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ISSN 1064–5624, Doklady Mathematics, 2006, Vol. 73, No. 1, pp. 6–10. © Pleiades Publishing, Inc., 2006. Original Russian Text © A.A. Voroshilov, A.A. Kilbas, 2006, published in Doklady Akademii Nauk, 2006, Vol. 406, No. 1, pp. 12–16.

MATHEMATICS

A Cauchy-Type Problem for the Diffusion-Wave Equation with Riemann–Liouville Partial Derivative A. A. Voroshilov and A. A. Kilbas Presented by Academician S.M. Nikol’skii April 13, 2005 Received May 26, 2005

DOI: 10.1134/S1064562406010029

Consider the linear differential equation α

( D 0+, t u ) ( x, t ) = λ ∆ x u ( x, t ), 2

x∈R , n

α 2 ∂ u ( x, t ) - , x ∈ R, t > 0; λ > 0. (5) ( D 0+, t u ) ( x, t ) = λ -------------------2 ∂x 2

t > 0, (1)

The interest to Eqs. (1) and (5) is caused by their numerous applications to solving diffusion problems of physics, mechanics, and other applied sciences [2, Sections 4.2.1 and 4.2.2; 3; 4] (historical information and a review of the related results are contained in survey [5, Ch. 7]).

α

where λ > 0; ( D 0+, t u)(x, t) is the partial Riemann–Liouville fractional derivative of order α > 0 of the function u(x, t) with respect to the second variable [1, p. 342], i.e., t

α ( D 0+, t u ) ( x,

α > 0;

1 u ( x , τ ) dτ ∂ n -, t ) = ⎛ -----⎞ --------------------- ---------------------------⎝ ∂t⎠ Γ ( n – α ) ( t – τ ) α – n + 1 0

∫

n = [ α ] + 1;

x∈R ;

In this paper, we solve Eq. (1) of order α > 0 with initial conditions

(2)

t > 0,

m

α–k

( D 0+, t u ) ( x, 0+ ) = f k ( x ),

and ∆x is the Laplace operator in the first variable m

x = (x1, x2, …, xm) ∈ Rm, i.e., ∆x =

k = 1, 2, …, n = – [ – α ]; x ∈ R .

∂ u --------2 . ∂x j j=1

∑

2

α–k

Here, ( D 0+, t u)(x, 0+) with n – 1 < α < n is understood as

∂u ( x, t ) 1 If α = 1 and α = 2, then ( D 0+, t u)(x, t) = ------------------- and ∂t

α–k

t → 0+

2

2 ∂u ------ = λ ∆ x u ( x, t ), ∂t

x∈R , n

t > 0,

k = 1, 2, …, n – 1, α–n

2 ∂ u -------2- = λ ∆ x u ( x, t ), ∂t

x∈R , n

t > 0.

n–α

( D 0+, t u ) ( x, 0+ ) = lim ( I 0+, t u ) ( x, t ); t → 0+

(7)

(8)

for α = n ∈ N, it equals

(3)

and for α = 2, it coincides with the wave equation 2

α–k

( D 0+, t u ) ( x, 0+ ) = lim ( D 0+, t u ) ( x, t ),

∂ u ( x, t ) - , respectively; thus, for α = 1, t) = -------------------2 ∂t Eq. (1) coincides with the heat (diffusion) equation 2 ( D 0+, t u)(x,

(6) m

∂ -u ( x, 0 ), 0+ ) = ----------n–k ∂t k = 1, 2, …, n, n–k

n–k ( D 0+, t u ) ( x,

(4)

(9)

n–α

For this reason, Eq. (1) is called the diffusion-wave equation [2, p. 146]. In particular, for m = 1, Eq. (1) takes the form

where ( I 0+, t u)(x, t) is the partial Riemann–Liouville fractional integral of order n – α [1, p. 341], that is, t

n–α ( I 0+, t u ) ( x,

Belarussian State University, pr. F. Skoriny 4, Minsk, 220050 Belarus e-mail: [email protected], [email protected]

1 u ( x , τ ) dτ -, t ) = --------------------- ---------------------------Γ ( n – α ) ( t – τ )1 – n + α 0

∫

( I 0+, t u ) ( x, t ) = u ( x, t ). 0

6

α < n,

(10)

A CAUCHY-TYPE PROBLEM FOR THE DIFFUSION-WAVE EQUATION

If α = n ∈ N, then, according to (2) and (9), problem (1), (6) takes the form of the Cauchy problem ∂ u ( x, t ) 2 -------------------- = λ ∆ x u ( x, t ), n ∂t n

∂ u ( x, 0 ) --------------------------- = f k ( x ), n–k ∂t n–k

x∈R , m

(11)

∫ u ( x, t )e

– st

dt,

x∈R ,

s ∈ C, (13)

m

0

∫ u ( x, t )e R

ix · σ

d x,

σ∈R , m

m

t > 0, (14)

and their inverse transforms with respect to s ∈ C and σ ∈ Rm: –1 ( L s u ) ( x,

1 t ) = -------2πi x∈R , m

γ + i∞

∫

e u ( x, s ) ds, st

γ – i∞

R

m

i

i=1

σ = (σ1, σ2, …, σm) ∈ Rm; γ ∈ R is a fixed real number. The properties of the direct and inverse Laplace and Fourier transforms can be found in, e.g., [6, Chapter II; 7, Chapter 16]. In particular, the operators in (15), (13) and (16), (14) are mutually invertible for sufficiently good functions u(x, t). Suppose that, for the functions fk(x) (k = 1, 2, …, n) in (6), the Fourier transforms (Fx fk)(σ) are defined. Applying Laplace transform (13) to both sides of Eq. (1) and taking into account the expression for the Laplace transform of partial Riemann–Liouville fractional derivative (2) [2, (2.248)] and initial conditions (6), we obtain n

∑s

k–1

f k ( x ) = λ ∆ x ( L t u ) ( x, s ). 2

k=1

Applying Fourier transform (14) to this equality and using the expression for the Fourier transform of the DOKLADY MATHEMATICS

=

∑σ

2 i .

