# On the Relationship between the Inhomogeneous Wave and Helmholtz Equations in a Fractional Setting

Authors: Mateo Dulce; Alexander Getmanenko
Journal: Abstract and Applied Analysis
 (2019)

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## Abstract

We study convergence of solutions of a space and time inhomogeneous fractional wave equation on the quarter-plane to the stationary regime described by solutions of the Helmholtz equation.


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## Body

## 1. Introduction and Main Result

Consider an inhomogeneous wave equation on the half-line with a time-periodic forcing term(1)uttx,t-uxxx,t=eiλtψx,t≥0,x≥0,where ψ(x) is, for simplicity, a C∞ function with compact support. Closely related to it is the one-dimensional Helmholtz equation(2)vxxx+λ2vx=-ψx.On the one hand, for any solution v(x) of the Helmholtz equation (2) the function v(x)eiλt solves the wave equation (1); on the other, there are pairs of solutions u(x,t) and v(x) of (1) and (2), respectively, such that u(x,t)-v(x)eiλt→0 as t→∞.Results of this type go back at least to [1] (for the wave equation in R3) and are known under the name of radiation principles. For an extensive treatment see the classical work [2]. The case of (1) will be recalled in Section 3.In this paper we study a very simple question that clarifies, to us, the role of the fractional analogue, (5) below, of the Helmholtz equation (2): to what extent does a similar result hold in fractional settings? For concreteness, we chose an inhomogeneous fractional wave equation on a half-line(3)Dxαux,t-DtβCux,t=εtfx,x,t≥0.Here Dxα denotes the Riemann-Liouville fractional derivative, DtβC denotes the Caputo fractional derivative defined in Section 2, and ε(t) is a fractional-harmonic function (see below), and 1<α,β<2. In various settings, (3) is a subject of active research; see [3] and related literature. The termfractional diffusion-wave equation is also used in the literature.By afractional-harmonic function ε(t) we mean in this context a solution of the fractional differential equation (FDE)(4)DtβCεt=Ωεt,where Ω∈C∖0 is a fixed constant. Taking a periodic ε(t), as in [4], does not seem to lead to a nice result.By [5, (3.1.34)], as ε(t) we can take any linear combination of Mittag-Leffler functions Eβ(Ωtβ) and tEβ,2(Ωtβ) defined in (14) below. This is quite natural since in the limiting case β=2, E2(-t2)=cos⁡t, and tEβ,2(-t2)=sin⁡t. We will work out only the case ε(t)=Eβ(Ωtβ); an arbitrary linear combination leads to similar results.Let nowv(x) be some function satisfying the fractional Helmholtz equation(5)Dxαvx-Ωvx=fx,x≥0.The classically known solutions of (5) are given in Theorem 4 below. Note that more recently related multidimensional equations have been solved, e.g., in [6] by the spectral method and in [7] in the form of an integral representation.Then(6)Dxαvxεt-DtβCvxεt=εtDxαvx-Ωvx=εtfx,and so the function(7)ux,t=vxEβΩtβsatisfies the fractional inhomogeneous wave equation (3).Conversely, do solutionsu(x,t) of (3) behave, for t→∞, as v(x)Eβ(Ωtβ)? Our main result shows that the answer is “not quite.”Before formulating the results, notice that, depending on the phase of the complex numberΩ, the function Eβ(Ωtβ) either is exponentially growing, or is bounded by a constant, or behaves as O(t-β) as t→∞; see [8, Theorems 1.3, 1.4]. Therefore any interesting result on the behavior of u(x,t)-v(x)Eβ(Ωtβ) for t→∞ ought to include information on asymptotic terms of order O(t-β) and smaller.The main results of this paper are as follows.Theorem 1.
Let1<β≤α<2, Ω∈C∖0, and ε(t)=Eβ(Ωtβ), and let f(x) be a C∞ function with compact support on [0,∞). Then there exist the following:
(a) A solutionu(x,t) of the fractional wave equation (3) satisfying u(x,0)=ut(x,0)=0, Dxα-2u(0+,t)=0
(b) A solutionv(x) of the fractional Helmholtz equation (5) satisfying Dxα-2v(0+)=0,
such that, ast→+∞,(8)ux,t-EβΩtβvx=sin⁡βππΓβCβ-1x1tβ+sin⁡β+β/αππΓβ+βαCβ-1+β/αx1tβ+β/α+sin⁡2βππΓ2βC2β-1x1t2β+O1tβ+2β/α,where the O symbol should be understood pointwise with respect to x and the functions Cβ-1, Cβ-1+β/α, and C2β-1 are given in (66), (67), and (68).Thus, ifΩ is such that Eβ(Ωtβ)=O(t-β), the right-hand side in (8) is of the same order as u(x,t).Theorem1 is proven in Section 4. The proof follows, with complications specific to fractional calculus, the outline of the case of the wave equation, Section 3, and of the paper [1].The caseβ>α is much less natural. Indeed, a prototype for this case would be a heat equation with aspace-harmonic source term and the limit as the space variable tends to infinity. We are not aware of any good result for such an equation which we would want to generalize to the fractional case. Nevertheless, if we impose a condition 3/2α>β>α, we are able to prove the following statement; see Section 5.Theorem 2.
Let1<α<β<min⁡{3α/2,2}, δ=α1-β-1, Ω∈C∖0, and ε(t)=EβΩtβ, and let fx be a C∞ function with compact support on [0,∞). Then there exist the following:
(a) A solutionu(x,t) of the fractional wave equation (3) satisfying u(x,0)=ut(x,0)=0,(9)ux,0=∫0xx-ξα-1Eα,αΩx-ξαfξdξ,utx,0=∫0xx-ξδ-1Eα,δΩx-ξαfξdξ,Dxα-2u0+,t=Dxα-1u0+,t=0;
(b) a solutionv(x) of the fractional Helmholtz equation (5) satisfying Dxα-2v(0+)=Dxα-1v(0+)=0,
such that, ast→+∞,(10)ux,t-EβΩtβvx=sin⁡βππΓβBβ-1x1tβ+sin⁡2βππΓ2βB2β-1x1t2β+O1t3β,where the O symbol should be understood pointwise with respect to x and the functions Bβ-1, B2β-1 are given in (89) and (90).The functionsu and v constructed in the proofs of the above theorems are such that the corresponding fractional derivatives exist in the sense of definitions (11) and (13) below, in which the integral is the Lebesgue integral and the derivatives are taken in the sense of elementary calculus. In this paper we do not address the questions of uniqueness of solutions and appropriate functional spaces where such uniqueness would hold; therefore we phrase our results in the language “there exists a solution...” rather than “the solution....”It would be curious to have an interpretation of the functionsCβ-1(x) and Bβ-1(x).


