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Electronic Journal of Differential Equations, Vol. 2003(2003), No. 103, pp. 1–8. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp) A CLASSICAL SOLUTION WEAKENED ON THE AXIS FOR A MIXED PROBLEM OF INHOMOGENEOUS HYPERBOLIC EQUATIONS RAID AL-MOMANI Abstract. The main purpose of this paper is to present sufficient conditions, on the forcing term of a mixed problem for three dimensional hyperbolic equations of any even order, for the existence of axially weakened classical solutions. 1. introduction p + + and G = {x ∈ R3 : 0 < r < R < ∞}. On the cylinder Let r = Q = G × [0, T ], we consider the mixed problem x21 x22 x23 m Y k=1 p ∂2 − a2k ∆ u + F (r, t; u) = f (r, t) ∂t2 (1.1) ∂ u = φp (r), 0 ≤ p ≤ 2m − 1 ∂tp t=0 ∂∆k u 1 k + ∆ u Γ = 0, 0 ≤ k ≤ m − 1 , ∂ν r (1.2) (1.3) ∂ is the partial derivative where Γ denotes the lateral surface of the cylinder Q, ∂ν with respect to the normal ν which is outward to the lateral surface Γ of the ∂2 ∂2 ∂2 cylinder Q, the reals a1 , . . . , an are different, ∆ = ∂x 2 + ∂x2 + ∂x2 is the Laplacian, ∆r ϕ = ∂2ϕ ∂r 2 1 + 2 r + ∂ϕ ∂r , 2 3 and F (r, t; u) = 3 X X ∂ 2p+1 ∆q u bp,q (r, t) xi ∂t2p ∂xi p+q<m i=1 (1.4) ∂ 2p+1 ∆q u ∂ 2p ∆q u + Cp+q (r, t) + d (r, t) . p,q ∂t2p+1 ∂t2p We assume that bp,q (R, t) = 0 (1.5) which is the consistency condition. We also assume that the coefficients bp,q , Cp,q , ∂C ∂d ∂b dp,q are continuous in Q and their derivatives ∂rp,q , ∂rp,q , ∂rp,q are bounded in Q. 2000 Mathematics Subject Classification. 35L70, 58J45. Key words and phrases. Weakened classical solutions, hyperbolic equations. c 2003 Texas State University-San Marcos. Submitted December 20, 2002. Published October 9, 2003. 1 2 RAID AL-MOMANI EJDE–2003/103 The central symmetry of the problem with respect to the variable x is reflected by the dependence of the data on it, in the invariance of the Laplacian with respect to rotation, and in the geometry of G. Before we state our results, let us dwell a moment on the existing literature concerning similar problems. First, we recall known facts concerning hyperbolic equations; as it is well known, hyperbolic equations do not keep smoothness. So, the space of continuous functions is badly adapted for investigating them; it is not suitable in the sense that for more than one space variable, we can not establish the agreement of necessary and sufficient conditions imposed on the initial functions for the existence of the classical solutions as it was shown by Sobolev [10] for the Cauchy problem of the wave equation. In contrast, Sobolev spaces are more adapted for investigating hyperbolic equations. But if the dimension is greater than the order of the operator, then solutions in Sobolev spaces may be discontinuous even at any point. Therefore, it arises the question of how to use the good quality of the spaces of continuous functions. Naturally, arises the question to localize the set of irregular points of the solutions. We will then make some compromise, we weaken the requirements on the