Applying inverse Laplace trans-

i=1

form (15) and Fourier transform (16), we obtain a solution u(x, t) to initial problem (1), (6); this is ⎛ –1 –1 u( x, t) = ⎜ L s F σ ⎝

k–1 ⎞ s ------------------------2(F x f k)(σ) ⎟ ( x, t). (18) α 2 s +λ σ ⎠ k=1 n

∑

Let us express solution (18) in terms of the MittagLeffler special function [8, 18.1(18)] ∞

E α, β ( z ) =

∑ j=0

j

z ------------------------ , α > 0, β > 0, z ∈ C, (19) Γ ( αj + β )

which is an entire function of z. The formula for the Laplace transform of this function (see, e.g., [1, (1.93)]) implies α–k

k–1

s 2 2 α -, E α, α – k + 1 ( – λ σ t ) ] ) ( s ) = -----------------------α 2 2 s + λ σ (20)

k = 1, 2, …, n; s ∈ C, σ ∈ R , λ > 0; λ σ s

– 2/α

< 1.

Applying the inverse Laplace and Fourier transforms to both sides of (17) and taking into account (20), we obtain a solution to problem (1), (6) in the form

t > 0.

∑ x σi for x = (x1, x2, …, xm) ∈ Rm and

s ( L t u ) ( x, s ) –

s ∈ C,

m

(16)

m

α

where

|σ|2

( Lt [ t

∫

x∈R ,

(15)

t > 0,

1 – iσ ⋅ x –1 ( F σ u ) ( x, t ) = -------------m- u ( σ, t )e dσ, ( 2π ) m

Here, x · σ =

σ∈R ,

(17)

m

∞

( F x u ) ( σ, t ) =

∑ m

m

k–1

s ------------------------2 ( F x f k ) ( σ ), α 2 s + λ σ k=1

( F x L t u ) ( σ, s ) =

k = 1, 2, …, n; x ∈ R . (12)

For this reason, by analogy, problem (1), (6) is called a Cauchy-type problem. To solve problem (1), (6), we apply, respectively, the Laplace and Fourier transforms of the function u(x, t) with respect to t > 0 and x ∈ Rm, which are ( L t u ) ( x, s ) =

operator ∆x: (Fx∆xu)(σ, s) = –|σ|2(Fxu)(σ, s), we come to the relation n

t > 0,

7

Vol. 73

No. 1

2006

n

u ( x, t ) =

∑

α–k

t -------------m- E α, α – k + 1 ( 2π ) m k=1

∫

R

2 α

× ( – λ σ t ) ( F x f k ) ( σ )e 2

– ix ⋅ σ

dσ.

(21)

Theorem 1. Suppose that, for the functions fk(x) with k = 1, 2, …, n, the Fourier transforms (Fx fk)(σ) are defined and the integrals on the right-hand side of (21) converge. Then, Cauchy-type problem (1), (6) is solvable, and its solution is given by (21). In particular, for m = 1, equality (21) gives a solution to problem (5), (6). Corollary 1. The solution to Cauchy problem (11), (12) is given by formula (21) with α = n provided that the integrals on the right-hand side of (21) converge. If the functions fk(x) with k = 1, 2, …, n are infinitely differentiable on Rm, then, substituting (19) into (21), interchanging integration and summation [which can

8

VOROSHILOV, KILBAS

be done because series (19) uniformly converges] and taking into account the equality 2, 0

H 2, 2 z

1 2 j – iσ ⋅ x ( ∆ f k ) ( x ) = -------------m- ( – σ ) ( F x f k ) ( σ )e dσ, ( 2π ) m

∫

j

⎛ a, 1---⎞ , ⎛ b, α ---⎞ ⎝ 2⎠ ⎝ 2 ⎠

R

j = 1, 2, …, for the jth powers of the Laplace operator

1 ( ∆x

j

1

j–1

n

u ( x, t ) =

∑

t

α–k

k=1

∞

∑ j=0

1 Γ ( c + τ )Γ ( d + τ/2 ) –τ = -------- --------------------------------------------------------z dτ. 2πi Γ ( a + τ/2 )Γ ( b + ατ/2 )

∫

= ∆x and

∆ x = ∆ x ∆ x for j = 2, 3, …), we obtain the representation of solution (21) in the form 2 α j

(λ t ) j ------------------------------------------ ( ∆ x f x ) ( x ). (22) Γ ( αj + α – k + 1 )

In particular, for m = 1, the solution to problem (5), (6) has the form n

u ( x, t ) =

∑ k=1

t

α–k

∞

∑ j=0

2 α j

(λ t ) (2 j) ------------------------------------------ f k ( x ). (23) Γ ( αj + α – k + 1 )

Corollary 2. The solution to Cauchy problem (11), (12) is given by formula (22) with α = n provided that the series on the right-hand side of (22) converge. If 0 < α < 2, then the solution to Eq. (1) with initial conditions (6) for 0 < α ≤ 1 and n = 1 or 1 < α < 2 and n = 2 can be expressed in terms of the so-called ç-function [9, Section 8.3; 10, Chapters 1 and 2]. The proof is based on the application of the inverse Fourier and Laplace transforms to (17) and on the expressions for the Fourier and Laplace transforms of the modified Bessel function Kv(z) of the third kind [11, 7.2(13)] and of the special ç-function

