CEJM 3(3) 2005 508–515
Radial-type complete solutions for a class of partial
differential equations
Ayşegül Çetinkaya1∗, Nuri Özalp2†
1
Department of Mathematics, Faculty of Arts and Sciences,
Gazi University,
40100, Kirsehir, Turkey
2
Department of Mathematics, Faculty of Sciences,
Ankara University,
Beşevler, 06100 Ankara, Turkey
Received 17 November 2004; accepted 1 June 2005
Abstract: We give some fundamental solutions of a class of iterated elliptic equations including
Laplace equation and its iterates.
c Central European Science Journals. All rights reserved.
Keywords: Complete solutions, iterated equations, Almansi’s expansion, Kelvin principle
MSC (2000): 35A08, 35C05, 35G99
1
Introduction
Much of the physical problems are solved in spherical or cylindrical domains. It means,
most of the time, that the solutions are symmetric functions with respect to a point or
with respect to an axis. That is why, when investigating the solutions, we see a frequent
use of the type of solutions in terms of a variable r defining a distance to a point. For some
of the research have been done for various type of problems, we refer to the references
[1-9]. Here in this study, we apply the idea to a class of linear partial differential equations
of second order and its iterates.
∗
†
Let us consider the class of equations
p n 2
X
1
∂u
γu
2∂ u
Lu =
xi 2 + αi xi
+ p =0
xi
∂xi
∂xi
r
i=1
E-mail: [email protected]
E-mail: [email protected]
(1)
A. Çetinkaya, N. Özalp / Central European Journal of Mathematics 3(3) 2005 508–515
509
where γ, αi (i = 1, 2, ..., n) are real parameters, p (> 0) is a real constant and r is defined
by
r p = xp1 + xp2 + ... + xpn .
(2)
The domain of the operator L is the set of all real valued functions u(x) of the class
C (D), where x = (x1 , x2 , ..., xn ) denotes points in Rn and D is a regularity domain of u
in Rn .
2
Almansi gave an expansion formula for the solutions of Laplace equation [1]. Lyakhov
and Ryzhkov, [5], obtained Almansi’s expansions for B- polyharmonic equation i.e. obtained the solutions of the equation
∆m
B f = 0,
where
∆B =
n
X
J=1
Bj +
N
X
∂2
,
2
∂x
i
i=n+1
Bj = ∂ 2 /∂x2j +
γj ∂
.
xj ∂x
Altın generalized the idea to a class of singular partial differential equations and obtained
a Lord Kelvin principle and radial type solutions for this class of equations [2,3,4,9]. In
[7], Özalp and Çetinkaya gave an expansion formula and Kelvin principle for the class of
equations given by (1). More precisely, they proved that if ui (x), i = 1, ..., k − 1 are any
k solutions of the equation (1), then the functions
w=
k−1
X
r jpui (x)
i,j=0
and
w=r
−φ
k−1
X
i=0
r ip ui
x
1
,
2
r
x2
xn ,
...,
r2
r2
are solutions of the iterated equation Lk u = 0 .
In [6], it is shown that, if u = f (r m ), f ∈ C 2 , then
r p L(u) =
′′
m2 v 2 f (v) + m (m − p + n(p − 1) +
Pn
i=1
′
αi ) vf (v) + γf (v) = 0
with v = r m . Since, this is an Euler equation, and the solutions of Euler equation are in
the form f (r m ) = r cm ,( where c is a root of the characteristic equation), we conclude that
equation (1) have solutions depending on powers of r m . We call these type of solutions as
r m − type solutions. Here, in this study we first give r m − type solutions for the iterates
of the equation (1) (i.e. for the equation Lk u = 0 ) and we generalize the idea to the
equations of the form
Lkq q ...Lk22 Lk11 u = 0
(3)
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A. Çetinkaya, N. Özalp / Central European Journal of Mathematics 3(3) 2005 508–515
where q, k1 , ..., kq are positive integers,
p n 2
X
∂
γv
1
(v)
2 ∂
Lv =
xi 2 + αi xi
+ p.
xi
∂xi
∂xi
r
i=1
(4)
(v)
Here, γv , αi (v = 1, 2, .., q; i = 1, 2, ...n) are real constants, and the operator Lkvv
denotes, as usual, the successive applications of the operator Lv onto itself, that is Lkvv u =
Lv (Lkvv −1 u).
2
rm type solutions for Lk u = 0
We borrow the following result from [7].
Lemma 2.1. For any real or complex parameter m,
Lk (r m ) =
k−1
Y
[(m − pj)(m − pj + 2ψ) + γ] r m−pk
(5)
j=0
where the integer k is the iteration number and
2ψ = −p + n(p − 1) +
n
X
αi
i=1
Now, we are ready to generate the r m type solutions of the iterated equation Lk u = 0.
Let
Φ(m) = m(m + 2ψ) + γ.
Then, equation (5) becomes
Lk (r m ) =
k−1
Y
Φ(m − pj)r m−pk .
j=0
On the other hand, for each j(= 0, 1, ...k − 1), the roots of the quadratic equations
Φ(m − pj) = 0 are

