Mathematical and Computer Modelling 49 (2009) 327–336
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A note on defining singular integral as distribution and partial differential equations with convolution term Adem Kılıçman ∗ , Hassan Eltayeb Department of Mathematics, University Malaysia Terengganu, 21030 Kuala Terengganu, Terengganu, Malaysia
article
a b s t r a c t
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Article history: Received 17 July 2007 Received in revised form 4 May 2008 Accepted 28 May 2008
In this study first we consider the singular integrals as generalized functions in two dimensions and then we solve the non-homogeneous wave equation with convolutional term by using the generalized functions as boundary conditions. © 2008 Elsevier Ltd. All rights reserved.
Keywords: Double Laplace transform Double inverse Laplace transform Generalized partial derivative
1. Introduction Integral transforms are extensively used in solving boundary value problems and integral equations. The problem related to partial differential equation commonly can be solved by using a special integral transform thus many authors solved the boundary value problems by using single Laplace transform. Furthermore, double Laplace transforms in classical sense for solving linear second-order partial differential equations were used by Estrin and Higgins [10], Brychkov et al. see [12], Sneddon et al. [5], and recently, in [4,2], Kılıçman and Hassan solved linear second-order partial differential equation with variable coefficient by using the double Laplace transform. In this paper, we first rewrite generalized partial derivatives by using double Laplace transform and transform the one-dimensional pseudo-functions to two-dimensional functions. Then, a non-homogeneous wave equation with variable coefficients is solved by applying the double Laplace transform. The non-homogeneous wave equationPhas P a double m n j i convolution term in the non-homogeneous part and variable coefficients are of the following form k(x, t ) = j=1 i =1 x t . Higgins defines the double Laplace transform and the first-order partial derivative as Lx Lt [f (x, s)] = F (p, s) =
∞
Z
e−px 0
Lx Lt
∞
Z
e−st f (x, t )dtdx, 0
∂ f (x, t ) = pF (p, s) − F (0, s) ∂x
and the double Laplace transform for second partial derivative with respect to x and t are given by
∂ 2 f ( x, t ) ∂ F (0, s) = p2 F (p, s) − pF (0, s) − ∂ 2x ∂x 2 ∂ f ( x, t ) ∂ F (p, 0) Lx Lt = s2 F (p, s) − sF (p, 0) − ∂ 2t ∂t respectively, where x, t > 0 and p, s complex value.
Lx Lt
∗
Corresponding author. E-mail addresses:
[email protected] (A. Kılıçman),
[email protected] (H. Eltayeb).
0895-7177/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2008.05.048
(1.1)
328
A. Kılıçman, H. Eltayeb / Mathematical and Computer Modelling 49 (2009) 327–336
Similarly, the double Laplace transform of a mixed partial derivative can be deduced from single Laplace transform as
Lx Lt
∂ 2 f (x, t ) = psF (p, s) − pF (p, 0) − sF (0, s) − F (0, 0). ∂ x∂ t
(1.2)
Throughout this paper we use the notations pf for the pseudo-function and FP for the Hadamard finite parts respectively. Now we consider D as the space of infinitely differentiable functions with compact support and D 0 the space of distributions defined on D . Then a distribution f ∈ D 0 is given by
Z
hf , φi =
f (x)φ(x)dx,
for all φ ∈ D , see details in [1]. Now if f is a singular distribution, Estrada and Fulling in [9], defined the singular distribution as follows F () =
Z
f (x)φ(x)dx = F0 () + F1 (), |x−x0 |>
where F0 () is known as Hadamard finite part and infinite part F1 () is also chosen so that lim→0 F0 () exists, since any finite −α piece of F0 could be moved to F1 without violating lim→0 F0 (). If Hadamard’s prescription is applied to f (x) = x H (x) H (x) xα
then defines a distribution as pf
for all α , where H is Heaviside function. The derivative of these distributions for
k ∈ Z+ is d
pf
dx
H ( x)
= −kpf
xk
H (x)
+
xk+1
R∞
(−1)k δ k (x) , k!
see [6,8].