L

m ----

------------⎛ ⎞ 2π 2 c 2 -, K m – 2 ( c x ) ⎟ ( σ ) = ⎛ ------⎞ -----------------⎜ Fx x 2 2 ⎝ ⎠ ------------c (25) c + σ ⎝ ⎠ 2 2–m

σ∈R . m

α/2

s According to (25) with c = -------- and the theorem about λ the Fourier convolution, relation (17) takes the form ( F x L t u ) ( σ, s ) ⎛ = ⎜F x ⎜ ⎝

α ( m – 2) k – 1 + --------------------4

2–m α ⎞ ------------s x --2-⎞ 2 ⎛ --------------------------K m – 2 -----s * x f k( x) ⎟ (σ). x m ⎟ ⎠ ------------- ⎝ λ ---2 2 k=1 ⎠ λ ( 2λπ ) Applying inverse Fourier transform (16) to this equality, we obtain ( L t u ) ( x, s ) n

∑

n

α ( m – 2) k – 1 + --------------------4

= 2 π

– 1/2

s

α

2–m

------------s x --2-⎞ 2 ⎛ ------------------------------s * x f k( x) (26) = K x m – 2 m ⎠ ------------- ⎝ λ ---2 2 k=1 λ ( 2λπ ) for x ∈ Rm and s ∈ C. Lemma 2. If 0 < α ≤ 1 and k = 1 or 1 < α < 2 and k = 1, 2, then

∑

⎛ 1 α(m – 2) α ⎛m ⎜ ----, ---⎞ , ⎛ 1 – k – ----------------------, ---⎞ α α(m – 2) – --– k – ---------------------⎝ ⎠ ⎝ ⎜ 4 2 2⎠ 4 2 2, 0 x 4 H 2, 2 -----t ⎜ Lt t λ ⎜ 1 ⎛ 1--- – m ⎛m ----, ---⎞ ---- – 1, 1⎞ ⎜ ⎝ ⎝ ⎠ 2⎠ 2 4 2 ⎝ m ---2

(24)

Here, a, b, c, d ∈ R and L is a special infinite contour on the right of the poles of the gamma-functions Γ(c + τ) and Γ(d + τ/2). Theorem 1.1 from [10] implies the existence of the function defined by (24) for 0 < α < 2; π a, b, c, d ∈ R; and z ≠ 0 with | arg ( z)| < (2 – α) --- . 4 Lemma 1. For m ∈ N and c > 0,

Theorem 2. Suppose that the functions fk(x), where k = 1, 2, …, n, are infinitely differentiable on Rm and the series on the right-hand side of (22) converge. Then, Cauchy-type problem (1), (6) is solvable, and its solution is given by formula (22).

1 ( c, 1 ), ⎛ d, ---⎞ ⎝ 2⎠

α(m – 2) k – 1 + ---------------------4

⎞ ⎟ ⎟ ⎟ (s) ⎟ ⎟ ⎠

α

x --2K m – 2 ⎛ -----s ⎞ . ⎠ ------------- ⎝ λ 2 DOKLADY MATHEMATICS

(27) Vol. 73

No. 1

2006

A CAUCHY-TYPE PROBLEM FOR THE DIFFUSION-WAVE EQUATION

Applying inverse Laplace transform (15) to (26) and taking into account (27), we obtain the following solution to problem (1), (6): –m

2–m ------------2

1 + ---2

------------2 k=1

2 x u ( x, t ) = -----------------------m m–1 λ

2, 0 × H 2, 2

π

n

∑t

α

x – --2-----t λ ⎛m ---- – 1, 1⎞ ⎝2 ⎠

α

where G 1 (x, t) is given by (31) and α

G 2 ( x, t ) =

α(m – 2) – k – ---------------------4

1 α(m – 2) α ⎛m ----, ---⎞ , ⎛ 1 – k – ----------------------, ---⎞ ⎝ 4 2⎠ ⎝ 2⎠ 4 1 ⎛ 1--- – m ----, ---⎞ ⎝ 2 4 2⎠

2–m ------------α(m – 2) 2 – 2 – ---------------------2 x 4 ------------------------t m m–1 1 + ---- ------------2 2 –m

λ

× * x f k ( x ), (28)

9

2, 0 H 2, 2

π

1 α(m – 2) α ⎛m ----, ---⎞ , ⎛ – 1 – ----------------------, ---⎞ α 2⎠ 4 x – --2- ⎝ 4 2⎠ ⎝ -----t λ 1 ⎛m ⎛ 1--- – m ---- – 1, 1⎞ ----, ---⎞ ⎝2 ⎠ ⎝ 2 4 2⎠

(34)

provided that the integral in (33) converges. where n = 1 for 0 < α ≤ 1 and n = 2 for 1 < α < 2. This and the Fourier convolution imply the following results. Theorem 3. If 0 < α ≤ 1, m ∈ N, and λ > 0, then the Cauchy-type problem α

2

α–1

x∈R

m

∫ R

α G 1 ( x,

t) =

2, 0

α

(30)

2 ∂ u ( x, t ) α -, ( D 0+, t u ) ( x, t ) = λ -------------------2 ∂x 2

2–m ------------α(m – 2) 2 – 1 – ---------------------2 x 4 ------------------------t m m–1 1 + ---- ------------2 2 –m

