Electronic Journal of Differential Equations, Vol. 2011(2011), No. 04, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLUTION TO THE TRIHARMONIC HEAT EQUATION
WANCHAK SATSANIT
Abstract. In this article, we study the equation
∂
u(x, t) − c2 ~ u(x, t) = 0
∂t
with initial condition u(x, 0) = f (x). Where x is in the Euclidean space Rn ,
~=
p
p+q
“X
∂ 2 ”3 “ X ∂ 2 ”3
+
∂x2i
∂x2j
i=1
j=p+1
with p + q = n, u(x, t) is an unknown function, (x, t) = (x1 , x2 , . . . , xn , t) ∈
Rn × (0, ∞), f (x) is a generalized function, and c is a positive constant. Under
suitable conditions on f and u, we obtain a unique solution. Note that for
q = 0, we have the triharmonic heat equation
∂
u(x, t) − c2 ∆3 u(x, t) = 0 .
∂t
1. Introduction
It is well known that the heat equation
∂
u(x, t) = c2 ∆u(x, t),
∂t
(1.1)
with the initial condition u(x, 0) = f (x), has solution
Z
|x − y|2 1
f (y)dy ,
(1.2)
u(x, t) =
exp
−
4c2 t
(4c2 πt)n/2 Rn
Pn ∂ 2
where (x, t) = (x1 , x2 , . . . , xn , t) ∈ Rn × (0, ∞), and ∆ = i=1 ∂x
2 is the Laplace
i
operator. It is also known that the solution can be written as the convolution
u(x, t) = E(x, t) ∗ f (x), where
E(x, t) =
|x|2 1
exp
−
,
4c2 t
(4c2 πt)n/2
(1.3)
which is called the heat kernel [1, pp. 208-209]. Here |x|2 = x21 + x22 + · · · + x2n and
t > 0.
2000 Mathematics Subject Classification. 46F10, 46F12.
Key words and phrases. Fourier transform; tempered distribution; diamond operator.
c
2011
Texas State University - San Marcos.
Submitted June 14, 2010. Published January 7, 2011.
1
2
W. SATSANIT
EJDE-2011/04
In 1996, Kananthai [3] introduced the Diamond operator
p
p+q
X
∂ 2 2 X ∂ 2 2
♦=
−
,
∂x2i
∂x2j
i=1
j=p+1
with p + q = n .
(1.4)
This operator can be written in the form ♦ = ∆ = ∆, where
∂2
∂2
∂2
+
+
·
·
·
+
∂x21
∂x22
∂x2n
(1.5)
∂2
∂2
∂2
∂2
∂2
∂2
+
+
·
·
·
+
−
−
−
·
·
·
−
∂x21
∂x22
∂x2p
∂x2p+1
∂x2p+2
∂x2p+q
(1.6)
∆=
is the Laplacian, and
=
is the ultra-hyperbolic operator. The Fourier transform and the elementary solution
of the Diamond operator has been studied; see for example [3]. Nonlaopon and
Kananthai [5] studied the equation
∂
u(x, t) = c2 u(x, t),
∂t
and obtain the ultra-hyperbolic heat kernel
Pp x2 − Pp+q x2 iq
j=p+1 j
i=1 i
exp −
,
E(x, t) =
4c2 t
(4c2 πt)n/2
√
where p + q = n, and i = −1.
The purpose of this work is to study the equation
∂
u(x, t) − c2 ~ u(x, t) = 0,
(1.7)
∂t
with the initial condition u(x, 0) = f (x), for x ∈ Rn . The operator is
~=
=
p
p+q
X
∂ 2 3 X ∂ 2 3
+
∂x2i
∂x2j
i=1
j=p+1
p
p
p+q
p
p+q
X
X
∂2
∂ 2 h X ∂ 2 2 X ∂ 2 X ∂ 2 −
+
∂x2i
∂x2j
∂x2i
∂x2i
∂x2j
i=1
i=1
j=p+1
j=p+1
i=1
+
p+q
X
∂ 2 2 i
∂x2j
j=p+1
3
= ∆(∆2 − (∆ + )(∆ − ))
4
3
1 3
= ♦ + ∆
4
4
where ∆, , ♦ are defined by (1.5), (1.6) and (1.4) respectively.
Here, p + q = n, u(x, t) is an unknown function, (x, t) = (x1 , x2 , . . . , xn , t) is in
Rn × (0, ∞), f (x) is a generalized function, and c is a positive constant. We obtain
a solution u(x, t) = E(x, t) ∗ f (x), where
Z
p
p+q
h
h X
3 X
3 i
i
1
2
2
2
exp
−
c
ξ
+
ξ
t
+
i(ξ,
x)
dξ ,
(1.8)
E(x, t) =
i
j
(2π)n Ω
i=1
j=p+1
and Ω ⊂ Rn is the spectrum of E(x, t) for any fixed t > 0.