3
The finite part of divergent integral 0 t − 2 φ(t )dt can also be defined as a singular distribution for all φ ∈ R+ , the Hadamard finite part FP of divergent integral may be given by ∞
Z FP
t
− 32
φ(t )dt =
b
Z
0
1
1
t − 2 ψ(t ) dt − 2φ(0)b− 2 ,
(1.3)
0 00
where supp φ ⊂ [0, b) and ψ(t ) = φ 0 (0) + 2t! φ (0) + · · · +
t n−1 n!
n
φ (ζ ), 0 < ζ < 1 for more details see [7].
2. Generalized partial derivatives and double Laplace transform In this section we are going to apply double Laplace transform to second-order generalized partial derivative. Single Laplace transform of delta function is given in [6,8,11] by Lt [δ(t − a)] =
∞
Z
e−st δ(t − a)dt = e−sa ,
(2.1)
0
and the derivative of delta function followed by Lt [δ (t − a)] = 0
∞
Z
e
δ (t − a)dt = −
−st 0
0
d dt
e
−st
= se−sa .
t =a
We extend the single Laplace transform to multiple Laplace transform as follows: Lx Lt [δ(t − a)δ(x − b)] =
∞
Z
e−px 0
∞
Z
e−st δ(t − a)δ(x − b)dtdx = e−sa−pb , 0
and also partial derivative with respect to x and t as
Lx Lt
2 ∂ ∂ ∂ δ(t − a) δ(x − b) = e−st −px = pse−sa−pb . ∂t ∂x ∂ x∂ t t =a,x=b
In general multiple Laplace transform of delta function in n-dimensional case is given by Ltn [δ(t1 − a1 )δ(t2 − a2 ) . . . δ(tn − an )] = e−s1 a1 −s2 a2 ···−sn an ,
(2.2)
where Ltn means multiple Laplace transform in n dimensional. Kanwal defined the classical derivative of a piecewise-defined function f (t ) =
g2 (t ), g1 (t ),
t >a t 0, g1 , g2 are continuously differentiable functions and its derivative defined as f 0 (t ) = g10 (t )H (a − t ) + g20 (t )H (t − a) for all t 6= a see [8]. Next, we extend Kanwal’s result from single variable to two variables as f ( x, t ) =
g2 (x, t ), g1 (x, t ),
x > a, t > b x < a, t < b.
(2.3)
The latter function can be written in the form of f (x, t ) = g1 (x, t )H (a − x)H (b − t ) + g2 (x, t )H (x − a)H (t − b)
(2.4)
where H Heaviside function a > 0 and b > 0, g1 , g2 are continuously differentiable function and the classical partial derivative with respect to t is given by ft =
∂ g 1 ( x, t ) ∂ g 2 ( x, t ) H (a − x)H (b − t ) + H (x − a)H (t − b). ∂t ∂t
(2.5)
If we take the partial derivative of (2.5) with respect to x, we obtain ftx =
∂ 2 g1 (x, t ) ∂ 2 g 2 ( x, t ) H (a − x)H (b − t ) + H (x − a)H (t − b) ∂t∂x ∂t∂x
(2.6)
since H 0 = δ and the second partial derivative with respect to x is given by
∂ 2 g1 (x, t ) ∂ 2 g2 (x, t ) H ( a − x ) H ( b − t ) + H (x − a)H (t − b), ∂ x2 ∂ x2 and the second-order partial derivative of f (x, t ) with respect to t is given by fxx =
ftt =
(2.7)
∂ 2 g2 (x, t ) ∂ 2 g1 (x, t ) H (a − x)H (b − t ) + H (x − a)H (t − b) 2 ∂t ∂t2
