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
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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)
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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
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Raid Al-Momani
Department of Mathematics, Yarmouk University, Irbid, Jordan
E-mail address: raid@yu.edu.jo