Particular solutions of a certain class of associated Cauchy-Euler fractional partial differential equations via fractional calculus

  • Shy-Der Lin1Email author,

    Affiliated with

    • Chia-Hung Lu1 and

      Affiliated with

      • Shou-Mei Su1

        Affiliated with

        Boundary Value Problems20132013:126

        DOI: 10.1186/1687-2770-2013-126

        Received: 30 January 2013

        Accepted: 1 May 2013

        Published: 16 May 2013

        Abstract

        In recent years, various operators of fractional calculus (that is, calculus of integrals and derivatives of arbitrary real or complex orders) have been investigated and applied in many remarkably diverse fields of science and engineering. Many authors have demonstrated the usefulness of fractional calculus in the derivation of particular solutions of a number of linear ordinary and partial differential equations of the second and higher orders. The purpose of this paper is to present a certain class of the explicit particular solutions of the associated Cauchy-Euler fractional partial differential equation of arbitrary real or complex orders and their applications as follows:

        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equa_HTML.gif

        where u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq1_HTML.gif; A, B, C, M, N, α and β are arbitrary constants.

        MSC: 26A33, 33C10, 34A05.

        Keywords

        fractional calculus differential equation partial differential equation generalized Leibniz rule analytic function index law linearity property principle value initial and boundary value

        Dedication

        Dedicated to Professor Hari M Srivastava.

        1 Introduction, definitions and preliminaries

        The subject of fractional calculus (that is, derivatives and integrals of any real or complex order) has gained importance and popularity during the past two decades or so, due mainly to its demonstrated applications in numerous seemingly diverse fields of science and engineering (cf. [115]). By applying the following definition of a fractional differential (that is, fractional derivative and fractional integral) of order ν R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq2_HTML.gif, many authors have obtained particular solutions of a number of families of homogeneous (as well as nonhomogeneous) linear fractional differ-integral equations.

        In this paper, we present a direct way to obtain explicit solutions of such types of the associated Cauchy-Euler fractional partial differential equation with initial and boundary values. The results are a coincidence that the solutions are obtained by the methods applying the Laplace transform with the residue theorem. In this paper, we present some useful definitions and preliminaries for the paper as follows.

        Definitions 1.1 (cf. [610])

        If the function f ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq3_HTML.gif is analytic and has no branch point inside and on C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq4_HTML.gif, where
        C : = { C , C + } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ1_HTML.gif
        (1.1)
        C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq5_HTML.gif is an integral curve along the cut joining the points z and + i T ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq6_HTML.gif, C + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq7_HTML.gif is an integral curve along the cut joining the points z and + i T ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq8_HTML.gif,
        f ν ( z ) = C f ν ( z ) : = Γ ( ν + 1 ) 2 π i C f ( ζ ) d ζ ( ζ z ) ν + 1 ( ν R / Z ; Z : = { 1 , 2 , 3 , } ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ2_HTML.gif
        (1.2)
        and
        f n ( z ) : = lim ν n { f ν ( z ) } ( n N : = { 1 , 2 , 3 , } ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ3_HTML.gif
        (1.3)
        where ζ z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq9_HTML.gif,
        π arg ( ζ z ) π for  C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ4_HTML.gif
        (1.4)
        and
        0 arg ( ζ z ) 2 π for  C + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ5_HTML.gif
        (1.5)
        then f ν ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq10_HTML.gif ( ν > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq11_HTML.gif) is said to be the fractional derivative of f ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq3_HTML.gif of order ν and f ν ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq10_HTML.gif ( ν < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq12_HTML.gif) is said to be the fractional integral of f ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq3_HTML.gif of order −ν, provided that
        | f ν ( z ) | < ( ν R ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ6_HTML.gif
        (1.6)

        First of all, we find it is worthwhile to recall here the following useful lemmas and properties associated with the fractional differ-integration which is defined above.

