Double Laplace Transform and Explicit Fractional Analogue of 2D Laplacian

Chapter 1 Double Laplace Transform and Explicit Fractional Analogue of 2D Laplacian

S. Rogosin and M. Dubatovskaya

1.1 Introduction

Fractional modeling and fractional differential equations becomes recently very important tool of applied mathematics (see, e.g. [GKMR], [KiSrTr], [SaKiMa]). By using fractional models and solving corresponding fractional differential equation one can understand the nature of nonstandard and anomalous processes in Physics, Engineering, Chemistry, Biology, Economics etc. (see, e.g., [Uch13]). One of the useful methods in the study of fractional differential equations is the integral transform method (see [DebBha], [Deb16]). The development of this method is an actual challenging problem. The aim of this paper is to study the properties of different type fractional analogs of 2D Laplacian by using double integral Laplace transform. The double Laplace transform has been investigated since late 1930s. An extended description of this transform and its properties was made by D.Bernstein (her PhD thesis has been published in [Bern41]). Some of the properties of the double Laplace transform are presented in the Sec. 1.2. Multi-dimensional fractional differential operators are not completely recognized by fractional society. An approach describing these operators in terms of Fourier images (see e.g. [SaKiMa, Ch. 5]) is more theoretical than practical. Recently, two attempts were made to ﬁnd an explicit form of the fractional analog of 2D Laplace operator (see [Yak10], [DalYak]), namely, these analogs have the following form

α,β 1+α 1+β 4 u(x,y) = D0+,xu(x,y) + D0+,yu(x,y), 0 < α,β < 1, (1.1)

S. Rogosin The Belarusian State University, Minsk, Belarus, e-mail: [email protected] M. Dubatovskaya The Belarusian State University, Minsk, Belarus, e-mail: [email protected]

1 2 S. Rogosin and M. Dubatovskaya

α,β α α β β 4−+u(x,y) = −D1−,xD0+,xu(x,y) − D1−,yD0+,yu(x,y), 0 < α,β < 1. (1.2) We consider here operator (1.1) and the skewed modiﬁcations of operator (1.2) for the unit square and for the quarter-plane, respectively (introduced in [Lip11])1

α,β;S α β α β 4−+ u(x,y) = −D0+,xD1−,yu(x,y) − D0+,yD1−,xu(x,y), 0 < α + β ≤ 2, (1.3)

α,β;Q α β α β 4−+ u(x,y) = −D0+,xD−,yu(x,y) − D0+,yD−,xu(x,y), 0 < α + β ≤ 2. (1.4) γ γ γ Here (D0+,xu)(x,y), (D1−,xu)(x,y), (D−,xu)(x,y) are the left- and right-sided Riemann-Liouville fractional derivatives and the Liouville fractional derivative de- ﬁned for arbitrary positive γ and n = [γ] + 1 (see, e.g., [KiSrTr]):

x 1 d n Z u(τ,y)dτ d n (Dγ u)(x,y) = = (In−γ u)(x,y), (1.5) 0+,x Γ (n − γ) dx (x − τ)γ−n+1 dx 0+,x 0

1 1 d n Z u(τ,y)dτ d n (Dγ u)(x,y) = − = − (In−γ u)(x,y), 1−,x Γ (n − γ) dx (τ − x)γ−n+1 dx 1−,x x (1.6) +∞ 1 d n Z u(τ,y)dτ d n (Dγ u)(x,y) = − = − (In−γ u)(x,y). −,x Γ (n − γ) dx (τ − x)γ−n+1 dx −,x x (1.7) Using the double Laplace transform applied to the above said fractional analogs of the 2D laplacian we show that the proper boundary conditions for the fractional Laplace equations have the form of the so-called Cauchy type conditions. This means that the proposed fractional analogs of the 2D laplacian are correct generalizations of the standard one-dimensional fractional differential operator (see [KiSrTr]). Main attention in this paper is paid to the operators (1.1) and (1.4). The study of another (and more cumbersome) form (1.3) is a subject of the forthcoming paper.