EJDE-2011/04
TRIHARMONIC HEAT EQUATION
3
Here E(x, t) is the elementary solution of (1.7), whose properties will be studied
in this article. If we put q = 0, then (1.7) reduces to the equation
∂
u(x, t) − c2 ∆3 u(x, t) = 0
∂t
which is related to the triharmaoic heat equation.
2. Preliminaries
Definition 2.1. Let f (x) ∈ L1 (Rn ), the space of integrable function in Rn . Then
the Fourier transform of f (x) is
Z
1
b
e−i(ξ,x) f (x) dx,
(2.1)
f (ξ) =
(2π)n/2 Rn
where ξ = (ξ1 , ξ2 , . . . , ξn ), x = (x1 , x2 , . . . , xn ) ∈ Rn , (ξ, x) = ξ1 x1 + ξ2 x2 + · · · +
ξn xn , and dx = dx1 dx2 . . . dxn . The inverse of Fourier transform is defined as
Z
1
ei(ξ,x) fb(ξ) dξ.
(2.2)
f (x) =
(2π)n/2 Rn
If f is a distribution with compact support by [6, Theorem 7.4-3], we can write
fb(ξ) =
1
hf (x), e−i(ξ,x) i.
(2π)n/2
(2.3)
Definition 2.2. The spectrum of the kernel E(x, t) in (1.6) is the bounded support
\t) for any fixed t > 0.
of the Fourier transform E(ξ,
Definition 2.3. Let ξ = (ξ1 , ξ2 , . . . , ξn ) be a point in Rn and let
2
2
2
Γ+ = {ξ ∈ Rn : ξ12 + ξ22 + · · · + ξp2 − ξp+1
− ξp+2
− · · · − ξp+q
> 0 and ξ1 > 0}
be the interior of the forward cone, and Γ+ denote the closure of Γ+ .
Let Ω be spectrum of E(x, t) defined by Definition 2.2 for any fixed t > 0, and
Ω ⊂ Γ+ . Let the Fourier transform of E(x, t) be

3 i
3 P
h
P
 1
p+q
p
2
2
2
−
c
t for ξ ∈ Γ+ ,
+
ξ
exp
ξ
n/2
j
i
j=p+1
i=1
\t) = (2π)
E(ξ,
0
for ξ ∈
/Γ .
+
(2.4)
Lemma 2.4. The Fourier transform of ~δ is
3
(−1)3
2
2
2
2
2
2
[(ξ
+
ξ
+
·
·
·
+
ξ
+ (ξp+1
+ ξp+2
+ · · · + ξp+q
)3 ]
F ~δ =
1
2
p
(2π)n/2
where F is defined by (2.1). Let the norm of ξ be kξk = (ξ12 + ξ22 + · · · + ξn2 )1/2 .
Then
M
kξk6 ,
|F ~ δ| ≤
(2π)n/2
where M is a positive constant. That is, F~ is bounded and continuous on the
space S 0 of the tempered distribution. Moreover, by (2.2),
~δ = F −1
1
2
2
2
[(ξ 2 + ξ22 + · · · + ξp2 )3 + (ξp+1
+ ξp+2
+ · · · + ξp+q
)3 ]
(2π)n/2 1
4
W. SATSANIT
EJDE-2011/04
Proof. By (2.3),
F ~δ =
=
=
=
=
1
h~δ, e−i(ξ,x) i
(2π)n/2
1
hδ, ~e−i(ξ,x) i
(2π)n/2
3
1
1 3 −i(ξ,x)
i
hδ,
♦
+
∆ e
4
4
(2π)n/2
1
3
1
1
hδ, ♦e−i(ξ,x) i +
hδ, ∆3 e−i(ξ,x) i
4
4
(2π)n/2
(2π)n/2
p
p+q
h X
2 X
2 i
1 D 3
3
2
2
−
(−1)
ξ
ξ
δ,
i
j
4
(2π)n/2
i=1
j=p+1
×
p
h X
ξi2
−
i=1
p+q
X
ξj2
i
e−i(ξ,x)
E
j=p+1
n
h X
i3
E
1 D 1
3
+
δ,
(−1)
ξi2
e−i(ξ,x)
n/2
4
(2π)
i=1
=
p+q
p
p+q
p
2 iih X
X
i
h X
2 X
1 h3
2
2
2
3
2
−
ξ
ξ
−
ξ
(−1)
ξ
j
i
j
i
(2π)n/2 4
j=p+1
i=1
j=p+1
i=1
n
+
h X i3 1 1
3
(−1)
ξi2
(2π)n/2 4
i=1
p+q
p
(−1)3 h X 2 3 X 2 3 i
+
ξj
ξ
(2π)n/2 i=1 i
j=p+1
3 i
3 (−1)3 h 2
2
2
2
2
2
=
+
ξ
+
ξ
+
·
·
·
+
ξ
.