        Lemma 1.1 (Linearity property)

        If the functions f ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq3_HTML.gif and g ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq13_HTML.gif are single-valued and analytic in some domain Ω C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq14_HTML.gif, then
        ( k 1 f + k 2 g ) ν = k 1 f ν + k 2 g ν ( ν R , z Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ7_HTML.gif
        (1.7)

        for any constants k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq15_HTML.gif and k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq16_HTML.gif.

        Lemma 1.2 (Index law)

        If the function f ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq3_HTML.gif is single-valued and analytic in some domain Ω C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq14_HTML.gif, then
        ( f μ ) ν = f μ + ν = ( f ν ) μ ( f μ 0 , f ν 0 , μ , ν R , z Ω ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ8_HTML.gif
        (1.8)

        Lemma 1.3 (Generalized Leibniz rule)

        If the functions f ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq3_HTML.gif and g ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq13_HTML.gif are single-valued and analytic in some domain Ω C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq14_HTML.gif, then
        ( f g ) ν = n = 0 ( ν n ) f ν n g n ( ν R , z Ω ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ9_HTML.gif
        (1.9)

        where g n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq17_HTML.gif is the ordinary derivative of g ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq13_HTML.gif of order n ( n N 0 : = N { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq18_HTML.gif), it being tacitly assumed (for simplicity) that g ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq13_HTML.gif is the polynomial part (if any) of the product f ( z ) g ( z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq19_HTML.gif.

        Lemma 1.4 (Cauchy’s residue theorem)

        Let Ω be a simple connected domain, and let C be a simple closed positively oriented contour that lies in Ω. If f is analytic inside C and on C, expect at the point z 1 , z 2 , , z n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq20_HTML.gif that lie inside C, then
        C f ( z ) d z = 2 π i k = 1 n Res [ f , z k ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equb_HTML.gif
        (I) If f has a simple pole at z 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq21_HTML.gif, then
        Res [ f , z 0 ] = lim z z 0 ( z z 0 ) f ( z ) ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equc_HTML.gif
        (II) If f has a pole of order k at z 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq21_HTML.gif, then
        Res [ f , z 0 ] = 1 ( k 1 ) ! lim z z 0 d k 1 d z k 1 ( z z 0 ) k f ( z ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equd_HTML.gif

        Property 1.1 (cf. [610])

        For a constant a,
        ( e a z ) ν = a ν e a z ( a 0 , ν R , z C ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ10_HTML.gif
        (1.10)

        Proof The proofs between ‘ν is not an integer’ and ‘ν is an integer’ are not coincident, so we mention the proof as follows.

        In case of | arg a | < π / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq22_HTML.gif, we have
        ( e a z ) v = C ( e a z ) v = Γ ( v + 1 ) 2 π i C e a ζ ( ζ z ) v + 1 d ζ ( put  ζ z = η  and  a η = ξ , | arg η | π ) = a v e a z Γ ( v + 1 ) 2 π i e i ϕ ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( ϕ = arg a ) = a v e a z Γ ( v + 1 ) 2 π i ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( for  | ϕ | < π 2 ) = a v e a z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Eque_HTML.gif

        for | arg a | < π / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq23_HTML.gif since ( 0 + ) ξ ( v + 1 ) e ξ d ξ = 2 π i Γ ( v + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq24_HTML.gif.

        In case of π / 2 < | arg a | π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq25_HTML.gif, we have
        ( e a z ) v = C + ( e a z ) v = Γ ( v + 1 ) 2 π i C + e a ζ ( ζ z ) v + 1 d ζ ( put  ζ z = η  and  a η = ξ , 0 arg η 2 π ) = a v e a z Γ ( v + 1 ) 2 π i e i ϕ ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( ϕ = arg a ) = a v e a z Γ ( v + 1 ) 2 π i ( 0 + ) ξ ( v + 1 ) e ξ d ξ ( for  π 2 < | ϕ | π ) = a v e a z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equf_HTML.gif