(2.8) −
for all x 6= a and t 6= b. If the generalized partial derivative of Eq. (2.4) with respect to x denoted by f x then: −
f x (x, t ) =
∂ g 1 ( x, t ) H (a − x)H (b − t ) − g1 (x, t )δ(a − x)H (b − t ) ∂x ∂ g 2 ( x, t ) H (x − a)H (t − b) + g2 (x, t )δ(x − a)H (t − b). + ∂x
(2.9)
Similarly, −
f xt (x, t ) =
∂ 2 g1 (x, t ) ∂ 2 g2 (x, t ) ∂ g2 (x, t ) H (a − x)H (b − t ) + H (x − a)H (t − b) + H (x − a)δ(t − b) ∂ x∂ t ∂ x∂ t ∂x ∂ g 1 ( x, t ) ∂ g 2 ( x, t ) ∂ g1 (x, t ) − H (a − x)δ(b − t ) + δ(x − a)H (t − b) − δ(a − x)H (b − t ) ∂x ∂t ∂t + g2 (x, t )δ(x − a)δ(t − b) + g1 (x, t )δ(a − x)δ(b − t ),
(2.10)
and −
f xx (x, t ) =
∂ 2 g1 (x, t ) ∂ 2 g 2 ( x, t ) ∂ g1 (x, t ) H (a − x)H (b − t ) + H (x − a)H (t − b) − 2 δ(a − x)H (b − t ) 2 ∂x ∂ x2 ∂x ∂ g 2 ( x, t ) ∂δ(a − x) ∂δ(x − a) +2 δ(x − a)H (t − b) + g1 (x, t ) H (b − t ) + g2 (x, t ) H (t − b) ∂x ∂x ∂x
(2.11)
and generalized second partial derivative with respect to t is given by −
f tt (x, t ) =
∂ 2 g1 (x, t ) ∂ 2 g2 (x, t ) ∂ g1 (x, t ) H (a − x)H (b − t ) + H (x − a)H (t − b) − 2 H (a − x)δ(b − t ) 2 2 ∂t ∂t ∂t ∂ g2 (x, t ) ∂δ(b − t ) ∂δ(t − b) +2 H (x − a)δ(t − b) + g1 (x, t )H (a − x) + g2 (x, t )H (x − a) . ∂t ∂t ∂t
(2.12)
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A. Kılıçman, H. Eltayeb / Mathematical and Computer Modelling 49 (2009) 327–336
The Laplace transform of Eq. (2.7) with respect to x follows that Lx [fxx ] = H (b − t )
Z
a
e−px
0
∂ 2 g1 (x, t ) dx + H (t − b) ∂ x2
∞
Z
e−px a
∂ 2 g2 (x, t ) dx ∂ x2
(2.13)
where the integration by parts is applied to the right-hand side of the equality to obtain
Z a ∂ g1 (0, t ) −px −pa 2 e g1 (x, t )dx Lx [fxx ] = H (b − t ) e − + pe g1 (a, t ) + H (b − t ) −pg1 (0, t ) + p ∂x ∂x 0 Z ∞ ∂ g 2 ( a, t ) e−px g2 (x, t )dx. (2.14) + H (t − b) −e−pa − pe−pa g2 (a, t ) + H (t − b)p2 ∂x a
−pa ∂ g1 (a, t )
By taking Laplace transform of the Eq. (2.14) with respect to t leads to the double Laplace transform as Lt Lx [fxx ] = p F (p, s) + pe 2
−pa
b
Z
e
−st
g1 (a, t )dt −
e
b
Z
e−st 0
−st
g2 (a, t )dt
b
0
+ e−pa
∞
Z
∂ g 1 ( a, t ) dt − ∂x
∞
Z
e−st
b
Z b Z b ∂ g2 (a, t ) ∂ g1 (0, t ) e−st e−st g1 (0, t )dt (2.15) dt − dt − p ∂x ∂x 0 0
since g1 , g2 ∈ C . In particular if a = 0, b = 0 and x, t > 0 in Eq. (2.15) it is easy to see that the Eq. (2.15) gives double Laplace transform of second-order partial derivative with respect to x in classical sense as 1
∂ g2 (0, s) − pg2 (0, s). ∂x
Lt Lx [fxx ] = p2 F (p, s) −
(2.16)
Similarly, the double Laplace transform of Eq. (2.8) with respect to x, t , is obtained Lx Lt [ftt ] = s2 F (p, s) + se−sb