## 2. Preliminaries

In this section we collect some definitions and results from fractional calculus in order to make the paper self-contained.TheRiemann-Liouville fractional derivative of order α>-1, α∉Z, of a function f(x) is defined ([5, §2.1] for α>0 and, in different notation, [9] for -1<α<0) as(11)Dαfx=1Γn-αddxn∫0xfξx-ξα-n+1dξ,n=α+1,x>0.An additional subscript x in Dxα will emphasize that the differentiation is with respect to x.Remark 3.
Note that the usual formula for differentiation under the integral sign(12)ddx∫0xφx,ξdξ=φx,x+∫0x∂∂xφx,ξdξis not applicable in (11) because the derivative of the integrand is not integrable at zero.TheCaputo fractional derivative of order β>0, β∉Z, of a function f(t) can be defined [5, §2.4] by(13)DβCft=1Γm-β∫0tfmτt-τβ-m+1dτ,m=β+1,t>0.An additional subscript t in DtβC will emphasize that the differentiation is with respect to t.Some analytical conditions need to be imposed on the functionf to guarantee the existence of its fractional derivatives; see [5].The followingMittag-Leffler functions treated in detail in [5, §1.8] and [8, §1.2] play the same role for FDEs as does the function ex in the theory of integer-order differential equations:(14)Eαx=∑k=0∞xkΓαk+1,Eα,βx=∑k=0∞xkΓαk+β.There is the following result on FDEs with the Riemann-Liouville derivative.Theorem 4 (see [5, §3.1], [9]).
Ifα>0, α∉Z, n=[α]+1, f(x) is an integrable function on [0,b) for some b, and λ∈C is a constant, then the initial-value problem(15)Dαyx-λyx=fx;Dα-ky0+=bk,k=1,…,n,has a unique solution(16)yx=∑k=1nbkxα-kEα,α-k+1λxα+∫0xx-ξα-1Eα,αλx-ξαfξdξ.This theorem indicates the correct way to set up the initial conditions for equations with the Riemann-Liouville fractional derivative. By contrast, [5, §3.5], one sets up the initial value problem for an FDE with Caputo derivative, say, DβCyt=ft,yt in the usual way: y(k)(0)=bk, k=0,…,[β].