ξ
+
ξ
+
·
·
·
+
ξ
p+1
p+2
p+q
1
2
p
(2π)n/2
=
Then
|F ~ δ|
3 1 2
2
2 3
2
2
2
=
ξ
+
ξ
+
·
·
·
+
ξ
+
ξ
+
ξ
+
·
·
·
+
ξ
1
2
p
p+1
p+2
p+q
(2π)n/2
1
≤
|ξ12 + · · · + ξn2 |(ξ12 + · · · + ξn2 )2 + (ξ12 + · · · + ξn2 )2 + (ξ12 + · · · + ξn2 )2 n/2
(2π)
M
kξk6 ,
≤
(2π)n/2
1/2
where kξk = ξ12 + ξ22 + · · · + ξn2
, ξi ∈ R (i = 1, 2, . . . , n). Hence we obtain F ~ δ
is bounded and continuous on the space S 0 of the tempered distribution.
Since F is a one-to-one transformation from the space S 0 of the tempered distribution to the real space R, by (2.2), we have
~δ = F −1
1
2
2
2
[(ξ 2 + ξ22 + · · · + ξp2 )3 + (ξp+1
+ ξp+2
+ · · · + ξp+q
)3 ].
(2π)n/2 1
This completes the proof.
EJDE-2011/04
TRIHARMONIC HEAT EQUATION
5
Lemma 2.5. Let
p
p+q
h X
∂ 2 3 X ∂ 2 3 i
∂
2
−c
+
,
L=
∂t
∂x2i
∂x2j
i=1
j=p+1
(2.5)
where
p
p+q
X
3
1
∂ 2 3 X ∂ 2 3
+
= ♦ + ∆3 ,
2
2
∂xi
∂xj
4
4
i=1
j=p+1
p + q = n, (x, t) = (x1 , x2 , . . . , xn , t) ∈ Rn × (0, ∞), and c is a positive constant.
Then
Z
p
p+q
h X
h
3 X
3 i
i
1
2
2
2
exp
−
c
ξ
+
E(x, t) =
ξ
t
+
i(ξ,
x)
dξ.
(2.6)
i
j
(2π)n Ω
i=1
j=p+1
is an elementary solution of (2.5).
Proof. Let E(x, t) be an elementary solution of operator L. Then
LE(x, t) = δ(x, t),
where δ is the Dirac-delta distribution. Thus
p
p+q
h X
∂
∂ 2 3 X ∂ 2 3 i
2
E(x, t) − c
+
E(x, t) = δ(x)δ(t).
∂t
∂x2i
∂x2j
i=1
j=p+1
Taking the Fourier transform on both sides of the equation, we obtain
p
p+q
h X
3 X
3 i
1
∂ \
\t) =
δ(t).
E(ξ, t) + c2
ξi2 +
ξj2
E(ξ,
∂t
(2π)n/2
i=1
j=p+1
Thus
p+q
p
3 i
h
X
3 X
2
2
\t) = H(t) exp − c2
E(ξ,
+
ξ
t
ξ
j
i
(2π)n/2
j=p+1
i=1
where H(t) is the Heaviside function. Since H(t) = 1 for t > 0. Therefore,
\t) =
E(ξ,
p
p+q
h
X
3 X
3 i
1
2
2
2
exp
−
c
ξ
+
ξ
t
i
j
(2π)n/2
i=1
j=p+1
which by (2.3), we obtain
E(x, t) =
1
(2π)n/2
Z
\t) dξ =
ei(ξ,x) E(ξ,
Rn
1
(2π)n/2
Z
\t) dξ
ei(ξ,x) E(ξ,
Ω
where Ω is the spectrum of E(x, t). Thus from (2.2),
E(x, t) =
1
(2π)n
Z
Ω
This completes the proof.
p
p+q
h
h X
3 X
3 i
i
exp − c2
ξi2 +
ξj2
t + i(ξ, x) dξ.