a
Z
e−px g1 (x, b)dx −
∞
Z
0
e−px g2 (x, b)dx
a
Z ∞ ∂ g2 (x, b) ∂ g1 (x, b) dx − e−px dx + e−sb e−px ∂t ∂t a 0 Z a Z a ∂ g ( x , 0 ) 1 − e−px dx − s e−px g1 (x, 0)dt . ∂t 0 0 Z
a
(2.17)
In particular if a = 0, b = 0 and x, t > 0 in Eq. (2.17) then it follows that the double Laplace transform of second-order partial derivative with respect to t in classical sense as Lx Lt [ftt ] = s2 F (p, s) − sg2 (p, 0) −
∂ g2 (p, 0) . ∂t
Finally, double Laplace transform of the mixed partial derivative of the Eq. (2.6) is obtained as follows Lt Lx [ftx ] = psF (p, s) + e−pa e−sb [g1 (a, b) + g2 (a, b)] + g1 (0, 0) − e−pa g1 (a, 0)
− e−sb g1 (0, b) + se−pa
b
Z
e−st g1 (a, t )dt − 0
+ pe
−sb
e
−px
e−st g2 (a, t )dt
b
a
Z
∞
Z
g1 (x, b)dx −
0
∞
Z
e
−px
g2 (x, b)dx − s
a
Z
b
−st
e
g1 (0, t )dt − p
0
Z
a
e−px g1 (x, 0)dx.
(2.18)
0
In particular if a = 0, b = 0 and x, t > 0 Eq. (2.18) becomes Lt Lx [ftx ] = psF (p, s) + g2 (0, 0) − s
∞
Z
−st
e
g2 (0, t )dt − p
0
Z
∞
e−px g2 (x, 0)dx
(2.19)
0
and the Eq. (2.19) can be written in the form of Lt Lx [ftx ] = psF (p, s) − sF (0, s) − pF (p, 0) + g2 (0, 0)
(2.20)
that is, the Eq. (2.20) is the double Laplace transform in classical sense for mixed partial derivative with respect to x and t. Example 1. Find the double Laplace transform for a regular generalized function f (x, t ) = H (t ) ⊗ H (x) ln(t ) ln(x)
(2.21)
A. Kılıçman, H. Eltayeb / Mathematical and Computer Modelling 49 (2009) 327–336
331
and the mixed derivatives
∂2 ∂2 [H (t ) ⊗ H (x) ln(t ) ln(x)] = pf f (x, t ) = ∂ x∂ t ∂ x∂ t
H (t ) ⊗ H (x)
xt
(2.22)
where H (x, t ) = H (t ) ⊗H (x) is a Heaviside function and ⊗ is a tensor product. The double Laplace transform with respect to x, t of (2.21) becomes Lx Lt [f (x, t )] =
∞
Z
e−px ln(x)
e−st ln(t )dtdx
0
0
=−
∞
Z
∞
Z
1 s
1
e−px ln(x) [γ + ln s] dx =
sp
0
γ 2 + ln(p) ln(s)
(2.23)
where γ is Euler’s constant see [8,3]. Thus, Lx Lt [f (x, t )] =
1
γ 2 + ln(p) ln(s)
sp
where Re s, p > 0.
The double Laplace transform of (2.22) with respect to x and t is obtained
Lx Lt
∂2 f (x, t ) = Lx Lt [H (t )H (x) ln(t ) ln(x)] ∂ x∂ t 1 2 = ps γ + ln(p) ln(s) = γ 2 + ln(p) ln(s). sp
Now let L be the space of test functions that Laplace transformable and the dual space of L is L0 , see [8], then a linear continuous functional over the space L of test functions is called a distribution of exponential growth. Example 2. Let us define a function xα + t β + = H (x) ⊗ H (t )xα t β where α, β 6= −1, −2, . . . then one can easily show that xα + t β + ∈ L0 and the double Laplace transform given by
Lx Lt
h
α
x +t
β
i +
∞
Z
α −px
=
∞
Z
x e 0
t β e−st dtdx.