## 3. Case of the Classical Inhomogeneous Wave Equation

In this section we recall the case of the usual wave equation in order to motivate the fractional version of the argument and especially the formulas (30)-(31).It is well-known (see, e.g., [10, Lecture 13]) that the function(17)ux,t=12∫0tds∫x-t-sx+t-sdyfy,sis a solution of the initial value problem (18)utt-uxx=fx,t,t≥0,-∞<x<∞;ux,0=0,utx,0=0.Define the Laplace transforms with respect to t:(19)u~x,γ=∫0∞e-tγux,tdt;f~x,γ=∫0∞e-tγfx,tdt.Then (17), after simple manipulations, yields(20)u~x,γ=12γeγx∫x∞dξe-γξf~ξ,γ+12γe-γx∫-∞xdηeγηf~η,γ.By the method of images we know that iff(x,t) is originally defined for x≥0 and extended to the negative axis by f(-x,t)=-f(x,t), then u(x,t) from (17) solves the boundary-value problem on the half-line:(21)utt-uxx=fx,t,t≥0,x≥0;ux,0=0,utx,0=0,u0,t=0.In this case (20) becomes(22)u~x,γ=eγx2γ∫x∞dξe-γξf~ξ,γ+e-γx2γ∫0xdξeγξf~ξ,γ-e-γx2γ∫0∞dξe-γξf~ξ,γ.If the forcing termf(x,t) in (21) is periodic in time,(23)fx,t=ψxeiλt,f~x,γ=1γ-iλψx,we get(24)u~x,γ=12γγ-iλeγx∫x∞dξe-γξψξ+e-γx∫0xdξeγξψξ-e-γx∫0∞dξe-γξψξ.On the other hand, the Helmholtz equation on the half-line(25)vxxx-ω2vx=φx,x≥0,v0=0has a solution(26)vx=-12ωeωx∫x∞dξe-ωξφξ-12ωe-ωx∫0xdξeωξφξ+12ωe-ωx∫0∞dξe-ωξφξ.Now let us understand the behavior ofu(x,t) for large t, pointwise with respect to x, assuming supp⁡ψ⊂[0,M]. If t>x+M, the Bromwich integral that expresses u(x,t) via (24) can be closed on the left, yielding that u(x,t) is simply a sum of residues:(27)ux,t=eγt12-iλ∫x∞dξψξ+∫0xdξψξ-∫0∞dξψξ+eiλt12iλeiλx∫x∞dξe-iλξψξ+e-iλx∫0xdξeiλξψξ-e-iλx∫0∞dξe-iλξψξ.Looking at (26) we recognize that (28)ux,t=eiλtvx,t>x+M,where v(x) is a solution of the Helmholtz equation above with ω=iλ and φ=-ψ: (29)vxxx+λ2vx=-ψx.