1.2 Double Laplace transforms. Deﬁnition and main properties

Double Laplace transforms is deﬁned by the following relation (see, e.g. [DebBha, p. 274]2)

1 One of the reasons to considered the operator in skewed form, i.e. to have both derivatives in x and y in both terms, is to enrich the set of eigenfunctions of the operator. 2 Similar deﬁnition with extra normalization was proposed in [DitPru]. 1 Double Laplace Transform and Fractional Laplacian 3

+∞ +∞ Z Z ˆ −px−qy fˆ(p,q) = (L2 f )(p,q) = (L2 f (x,y))(p,q) = f (x,y)e dxdy. (1.8) 0 0 We present here the following main properties of the double Laplace transforms following [DebBha].

• (linearity) (L2(a f + bg))(p,q) = a(L2 f )(p,q) + b(L2g)(p,q), a,b ∈ R; • (scaling) 1 p q (L f (amx,bny))(p,q) = (L f (x,y)) , , 2 ambn 2 am bn

a,b > 0, m,n ∈ N; −ax−by • (ﬁrst shifting property) L2e f (x,y) (p,q) = (L2 f )(p + a,q + b); • (second shifting property)

−ap−bq (L2 f (x − a,y − b)H(x − a)H(y − b))(p,q) = e (L2 f )(p,q), a,b > 0,

H is the Heaviside function; • (transform of functions with special dependence on the arguments)

1 (L2 f (x + y))(p,q) = p−q [(L f )(p) − (L f )(q)]; 1 (L2 f (x − y))(p,q) = p+q [(L f )(p) + (L f )(q)] f is even, 1 = p+q [(L f )(p) − (L f )(q)] f is odd;

• (transform of the truncation of functions)

1 (L2 f (x)H(x − y))(p,q) = q [(L f )(p) − (L f )(p + q)]; 1 (L2 f (x)H(y − x))(p,q) = q (L f )(p + q); 1 (L2 f (x)H(x + y))(p,q) = q (L f )(p);

• (differential properties)

∂ f L2 ∂x (x,y) (p,q) = p(L2 f (x,y))(p,q) − (Ly f (0,y))(q); ∂ f L2 ∂y (x,y) (p,q) = q(L2 f (x,y))(p,q) − (Lx f (x,0))(p); ∂ 2 f L2 ∂x∂y (x,y) (p,q) = p(L2 f (x,y))(p,q) + q(L2 f (x,y))(p,q) − (Ly f (0,y))(q) − (Lx f (x,0))(p) + f (0,0); 2 ∂ f (x,y) (p,q) = p2 ( f (x,y))(p,q) − p( f (0,y))(q) L2 ∂x2 L2 Ly ∂ f − Ly ∂x (0,y) (q) + f (0,0);

• (convolution) (L2 f ∗ g)(p,q) = (L2 f (x,y))(p,q) · (L2g(x,y))(p,q), where f ∗ x y g is the double convolution: ( f ∗ g)(x,y) = R R f (x −t,y − s)g(t,s)dt ds; 0 0 • (inverse transform) 4 S. Rogosin and M. Dubatovskaya

c+i∞ d+i∞ Z Z −1 ˆ 1 px 1 qy ˆ f (x,y) = L fˆ(p,q) (x,y) = e dp e fˆ(p,q)dq. (1.9) 2 2πi 2πi c−i∞ d−i∞

The last formula is valid if the direct double Laplace transform fˆ(p,q) is an analytic function in the direct product of two semi-planes, namely in the domain (p,q) ∈ C2 : Re p > c, Req > d , where c,d are some real numbers. All other formulas are valid provided existence of the both of their sides.

Remark 1. Double integral transform in the Laplace-Carson form (see [DitPru])

+∞ +∞ Z Z ¯ −px−qy f¯(p,q) = (C2 f )(p,q) = (C2 f (x,y))(p,q) = pq f (x,y)e dxdy (1.10) 0 0 has similar properties. In particular, the scaling property has the following form (which is suitable for the fractal theory, see [NigBal13]) p q (C f (amx,bny))(p,q) = (C f (x,y)) , , 2 2 am bn a,b > 0, m,n ∈ N. Few examples illustrating the above formulas. −kx−ly Example 1. f1(x,y) = e .

+∞ +∞ Z Z −(p+k)x −(q+l)y (L2 f1)(p,q) = e dx e dy = 0 0

1 +∞ 1 +∞ 1 1 − e−(p+k)x − e−(q+l)y = . p + k x=0 q + l y=0 p + k q + l

Example 2. f2(x,y) = H(x − y).