0
By letting u = px and v = st for p, t > 0, it follows that Lx Lt
h
i
xα + t β +
pα+1 sβ+1 1
=
∞
Z
1
=
pα+1 sβ+1
uα e−u
∞
Z
0
v β e−v dv du,
0
Γ (α + 1)Γ (β + 1).
(2.24)
In particular if α, β = 0 the Eq. (2.24) becomes Lx Lt [H (x) ⊗ H (t )] =
∞
Z
e−px
∞
Z
0
e−st dtdx = 0
1 ps
.
3. Distribution defined by divergent integrals In this section, the idea for one-dimensional pseudo-function is extended to two-dimensional pseudo-function that is of the following form f ( x, y ) =
=
x−n y−n , 0,
x, y > 0, x, y < 0,
H ( x, y ) xn y n
,
(3.1)
where n is positive integer and H (x, y) =
n
1, x, y > 0 0, x, y < 0
in the form of tensor product as H (x, y) = H (x)⊗ H (y). We first consider
a simple distribution for n = 1
hf (x, y), φ(x, y)i =
∞
Z
−∞ Z ∞
= 0
∞
H (x) ⊗ H (y)
−∞
xy
Z 1 y
∞
Z 0
1 x
φ(x, y)dxdy,
φ(x, y) dx dy.
(3.2)
332
A. Kılıçman, H. Eltayeb / Mathematical and Computer Modelling 49 (2009) 327–336
where φ(x, y) is a test function(infinitely differentiable) on R2 and has a compact support on R2+ . The Taylor series expansion of φ(x, y) around (0; 0) is
φ(x, y) = φ(0, 0) + yφy (0, 0) + xφx (0, 0) + xyψ(x, y)
(3.3)
where ψ(x, y) is given by 1
ψ(x, y) =
2
1
xy−1 φxx (0, 0) + φxy (0, 0) +
2
yx−1 φyy (0, 0) + · · · +
xn−1−k yk−1 ∂ n φ(t , t )
(n − k)!k! ∂ xn−k ∂ yk
for 0 < t < 1.
(3.4)
Since the inner integral in (3.2) has a singularity at x = 0, an improper integral form of the integral for ε > 0 is introduced as ∞
Z
1
a
Z
ε→0 ε
x
0
φ(x, y)dx = lim
1 x
φ(x, y)dx
= lim φ(0, 0) ln a − φ(0, 0) ln ε + yφy (0, 0) ln a ε→0
+ lim −yφy (0, 0) ln ε + aφx (0, 0) − εφx (0, 0) + lim ε→0
Z
a
yψ(x, y)dx.
ε→0 ε
(3.5)
Similarly, for the outer integral in (3.2), an improper integral form of the integral for ε > 0 is introduced as ∞
Z
y
0
∞
Z
1
1 x
0
φ(x, y)dx dy = lim lim
b
Z
1
ε→0 β→0 β
y b
Z
1
+ lim lim
ε→0 β→0 β
−yφy (0, 0) ln ε + aφx (0, 0) − εφx (0, 0) dy
y b
Z
φ(0, 0) ln a − φ(0, 0) ln ε + yφy (0, 0) ln a dy
a
Z
ψ(x, y) dx dy.
+ lim lim
β→0 ε→0 β
ε
(3.6)
Substituting (3.5) and (3.6) into (3.2) gives rise ∞
Z
∞
Z
−∞
H (x) ⊗ H (y) xy
−∞
φ(x, y) dx dy =
Z
∞
Z
∞
−∞ Z ∞
−∞ Z ∞
−∞
−∞
=
H (x) ⊗ H (y) xy
φ(x, y) dx dy
φxy (x, y)H (x) ⊗ H (y) ln(x) ln(y) dx dy.