p
(1)


m
=
pj
−
ψ
+
ψ2 − γ

j


(6)



p

 m(2) = pj − ψ − ψ 2 − γ.
j
Thus, we can rewrite (5) as
k
m
L (r ) =
k−1
Yh
j=0
i
(1)
(2)
(m − mj )(m − mj ) r m−pk .
(7)
A. Çetinkaya, N. Özalp / Central European Journal of Mathematics 3(3) 2005 508–515
511
From (7), we conclude that
(1)
u = r mj ,
(2)
u = r mj
(j = 1, 2, ..., k − 1),
are solutions of Lk u = 0. Since the operator L is linear, by the superposition principle,
the function
k−1
X
(1)
(2)
(1)
(2)
u=
Aj r mj + Aj r mj
j=0
(1)
(2)
is a complete solution of the iterated equation Lk u = 0. Here Aj and Aj are arbitrary
constants. To generate the real valued solutions, we have three cases for the roots of the
quadratic equations Φ(m − pj) = 0 for each fixed j.
(1)
ψ 2 − γ > 0, i.e. mj
(i)
(2)
and mj
(1)
are two different real roots. In this case, r mj and
(2)
r mj are real valued linearly independent solutions of Lk u = 0.
(1)
(2)
(ii) ψ 2 − γ = 0, i.e. mj = mj = pj − ψ is a multiple real root. In this case, from
the theory of elementary differential equations we know that r pj−ψ and r pj−ψ ln r are real
valued linearly independent solutions of Lk u = 0.
(iii)
ψ 2 − γ < 0, i.e. the roots are complex conjugates. Once
p again from the
pj−ψ
theory of p
elementary differential equations we know that r
cos( γ − ψ 2 ln r) and
r pj−ψ cos( γ − ψ 2 ln r) are real valued linearly independent solutions of Lk u = 0.
Now let us define three auxiliary functions as


 1, if ψ 2 − γ > 0
Λ1 =
,

 0, else


 1, if ψ 2 − γ = 0
Λ2 =
,

 0, else


 1, if ψ 2 − γ < 0
Λ3 =
.

 0, else
Hence, we can give the following result:
Theorem 2.2. Real valued r m type solutions of the iterated equation Lk u = 0 is given
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A. Çetinkaya, N. Özalp / Central European Journal of Mathematics 3(3) 2005 508–515
by the formula
u=
k−1
X
j=0
+
(1)
(2)
(1)
(2)
Λ1 Aj r mj + Aj r mj
k−1
X
j=0
+
k−1
X
(1)
(2)
Λ2 r pj−ψ Bj + Bj ln r
Λ3 r
pj−ψ
j=0
(i)
(i)
(i)
where Aj , Bj , Cj
3
r
m
(8)
hp
i
hp
i
(1)
(2)
2
2
Cj cos
γ − ψ ln r + Cj sin
γ − ψ ln r
(i = 1, 2) are arbitrary constants.
kq
k2 k1
type solutions for Lq ...L2 L1 u = 0
Now let v and q be positive integers, 1 ≤ v ≤ q,
2ψv = −p + n(p − 1) +
n
X
(v)
αi
(9)
Φv (m) = m(m + 2ψv ) + γv
(10)
i=1
and
Then, for any positive integer kv , from (5) we have
Lkvv (r m )
=
kY
v −1
Φv (m − pj)r m−pkv
j=0
=
kY
v −1
[(m − pj)(m − pj + 2ψv ) + γv ] r m−pkv
(11)
j=0
Lemma 3.1. For any positive integer q, the following equality holds.
" q k −1
!#
v−1
v
YY
X
Pq
m
k2 k1
kq
Φv m − p(j +
kl )
r m−p( l=1 kl ) .
Lq ...L2 L1 (r ) =
v=1 j=0
Here, we assume that
P0
l=1
l=1
kl = 0.
Proof. We use induction argument on q. For q = 1, (12) is reduced to
Lk11 (r m )
=
kY
1 −1
j=0
Φ1 (m − pj)r m−pk1
(12)
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513
which gives (11) for v = 1. Now assume that (12) holds for q − 1. Now, for q, we get
kq−1
Lkq q ...Lk22 Lk11 (r m ) = Lkq q Lq−1
...Lk22 Lk11 (r m )
= Lkq q
"q−1 k −1
v
YY
Φv
m − p(j +
v=1 j=0
=
"q−1 k −1
v
YY
Φv
v−1
X
!
kl ) r
l=1
m − p(j +
v=1 j=0
v−1
X
kl )
l=1
!#
P
m−p( q−1
l=1 kl )
P q−1
Lkq q (r m−p(
l=1
#
kl )
)
kl )
)
On the other hand, from (11) we have
kq −1
P q−1
Lkq q (r m−p( l=1 kl ) )
=
Y
Φq (m − p(
j=0
q−1
X
P q−1
kl ) − pj)r m−p(
l=1
kl )−pkq
l=1
.
Thus,
"q−1 k −1
v
YY
Φv
Lkq q ...Lk22 Lk11 (r m ) =
m − p(j +
v=1 j=0
=
"q−1 k −1
v
YY
Φv
m − p(j +
Φq
m − p(
j=0
=
" q k −1
v
YY
v−1
X
kl )
!#
l=1
kq −1
×
kl )
!#
l=1
v=1 j=0
Y
v−1
X
q−1
X
kl ) − pj
l=1
Φv
m − p(j +
v=1 j=0
v−1
X
l=1
!
kl )
P q−1
Lkq q (r m−p(
P q−1
r m−p(
!#
l=1
kl )−pkq
Pq
r m−p(
l=1
l=1
kl )
.
Hence the proof is completed.
Since for any fixed v, the roots of the quadratic equation
!
v−1
X
Φv m − p(j +
kl ) = 0
l=1
are