sought for solutions; As, in our situation, they are classical except on the axis of the cylinder, we will impose some conditions there, and call the new defined solutions, weakened on the axis. Classical solution of mixed problems for hyperbolic equations has been the subject of extensive research. However, their common drawback, is that the sufficient conditions, imposed on the initial functions for the existence of a classical solution, disagree with the necessary ones, see, for example, the works of Steklov [11] and Petrovsky [9]. This limitation has been eliminated for one-dimensional hyperbolic equations; an agreement between necessary and sufficient conditions for the initial functions has been established in the case of mixed problems for the telegraph equation by Chernyatin [5,6] and Egorov [7]. Baranowskaya [2] extended this result to more general one-dimensional hyperbolic equations of second order. Also, it has been extended to one-dimensional hyperbolic equations of any even order in the work of Koku and Yurchuk [4]. For multidimensional hyperbolic equations, the situation is much more delicate. All the known sufficient conditions for the existence of classical solutions contain additional conditions for the initial functions. Moreover, Sobolev [10] proved that the requirements for the initial-value (Cauchy) problem for the wave equation could not be reduced by one derivative. The solution at the point r = 0 requires that Φ ∈ C 3 and Ψ ∈ C 2 in the case of central symmetry. Besov [3] arrived at similar results in terms of what is called now Besov spaces. An agreement between necessary and sufficient conditions for the initial functions has been established in the case of mixed problems for homogeneous three-dimensional hyperbolic equations of any even order in the case of central symmetry by Al-Momani [1] and Al-Momani and Yurchuk [12]. EJDE–2003/103 A CLASSICAL SOLUTION WEAKENED ON THE AXIS 3 2m By C{|x|=0} (Q) we denote the set of functions u ∈ C 1 (Q) ∩ C 2m (Q\{0} × [0, T ]) satisfying the conditions ∂ p ∆n u ∈ C 1 (Q), p + 2n < 2m − 1 ∂tp ∂ p+2 ∆n u ∈ C(Q), p + 2n < 2m − 2 ∂tp ∂r2 3 X ∂ p+1 ∆n u(x, t) lim xi = 0, p + 2n = 2m − 1 r→0 ∂tp ∂xi i=1 lim r r→0 ∂ p ∆n u(x, t) , ∂tp p + 2n = 2m . (1.6) (1.7) (1.8) (1.9) Definition. The weakened on the axis r = 0 classical solution in Q of the problem 2m (1.1)-(1.3) is a function in C{|x|=0} (Q), which satisfies equation (1.1) pointwise in Q\{0} × [0, T ] and satisfies the conditions (1.2), (1.3) in the usual sense. Next, we state a result from [1] which provides the existence of the weakened on the axis classical solution in Q of our problem (1.1)-(1.3) with f (r, t) = 0 in (1.1). Theorem 1.1. Let the coefficients bp,q , Cp,q , dp,q be continuous in Q and have bounded first partial derivatives in Q with respect to r and the condition (1.5) holds. The necessary and sufficient conditions for the existence of the weakened on the axis classical solution in Q of the problem m Y k=1 ∂2 − a2k ∆ u + F (r, t; u) = 0 ∂t2 with (1.2) and (1.3) are the following