(3.7)
The right-hand side of (3.7) yields to the Hadamard finite part of the integral as ∞
Z
∞
Z
FP −∞
H (x) ⊗ H (y) xy
−∞
φ(x, y)dxdy =
∞
Z
−∞
Z
∞
φxy (x, y)H (x) ⊗ H (y) ln(x) ln(y) dx dy.
(3.8)
−∞
Hence,
∂2 ln(x) ln(y). xy ∂ x∂ y Next we study the pseudo-function for n = 2, in (3.1), (see [8]) 0, x, y < 0, f (x, y) = x−2 y−2 , x, y > 0,
pf
H (x) ⊗ H (y)
=
H (x, y) x2 y2
=
.
(3.9)
Let us now examine the above function Z ∞Z ∞ H (x, y) hf (x, y), φ(x, y)i = φ(x, y)dxdy, 2 2 −∞ −∞ x y
Z = 0
∞
1 y2
∞
Z
x
0
φ( x , y ) dx dy. 2
1
(3.10)
By similar way we obtain Hadamard finite part as follows
Z
∞ −∞
Z
∞ −∞
H (x, y) x2 y 2
φ(x, y)dxdy =
Z
∞
−∞ Z ∞
Z
∞
φxy (x, y)
−∞ Z ∞
= −∞
−∞
φxy (x, y)
H (x, y)
xy
H (x, y) xy
dxdy + φxy (0, 0),
Z
∞
Z
∞
φ(x, y)δxy (x, y)dxdy.
dxdy + −∞
−∞
A. Kılıçman, H. Eltayeb / Mathematical and Computer Modelling 49 (2009) 327–336
333
Finally the relation
pf
H (x, y)
x2 y 2
∂2 H (x, y) = pf + δxy (x, y). ∂ x∂ y xy
(3.11)
We can continue the above analysis to generalized equation (3.11) as
pf
H (x, y)
(xy)m+1
∂2 = pf ∂ x∂ y
H (x, y)
m2 (xy)
+
m
1
2
m!
∂ 2m δ(x, y), ∂ xm ∂ y m
m ≥ 1.
(3.12)
Thus we can generalize a distributional derivative from one-dimensional pseudo-function to two-dimensional case as follows H ( x, y ) h(x, y) = α+1 β+1 , x y
where 0 < α, β < 1
(3.13)
see [2,6,12]. Take any arbitrary test function φ(x, y) and consider h having singularity at x, y = 0 where H (x, y) defined by tensor product as above then let
φ(x, y) = φ(0, 0) + xφx (0, 0) + yφy (0, 0) + xyψ(x, y)
(3.14)
where the last term on the right-hand side of Eq. (3.14) defined by
ψ(x, y) =
1 2
xy−1 φxx (0, 0) + φxy (0, 0) +
1 2
yx−1 φyy (0, 0) + · · · +
xn−1−k yk−1 ∂ n φ(t , t )
(n − k)!k! ∂ xn−k ∂ yk
,
0 < t < 1.
(3.15)
Then the distributional derivative of Eq. (3.13) defined by
hf (x, y), φ(x, y)i =
∞
Z 0
φ(x, y)
∞
Z
Z
xα+1 yβ+1
0
dxdy = lim
ε→0 γ →0
γ
b
a
Z
1 yβ+1
ε
φ(x, y) xα+1
dx dy.
(3.16)
Now if we want to calculate the integral inside bracket on the right-hand side of Eq. (3.16) we can have a
Z
φ(x, y) xα+1
ε
1 1 1 1 dx = −φ(0, 0) a−α + φ(0, 0) ε −α − φx (0, 0) a−α+1 + φx (0, 0) ε −α+1
α
α
α
1 1 − yφy (0, 0) a−α + yφy (0, 0) ε −α − φx (0, 0)
α
+ φx (0, 0)
1
α(α − 1)
ε −α+1 +
α Z a
yψ(x, y) xα
ε
α
1
α(α − 1)
a−α+1
dx.