## 4. Caseβ≤α

This section contains the proof of Theorem1. We begin by constructing a solution u(x,t) of the inhomogeneous fractional wave equation using the method of Laplace transform. Later on b(γ) in Lemma 5 will be specialized to the Laplace transform of a Mittag-Leffler function.Lemma 5.
Let
(a)f(x) be a continuous function with compact support
(b)b(γ) be an analytic function defined in C∖D(0,R)∪(-∞,0] estimated as O(γ-1) as γ→∞ in that region
(c)∫c-i∞c+i∞eγtb(γ)dγ be defined for t=0 in the sense of principal value
Then the function(30)ux,t=12πi∫c-i∞c+i∞etγFx,γdγ,t≥0,where(31)Fx,γ=bγ∫0xdξfξx-ξα-1Eα,αγβx-ξα-e-ξγβ/αxα-1Eα,αγβxα-bγ∫x∞dξe-ξγβ/αxα-1Eα,αγβxαfξfor x≥0 solves the FDE(32)Dxαux,t-DtβCux,t=fx12πi∫c-i∞c+i∞eγtbγdγ.Moreover, u(x,t) satisfies the boundary condition(33)Dxα-2u0+,t=0and(34)ux,0=utx,0=0.Remark that sincexE2,2(x2)=sinh⁡x, the function (31) is a fractional generalization of (22). While the statement of the lemma is quite natural on the algebraic level, various analytic justifications need to be carried out.Proof of the Lemma.
Denote(35)F1x,γ=∫0xdξfξx-ξα-1Eα,αγβx-ξα-e-ξγβ/αxα-1Eα,αγβxα,(36)F2x,γ=∫x∞dξe-ξγβ/αxα-1Eα,αγβxαfξ.
Let us first show that the integral (30) converges.
According to [8, Th.1.3], for γ→∞, Rγ≥0, (37)x-ξα-1Eα,αγβx-ξα~1αγβ1-α/αeγβ/αx-ξ-∑k=2∞γ-βkx-ξ1-kα-1Γα-αk;e-ξγβ/αxα-1Eα,αγβxα~e-ξγβ/α1αγβ1-α/αeγβ/αx-e-ξγβ/α∑k=2∞γ-kβx1-kα-1Γα-αkand so the bracket in (35) has an asymptotic expansion(38)e-ξγβ/α∑k=2∞γ-kβx1-kα-1Γα-αk-∑k=2∞γ-βkx-ξ1-kα-1Γα-αk.Using that Rγβ/α>0 for Rγ≥0 and that ξ≥0, we see that (38) is O(γ-2β) as γ→∞, Rγ≥0. Thus the integral ∫c-i∞c+i∞eγtb(γ)F1(x,γ)dγ converges.
Concerning theF2 term, integrate (36) by parts to get(39)F2x,γ=γ-β/αEα,αγβx-e-xγβ/αfx+∫x∞e-ξγβ/αf′ξdξ.Analogously to (37), we see that the right-hand side of (39) is O(γ-β/α·γβ(1-α)/α)=O(γ-β) for γ→∞, Rγ≥0. As b(γ)=O(γ-1) by assumptions of the lemma, the integral ∫c-i∞c+i∞eγtb(γ)F2(x,γ)dγ also converges.
Thus the definition ofu(x,t) by (30) makes sense.
Ift=0, we can close the integration contour in (30) on the right and obtain u(x,0)=0. The decay of the integrand in (30) for γ→∞ is sufficiently fast to allow the differentiation under the integral sign; putting t=0 in the integral(40)utx,t=12πi∫c-i∞c+i∞etγFx,γ·γdγand again closing the contour on the right, we obtain ut(x,0)=0.
Fort>0 the Bromwich integration contour in (30) can be replaced with the integration contour μ shown on Figure 1: (41)ux,t=12πi∫μetγFx,γdγwhere now(42)Fx,γ=bγ∫0xfξx-ξα-1Eα,αγβx-ξαdξ+xα-1Eα,αγβxα∫0∞e-ξγβ/αfξdξ.