+∞ +∞ +∞ x Z Z Z Z −px −qy −px −qy (L2 f2)(p,q) = e dx e H(x − y)dy = e dx e dy = 0 0 0 0

+∞ 1 Z 1 1 1 +∞ 1 = e−px(1 − e−qx)dx = − e−px + e−(p+q)x = . q q p p + q x=0 p(p + q) 0 Remark 2. There exists a problem with application of the inverse double Laplace transform. To describe this problem let us consider a simple situation, namely, when the double Laplace image is a rational function of two complex variables p and q:

R(p,q) (L f )(p,q) = . 2 S(p,q) 1 Double Laplace Transform and Fractional Laplacian 5

Let the denominator is represented in the form

m n S(p,q) = C0 ∏(p − ak)∏(q − bl), (1.11) k=1 l=1 where ak, bl are certain complex numbers (not necessarily different). Then to re- cover an initial function f (x,y) one needs to apply Residue Theorem in two semi- planes of variables p and q, respectively. But, if the polynomial S(p,q) is irreducible, i.e. there exist no nontrivial poly- nomials S1(p,q), S2(p,q) of the above described form (1.11), such that S(p,q) = S1(p,q)S2(p,q), then such technique cannot be applied. Simplest example of the irreducible polynomial is S(p,q) = p2 + q2 − 1.

Let us repeat a simple example of application of the double Laplace transform to solving of an initial-boundary problem for a second order differential equation in the quarter-plane (see [DebBha, p. 278]). Example 3. Let us consider the wave equation

2 c uxx = utt , x ≥ 0, t > 0, (1.12) under the following initial conditions

u(x,0) = f (x), ut (x,0) = g(x), (1.13) the following boundary conditions

u(0,t) = 0, ux(0,t) = 0. (1.14)

Assume that the given functions f (x) and g(x) are suitable to apply the below technique (e.g. are absolutely integrable in x ∈ R+). Let us apply the double Laplace transform uˆ(p,q) = (L2u(x,t))(p,q) to the equation (1.12):

2 c (L2uxx(x,t))(p,q) = (L2utt (x,t))(p,q).

By using the differential properties of the double Laplace transform we have

2 2 c p uˆ(p,q) − p(L2u(0,t))(q) − (L2ux(0,t))(q) + u(0,0) =

2 = p uˆ(p,q) − q(L2u(x,0))(p) − (L2ut (x,0))(p) + u(0,0) . Applying conditions (1.13), (1.14) we arrive at the following relation

c2 p2 − q2uˆ(p,q) = −q fˆ(p) + gˆ(p) or 6 S. Rogosin and M. Dubatovskaya

q fˆ(p) + gˆ(p) uˆ(p,q) = . (1.15) q2 − c2 p2 The direct calculation shows that with a > cRe p we have

a+i∞ 1 Z q fˆ(p) + gˆ(p) 1 eqt dq = fˆ(p)cosh cpt + gˆ(p)sinhcpt. 2πi q2 − c2 p2 cp a−i∞ Thus the inverse double Laplace transform (1.9) applied to (1.15) gives for any b > 0

b+i∞ 1 Z 1 u(x,t) = fˆ(p)cosh cpt + gˆ(p)sinhcpt epxdp = 2πi cp b−i∞

 x−ct x+ct  1 1 Z Z = [ f (x + ct) + f (x − ct)] +  g(τ)dτ − g(τ)dτ. 2 2c 0 0 Therefore, the solution to the initial-boundary value problem (1.13)-(1.14) for equation (1.12) have the following form (known in the literature d’Alembert formula for the solution of 2D wave equation)

 x+ct  1 1 Z u(x,t) = [ f (x + ct) + f (x − ct)] +  g(τ)dτ. (1.16) 2 2c x−ct

1.3 Application of the double Laplace transform to explicit fractional laplacian

In this section we present the values of the double Laplace transform of the fractional analogs (1.1) and (1.4) of the 2D laplacian.