        Therefore we have Property 1.1 for arbitrary a 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq26_HTML.gif. □

        Property 1.2 For a constant a,
        ( e a z ) ν = e i π ν a ν e a z ( a 0 , ν R , z C ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ11_HTML.gif
        (1.11)
        Property 1.3 For a constant a,
        ( z a ) ν = e i π ν Γ ( ν a ) Γ ( a ) z a ν ( ν R , z C , | Γ ( ν a ) Γ ( a ) | < ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ12_HTML.gif
        (1.12)

        Property 1.4 (cf. [2, 16, 17])

        The fractional derivative of a causal function f ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq27_HTML.gif is defined by
        d α d t α f ( t ) : = { f ( n ) ( t ) if α = n N , 1 Γ ( n α ) 0 t f ( n ) ( τ ) ( t τ ) α + 1 n d τ if n 1 < α < n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ13_HTML.gif
        (1.13)

        where f ( n ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq28_HTML.gif denotes the ordinary derivative of order n and Γ is the gamma function.

        The Laplace transform of a function f ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq27_HTML.gif is denoted as
        L { f ( t ) } ( s ) = 0 + e s t f ( t ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ14_HTML.gif
        (1.14)
        where s is the Laplace complex parameter. We recall from the fundamental formula (cf. [16])
        L { d α d t α f ( t ) } ( s ) = s α L { f ( t ) } ( s ) k = 1 n s α k f ( k 1 ) ( 0 ) , n 1 < α n , n N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ15_HTML.gif
        (1.15)

        2 Main results

        Theorem 2.1 The fractional partial differential equation
        A x 2 2 u x 2 + B x u x + C u = M α u t α + N β u t β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ16_HTML.gif
        (2.1)

        with u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq1_HTML.gif, n 1 < α , β n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq29_HTML.gif, n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq30_HTML.gif and A (≠0), B, C, M, N are constants, has its solutions of the form given by

        (a)
        u ( x , t ) = e λ t [ k 1 x A B + ( A B ) 2 4 A ( C M λ α N λ β ) 2 A + k 2 x A B ( A B ) 2 4 A ( C M λ α N λ β ) 2 A ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ17_HTML.gif
        (2.2)

        when the discriminant ( A B ) 2 4 A ( C M λ α N λ β ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq31_HTML.gif;

        (b)
        u ( x , t ) = e λ t [ k 1 x m + k 2 x m ln x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equg_HTML.gif

        when the discriminant ( A B ) 2 4 A ( C M λ α N λ β ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq32_HTML.gif, and the roots m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq33_HTML.gif, m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq34_HTML.gif of Equation (2.5) are repeated; that is, m 1 = m 2 = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq35_HTML.gif;

        (c)
        u ( x , t ) = e λ t [ k 1 x a + b i + k 2 x a b i ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equh_HTML.gif

        when the discriminant ( A B ) 2 4 A ( C M λ α N λ β ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq36_HTML.gif, and a + b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq37_HTML.gif, a b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq38_HTML.gif are the conjugate pair roots of Equation (2.5).

        Proof Suppose that u ( x , t ) = x m e λ t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq39_HTML.gif. We have
        u x = m x m 1 e λ t , 2 u x 2 = m ( m 1 ) x m 1 e λ t , α u t α = λ α x m e λ t , and β u t β = λ β x m e λ t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ18_HTML.gif
        (2.3)
        So that the given equation (2.1) becomes
        A m ( m 1 ) x m e λ t + B m x m e λ t + C x m e λ t M λ α x m e λ t N λ β x m e λ t = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ19_HTML.gif
        (2.4)
        Equation (2.4) leads to the auxiliary equation
        A m ( m 1 ) + B m + C M λ α N λ β = 0 ( or  A m 2 ( A B ) m + ( C M λ α N λ β ) = 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ20_HTML.gif
        (2.5)
        That is,
        m 1 = A B + ( A B ) 2 4 A ( C M λ α N λ β ) 2 A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equi_HTML.gif
        and
        m 2 = A B ( A B ) 2 4 A ( C M λ α N λ β ) 2 A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equj_HTML.gif

        are the two roots of the auxiliary equation (2.5). Thus, u ( x , t ) = i = 1 2 k i x m i e λ t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq40_HTML.gif is a solution of the fractional partial differential equation (2.1) whenever m i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq41_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq42_HTML.gif) is a solution of the auxiliary equation (2.5).