α–1 ( D 0+, t u ) ( x,

π

1 α(m – 2) α ⎛m ----, ---⎞ , ⎛ – ----------------------, ---⎞ α 2⎠ 4 x – --2- ⎝ 4 2⎠ ⎝ -----t , λ 1 1 m m ⎛ --- – ----, ---⎞ ⎛ ---- – 1, 1⎞ ⎝ 2 4 2⎠ ⎝2 ⎠

α–1 ( D 0+, t u ) ( x, α–2

( D 0+, t u ) ( x, 0+ ) = f 2 ( x ),

x∈R

m

(32)

α

α

m

Vol. 73

No. 1

α

1 --2- – 2 x – --2α α t ) = ------t ϕ ⎛ – ---, --- – 1 ; – -----t ⎞ . ⎝ 2 2 2λ λ ⎠

(38)

∫ G ( x – τ, t ) f ( τ ) dτ, R

– τ, t ) f 1 ( τ ) + G 2 ( x – τ, t ) f 2 ( τ ) ] dτ, (33)

DOKLADY MATHEMATICS

(37)

Corollary 4. If 1 < α < 2 and λ > 0, then the Cauchytype problem for Eq. (5) with initial conditions (32) (where m = 1) is solvable, and its solution is given by α (33) with m = 1, where G 1 (x, t) is defined by (37) and

u ( x, t ) =

u ( x, t )

R

α

1 --2- – 1 α α x – --2α G 1 ( x, t ) = ------t ϕ ⎛ – ---, ---; – -----t ⎞ . ⎝ 2 2 λ ⎠ 2λ

For α = 1, Theorem 3 implies the known solution

is solvable, and its solution has the form

α 1 (x

0+ ) = f ( x ), x ∈ R, t > 0

α

(31)

α G 2 ( x,

0+ ) = f 1 ( x ),

(36)

is solvable, and its solution is given by (30) with m = 1, where

Theorem 4. If 1 < α < 2, m ∈ N, and λ > 0, then the Cauchy-type problem for Eq. (1) with initial conditions

∫ [G

Corollary 3. If 0 < α ≤ 1 and λ > 0, then the Cauchytype problem

m

provided that the integral in (30) converges.

=

j

z ---------------------------, a > 0, b > 0, z ∈ C, (35) j!Γ ( aj + b )

and Theorems 3 and 4 imply the corresponding results for Eq. (5).

G 1 ( x – τ, t ) f ( τ ) dτ,

λ

× H 2, 2

∑ j=0

(29)

is solvable, and its solution has the form u ( x, t ) =

∞

ϕ ( a, b; z ) =

( D 0+, t u ) ( x, t ) = λ ( ∆ x u ) ( x, t ), ( D 0+, t u ) ( x, 0+ ) = d ( x ),

2, 0

For m = 1, the H 2, 2 -functions in (31) and (34) reduce to the special Wright function [8, 18.1(21)]

2006

m

x m – ---- – ---------2 4λ t 2

1 G ( x, t ) = ---------------------m- t e ( 2λ π )

(39)

10

VOROSHILOV, KILBAS

REFERENCES

to the Cauchy problem for heat equation (3): ∂u ( x, t ) 2 ------------------- = λ ( ∆ x u ) ( x, t ), ∂t x∈R ,

u ( x, 0 ) = f ( x ),

(40)

t > 0.

m

In particular, for m = 1, the Cauchy problem ∂u ( x, t ) 2 ∂ u ( x, t ) -, ------------------- = λ -------------------2 ∂t ∂x 2

x ∈ R,

u ( x, 0 ) = f ( x ),

2. I. Podlubny, Fractional Differential Equations (Academic, San Diego, 1999).

(41)

t>0

3. A. Carpinteri and F. Mainardi, Fractals and Fractional Calculus in Continuum Mechanics (Springer, Wien, 1997). 4. R. Hilter, Applications of Fractional Calculus in Physics (World Sci., Singapore, 2000).

has the solution ∞

u ( x, t ) =

1. S. G. Samko, A. A. Kilbas, and O. I. Marichev, Fractional Integrals and Derivatives and Some of Their Applications (Nauka i Tekhnika, Minsk, 1987) [in Russian].

5. A. A. Kilbas and J. J. Trujillo, Appl. Anal. 81 (2), 435– 493 (2002).

∫ G ( x – τ, t ) f ( τ ) dτ,

–∞

x 1 – --- – ---------2 2 4λ t

1 G ( x, t ) = -------------- t e 2λ π

(42) .

ACKNOWLEDGMENTS The authors thank the participants of the seminar of Academician S.M. Nikol’skii, Corresponding Member of the Russian Academy of Sciences O.V. Besov, and Corresponding Member of the Russian Academy of L.D. Kudryavtsev at the function theory division of Steklov Institute of Mathematics, Russian Academy of Sciences, for useful discussions. This work was supported by the Foundation for Fundamental Research of Republic Belarus (project no. F05MS-050).

6. V. A. Ditkin and A. P. Prudnikov, Integral Transforms and Operational Calculus (Pergamon, Oxford, 1966; Nauka, Moscow, 1974). 7. S. M. Nikol’skii, Course in Mathematical Analysis (Nauka, Moscow, 1983) [in Russian]. 8. Higher Transcendental Functions (Bateman Manuscript Project), Ed. by A. Erdelyi (McGraw-Hill, New York, 1955; Nauka, Moscow, 1967), Vol. 3. 9. A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, Integrals and Series, Vol. 3: Supplementary Chapters (Nauka, Moscow, 1986) [in Russian]. 10. A. A. Kilbas and M. Saigo, H-Transforms. Theory and Applications (Chapman & Hall, Boca Raton, 2004). 11. Higher Transcendental Functions (Bateman Manuscript Project), Ed. by A. Erdelyi (McGraw-Hill, New York, 1953; Nauka, Moscow, 1966), Vol. 2.