1.3.1 Power type fractional laplacian 4α,β

Let us apply the double Laplace transform to the fractional laplacian (1.1). It is 1+α sufﬁcient to calculate L2D0+,xu(x,y) (p,q) with 0 < α < 1:

+∞ +∞  x  Z Z 2 Z 1+α −qy −px d 1 u(t,y)dt L2D u(x,y) (p,q) = e dy e  dx. 0+,x dx2 Γ (1 − α) (x −t)α 0 0 0 (1.17) Denote for shortness 1 Double Laplace Transform and Fractional Laplacian 7

x 1 Z u(t,y)dt ϕ(x,y) := = I1−α u(x,y). Γ (1 − α) (x −t)α 0+,x 0 Then by applying the differentiation property of the Laplace transform to the inter- nal integral in (1.17) we have

2 d 2 dϕ Lx ϕ(x,y) (p) = p (Lxϕ(x,y))(p) − pϕ(0,y) − (0,y) = dx2 dx

2 1−α α = p (Lxϕ(x,y))(p) − p(I0+,x u(x,y))(0) − (D0+,xu(x,y))(0). Further, we use the formula for the Laplace transform of the fractional integral [SaKiMa, p. 140] 1−α −1+α LxI0+,x φ = p (Lxφ). (1.18) Hence

+∞  x  Z 2 Z 1+α −px d 1 u(t,y)dt LxD u(x,y) (p) = e  dx = (1.19) 0+,x dx2 Γ (1 − α) (x −t)α 0 0

1+α 1−α α = p (Lxu(x,y))(p) − p(I0+,x u(x,y))(0) − (D0+,xu(x,y))(0). Applying the external integral in (1.17) and using the symmetry of the formula (1.1) we obtain the following result. Proposition 1. The double Laplace transform of the fractional laplacian (1.1) is represented in the following form

α,β 1+α 1+β L24 u(x,y) (p,q) = p + q (L2u(x,y))(p,q)− (1.20)

1−α 1−β −p Ly I0+,x u(x,y) (q) − q Lx I0+,yu(x,y) (p)− x=0 y=0 α β − Ly D0+,xu(x,y) (q) − Lx D0+,yu(x,y) (p). x=0 y=0 Remark 3. It follows from [SaKiMa, Thm. 3.1, Thm. 3.6] that if the function u is either Hölder continuous in both variables, or Lr-integrable on R+ × R+ with sufﬁ- ciently large r, then both terms in the second line of (1.20) vanishes.

α,β;Q 1.3.2 Skewed fractional laplacian 4−+ in the quarter plane

In this subsection we determine the double Laplace transform of the skewed fractional laplacian in the quarter plane (1.4). It is natural (see [DalYak]) to consider such construction in three cases: 8 S. Rogosin and M. Dubatovskaya

(a) 0 < α,β < 1; (b) 0 < α < 1, 1 < β < 2; α + β = 2; (c) 1 < α < 2, 0 < β < 1; α + β = 2. α,β;Q Case (a). Due to the symmetry of the operator 4−+ it suffices to transform the first term in (1.4): α β d1(x,y) := D0+,xD−,y. Note as in previous subsection that for any 0 < γ < 1 and sufficiently smooth φ

z 1 d Z φ(τ)dτ d (Dγ φ)(z) = = (I1−γ φ)(z), (1.21) 0+,z Γ (1 − γ) dz (z − τ)γ dz 0+,z 0

+∞ 1 d Z φ(τ)dτ d (Dγ φ)(z) = − = − (I1−γ φ)(z). (1.22) −,z Γ (1 − γ) dz (τ − z)γ dz −,z z 1−γ Besides, the right-sided fractional integral I−,z is related to the left-sided fractional integral by two formulas (see [SaKiMa, formulas (11.27), (11.29)])

γ γ γ −γ γ (I−,zψ)(z) = cos πγ (I0+,zψ)(z) + sin πγ (I0+,zx St ψ)(z), (1.23)

γ γ γ (I−,zψ)(z) = cos πγ (I0+,zψ)(z) + sin πγ (SI0+,xψ)(z), (1.24) where S is the singular integral operator over semi-axis and compositions are deﬁned as follows

z +∞ 1 Z dx 1 Z tγ ψ(t)dt (Iγ x−γ Stγ ψ)(z) = , 0+,z Γ (γ) (z − x)1−γ πxγ t − x 0 0