By definition (11), since 1<α<2,(43)Dxαux,t=d2dx21Γ2-α∫0xuη,tx-η1-αdη=d2dx21Γ2-α∫0xdηx-η1-α12πi∫μdγeγtFη,γ.The assumptions of Fubini’s theorem are clearly satisfied for interchanging the order of integration with respect to dη and dγ (while keeping x fixed); thus(44)=d2dx21Γ2-α12πi∫μdγeγt∫0xdηx-η1-αFη,γ.
Claim(45) 44 = 1 Γ 2 - α 1 2 π i ∫ μ d γ e γ t d 2 d x 2 ∫ 0 x d η x - η 1 - α F η , γ .
Proof of the Claim. We have to work around the obstacle mentioned in Remark 3. We need to show that the first and the second x-derivatives in the integrand of (44) are integrable with respect to dγeγt.
Step A. The function(46)d2dx21Γ2-α∫0xdηx-η1-αFη,γ=DxαFη,γequals by Theorem 4(47)γβFx,γ+fxwhich is integrable along μ with respect to the measure eγtdγ, uniformly for x in compact sets.
Step B. Using the beta-integral, for x>ξ we find(48)∫η=ξη=xx-η1-αη-ξkα-1dη=Γ2-αΓkαΓ2+k-1αx-ξk-1α+1.Therefore, for each fixed value of γ∈C and x≥0,(49)∫0xdηx-η1-α∫0ηfξη-ξα-1Eα,αγβη-ξdξ=∫0xdηx-η1-α∫0ηfξ∑k=0∞γβkη-ξk+1α-1Γkα+αdξ(since the series converges uniformly with respect to the integration variables)(50)=∫0xdξfξ∑k=0∞Γ2-αx-ξkα+1γβkΓ2+kα=Γ2-α∫0xdξfξ·x-ξEα,2γβx-ξα.The integrand is regular enough to differentiate under the integral sign:(51)∂∂x50=∫0xdξfξ∑k=0∞Γ2-αx-ξkαγβkΓ1+kα=Γ2-α∫0xdξfξEαγβx-ξα,which is integrable along μ with respect to dγb(γ)eγt.
Step C. Similarly, using (48), (52)∫0xdηx-η1-αηα-1Eα,αγβηα=∫0xdηx-η1-α∑k=0∞γβkηk+1α-1Γkα+α=∑k=0∞Γ2-αxkα+1γβkΓ2+kα=Γ2-αxEα,2γβxα,and ∂/∂x (52) is also integrable along μ with respect to dγb(γ)eγt.
Step D. Adding the results of Steps B and C we conclude that also(53)ddx∫0xdηx-η1-αFη,yis integrable along μ with respect to eγtdγ, uniformly with respect to x on compact sets. Therefore two consecutive differentiations with respect to x can be carried out under the integral sign in (44) and the claim is proven.
Resuming the Proof of Lemma5. By the result of the claim and by Step A in its proof, we have(54)Dxαux,t=12πi∫μdγeγtγβFx,γ+fx.
By definition (13),(55)DtβCux,t=1Γ2-β∫0td2dτ2ux,τt-τ1-βdτand similarly to [5, (5.3.3)](56)DtβC∫μdγeγtFx,γ=∫μdγγβeγtFx,γ.Here we used the fact that convolutions interact with the Laplace integral in the usual way even if the contour of the Laplace integral is not rectilinear; see [11, Pré I.5].
Collecting the terms from (54) and (56), we obtain (32).