DOKLADY MATHEMATICS

Vol. 73

No. 1

2006

MATHEMATICS

A Cauchy-Type Problem for the Diffusion-Wave Equation with Riemann–Liouville Partial Derivative A. A. Voroshilov and A. A. Kilbas Presented by Academician S.M. Nikol’skii April 13, 2005 Received May 26, 2005

DOI: 10.1134/S1064562406010029

Consider the linear differential equation α

( D 0+, t u ) ( x, t ) = λ ∆ x u ( x, t ), 2

x∈R , n

α 2 ∂ u ( x, t ) - , x ∈ R, t > 0; λ > 0. (5) ( D 0+, t u ) ( x, t ) = λ -------------------2 ∂x 2

t > 0, (1)

The interest to Eqs. (1) and (5) is caused by their numerous applications to solving diffusion problems of physics, mechanics, and other applied sciences [2, Sections 4.2.1 and 4.2.2; 3; 4] (historical information and a review of the related results are contained in survey [5, Ch. 7]).

α

where λ > 0; ( D 0+, t u)(x, t) is the partial Riemann–Liouville fractional derivative of order α > 0 of the function u(x, t) with respect to the second variable [1, p. 342], i.e., t

α ( D 0+, t u ) ( x,

α > 0;

1 u ( x , τ ) dτ ∂ n -, t ) = ⎛ -----⎞ --------------------- ---------------------------⎝ ∂t⎠ Γ ( n – α ) ( t – τ ) α – n + 1 0

∫

n = [ α ] + 1;

x∈R ;

In this paper, we solve Eq. (1) of order α > 0 with initial conditions

(2)

t > 0,

m

α–k

( D 0+, t u ) ( x, 0+ ) = f k ( x ),

and ∆x is the Laplace operator in the first variable m

x = (x1, x2, …, xm) ∈ Rm, i.e., ∆x =

k = 1, 2, …, n = – [ – α ]; x ∈ R .

∂ u --------2 . ∂x j j=1

∑

2

α–k

Here, ( D 0+, t u)(x, 0+) with n – 1 < α < n is understood as

∂u ( x, t ) 1 If α = 1 and α = 2, then ( D 0+, t u)(x, t) = ------------------- and ∂t

α–k

t → 0+

2

2 ∂u ------ = λ ∆ x u ( x, t ), ∂t

x∈R , n

t > 0,

k = 1, 2, …, n – 1, α–n

2 ∂ u -------2- = λ ∆ x u ( x, t ), ∂t

x∈R , n

t > 0.

n–α

( D 0+, t u ) ( x, 0+ ) = lim ( I 0+, t u ) ( x, t ); t → 0+

(7)

(8)

for α = n ∈ N, it equals

(3)

and for α = 2, it coincides with the wave equation 2

α–k

( D 0+, t u ) ( x, 0+ ) = lim ( D 0+, t u ) ( x, t ),

∂ u ( x, t ) - , respectively; thus, for α = 1, t) = -------------------2 ∂t Eq. (1) coincides with the heat (diffusion) equation 2 ( D 0+, t u)(x,

(6) m

∂ -u ( x, 0 ), 0+ ) = ----------n–k ∂t k = 1, 2, …, n, n–k

n–k ( D 0+, t u ) ( x,

(4)

(9)

n–α

For this reason, Eq. (1) is called the diffusion-wave equation [2, p. 146]. In particular, for m = 1, Eq. (1) takes the form

where ( I 0+, t u)(x, t) is the partial Riemann–Liouville fractional integral of order n – α [1, p. 341], that is, t

n–α ( I 0+, t u ) ( x,

Belarussian State University, pr. F. Skoriny 4, Minsk, 220050 Belarus e-mail: [email protected], [email protected]

1 u ( x , τ ) dτ -, t ) = --------------------- ---------------------------Γ ( n – α ) ( t – τ )1 – n + α 0

∫

( I 0+, t u ) ( x, t ) = u ( x, t ). 0

6

α < n,

(10)

A CAUCHY-TYPE PROBLEM FOR THE DIFFUSION-WAVE EQUATION

If α = n ∈ N, then, according to (2) and (9), problem (1), (6) takes the form of the Cauchy problem ∂ u ( x, t ) 2 -------------------- = λ ∆ x u ( x, t ), n ∂t n

∂ u ( x, 0 ) --------------------------- = f k ( x ), n–k ∂t n–k

x∈R , m

(11)

∫ u ( x, t )e

– st

dt,

x∈R ,

s ∈ C, (13)

m

0

∫ u ( x, t )e R

ix · σ

d x,

σ∈R , m

m

t > 0, (14)

and their inverse transforms with respect to s ∈ C and σ ∈ Rm: –1 ( L s u ) ( x,

1 t ) = -------2πi x∈R , m

γ + i∞

∫

e u ( x, s ) ds, st

γ – i∞

R

m

i

i=1

σ = (σ1, σ2, …, σm) ∈ Rm; γ ∈ R is a fixed real number. The properties of the direct and inverse Laplace and Fourier transforms can be found in, e.g., [6, Chapter II; 7, Chapter 16]. In particular, the operators in (15), (13) and (16), (14) are mutually invertible for sufficiently good functions u(x, t). Suppose that, for the functions fk(x) (k = 1, 2, …, n) in (6), the Fourier transforms (Fx fk)(σ) are defined. Applying Laplace transform (13) to both sides of Eq. (1) and taking into account the expression for the Laplace transform of partial Riemann–Liouville fractional derivative (2) [2, (2.248)] and initial conditions (6), we obtain n

∑s

k–1

f k ( x ) = λ ∆ x ( L t u ) ( x, s ). 2

k=1

Applying Fourier transform (14) to this equality and using the expression for the Fourier transform of the DOKLADY MATHEMATICS

=

∑σ

2 i .