+∞ x 1 Z dx 1 Z ψ(t)dt (SIγ ψ)(z) = . 0+,x π (x − z) Γ (γ) (x −t)1−γ 0 0

It follows from (1.21), (1.22), (1.23), that the operator d1(x,y) possesses the following representation d d d (x,y) = − I1−α I1−β = (1.25) 1 dx 0+,x dy −,y d d h i I1−α cos πβ I1−β − sin πβ I1−β ω(t,τ) , dx 0+,x dy 0+,y 0+,y where ω(x,y) := y−1+β (Sr1−β u(x,r))(y). (1.26) Denoting d ϕ (x,y) := (I1−β u(x,t))(y) 1 dy 0+,y 1 Double Laplace Transform and Fractional Laplacian 9 we get by using Laplace transform properties d 1−α h α 1−α i L I ϕ (x,y) (p,q) = Ly (p Lx ϕ (x,y))(p) − (I ϕ (t,y))(0) (q) = 2 dx 0+,x 1 1 0+,x 1 α d 1−β 1−α = p Lx Ly (I u(x,τ))(y) (q) (p) − Ly(I ϕ (t,y))(0) (q) = dy 0+,y 0+,x 1 α β α 1−β d 1−β = p q (L u(x,y))(p,q)− p Lx(I u(x,t))(0) (p)− Ly (I ϕ (0,t))(y) (q), 2 0+,y dy 0+,y 2 where 1−α ϕ2(0,t) = (I0+,x u(τ,t))(0). Since d 1−β β 1−β Ly (I ϕ (0,t))(y) (q) = q (Lyϕ (0,y))(q) − (I ϕ (0,t))(0), dy 0+,y 2 2 0+,y 2 and   y  x  Z Z 1−β 1 1  dτ u(t,τ)dt  (I ϕ2(0,τ))(0) =     = cα,β;u, 0+,y Γ (1 − β) Γ (1 − α)  (y − τ)β (x −t)α  0 0 x=0 y=0 then d d 1−β L I1−α (I u(x,τ))(y) (p,q) = pα qβ (L u(x,y))(p,q)− (1.27) 2 dx 0+,x dy 0+,y 2 α 1−β β 1−α −p Lx I u(x,τ) (p) − q Ly I u(t,y) (q) − c . 0+,y y=0 0+,x x=0 α,β;u Similar to the previous calculation we have d d 1−β L I1−α (I ω(x,τ))(y) (p,q) = pα qβ (L ω(x,y))(p,q)− (1.28) 2 dx 0+,x dy 0+,y 2 α 1−β β 1−α −p Lx I ω(x,τ) (p) − q Ly I ω(t,y) (q) − c , 0+,y y=0 0+,x x=0 α,β;ω where   y  x  1 1 Z dt Z ω(t,τ)dt  cα,β;ω =     . Γ (1 − β) Γ (1 − α)  (y − τ)β (x −t)α  0 0 x=0 y=0 10 S. Rogosin and M. Dubatovskaya

α,β;Q It follows from (1.23) that the operator 4−+ has the representation: d d d d 4α,β;Qu(x,y) = cos πβ I1−α I1−β u(x,y) + I1−α I1−β u(x,y) − −+ dx 0+,x dy 0+,y dy 0+,y dx 0+,x (1.29) d d d d −sin πβ I1−α I1−β ω(x,y) + I1−α I1−β ω(x,y) dx 0+,x dy 0+,y dy 0+,y dx 0+,x where ω(x,y) is deﬁned in (1.26). Then, by (1.27) and (1.28) we get the following result. Proposition 2. Let 0 < α,β < 1. Let ω(x,y) be deﬁned in (1.26). Then the double Laplace transform of the skewed fractional laplacian (1.4) is represented in the following form

α,β;Q α β β α L24−+ u(x,y) (p,q) = p q + p q (L2 (cos πβu(x,y) − sin πβω(x,y)))(p,q)− (1.30) α 1−β −p Lx I0+,y (cos πβu(x,τ) − sin πβω(x,τ)) (p)− y=0 β 1−α −p Lx I0+,y (cos πβu(x,τ) − sin πβω(x,τ)) (p)− y=0 α 1−β −q Ly I0+,x (cos πβu(t,y) − sin πβω(t,y)) (q)− x=0 β 1−α −q Lx I0+,x (cos πβu(t,y) − sin πβω(t,y)) (p)− x=0