Finally, let us computeDxα-2ux,tx=0+. With the same analytical details as above,(57)Dxα-2ux,tx=0+=∫μeγtDxα-2Fx,γx=0+dγ=0,using Theorem 4.Figure 1
The contourμ in the γ-plane.Let us now subtract from the solution of the form (30) the solution of the type (7) and estimate the difference. Since we want ε(t)=Eβ(Ωtβ) in (3), we take in (32) (58)bγ=L.T.EβΩtβ=γβ-1γβ-Ω.We will treatΩ as belonging to the cut complex plane C∖(-∞,0]; and let, for definiteness, Ω1/α be the principal branch, i.e., Ω1/α>0 if Ω>0. If arg⁡Ω∈-2-βπ,2-βπ then sβ-Ω has only one root s=r1 in the cut complex plane C∖(-∞;0]; otherwise it has two roots s=r1 and s=r2. If arg⁡Ω∈[-πβ/2,πβ/2], one of the roots rκ is located in the left half-plane Rs>0.As a solution of the Helmholtz equation (5) satisfying Dα-2v(0+)=0 we take(59)vx=∫0xdξfξx-ξα-1Eα,αΩx-ξα-e-ξΩ1/αxα-1Eα,αΩxα-∫x∞dξe-ξΩ1/αxα-1Eα,αΩxαfξ.The fact that v(x) is indeed a solution follows from Theorem 4; the form of v(x) was derived similarly to (31) by imitating (26) in the case of the wave equation. As u(x,t) we take the solution of the fractional wave equation constructed in Lemma 5. With these choices,(60)ux,t-EβΩtβvx=12πi∫μdγeγtγβ-1γβ-Ω×∫0xx-ξα-1Eα,αγβx-ξα-Eα,αΩx-ξαfξdξ-∫0∞e-ξγβ/αEα,αγβxα-e-ξΩ1/αEα,αΩxαfξxα-1dξ.The integrand of (60) is analytic at rκ, κ=1 or κ=1,2, as the case may be. Therefore the only contribution to the integral comes from the discontinuity along the cut (-∞,0]. To the jump of the integrand along that cut we apply the following.Lemma 6 (generalized Watson’s lemma, [12, p.22]).
Consider the integral(61)Gz=∫0∞eiygte-ztdtin the complex domain, where y∈R, and the path of integration is the straight line joining t=0 to t=∞eiy. Suppose that the integral G(z) exists for some fixed z=z0 and that, as t→0 along arg⁡t=y,(62)gt~∑n=0∞antλn-1,where λ0>0 and λn+1>λn. Then (63)Gz~∑n=0∞anΓλnz-λnas z→∞ in arg⁡zeiy≤π/2-Δ for any Δ in the interval 0<Δ≤π/2.An elementary calculation with power series shows that(64)γβ-1γβ-Ω∫0xx-ξα-1Eα,αγβx-ξα-Eα,αΩx-ξαfξdξ-∫0∞e-ξγβ/αEα,αγβxα-e-ξΩ1/αEα,αΩxαfξxα-1dξ=Cβ-1xγβ-1+Cβ-1+β/αxγβ-1+β/α+C2β-1γ2β-1+Oγβ-1+2β/α,where in the ordering of the terms we remembered that 1/2<β/α≤1 and 1<β<2 and where(65)Cβ-1x=∫0xx-ξ2α-1Eα,2αΩx-ξαfξdξ(66)+xα-1Ω∫0∞1Γα-e-ξΩ1/αEα,αΩxαfξdξ;(67)Cβ-1+β/αx=-xα-1ΩΓα∫0∞ξfξdξ;(68)C2β-1x=1ΩCβ-1-1ΩΓ2α∫0xx-ξ2α-1-x2α-1fξdξ.Now (8) follows from Lemma 6 and the formula(69)12πi∫μγq-1eγtdγ=sin⁡qππΓq·1tq.The proof of Theorem 1 is complete.