(3.17)
Then by using the integration by parts for the Eq. (3.17) with respect to y we obtain b
Z
yβ+1
γ
a
Z
1
ε
φ(x, y) xα+1
dx dy =
b
1 −α 1 −α −φ( 0 , 0 ) a + φ( 0 , 0 ) ε dy β+1 α α γ y Z b 1 1 −α+1 1 −α+1 + −φ ( 0 , 0 ) a + φ ( 0 , 0 ) ε dy x x β+1 α α γ y Z b 1 1 −α 1 −α + − y φ ( 0 , 0 ) a + y φ ( 0 , 0 ) ε dy y y β+1 α α γ y Z b 1 1 −α+1 −φx (0, 0) + a dy β+1 α(α − 1) γ y Z b Z bZ a 1 1 ψ(x, y) −α+1 + φ ( 0 , 0 ) ε dy + dxdy. x β+1 α(α − 1) xα yβ γ y γ ε Z
1
(3.18)
We calculate the integral on the right-hand side of Eq. (3.18) and take the limit as γ → 0, ε → 0 then the Hadamard finite part of Eq. (3.16) follows
Z
∞
Z
∞
FP −∞
−∞
φ(x, y) xα+1 y
φ(x, y)dxdy = β+1
1
αβ
Z
∞ −∞
Z
∞
φxy (x, y) H (x, y)x−α y−β dxdy −∞
334
A. Kılıçman, H. Eltayeb / Mathematical and Computer Modelling 49 (2009) 327–336
now one can easily check the convergence of the last integral of Eq. (3.18). Now we can show that the last integral of Eq. (3.18) defines distribution. The linearity of the function is obvious. We show the continuity, we note that b
Z
lim lim ε→0 γ →0
a
Z
ψ(x, y)
ε
γ
b
Z
dxdy = xα yβ
φ(x, y) − φ(0, 0) − xφx (0, 0) − yφy (0, 0) xα+1 yβ+1
0
0
Z ≤
a
Z
a
bZ
1
dxdy xα+1 yβ+1
0
0
Z yZ
≤ sup |φuv (u, v)|
a−α b−β
αβ
0≤u≤a 0≤v≤b
0
0
x
dxdy
φuv (u, v)dudv
.
Therefore if {φnm } is a sequence in two-dimensional form then it converges in D(R2+ ) to zero, then φnm (0, 0) → 0 and
RbRa 0
ψnm (x,y) xα y β
0
n
Thus FP
αβ
dxdy → 0 as n, m → ∞. o∞ R ∞ R ∞ φnm (x,y)
∞
Z
Z
−∞
dxdy
xα yβ
0
0
∞
φ(x, y)pf
n,m=1
converges to zero therefore Eq. (3.16) defines the following distribution
H (x, y)
xα+1 yβ+1
−∞
∞
Z
∞
Z
φxy (x, y)(H (x, y)x−α y−β )dxdy
dxdy = −∞
(3.19)
−∞
then we have
φ(x, y), αβ pf
H (x, y)
xα+1 yβ+1
= φxy (x, y), (H (x, y)x−α y−β )
which means that
∂2 (H (x, y)x−α y−β ) = αβ pf ∂ x∂ y
H ( x, y )
xα+1 yβ+1
.
Example 3. Consider the function H (x, y) h(x, y) = α+2 β+2 , x y
where 0 < α, β < 1
(3.20)
as before let 1
1
1
2
2
6
φ(x, y) = φ(0, 0) + yφy (0, 0) + xφx (0, 0) + x2 φxx (0, 0) + xyφxy (0, 0) + y2 φyy (0, 0) + x3 φxxx (0, 0) 1
1
1
+ x yφxxy (0, 0) + xy φxyy (0, 0) + y φyyy (0, 0) + x y ψ(x, y) 2
2
2
2
3
2 2
6
where
ψ(x, y) =
1 2 −2 1 1 x y φxxxxx (0, 0) + xy−1 φxxxy (0, 0) + φxxyy (0, 0) 24 6 4 1 1 xn−k yk n ∂ n φ(t , t ) , + x−1 yφxyyy (0, 0) + y2 x−2 φyyyy (0, 0) + · · · + 6 24 n! k ∂ xn−k ∂ yk
0
