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Open Access Published by De Gruyter Open Access February 8, 2018

Singular Cauchy problem for the general Euler-Poisson-Darboux equation

General Euler-Poisson-Darboux equation

E.L. Shishkina
From the journal Open Mathematics

Abstract

In this paper we obtain the solution of the singular Cauchy problem for the Euler-Poisson-Darboux equation when differential Bessel operator acts by each variable.

MSC 2010: 26A33; 44A15

1 Introduction

The classical Euler-Poisson-Darboux equation has the form

i=1n2uxi2=2ut2+ktut,u=u(x,t),xRn,t>0,<k<.(1)

The operator acting by t in (1) is called the Bessel operator. For the Bessel operator we use the notation (see. [1], p. 3)

(Bk)t=2t2+ktt.

The Euler-Poisson-Darboux equation for n = 1 appears in Euler’s work (see [2], p. 227). Further Euler’s case of (1) was studied by Poisson in [3], Riemann in [4] and Darboux in [5] (for the history of this issue see also in [6], p. 532 and [7], p. 527). The generalization of it was studied in [8]. When n ≥ 1 the equation (1) was considered, for example, in [9, 10]. The Euler-Poisson-Darboux equation appears in different physics and mechanics problems (see [11, 12, 13, 14, 15]). In [16] (see also [17], p. 243) and in [18] there were different approaches to the solution of the Cauchy problem for the general Euler-Poisson-Darboux equation

i=1n2uxi2+γixiuxi=2ut2+ktut,0<γi,i=1,...,n,k>0(2)

with the initials conditions

u(x,0)=f(x),ut|t=0=0.(3)

The Cauchy problem with the nonequal to zero first derivative by t of u for the (2) (and for (1)) is incorrect. However, if we use the special type of the initial conditions containing the nonequal to zero first derivative by t of u then such Cauchy problem for the (2) will by solvable. Following [17] and [19] we will use the term singular Cauchy problem in this case. The abstract Euler-Poisson-Darboux equation (when in the left hand of (2) an arbitrary closed linear operator is presented) was studied in [20, 21, 22].

In this article we consider the solution of the problem (2)-(3) when −∞ < k < +∞ and its properties. Besides this, we get the formula for the connection of solution of the problem (2)-(3) and solution of a simpler problem. Also using the solution of the problem (2)-(3) we obtain solution of the singular Cauchy problem for the equation (2) when k < 1 with the conditions

u(x,0)=0,limt0tkut=φ(x).(4)

2 Property of general Euler-Poisson-Darboux equations’ solutions

In this section we give some necessary definitions and obtain two fundamental recursion formulas for solution of (2).

Let

R+n={x=(x1,,xn)Rn,x1>0,,xn>0}

and Ω is open set in ℝn which is symmetric correspondingly to each hyperplane xi=0, i=1, …, n, Ω+ = ΩR+n and Ω+ = ΩR¯+n where

R¯+n={x=(x1,,xn)Rn,x10,,xn0}.

We have Ω+R+n and Ω+R¯+n. Consider the set Cm(Ω+), m ≥ 1, consisting of differentiable functions on Ω+ by order m. Let Cm(Ω+) be the set of functions from Cm(Ω+) such that all their derivatives by xi for all i = 1, …, n are continuous up to the xi=0. Class Cevm(Ω¯+) consists of functions from Cm(Ω+) such that 2k+1fxi2k+1|x=0=0 for all non-negative integers km12 and all xi, i = 1, …, n (see [1], p. 21). A multi-index γ=(γ1, …, γn) consists of fixed positive numbers γi > 0, i=1, …, n and |γ|=γ1+…+γn.

We consider the multidimensional Euler-Poisson-Darboux equation wherein the Bessel operator acts in each of the variables:

(γ)xu=(Bk)tu,<k<,u=uk(x,t),xR+n,t>0,(5)

where

(γ)x=γ=i=1n(Bγ1)xi=i=1n2xi2+γixix,(6)
(Bk)t=2t2+ktt,kR.

Equation (5) we will call the general Euler-Poisson-Darboux equation.

Statement 2.1

Let uk = uk (x,t) denote the solution of(5)when the next two fundamental recursion formulas hold

uk=t1ku2k,(7)
utk=tuk+2.(8)

Proof

Following [23] we prove (7). Putting w = tk−1v, v = uk we have

wt=(k1)tk2v+tk1vt=k1tw+tk1vt,wtt=(k1)(k2)tk3v+(k1)tk2vt+(k1)tk2vt+tk1vtt==(k1)(k2)t2w+2(k1)tk2vt+tk1vtt,2ktwt=(k1)(k2)t2w+(2k)tk2vt,wtt+2ktwt=2(k1)tk2vt+tk1vtt+(2k)tk2vt=tk1vtt+ktvt

or

wtt+2ktwt=tk1vtt+ktvt.(9)

If w = tk−1v satisfies the equation

Δγw=wtt+2ktwt,

then using (9) we get

tk1Δγv=tk1vtt+ktvt

which means that v satisfies the equation

Δγv=vtt+ktvt.