−cα,β;u − cα,β;ω − cβ,α;u − cβ,α;ω , where constants cα,β;u,cα,β;ω are defined in (1.27) and (1.28), respectively, and constants cβ,α;u,cβ,α;ω are defined similarly by interchanging parameters in fractional integrals in (1.27) and (1.28). Case (b). Here we consider the double Laplace transform of the skewed frac- α,β;Q tional laplacian 4−+ with parameters 0 < α < 1,1 < β < 2. The remaining case (c) can be obtained by interchanging α and β. As before it suffices to apply the double Laplace transform only to the first term α,β;Q of 4−+ :

d d2 d˜ (x,y) = −Dα Dβ u(x,y) = − I1−α I2−β u(t,y) (x,y). 1 0+,x −,y dx 0+,x dy2 −,y

Using the relation between right- and left-sided fractional integrals (1.23) we repre- sent d˜1(x,y) in the form

d d2 h i d˜ (x,y) = − I1−α cos πβI2−β u(t,τ) − sin πβI2−β τβ−2Sr2−β u(t,r) (x,y) = 1 dx 0+,x dy2 0+,y 0+,y 1 Double Laplace Transform and Fractional Laplacian 11

d d2 d d2 = −cos πβ I1−α I2−β u(t,τ) (x,y)+sin πβ I1−α I2−β ω˜ (t,τ) (x,y) =: dx 0+,x dy2 0+,y dx 0+,x dy2 0+,y

=: d˜11(x,y) + d˜12(x,y). where ω˜ (t,τ) = τβ−2Sr2−β u(r,τ). (1.31)

2 d 1−α d 2−β L d˜ (x,y) (p,q) = −cos πβLy Lx I I u(t,τ) (x,y) (p,q) = 2 11 dx 0+,x dy2 0+,y d 1−α = −cos πβLy Lx I ϕ˜ (t,y) (x,y) (p,q), dx 0+,x 1 where d2 ϕ˜ (τ,y) = I2−β u(t,τ) (y). 1 dy2 0+,y Hence α 1−α L d˜ (x,y) (p,q) = −cos πβLy p (Lxϕ˜ (x,y))(p) − I ϕ˜ (t,y) (q). 2 11 1 0+,x 1 x=0 (1.32) If the the function u(x,y) is such that the operators Ly and Lx are interchangeable, then α α Ly (p (Lxϕ˜1(x,y))(p))(q) = p (LxLy (ϕ˜1(x,y))(p,q) = 2 α d 2−β = p LxLy I u(t,τ) (p,q) = dy2 0+,y α β α 2−β β−1 = p q (L2u(x,y))(p,q)− p qLx I0+,yu(x,τ) + Lx D0+,yu(x,τ) (p). y=0 y=0 Besides, if the fractional integrals in the second term in (1.32), then

2 1−α d 2−β 1−α − LyI0+,x ϕ˜1(t,y) (q) = − Ly I0+,y I0+,x u(t,τ) (q) = x=0 dy2 x=0 β 1−α 2−β 1−α = −q Ly I0+,x u(t,y) (q) + q I0+,y I0+,x u(t,τ) + x=0 x=0 y=0 β−1 1−α + D0+,y I0+,x u(t,τ) . x=0 y=0 Analogously,

2 d 1−α d 2−β L d˜ (x,y) (p,q) = sin πβLy Lx I I ω˜ (t,τ) (x,y) (p,q). 2 12 dx 0+,x dy2 0+,y 12 S. Rogosin and M. Dubatovskaya

Under similar conditions as before we have α 1−α L d˜ (x,y) (p,q) = sin πβLy p (Lxϕ˜ (x,y))(p) − I ϕ˜ (t,y) (q), 2 12 2 0+,x 2 x=0 (1.33) where d2 ϕ˜ (τ,y) = I2−β ω˜ (t,τ) (y). 2 dy2 0+,y