i=1
j=p+1
6
W. SATSANIT
EJDE-2011/04
3. Main Results
Theorem 3.1. Consider the equation
∂
u(x, t) − c2 ~ u(x, t) = 0
∂t
(3.1)
u(x, 0) = f (x)
(3.2)
with initial condition
and the operator
~=
=
p+q
p
X
∂ 2 3 X ∂ 2 3
+
∂x2i
∂x2j
j=p+1
i=1
p
p
p
p+q
p+q
X
X
∂2
∂ 2 h X ∂ 2 2 X ∂ 2 X ∂ 2 −
+
∂x2i
∂x2j
∂x2i
∂x2i
∂x2j
i=1
j=p+1
i=1
i=1
j=p+1
+
=
p+q
X
∂ 2 2 i
∂x2j
j=p+1
3
1
♦ + ∆3
4
4
where p + q = n, k is a positive integer, u(x, t) is an unknown function for (x, t) =
(x1 , x2 , . . . , xn , t) ∈ Rn × (0, ∞), f (x) is a given generalized function, and c is a
positive constant. Then
u(x, t) = E(x, t) ∗ f (x)
as a solution of (3.1)-(3.2), where E(x, t) is given by (2.5).
Proof. Taking the Fourier transform on both sides of (3.1), we obtain
p+q
p
3 i
h X
3 X
∂
2
2
ξj2
u
b(ξ, t) = 0,
u
b(ξ, t) + c
ξi +
∂t
j=p+1
i=1
(see Lemma 2.4). Thus
p
p+q
h
X
3 X
3 i
u
b(ξ, t) = K(ξ) exp − c2
ξi2 +
ξj2
t
i=1
(3.3)
j=p+1
where K(ξ) is constant and u
b(ξ, 0) = K(ξ). By (3.2) we have
Z
1
K(ξ) = u
b(ξ, 0) = fb(ξ) =
e−i(ξ,x) f (x) dx
(2π)n/2 Rn
(3.4)
and by the inversion in (2.2), (3.3) and (3.4) we obtain
u(x, t)
Z
1
=
ei(ξ,x) u
b(ξ, t) dξ
(2π)n/2 Rn
Z Z
p
p+q
h
X
3 X
3 i
1
i(ξ,x) −i(ξ,y)
2
2
e
e
f
(y)
exp
−
c
ξ
+
ξj2
t dy dξ.
=
i
n
(2π) Rn Rn
i=1
j=p+1
EJDE-2011/04
TRIHARMONIC HEAT EQUATION
7
Thus
1
(2π)n
Z
1
u(x, t) =
(2π)n
Z
u(x, t) =
Z
Rn
Rn
p
p+q
X
h
3 X
3 i
ei(ξ,x−y) exp − c2
ξi2 +
ξj2
t f (y) dy dξ
i=1
j=p+1
or
Z
h
2
exp −c
Rn
p
h X
Rn
ξi2
p+q
3 X
3 i
i
+
t+i(ξ, x−y) f (y) dy dξ.
ξj2
i=1
j=p+1
(3.5)
Set
E(x, t) =
1
(2π)n
Z
Rn
p
p+q
h X
h
3 X
3 i
i
exp − c2
ξi2 +
ξj2
t + i(ξ, x) dξ.
i=1
We choose Ω ⊂ Rn be the spectrum of E(x, t) and by (2.5), we have
Z
p+q
p
3 i
i
h
h X
3 X
1
2
2
2
+
ξ
t
+
i(ξ,
x)
dξ
exp
−
c
ξ
E(x, t) =
j
i
(2π)n Rn
j=p+1
i=1
1
=
(2π)n
Z
h
2
exp − c
p
h X
Ω
(3.6)
j=p+1
ξi2
3
+
p+q
X
ξj2
3 i
(3.7)
i
t + i(ξ, x) dξ.
j=p+1
i=1
Thus (3.5) can be written in the convolution form
u(x, t) = E(x, t) ∗ f (x).
Since E(x, t) exists,
1
lim E(x, t) =
t→0
(2π)n
Z
i(ξ,x)
e
Ω
1
dξ =
(2π)n
Z
ei(ξ,x) dξ = δ(x),
(3.8)
Rn
for x ∈ Rn ; see [3, Eq. (10.2.19b)]. Thus for the solution u(x, t) = E(x, t) ∗ f (x) of
(3.1),
lim u(x, t) = u(x, 0) = δ ∗ f (x) = f (x)
t→0
which satisfies (3.2).
Theorem 3.2. The kernel E(x, t) defined by (3.7) has the following properties:
(1) E(x, t) ∈ C ∞ for x ∈ Rn and t > 0, the space of function with infinitely
many continuous derivatives.
(2) For t > 0,
p
p+q
∂
h X
∂ 2 3 X ∂ 2 3 i
− c2
+
E(x, t) = 0 .