Applying inverse Laplace trans-

i=1

form (15) and Fourier transform (16), we obtain a solution u(x, t) to initial problem (1), (6); this is ⎛ –1 –1 u( x, t) = ⎜ L s F σ ⎝

k–1 ⎞ s ------------------------2(F x f k)(σ) ⎟ ( x, t). (18) α 2 s +λ σ ⎠ k=1 n

∑

Let us express solution (18) in terms of the MittagLeffler special function [8, 18.1(18)] ∞

E α, β ( z ) =

∑ j=0

j

z ------------------------ , α > 0, β > 0, z ∈ C, (19) Γ ( αj + β )

which is an entire function of z. The formula for the Laplace transform of this function (see, e.g., [1, (1.93)]) implies α–k

k–1

s 2 2 α -, E α, α – k + 1 ( – λ σ t ) ] ) ( s ) = -----------------------α 2 2 s + λ σ (20)

k = 1, 2, …, n; s ∈ C, σ ∈ R , λ > 0; λ σ s

– 2/α

< 1.

Applying the inverse Laplace and Fourier transforms to both sides of (17) and taking into account (20), we obtain a solution to problem (1), (6) in the form

t > 0.

∑ x σi for x = (x1, x2, …, xm) ∈ Rm and

s ( L t u ) ( x, s ) –

s ∈ C,

m

(16)

m

α

where

|σ|2

( Lt [ t

∫

x∈R ,

(15)

t > 0,

1 – iσ ⋅ x –1 ( F σ u ) ( x, t ) = -------------m- u ( σ, t )e dσ, ( 2π ) m

Here, x · σ =

σ∈R ,

(17)

m

∞

( F x u ) ( σ, t ) =

∑ m

m

k–1

s ------------------------2 ( F x f k ) ( σ ), α 2 s + λ σ k=1

( F x L t u ) ( σ, s ) =

k = 1, 2, …, n; x ∈ R . (12)

For this reason, by analogy, problem (1), (6) is called a Cauchy-type problem. To solve problem (1), (6), we apply, respectively, the Laplace and Fourier transforms of the function u(x, t) with respect to t > 0 and x ∈ Rm, which are ( L t u ) ( x, s ) =

operator ∆x: (Fx∆xu)(σ, s) = –|σ|2(Fxu)(σ, s), we come to the relation n

t > 0,

7

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2006

n

u ( x, t ) =

∑

α–k

t -------------m- E α, α – k + 1 ( 2π ) m k=1

∫

R

2 α

× ( – λ σ t ) ( F x f k ) ( σ )e 2

– ix ⋅ σ

dσ.

(21)

Theorem 1. Suppose that, for the functions fk(x) with k = 1, 2, …, n, the Fourier transforms (Fx fk)(σ) are defined and the integrals on the right-hand side of (21) converge. Then, Cauchy-type problem (1), (6) is solvable, and its solution is given by (21). In particular, for m = 1, equality (21) gives a solution to problem (5), (6). Corollary 1. The solution to Cauchy problem (11), (12) is given by formula (21) with α = n provided that the integrals on the right-hand side of (21) converge. If the functions fk(x) with k = 1, 2, …, n are infinitely differentiable on Rm, then, substituting (19) into (21), interchanging integration and summation [which can

8

VOROSHILOV, KILBAS

be done because series (19) uniformly converges] and taking into account the equality 2, 0

H 2, 2 z

1 2 j – iσ ⋅ x ( ∆ f k ) ( x ) = -------------m- ( – σ ) ( F x f k ) ( σ )e dσ, ( 2π ) m

∫

j

⎛ a, 1---⎞ , ⎛ b, α ---⎞ ⎝ 2⎠ ⎝ 2 ⎠

R

j = 1, 2, …, for the jth powers of the Laplace operator

1 ( ∆x

j

1

j–1

n

u ( x, t ) =

∑

t

α–k

k=1

∞

∑ j=0

1 Γ ( c + τ )Γ ( d + τ/2 ) –τ = -------- --------------------------------------------------------z dτ. 2πi Γ ( a + τ/2 )Γ ( b + ατ/2 )

∫

= ∆x and

∆ x = ∆ x ∆ x for j = 2, 3, …), we obtain the representation of solution (21) in the form 2 α j

(λ t ) j ------------------------------------------ ( ∆ x f x ) ( x ). (22) Γ ( αj + α – k + 1 )

In particular, for m = 1, the solution to problem (5), (6) has the form n

u ( x, t ) =

∑ k=1

t

α–k

∞

∑ j=0

2 α j

(λ t ) (2 j) ------------------------------------------ f k ( x ). (23) Γ ( αj + α – k + 1 )

Corollary 2. The solution to Cauchy problem (11), (12) is given by formula (22) with α = n provided that the series on the right-hand side of (22) converge. If 0 < α < 2, then the solution to Eq. (1) with initial conditions (6) for 0 < α ≤ 1 and n = 1 or 1 < α < 2 and n = 2 can be expressed in terms of the so-called ç-function [9, Section 8.3; 10, Chapters 1 and 2]. The proof is based on the application of the inverse Fourier and Laplace transforms to (17) and on the expressions for the Fourier and Laplace transforms of the modified Bessel function Kv(z) of the third kind [11, 7.2(13)] and of the special ç-function