        There are three different cases to be considered, depending on whether the roots of this quadratic equation (2.5) are distinct real roots, equal real roots (repeated real roots), or complex roots (roots appear as a conjugate pair). The three cases are due to the discriminant of the coefficients ( A B ) 2 4 A ( C M λ α N λ β ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq43_HTML.gif.

        • Case I: Distinct real roots (when ( A B ) 2 4 A ( C M λ α N λ β ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq44_HTML.gif).

        Let m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq33_HTML.gif and m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq34_HTML.gif denote the real roots of Equation (2.5) such that m 1 m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq45_HTML.gif. Then the general solution of Equation (2.1) is
        u ( x , t ) = e λ t ( k 1 x m 1 + k 2 x m 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equk_HTML.gif

        where k i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq46_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq42_HTML.gif) are constants.

        • Case II: Repeated real roots (when ( A B ) 2 4 A ( C M λ α N λ β ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq47_HTML.gif).

        If the roots of Equation (2.5) are repeated, that is, m 1 = m 2 = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq35_HTML.gif, then the general solution of Equation (2.1) is
        u ( x , t ) = e λ t ( k 1 x m + k 2 x m ln x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equl_HTML.gif

        where k i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq46_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq48_HTML.gif) are constants.

        • Case III: Conjugate complex roots (when ( A B ) 2 4 A ( C M λ α N λ β ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq49_HTML.gif).

        If the roots of Equation (2.5) are the conjugate pair m 1 = a + b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq50_HTML.gif and m 2 = a b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq51_HTML.gif, then a solution of Equation (2.1) is
        u ( x , t ) = e λ t ( k 1 x a + b i + k 2 x a b i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equm_HTML.gif

        where k i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq46_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq48_HTML.gif) are constants.

        In general,
        u ( x , t ) = e λ t [ k 1 x A B + ( A B ) 2 4 A ( C M λ α N λ β ) 2 A + k 2 x A B ( A B ) 2 4 A ( C M λ α N λ β ) 2 A ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equn_HTML.gif

        forms a fundamental solution, where k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq15_HTML.gif, k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq16_HTML.gif and λ are constants. □

        Theorem 2.2 The fractional partial differential equation
        A 2 u x 2 + B u x + C u = M α u t α + N β u t β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ21_HTML.gif
        (2.6)
        with u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq1_HTML.gif, n 1 < α , β n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq29_HTML.gif, n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq30_HTML.gif and A (≠0), B, C, M, N are constants, has its solutions of the form given by(a′)
        u ( x , t ) = e λ t [ k 1 e ( B + B 2 4 A ( C M λ α N λ β ) 2 A ) x + k 2 e ( B B 2 4 A ( C M λ α N λ β ) 2 A ) x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ22_HTML.gif
        (2.7)
        when the discriminant B 2 4 A ( C M λ α N λ β ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq52_HTML.gif;(b′)
        u ( x , t ) = e λ t [ k 1 e m x + k 2 x e m x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equo_HTML.gif
        when the discriminant B 2 4 A ( C M λ α N λ β ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq53_HTML.gif, and the roots m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq33_HTML.gif, m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq34_HTML.gif of Equation (2.10) are repeated; that is, m 1 = m 2 = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq35_HTML.gif;(c′)
        u ( x , t ) = e λ t [ k 1 x ( a + b i ) x + k 2 x ( a b i ) x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equp_HTML.gif

        when the discriminant B 2 4 A ( C M λ α N λ β ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq54_HTML.gif, and a + b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq37_HTML.gif, a b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq38_HTML.gif are the conjugate pair roots of Equation (2.10).

