (5) (a k a r ) p l From (5) we obtain THEOREM 6. The solutions of the equation (3) have not a logarithmic singularities if and only if k r q r+1 q k (q k + 1) l=0 (q r l) α lr = 0. (6) (a k a r ) qr+1 l If s = 1 the equation (3) is none other tahn Euler equation and has the common solution µ = C 1 (y 1 a 1 ) q1+1 + C (y 1 a 1 ) q1 (q 1 > 0 is integer) which does not contain the logarithmic singularity for each value of a 1. For s = it follows from (5) and (6), that the common solution of the equation (3) has the logarithmic singularities for any values of a 1 and a. If s >, as is shown in for example (14), there are some distribution of the roots a i, for which the common solution of (3) has no the logarithmic singularities. R E F E R E N C E S 10

11 1. J. HADAMARD, Lectures on Cauchy s Problem in Linear Partial Differential Equations, New Haven, Yale University Press, N. H. IBRAGIMOV, Transformation Groups in Mathematical Physics, Nauka, Moscow, 1983 (in Russian). 3. P. GÜNTHER, Huygens Principle and Hyperbolic Equations, Acad. Press, Boston, CH. HUYGENS, Traite de la Lumiere. Van de Aa, Leyden, (English translation: Treaties of Light, 1960, reprint by Dover Publications, N.-Y., 196). 5. A. O. OGANESIAN, Generalization of Lagnese-Stellmacher s method on a class of equations satisfying Huygens principle in Russian], Izv. Akad. Nauk Armenii, Matematika English translation: Journal of Contemporary Math. Anal. (Armenian Academy of Sciences)], vol. 5, no. 5, pp , N. H. IBRAGIMOV, A. O. OGANESIAN, Hierarchy of Huygens equations in the spaces with a nontrivial conformal group in Russian], Uspekhi Math. Nauk, 46, no. 3, pp , J. E. LAGNESE, The Structure of a Class of Huygens Operators. J. Math. Mech., 18 (1969) No. 1, p WEISS, M. TABOR, G. CARNEVALE, The Painleve Property for Partial Differential Equations. J. Math. Phys., 4 (1983), p V. V. GOLUBEV, Lectures on Analitical Theory of Differential Equations (in Russian), M.,

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