L

m ----

------------⎛ ⎞ 2π 2 c 2 -, K m – 2 ( c x ) ⎟ ( σ ) = ⎛ ------⎞ -----------------⎜ Fx x 2 2 ⎝ ⎠ ------------c (25) c + σ ⎝ ⎠ 2 2–m

σ∈R . m

α/2

s According to (25) with c = -------- and the theorem about λ the Fourier convolution, relation (17) takes the form ( F x L t u ) ( σ, s ) ⎛ = ⎜F x ⎜ ⎝

α ( m – 2) k – 1 + --------------------4

2–m α ⎞ ------------s x --2-⎞ 2 ⎛ --------------------------K m – 2 -----s * x f k( x) ⎟ (σ). x m ⎟ ⎠ ------------- ⎝ λ ---2 2 k=1 ⎠ λ ( 2λπ ) Applying inverse Fourier transform (16) to this equality, we obtain ( L t u ) ( x, s ) n

∑

n

α ( m – 2) k – 1 + --------------------4

= 2 π

– 1/2

s

α

2–m

------------s x --2-⎞ 2 ⎛ ------------------------------s * x f k( x) (26) = K x m – 2 m ⎠ ------------- ⎝ λ ---2 2 k=1 λ ( 2λπ ) for x ∈ Rm and s ∈ C. Lemma 2. If 0 < α ≤ 1 and k = 1 or 1 < α < 2 and k = 1, 2, then

∑

⎛ 1 α(m – 2) α ⎛m ⎜ ----, ---⎞ , ⎛ 1 – k – ----------------------, ---⎞ α α(m – 2) – --– k – ---------------------⎝ ⎠ ⎝ ⎜ 4 2 2⎠ 4 2 2, 0 x 4 H 2, 2 -----t ⎜ Lt t λ ⎜ 1 ⎛ 1--- – m ⎛m ----, ---⎞ ---- – 1, 1⎞ ⎜ ⎝ ⎝ ⎠ 2⎠ 2 4 2 ⎝ m ---2

(24)

Here, a, b, c, d ∈ R and L is a special infinite contour on the right of the poles of the gamma-functions Γ(c + τ) and Γ(d + τ/2). Theorem 1.1 from [10] implies the existence of the function defined by (24) for 0 < α < 2; π a, b, c, d ∈ R; and z ≠ 0 with | arg ( z)| < (2 – α) --- . 4 Lemma 1. For m ∈ N and c > 0,

Theorem 2. Suppose that the functions fk(x), where k = 1, 2, …, n, are infinitely differentiable on Rm and the series on the right-hand side of (22) converge. Then, Cauchy-type problem (1), (6) is solvable, and its solution is given by formula (22).

1 ( c, 1 ), ⎛ d, ---⎞ ⎝ 2⎠

α(m – 2) k – 1 + ---------------------4

⎞ ⎟ ⎟ ⎟ (s) ⎟ ⎟ ⎠

α

x --2K m – 2 ⎛ -----s ⎞ . ⎠ ------------- ⎝ λ 2 DOKLADY MATHEMATICS

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No. 1

2006

A CAUCHY-TYPE PROBLEM FOR THE DIFFUSION-WAVE EQUATION

Applying inverse Laplace transform (15) to (26) and taking into account (27), we obtain the following solution to problem (1), (6): –m

2–m ------------2

1 + ---2

------------2 k=1

2 x u ( x, t ) = -----------------------m m–1 λ

2, 0 × H 2, 2

π

n

∑t

α

x – --2-----t λ ⎛m ---- – 1, 1⎞ ⎝2 ⎠

α

where G 1 (x, t) is given by (31) and α

G 2 ( x, t ) =

α(m – 2) – k – ---------------------4

1 α(m – 2) α ⎛m ----, ---⎞ , ⎛ 1 – k – ----------------------, ---⎞ ⎝ 4 2⎠ ⎝ 2⎠ 4 1 ⎛ 1--- – m ----, ---⎞ ⎝ 2 4 2⎠

2–m ------------α(m – 2) 2 – 2 – ---------------------2 x 4 ------------------------t m m–1 1 + ---- ------------2 2 –m

λ

× * x f k ( x ), (28)

9

2, 0 H 2, 2

π

1 α(m – 2) α ⎛m ----, ---⎞ , ⎛ – 1 – ----------------------, ---⎞ α 2⎠ 4 x – --2- ⎝ 4 2⎠ ⎝ -----t λ 1 ⎛m ⎛ 1--- – m ---- – 1, 1⎞ ----, ---⎞ ⎝2 ⎠ ⎝ 2 4 2⎠

(34)

provided that the integral in (33) converges. where n = 1 for 0 < α ≤ 1 and n = 2 for 1 < α < 2. This and the Fourier convolution imply the following results. Theorem 3. If 0 < α ≤ 1, m ∈ N, and λ > 0, then the Cauchy-type problem α

2

α–1

x∈R

m

∫ R

α G 1 ( x,

t) =

2, 0

α

(30)

2 ∂ u ( x, t ) α -, ( D 0+, t u ) ( x, t ) = λ -------------------2 ∂x 2

2–m ------------α(m – 2) 2 – 1 – ---------------------2 x 4 ------------------------t m m–1 1 + ---- ------------2 2 –m

α–1 ( D 0+, t u ) ( x,

π

1 α(m – 2) α ⎛m ----, ---⎞ , ⎛ – ----------------------, ---⎞ α 2⎠ 4 x – --2- ⎝ 4 2⎠ ⎝ -----t , λ 1 1 m m ⎛ --- – ----, ---⎞ ⎛ ---- – 1, 1⎞ ⎝ 2 4 2⎠ ⎝2 ⎠