conditions on the initial functions φp : φp ∈ C 2m−p (0, R] (1.10) ∆nr φp ∈ C 2 [0, R], p + 2n < 2m − 2 (1.11) ∆nr φp p + 2n < 2m − 1 (1.12) 1 ∈ C (0, R], d n ∆ φp (r) = 0, p + 2n < 2m − 1 dr r lim r∆nr φp (r) = 0, p + 2n = 2m lim r r→0 r→0 d n 1 ∆r φp (r) + ∆nr φp (r) r=R = 0, dr r p + 2n < 2m (1.13) (1.14) (1.15) 2. Main result Theorem 2.1. Sufficient conditions for the existence of the weakened on the axis classical solution in Q of problem (1.1)-(1.3) are given by (1.10)-(1.15) on the initial functions φp , and the following conditions on the function f : f ∈ C(Q\{0} × [0, T ]), (2.1) lim rf (r, t) = 0, 0 ≤ t ≤ T r→0 Z R ∂f (r, t) 2 r2 | | dr ∈ C[0, T ]. ∂r 0 (2.2) (2.3) 4 RAID AL-MOMANI EJDE–2003/103 Proof. We transform (1.1)-(1.3) to the spherical coordinates and set ν(r, t) = ru(r, t) and use the equality ν(r, t) 1 ∂ 2 ν(r, t) = . r r ∂r2 Then the function ν(r, t) satisfies ∆r m Y 2 ∂2 2 ∂ − a ν + F (r, t; ν) = f1 (r, t) k ∂t2 ∂r2 k=1 (2.4) with the initial and boundary conditions ∂pu |t=0 = Φp (r), Φp (r) = rφp (r), 0 ≤ p ≤ 2m − 1 ∂tp ∂ 2k ν ∂ 2k+1 ν |r=0 = |r=R = 0, 0 ≤ k ≤ m − 1 , 2k ∂r ∂r2k+1 (2.5) (2.6) where F (r, t; ν) = X bp,q (r, t)r p+q<m ∂ 2p+2q+1 ν ∂t2p ∂r2q+1 ∂ 2p+2q ν ∂ 2p+2q+1 ν + d (r, t) − b (r, t) p,q p,q ∂t2p+1 ∂r2q ∂t2p ∂r2q and the functions Φp (r) = rφp (r) satisfy the conditions + Cp,q (r, t) Φp ∈ C 2m−p [0, R], Φ(2k+1) (R) p Φ(2n) (0) = 0, p 0 ≤ 2n ≤ 2m − p 0 ≤ 2k + 1 ≤ 2m − p . = 0, The function f1 (r, t) = rf (r, t) satisfies the conditions f1 ∈ C ([0, R] × [0, T ]) , f1 (0, t) = 0 Z R ∂f1 (r, t) 2 | dr ∈ C[0, T ] . | ∂r 0 As is shown in [4], under the conditions (2.1)-(2.3), problem (2.4)-(2.6) has a classical solution ν ∈ C 2m (Q) which satisfies ν(r, t) (−i) 1 X pn X (−i) = ak αi,k Φi (r + ak t) + (−1)i Φi (r − ak t) 2 0≤i≤2m−1 1≤k≤m Z t (1−2m) (1−2m) − α2m−1,k F (r + ak (t − τ ), τ ; ν) − F (r + ak (t − τ ), τ ; ν) dτ 0 Z t (1−2m) o (1−2m) + α2m−1,k f (r + ak (t − τ ), τ ) − f 1 (r − ak (t − τ ), τ ) dτ . 0 (2.7) Here, f (−k) r Z (r, τ ) = 0 (k) f 1 (r, F ) = (r − ζ)k−1 f (ζ, τ )dτ (k − 1)! 1 ∂k f (r, τ ), ∂rk 1 k≥0 (2.8) (2.9) EJDE–2003/103 A CLASSICAL SOLUTION WEAKENED ON THE AXIS 5 we denote by f 1 the 4R-periodic continuation with respect to r of the function f1 (r, t) from Q to R1 × [0, T ] which is constructed in the same way as Φ and F in [1]. Let ( 1, if i = k αi,k = 0, if i 6= k . Using (2.7), we will show that the function u(x, t) = ν(r,t) is the weakened on the r axis classical solution in Q of the problem (1.1)-(1.3). By substitution we can check that the function u satisfies at r = 0 the equation (1.1), initial conditions (1.2) and boundary conditions (1.3). Therefore, to complete the proof of our theorem it remains to prove that ν(r, t) , r u(x, t) = 2m u ∈ C{r=0} (Q) since ν ∈ C 2m (Q) the function u = νr ∈ C 2m (Q\{0} × [0, T ]). Now we must show that the function u satisfies the conditions (1.6)-(1.9). ∂ p+q ν Equation (2.7) implies that ∂t p ∂r q satisfies ∂ p+q ν 1 X pn = ak ∂tp ∂rq 2 1≤k≤m + − + (p+q−i) αi,k Φi (r + ak t) 0≤i≤2m−1 (p+q−i) (−1) Φi (r − ak t) Z t (p+q+1−2m) α2m−1,k F (r + ak (t − τ ), τ ; ν) 0 (p+q+1−2m) (−1)p+1 F (r − ak (t − τ ), τ ; ν) dτ p+i Z + α2m−1,k + X t (p+q+1−2m) [f 1 0 (p+q+1−2m) (r (−1)p+1 f 1 (r + ak (t − τ ), τ ) o − ak (t − τ ), τ )]dτ . The continuations Φi , ν, F and f 1 satisfy the following conditions: (p+q−i) Φi (r ± ak t) = 2m−p−q X j=0 (p+q−i+j) Φi (±ak t) rj + θ(r2m−p−q ) j! (2.10) where θ(r2m−p−q ) means a function g(r) such that limr→0 g(r)/(r2m−p−q ) = 0, Z t (p+q+1−2m) F (r ± ak (t − τ ), τ ; ν)dτ 0 = 2m−p−q X Z t F p+q+1−2m+j 0 j=0 rj (±ak (t − τ ), τ ; ν)dτ + θ(r2m−p−q ) , j! (2.11) and 1 Z f 1 ∈ C(R × [0, T ]), 0 t f 1 (r ± ak (t − τ ), τ )dτ ∈ C 1 (R1 × [0, T ]) . (2.12) 6 RAID AL-MOMANI EJDE–2003/103 By the Taylor formula with remainder in Peano form [8], Z t (p+q+1−2m) [f 1 (r ± ak (t − τ ), τ )dτ 0 = 2m−p−q X Z t (p+q+1−2m) f1 (±ak (t 0 j=0 rj − τ ), τ ) + θ(r2m−p−q ) j! (2.13) and (−k) f1 (−k) (r, τ ) = (−1)1+k f 1 (r, t). (2.14) From (2.3) and xi ∂u(r, t) ∂u(x, t) = ∂xi r ∂r and on the basis of (2.10)-(2.14), we have 2m−2n−p h X ∂ p ∆n u 1 X pn X (2n+p−i+j) j = a (1 − (−1) ) αi,k Φi (ak t) k ∂tp 2r j=0 0≤i≤2m−1 1≤k≤m Z t (2n+p+1−2m+j) − α2m−1,k F (ak (t − τ ), τ ; ν)dτ 0 Z t i rj o (2n+p+1−2m+j) f1 (ak (t − τ ), τ )dτ + θ(r2m−2n−p ) , + α2m−1,k j! 0 (2.15) xi ∂ p+1 ∆n u = 2 ∂tp ∂xi 2r h × p X apk n 2m−2n−p−1 X (1 + (−1)j )( j=0 1≤k≤m 1 1 − ) j! (j + 1)! (2n+p+1−i+j) αi,k Φi (ak t) X 0≤i≤2m−1 Z t − α2m−1,k F (2n+p+2−2m+j) (ak (t − τ ), τ ; ν)dτ 0 Z + α2m−1,k t (2n+p+2−2m+j) f1 (ak (t − τ ), τ )dτ 0 i rj o j! + θ(r2m−2n−p−1 ) (2.16) Equalities (2.15) and (2.16) show that X ph ∂ p ∆n u = ak r→0 ∂tp X lim 1≤k≤m (2n+p−i+1) αi,k Φi (ak t) 0≤i≤2m−1 Z t − α2m−1,k F (2n+p+2−2m) (ak (t − τ ), τ, ν)dτ 0 Z + α2m−1,k t (2n+p+2−2m) f1 (ak (t − τ ), τ )dτ 0 and ∂ p+1 ∆n u = 0. r→0 ∂tp ∂xi lim i (2.17) EJDE–2003/103 A CLASSICAL SOLUTION WEAKENED ON THE AXIS 7 Therefore, condition (1.6) holds. Equation (2.17) also implies that 3 1 X ∂ p+1 ∆n u 1 ∂ p+1 ∆n u lim = lim xi r→0 r 2 r→0 r ∂tp ∂r ∂tp ∂xi i=1 X 1 X p (2n+p+3−i) = ak [ αi,k Φi (ak t) 3 0≤i≤2m−1 1≤k≤m Z t (2n+p+4−2m) F (ak (t − τ ), τ, ν)dτ − α2m−1,k 0 Z t i (2n+p+4−2m) + α2m−1,k f1 (ak (t − τ ), τ )dτ . 0 Replacing n by n + 1 in equations (2.17) and (1.4) and taking into account that ∂ p+2 ∆n u ∂ p ∆n+1 u 2 ∂ p+1 ∆n u = − ∂tp ∂r2 ∂tp r ∂tp ∂r we obtain X ∂ p+2 ∆n u 1 X p (2n+p+3−i) = ak [ αi,k Φi (ak t) r→0 ∂tp ∂r 2 3 0≤i≤2m−1 1≤k≤m Z t (2n+p+4−2m) − α2m−1,k F (ak (t − τ ), τ, ν)dτ 0 Z t i (2n+p+4−2m) f1 (ak (t − τ ), τ )dτ + α2m−1,k lim 0 and consequently condition (1.7) holds. From (2.16) with p + 2n = 2m − 1, we have 3 X xi i=1 ∂ p+1 ∆n u = θ(ro ) ∂tp ∂xi (2.18) and from (2.15) with p + 2n = 2m, we have ∂ p ∆n u = θ(ro ) . (2.19) ∂tp Since limr→0 θ(ro ) = 0, so passing in (2.18) and (2.19) to the limit as r → 0, we show that the conditions (1.8) and (1.9) consequently hold. r References [1] R. 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Raid Al-Momani Department of Mathematics, Yarmouk University, Irbid, Jordan E-mail address: raid@yu.edu.jo