p
Pv−1
(1)


m
=
p(j
+
k
)
−
ψ
+
ψv 2 − γv

l
v
j,v
l=1


we can rewrite (12) as
(13)



p

 m(2) = p(j + Pv−1 kl ) − ψv − ψv 2 − γv
j,v
l=1
" q k −1
#
v
YY
Pq
(1)
(2)
Lkq q ...Lk22 Lk11 (r m ) =
(m − mj,v )((m − mj,v ) r m−p( l=1 kl ) .
v=1 j=0
(14)
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Hence, we see that for each integer v, the functions
(1)
(2)
u = r mj,v , u = r mj,v
(j = 1, 2, ..., kv − 1),
kq
k2 k1
are solutions of the iterated equations Lq ...L2 L1 u = 0. Since the operator Lv is
linear for each v, by the superposition principle, the function
u=
kX
v −1
(1)
(1)
(2)
(2)
Aj,v r mj,v + Aj,v r mj,v
j=0
k
(1)
(2)
is a complete solution of the iterated equation Lq q ...Lk22 Lk11 u = 0. Here Aj,v and Aj,v
are arbitrary constants. Now let us define three auxiliary functions as


 1, if ψ 2 − γv > 0
v
Λ1,v =
,

 0, else


 1, if ψ 2 − γv = 0
v
Λ2,v =
,

 0, else


 1, if ψ 2 − γv < 0
v
Λ3,v =
.

 0, else
Hence, by generating the real valued solutions with similar arguments of the case Lk u = 0,
we can give the following result:
k
Theorem 3.2. Real valued r m type solutions of the iterated equation Lq q ...Lk22 Lk11 u =
0 is given by the formula
u=
q k−1
X
X
Λ1,v
v=1 j=0
+
q k−1
X
X
(2)
(1)
(2)
(1)
Aj r mj,v + Aj r mj,v
Λ2,v r p(j+
P v−1
kl )−ψv
Bj + Bj ln r
Λ3,v r p(j+
P v−1
kl )−ψv
Cj cos
v=1 j=0
+
q k−1
X
X
v=1 j=0
(i)
(i)
(i)
where Aj , Bj , Cj
4
l=1
l=1
(1)
(1)
(2)
(15)
hp
i
hp
i
(2)
γv − ψv2 ln r + Cj sin
γv − ψv2 ln r
(i = 1, 2) are arbitrary constants.
Conclusion
Theorem 3.2 gives the complete solutions of the iterated equations (3) which includes
a wide range of elliptic type equations with equidimensional equations. Thus, we can
compute the analytical (symbolic) solutions of (3) by using (15) in a suitable algorithm.
A. Çetinkaya, N. Özalp / Central European Journal of Mathematics 3(3) 2005 508–515
515
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