α–1 ( D 0+, t u ) ( x, α–2

( D 0+, t u ) ( x, 0+ ) = f 2 ( x ),

x∈R

m

(32)

α

α

m

Vol. 73

No. 1

α

1 --2- – 2 x – --2α α t ) = ------t ϕ ⎛ – ---, --- – 1 ; – -----t ⎞ . ⎝ 2 2 2λ λ ⎠

(38)

∫ G ( x – τ, t ) f ( τ ) dτ, R

– τ, t ) f 1 ( τ ) + G 2 ( x – τ, t ) f 2 ( τ ) ] dτ, (33)

DOKLADY MATHEMATICS

(37)

Corollary 4. If 1 < α < 2 and λ > 0, then the Cauchytype problem for Eq. (5) with initial conditions (32) (where m = 1) is solvable, and its solution is given by α (33) with m = 1, where G 1 (x, t) is defined by (37) and

u ( x, t ) =

u ( x, t )

R

α

1 --2- – 1 α α x – --2α G 1 ( x, t ) = ------t ϕ ⎛ – ---, ---; – -----t ⎞ . ⎝ 2 2 λ ⎠ 2λ

For α = 1, Theorem 3 implies the known solution

is solvable, and its solution has the form

α 1 (x

0+ ) = f ( x ), x ∈ R, t > 0

α

(31)

α G 2 ( x,

0+ ) = f 1 ( x ),

(36)

is solvable, and its solution is given by (30) with m = 1, where

Theorem 4. If 1 < α < 2, m ∈ N, and λ > 0, then the Cauchy-type problem for Eq. (1) with initial conditions

∫ [G

Corollary 3. If 0 < α ≤ 1 and λ > 0, then the Cauchytype problem

m

provided that the integral in (30) converges.

=

j

z ---------------------------, a > 0, b > 0, z ∈ C, (35) j!Γ ( aj + b )

and Theorems 3 and 4 imply the corresponding results for Eq. (5).

G 1 ( x – τ, t ) f ( τ ) dτ,

λ

× H 2, 2

∑ j=0

(29)

is solvable, and its solution has the form u ( x, t ) =

∞

ϕ ( a, b; z ) =

( D 0+, t u ) ( x, t ) = λ ( ∆ x u ) ( x, t ), ( D 0+, t u ) ( x, 0+ ) = d ( x ),

2, 0

For m = 1, the H 2, 2 -functions in (31) and (34) reduce to the special Wright function [8, 18.1(21)]

2006

m

x m – ---- – ---------2 4λ t 2

1 G ( x, t ) = ---------------------m- t e ( 2λ π )

(39)

10

VOROSHILOV, KILBAS

REFERENCES

to the Cauchy problem for heat equation (3): ∂u ( x, t ) 2 ------------------- = λ ( ∆ x u ) ( x, t ), ∂t x∈R ,

u ( x, 0 ) = f ( x ),

(40)

t > 0.

m

In particular, for m = 1, the Cauchy problem ∂u ( x, t ) 2 ∂ u ( x, t ) -, ------------------- = λ -------------------2 ∂t ∂x 2

x ∈ R,

u ( x, 0 ) = f ( x ),

2. I. Podlubny, Fractional Differential Equations (Academic, San Diego, 1999).

(41)

t>0

3. A. Carpinteri and F. Mainardi, Fractals and Fractional Calculus in Continuum Mechanics (Springer, Wien, 1997). 4. R. Hilter, Applications of Fractional Calculus in Physics (World Sci., Singapore, 2000).

has the solution ∞

u ( x, t ) =

1. S. G. Samko, A. A. Kilbas, and O. I. Marichev, Fractional Integrals and Derivatives and Some of Their Applications (Nauka i Tekhnika, Minsk, 1987) [in Russian].

5. A. A. Kilbas and J. J. Trujillo, Appl. Anal. 81 (2), 435– 493 (2002).

∫ G ( x – τ, t ) f ( τ ) dτ,

–∞

x 1 – --- – ---------2 2 4λ t

1 G ( x, t ) = -------------- t e 2λ π

(42) .

ACKNOWLEDGMENTS The authors thank the participants of the seminar of Academician S.M. Nikol’skii, Corresponding Member of the Russian Academy of Sciences O.V. Besov, and Corresponding Member of the Russian Academy of L.D. Kudryavtsev at the function theory division of Steklov Institute of Mathematics, Russian Academy of Sciences, for useful discussions. This work was supported by the Foundation for Fundamental Research of Republic Belarus (project no. F05MS-050).

6. V. A. Ditkin and A. P. Prudnikov, Integral Transforms and Operational Calculus (Pergamon, Oxford, 1966; Nauka, Moscow, 1974). 7. S. M. Nikol’skii, Course in Mathematical Analysis (Nauka, Moscow, 1983) [in Russian]. 8. Higher Transcendental Functions (Bateman Manuscript Project), Ed. by A. Erdelyi (McGraw-Hill, New York, 1955; Nauka, Moscow, 1967), Vol. 3. 9. A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, Integrals and Series, Vol. 3: Supplementary Chapters (Nauka, Moscow, 1986) [in Russian]. 10. A. A. Kilbas and M. Saigo, H-Transforms. Theory and Applications (Chapman & Hall, Boca Raton, 2004). 11. Higher Transcendental Functions (Bateman Manuscript Project), Ed. by A. Erdelyi (McGraw-Hill, New York, 1953; Nauka, Moscow, 1966), Vol. 2.

DOKLADY MATHEMATICS

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2006

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