        Proof The similarity between the forms of solutions of Equation (2.1) and solutions of a linear equation with constant coefficients of Equation (2.6) is not just a coincidence.

        Suppose that u ( x , t ) = e m x e λ t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq55_HTML.gif. We have
        u x = m e m x e λ t , 2 u x 2 = m 2 e m x e λ t , α u t α = λ α e m x e λ t , and β u t β = λ β e m x e λ t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ23_HTML.gif
        (2.8)
        So that the given equation (2.6) becomes
        A m 2 e m x e λ t + B m e m x e λ t + C e m x e λ t M λ α e m x e λ t N λ β e m x e λ t = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ24_HTML.gif
        (2.9)
        If m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq33_HTML.gif and m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq34_HTML.gif are the two roots of the auxiliary equation
        A m 2 + B m + C M λ α N λ β = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ25_HTML.gif
        (2.10)
        then
        m = m 1 = B + B 2 4 A ( C M λ α N λ β ) 2 A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equq_HTML.gif
        and
        m = m 2 = B B 2 4 A ( C M λ α N λ β ) 2 A . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equr_HTML.gif

        The analysis of three cases is similar to Theorem 2.1, we can obtain each solution of the forms as follows:

        u ( x , t ) = e λ t ( k 1 e m 1 x + k 2 e m 2 x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq56_HTML.gif with m 1 m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq45_HTML.gif, two distinct real roots,

        u ( x , t ) = e λ t ( k 1 e m x + k 2 x e m x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq57_HTML.gif with m 1 = m 2 = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq58_HTML.gif, repeated real roots, and

        u ( x , t ) = e λ t ( k 1 e ( a + i b ) x + k 2 e ( a i b ) x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq59_HTML.gif with the conjugate complex roots.

         □

        Remark The constant λ in Equations (2.2) and (2.7) can be solved directly by constant initial value and constant boundary values (or by the numerical methods).

        Corollary 2.1 The fractional partial differential equation
        A x 2 2 u x 2 + B x u x + C = M α u t α , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ26_HTML.gif
        (2.11)
        with u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq1_HTML.gif, n 1 < α n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq60_HTML.gif, n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq30_HTML.gif and A (≠0), B, C, M are constants, has its solutions of the form given by(a″)
        u ( x , t ) = e λ t [ k 1 x A B + ( A B ) 2 4 A ( C M λ α ) 2 A + k 2 x A B ( A B ) 2 4 A ( C M λ α ) 2 A ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ27_HTML.gif
        (2.12)
        when the discriminant ( A B ) 2 4 A ( C M λ α ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq61_HTML.gif;(b″)
        u ( x , t ) = e λ t [ k 1 x m + k 2 x m ln x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equs_HTML.gif
        when the discriminant ( A B ) 2 4 A ( C M λ α ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq62_HTML.gif, and the roots m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq33_HTML.gif, m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq34_HTML.gif of Equation (2.5) with N = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq63_HTML.gif are repeated; that is, m 1 = m 2 = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq35_HTML.gif;(c″)
        u ( x , t ) = e λ t [ k 1 e a + b i + k 2 e a b i ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equt_HTML.gif

        when the discriminant ( A B ) 2 4 A ( C M λ α ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq64_HTML.gif, and a + b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq37_HTML.gif, a b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq38_HTML.gif are the conjugate pair roots of Equation (2.5) with N = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq63_HTML.gif.