If the the function u(x,y) is such that LyLxϕ˜2(x,y) = LxLyϕ˜2(x,y), then

α α Ly (p (Lxϕ˜2(x,y))(p))(q) = p (LxLy (ϕ˜2(x,y))(p,q) = α β α 2−β β−1 = p q (L2ω˜ (x,y))(p,q)− p qLx I0+,yω˜ (x,τ) + Lx D0+,yω˜ (x,τ) (p). y=0 y=0 Besides, if the fractional integrals in the second term in (1.33), then

2 1−α d 2−β 1−α − LyI0+,x ϕ˜2(t,y) (q) = − Ly I0+,y I0+,x ω˜ (t,τ) (q) = x=0 dy2 x=0 β 1−α 2−β 1−α = −q Ly I0+,x ω˜ (t,y) (q) + q I0+,y I0+,x ω˜ (t,τ) + x=0 x=0 y=0 β−1 1−α + D0+,y I0+,x ω˜ (t,τ) . x=0 y=0 α,β;Q The results for the second term of 4−+ :

d d2 d˜ (x,y) = −Dα Dβ u(x,y) = − I1−α I2−β u(t,τ) (x,y) = 2 0+,y −,x dy 0+,y dx2 −,x

d d2 d d2 = −cos πβ I1−α I2−β u(t,τ) (x,y)+sin πβ I1−α I2−β ω˜ (t,τ) (x,y) =: dy 0+,y dx2 0+,x dy 0+,y dx2 0+,x

=: d˜21(x,y) + d˜22(x,y) can be obtained simply by interchanging variables x and y. Combining the above results we obtain the following Proposition 3. Let 0 < α < 1,1 < β < 2. Let the above conditions are satisﬁed. Let

ζ(x,y) := −cos πβu(x,y) + sin πβω˜ (x,y), where ω˜ is deﬁned in (1.31). Then the double Laplace transform of the skewed α,β;Q fractional laplacian 4−+ has the following form

α,β;Q α β β α L24−+ u(x,y) (p,q) = (p q + p q )(L2ζ(x,y))(p,q)− (1.34) 1 Double Laplace Transform and Fractional Laplacian 13 α 2−β α 2−β −p q Lx I0+,yζ(x,τ) (p) − q p Ly I0+,xζ(t,y) (q)− y=0 x=0 α β−1 α β−1 −p Lx D0+,yζ(x,τ) (p) − q Ly D0+,xζ(t,y) (q)− y=0 x=0 β 1−α β 1−α −q Ly I0+,x ζ(t,y) (q) − p Lx I0+,y ζ(x,τ) (p)− x=0 y=0 2−β 1−α 2−β 1−α q I0+,y I0+,x ζ(t,τ) + p I0+,x I0+,y ζ(t,τ) + x=0 y=0 y=0 x=0 β−1 1−α β−1 1−α D0+,y I0+,x ζ(t,τ) + D0+,x I0+,y ζ(t,τ) . x=0 y=0 y=0 x=0 Remark 4. If the function u(x,y) is smooth enough (e.g. Hölder continuous in both variables), then all terms containing values of fractional integrals at the initial point vanishes. In this case in formula (1.34) remains only ﬁrst and third lines. Analo- gously, in this case formula (1.30) contains only ﬁrst line.

Discussion and Outlook

In this paper we apply the double Laplace transform to some explicit fractional analogs of the 2D laplacian. Using the obtained results we propose to use the Cauchy type boundary conditions for considered fractional Laplace equation. These conditions differ from the standard conditions for the Laplace equations and they have the nonlocal nature. Thus, such problems describe the processes with nonlocal be- haviour. In the forthcoming paper we suppose to fix proper functional spaces and to find corresponding solvability conditions of the Cauchy type problem for fractional Laplace equation as well as to determine the set of eigenfunctions and the fundamental solution. Since it is known (see [SaKiMa]) that there exist no unique fractional analog of the Laplace operator, then the question remains to find addi- tional properties of the proposed explicit fractional analogs of the Laplace operator.

Acknowledgement

The work is partially supported by the People Programme (Marie Curie Actions) of the European Union’s Seventh Framework Programme FP7/2007- 2013/ under REA grant agreement PIRSES-GA-2013-610547 - TAMER. The authors are grateful to an anonymous referee for attentive reading of the paper and making important remarks improving the presentation. 14 S. Rogosin and M. Dubatovskaya References

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