## 5. Case3/2α>β>α

In this section we prove Theorem2, following the model of the previous section. The condition 3/2α>β is imposed to make the function Eα,α(γβc), for c>0, decrease as γ→∞, π≤arg⁡γ≤π/2; see [8, Theorems 1.3, 1.4].Lemma 7.
Letf(x) be a continuous function with compact support.
Letb(γ) be an analytic function defined in C∖D(0,R)∪(-∞,0] and bounded in that region; moreover ∫c-i∞c+i∞eγtb(γ)dγ is defined for t=0 in the sense of principal value. Then the function(70)ux,t=12πi∫c-i∞c+i∞etγHx,γdγ,t≥0,where(71)Hx,γ=bγ∫0xdξfξx-ξα-1Eα,αγβx-ξαfor x≥0, solves the FDE(72)Dxαux,t-DtβCux,t=fx12πi∫c-i∞c+i∞eγtbγdγ.Moreover, u(x,t) satisfies the boundary conditions(73)Dxα-2u0+,t=Dxα-1u0+,t=0.Proof.
SinceEα,αγβx-ξα=Oγ-2β as γ→±i∞ and β>1, in the integral (74)ux,t=12πi∫c-i∞c+i∞dt∫0xdξetγbγx-ξα-1Eα,αγβx-ξαfξthe integrations can be carried out in arbitrary order; also, similarly to the proof of Lemma 5, one can perform fractional differentiations under the integral sign even without modifying the integration contour.Next we assume thatb(γ)=L.T.(Eβ(Ωtβ))=γβ-1/(γβ-Ω) and work out the initial conditions of u(x,t) from (70) for t=0; namely, we compute(75)ux,0=12πi∫0xdξ∫c-i∞c+i∞dγx-ξα-1Eα,αγβx-ξαγβ-1γβ-Ωfξand(76)utx,0=12πi∫0xdξ∫c-i∞c+i∞dγγx-ξα-1Eα,αγβx-ξαγβ-1γβ-Ωfξ.In the inner integral of (75) make the substitution s=γβ and obtain the integral(77)12πi∫L′Eα,αsx-ξαdss-Ω,where the integration contour L′ is shown on Figure 2.Figure 2
The contourL′ in the s-plane.SinceEα,α(sx-ξα)=O(s-2) as s→∞, arg⁡s>πα/2, we can close the contour L′ and calculate the integral (77) using the residue at s=Ω:(78)12πi∮Eα,αsx-ξαdss-ω=Eα,αΩx-ξα.Similarly, the inner integral of (76) becomes(79)12πi∫L′s1/βEα,αsx-ξαdss-Ω,where we now introduce a cut (-∞;0] in the s-plane. Let πα/2<μ<πβ/2; then the integral representation [8, (1.126)] implies(80)79=12πi∫L′dss1/β12πiα∫Γdζexp⁡ζ1/αζ1/α-1ζ-sx-ξα1s-Ω,where the contour Γ is as on Figure 3 and avoids a large enough circle around the origin so as not to intersect the similarity image max⁡{1,x-ξα}·L′ of the contour L′.Figure 3
The contourΓ.Interchanging the order of integration in (80) (legal because β>1), we obtain(81)12πi2α∫Γdζexp⁡ζ1/αζ1/α-1∫L′dss1/βζ-sx-ξαs-Ω.Closing the contourL′ on the right (which is possible for β>1) we pick up the residue at s=ζ/x-ξα, and (76) becomes (82)-12πiαx-ξα/β∫Γdζexp⁡ζ1/αζ1/β+1/α-1ζ-Ωx-ξα.Let δ=α-α/β; then in the numerator ζ(1/β)+(1/α)-1=ζ(1-δ)/α, and the integral equals (see [8, (1.130)])(83)-1x-ξα/βEα,δΩx-ξα.Coming back to (76), we obtain(84)76=-∫0xdξx-ξδ-1Eα,δΩx-ξαfξ.Putting together (78) and (84), we have proven the following.Lemma 8.
Iff(x) is a smooth function with compact support, the solution u(x,t) from Lemma 7 with β(γ)=γβ-1/(γβ-Ω) satisfies (85)ux,0=∫0xx-ξα-1Eα,αΩx-ξαfξdξ,utx,0=∫0xx-ξδ-1Eα,δΩx-ξαfξdξ,where δ=α(1-β-1).Let(86)vx=∫0xdξfξx-ξα-1Eα,αΩx-ξα.By Theorem 4, v(x) satisfies the Helmholtz equation (5) and the boundary conditions Dα-2v(0+)=Dα-1v(0+)=0. With the same choice of the contour μ as in the previous section,(87)ux,t-EβΩtβvx=12πi∫μdγeγtγβ-1γβ-Ω×∫0xx-ξα-1Eα,αγβx-ξα-Eα,αΩx-ξαfξdξ,and the only contribution to the integral on the right-hand side of (87) comes from the cut (-∞,0]. We compute that(88)γβ-1γβ-Ω∫0xx-ξα-1Eα,αγβx-ξα-Eα,αΩx-ξαfξdξ=Bβ-1xγβ-1+B2β-1xγβ-2+Oγ3β-3,where(89)Bγ-1=∫0xx-ξ2α-1Eα,2αΩx-ξαfξdξ;(90)B2γ-1=1ΩBγ-1-1Γ2αΩ∫0xx-ξ2α-1fξdξ.Using Lemma 6 and formula (69), we obtain (10). The proof of Theorem 2 is complete.



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