        Corollary 2.2 The fractional partial differential equation
        A 2 u x 2 + B u x + C u = M α u t α , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ28_HTML.gif
        (2.13)

        with u = u ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq1_HTML.gif, n 1 < α n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq60_HTML.gif, n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq30_HTML.gif and A (≠0), B, C, M are constants, has its solutions of the form given by

        ( a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq65_HTML.gif)
        u ( x , t ) = e λ t [ k 1 e ( B + B 2 4 A ( C M λ α ) 2 A ) x + k 2 e ( B B 2 4 A ( C M λ α ) 2 A ) x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ29_HTML.gif
        (2.14)

        when the discriminant B 2 4 A ( C M λ α ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq66_HTML.gif;

        ( b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq67_HTML.gif)
        u ( x , t ) = e λ t [ k 1 e m x + k 2 x e m x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equu_HTML.gif

        when the discriminant B 2 4 A ( C M λ α ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq68_HTML.gif, and the roots m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq33_HTML.gif, m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq34_HTML.gif of Equation (2.10) with N = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq63_HTML.gif are repeated; that is, m 1 = m 2 = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq35_HTML.gif;

        ( c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq69_HTML.gif)
        u ( x , t ) = e λ t [ k 1 e ( a + b i ) x + k 2 e ( a b i ) x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equv_HTML.gif

        when the discriminant B 2 4 A ( C M λ α ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq70_HTML.gif, and a + b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq37_HTML.gif, a b i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq38_HTML.gif are the conjugate pair roots of Equation (2.10) with N = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq63_HTML.gif.

        3 Examples

        Example 3.1 If the two-dimensional harmonic equation 1 2 u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq71_HTML.gif is transformed to plane polar coordinates r and θ, defined by x = r cos θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq72_HTML.gif, y = r sin θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq73_HTML.gif, it takes the form
        2 u r 2 + 1 r u r + 1 r 2 2 u θ 2 = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equ30_HTML.gif
        (3.1)
        then it has solutions of the form
        u ( r , θ ) = ( k 1 r i λ + k 2 r i λ ) e λ θ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equw_HTML.gif

        where k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq15_HTML.gif, k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq16_HTML.gif and λ are constants.

        Solution Equation (3.1) is coincident to
        r 2 2 u r 2 + r u r + 2 u θ 2 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equx_HTML.gif
        We have the solution
        u ( r , θ ) = ( k 1 r i λ + k 2 r i λ ) e λ θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equy_HTML.gif

        by taking A = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq74_HTML.gif, B = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq75_HTML.gif, C = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq76_HTML.gif, M = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq77_HTML.gif, N = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq63_HTML.gif and α = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq78_HTML.gif in Theorem 2.1. □

        Example 3.2 The fractional partial differential equation
        2 u x 2 + u x = α u t α with  n 1 < α n , n N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equz_HTML.gif
        Solution Putting A = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq74_HTML.gif, B = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq75_HTML.gif, C = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq76_HTML.gif and M = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq79_HTML.gif in Corollary 2.2, we obtain the solution
        u ( x , t ) = e λ t [ c 1 e ( 1 + 1 + 4 λ α 2 ) x + c 2 e ( 1 1 + 4 λ α 2 ) x ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equaa_HTML.gif

        where the discriminant 1 + 4 λ α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq80_HTML.gif.

        If 1 + 4 λ α = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq81_HTML.gif,
        u ( x , t ) = e ( 1 2 x + 1 64 t ) ( c 1 + c 2 x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equab_HTML.gif

        The analysis of the case 1 + 4 λ α < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq82_HTML.gif is similar to Theorem 2.2. □