∂t
∂x2i
∂x2j
i=1
j=p+1
(3) E(x, t) > 0 for t > 0.
(4) For t > 0,
|E(x, t)| ≤
22−n M (t)
,
π n/2 Γ( p2 )Γ( 2q )
where M (t) is a function of t in the spectrum Ω, and Γ denotes the Gamma
function. Thus E(x, t) is bounded for any fixed t > 0.
(5) limt→0 E(x, t) = δ.
8
W. SATSANIT
EJDE-2011/04
Proof. (1) From (3.7),
Z
p
p+q
h X
h
3 X
3 i
i
∂n
∂n
1
2
2
E(x,
t)
=
exp
−
c
ξ
+
ξj2
t + i(ξ, x) dξ.
i
n
n
n
∂x
(2π) Ω ∂x
i=1
j=p+1
Thus E(x, t) ∈ C ∞ for x ∈ Rn and t > 0.
(2) By a computation,
p
p+q
h X
3 X
3 i
∂
2
2
E(x, t) = 0.
−c
ξi +
ξj2
∂t
i=1
j=p+1
(3) E(x, t) > 0 for t > 0 is obvious by (3.7).
(4) We have
Z
p
p+q
h
h X
3 X
3 i
i
1
2
2
2
E(x, t) =
+
exp
−
c
t
+
i(ξ,
x)
dξ,
ξ
ξ
i
j
(2π)n Ω
i=1
j=p+1
|E(x, t)| ≤
1
(2π)n
Z
p+q
p
3 i
h
X
3 X
ξj2
dξ.
exp − c2
ξi2 +
Ω
i=1
j=p+1
By changing to bipolar coordinates
ξ1 = rω1 ,
ξ2 = rω2 ,
...,
ξp = rωp ,
ξp+1 = sωp+1 , ξp+2 = sωp+2 , . . . , ξp+q = sωp+q ,
Pp
Pp+q
where i=1 ωi2 = 1 and j=p+1 ωj2 = 1. Thus
Z
1
exp[−c2 (s6 + r6 )t]rp−1 sq−1 dr ds dΩp dΩq
|E(x, t)| ≤
(2π)n Ω
where dξ = rp−1 sq−1 dr ds dΩp dΩq , dΩp and Ωq are the elements of surface area of
the unit sphere in Rp and Rq respectively. Since Ω ⊂ Rn is the spectrum of E(x, t)
and we suppose 0 ≤ r ≤ R and 0 ≤ s ≤ T where R and T are constants. Thus we
obtain
Z Z
Ωp Ωq R T
2 6
6
|E(x, t)| ≤
exp[−c
(s
+
r
t]rp−1 sq−1 ds dr
(2π)n 0 0
Ωp Ωq
=
M (t)
(2π)n
22−n M (t)
= n/2 p
Γ( 2 )Γ( 2q )
π
for any fixed t > 0 in the spectrum Ω, where
Z RZ T
M (t) =
exp[−c2 (s6 + r6 )t]rp−1 sq−1 ds dr
0
0
p/2
is a function of t, Ωp = 2π /Γ( p2 ) and Ωq = 2π p/2 /Γ( 2q ). Thus, for any fixed
t > 0, E(x, t) is bounded.
(5) This statement follows from (3.8).
Acknowledgements. The authors would like to thank The Thailand Research
Fund and Graduate School, Maejo University, Chiang Mai, Thailand for financial
support and also Prof. Amnuay Kananthai Department of Mathematics, Chiang
Mai University for many helpful of discussion.
EJDE-2011/04
TRIHARMONIC HEAT EQUATION
9
References
[1] R. Haberman; Elementary Applied Partial Differential Equations, 2-nd Edition, Prentice-Hall
International, Inc. (1983).
[2] F. John; Partial Differential Equations, 4-th Edition, Springer-Verlag, New York, (1982).
[3] A. Kananthai; On the Fourier Transform of the Diamond Kernel of Marcel Riesz, Applied
Mathematics and Computation 101:151-158 (1999).
[4] A. Kananthai; On the Solution of the n-Dimensional Diamond Operator, Applied Mathematics and Computational 88:27-37 (1997).
[5] K. Nonlaopon, A. Kananthai; On the Ultra-hyperbolic heat kernel, Applied Mathematics Vol.
13 No. 2 (2003), 215-225.
[6] A. H. Zemanian; Distribution Theory and Transform Analysis, McGraw-Hill, New York,
1965.
Wanchak Satsanit
Department of Mathematics, Faculty of Science, Maejo University, Chiangmai, 50290,
Thailand
E-mail address: aunphue@live.com