        Example 3.3 The fractional partial differential equation
        x 2 2 u x 2 x u x = 2 1 2 u t 1 2 with  u ( 1 , t ) = u ( e , t ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equac_HTML.gif
        Solution Putting A = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq74_HTML.gif, B = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq83_HTML.gif, C = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq76_HTML.gif and M = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq84_HTML.gif and α = 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq85_HTML.gif in Corollary 2.1, we obtain the solution
        u ( x , t ) = x e λ t [ k 1 cos 4 × 2 λ 1 2 4 2 ln x + k 2 sin 4 × 2 λ 1 2 4 2 ln x ] = x e λ t [ k 1 cos 2 λ 1 2 1 ln x + k 2 sin 2 λ 1 2 1 ln x ] , u ( 1 , t ) = k 1 e λ t = 0 implies k 1 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equad_HTML.gif
        Then
        u ( x , t ) = k 2 x e λ t sin 2 λ 1 2 1 ln x , u ( e , t ) = k 2 e λ t + 1 sin 2 λ 1 2 1 = 0 implies that 2 λ 1 2 1 = n π , n Z . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equae_HTML.gif

        That is, λ = ( n 2 π 2 + 1 ) 2 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq86_HTML.gif.

        Thus, u ( x , t ) = k 2 x e ( n 2 π 2 + 1 ) 2 4 t sin n π ln x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq87_HTML.gif, n Z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq88_HTML.gif.

        If the discriminant 2 λ 1 2 1 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq89_HTML.gif, the solution is trivial. If the discriminant 2 λ 1 2 1 < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq90_HTML.gif, then the solution is
        u ( x , t ) = k 1 e ( n 2 + π 2 ) 2 t 4 ( x 1 + i n π x 1 i n π ) , n N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equaf_HTML.gif

         □

        Example 3.4 The fractional partial differential equation
        2 3 u t 2 3 = 1 4 2 u x 2 with  u ( 0 , t ) = e 2 t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equag_HTML.gif
        Solution Putting α = 2 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq91_HTML.gif, A = 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq92_HTML.gif, B = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq93_HTML.gif, C = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq76_HTML.gif and M = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq79_HTML.gif in Corollary 2.2, the discriminant is λ 2 3 = ( λ 1 3 ) 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq94_HTML.gif, but λ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq95_HTML.gif leads to a contradiction, hence there are different real roots m 1 = 2 λ 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq96_HTML.gif and m 2 = 2 λ 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq97_HTML.gif, so that we have
        u ( x , t ) = e λ t ( k 1 e 2 λ 1 3 x + k 2 e 2 λ 1 3 x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equah_HTML.gif
        By the boundary condition u ( 0 , t ) = e 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq98_HTML.gif, we obtain λ = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq99_HTML.gif and k 1 + k 2 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq100_HTML.gif. So,
        u ( x , t ) = e 2 t e 2 2 3 x 2 k 1 e 2 t sinh ( 2 2 3 x ) = e 2 t e 2 2 3 x + 2 k 2 e 2 t sinh ( 2 2 3 x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equai_HTML.gif
        and the particular solution is
        u ( x , t ) = e 2 t e 2 2 3 x ( or  u ( x , t ) = e 2 t e 2 2 3 x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equaj_HTML.gif
        If we apply the Laplace transform to ϕ ( t ) = u ( 0 , t ) = e 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq101_HTML.gif, then L { ϕ ( t ) } ( s ) = 1 s + 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq102_HTML.gif and u ˜ ( x , s ) = 1 s + 2 e 2 x s 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_IEq103_HTML.gif. Using the residue theorem,
        u ( x , t ) = 1 2 π i C e s t 1 s + 2 e 2 x s 1 3 d s = 1 2 π i 2 π i lim s 2 ( s + 2 ) e 2 x s 1 3 e s t s + 2 = e 2 t e 2 2 3 x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-126/MediaObjects/13661_2013_Article_380_Equak_HTML.gif

        The solution obtained by the method of Laplace transform and the residue theorem is a coincidence, which is our result above. □

        Declarations

        Acknowledgements

        The authors are deeply appreciative of the comments and suggestions offered by the referees for improving the quality and rigor of this paper. The present investigation was supported, in part, by the National Science Council of the Republic of China under Grant NSC-101-2115-M-033-002.

        Authors’ Affiliations

        (1)
        Department of Applied Mathematics, Chung Yuan Christian University

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