Denoting w = u2−k we obtain (7).

Now we prove the (8). Let tw = vt, v = uk. We obtain

wt=1t2vt+1tvtt,wtt=2t3vt2t2vtt+1tvttt.

We find now k+2twt:

k+2twt=k+2t3vt+k+2t2vtt.

Then we get

wtt+k+2twt=2t3vt2t2vtt+1tvtttk+2t3vt+k+2t2vtt==1tvtttkt3vt+kt2vtt=1tvtttkt2vt+ktvtt=1ttvtt+ktvt

or

wtt+k+2twt=1ttvtt+ktvt.(10)

□

Recursion formulas (7) and (8) allow us to obtain, from a solution uk of equation (5), the solutions of the same equation with the parameter k+2 and 2 − k, respectively. Both formulas are proved for Euler-Poisson-Darboux equation 2ut2+ktutu=0.

3 Weighted spherical mean and the first Cauchy problem for the general Euler-Poisson-Darboux equation

Here we present the solutions of the problem (2)-(3) for different values of k for which we obtain solution of (2)-(4) in the next section, and get formula for the connection of solution of problem (2)-(3) and solution of simpler problem when k = 0 in (2).

In R+n we will use multidimensional generalized translation corresponding to multi-index γ:

γTt=γ1Tx1t1...γnTxntn,

where each γiTxiτi is defined by the formula (see [24])

γiTxiτif(x)=Γγi+12Γγi2Γ120πf(x1,...,xi1,xi2+τi22xiτicosαi,xi+1,...,xn)sinγi1αidαi.

The below-considered weighted spherical mean generated by a multidimensional generalized translation γTt has the form (see [25])

Mfγ(x;r)=1|S1+(n)|γS1+(n)γTxrθf(x)θγdS,(11)

where θγ=i=1nθiγi,S1+(n)={θ:|θ|=1,θR+n} and the coefficient |S1+(n)|γ is computed by the formula

|S1+(n)|γ=S1+(n)i=1nxiγidS(y)=i=1nΓγi+122n1Γn+|γ|2(12)

(see [26], p. 20, formula (1.2.5) in which we should put N=n). Construction of a multidimensional generalized translation and the weighted spherical mean are transmutation operators (see [27]).

Theorems 3.1-3.4 have been proved in [28]. We give formulations of these theorems here because they will be needed in the next section.

Theorem 3.1

The weighted spherical mean of fCev2satisfies the general equation Euler–Poisson–Darboux equation

(Δγ)xMfγ(x;t)=(Bk)tMfγ(x;t),k=n+|γ|1(13)

and the conditions

Mfγ(x;0)=f,(Mfγ)t(x;0)=0.(14)

This theorem has been proved in [25]).

We give theorems on the solution of the Cauchy problem for the general Euler–Poisson–Darboux equation for the remaining values of k.

(Δγ)xu=(Bk)tu,u=uk(x,t),xR+n,t>0,(15)
uk(x,0)=f(x),utk(x,0)=0.(16)

Theorem 3.2

Let fCev2. Then for the case k > n+|γ | − 1 the solution of(15)(16)is unique and given by

uk(x,t)=2nΓk+12Γkn|γ|+12i=1nΓγi+12B1+(n)[γTtyf(x)](1|y|2)kn|γ|12yγdy.(17)

Using weighted spherical mean we can write

uk(x,t)=2t1kΓk+12Γkn|γ|+12Γn+|γ|20t(t2r2)kn|γ|12rn+|γ|1Mfγ(x;r)dr.(18)

Theorem 3.3

IffCevn+|γ|k2+2then the solution of(15)(16)for k < n+|γ | − 1, k ≠ −1,−3,−5,…

uk(x,t)=t1kttm(tk+2m1uk+2m(x,t)),(19)

where m is a minimum integer such thatmn+|γ|k12anduk+2m(x,t)is the solution of the Cauchy problem

(Bk+2m)tuk+2m(x,t)=(Δγ)xuk+2m(x,t),(20)
uk+2m(x,0)=f(x)(k+1)(k+3)...(k+2m1),utk+2m(x,0)=0.(21)

The solution of(15)(16)is unique for k ≥ 0 and not unique for negative k.

Theorem 3.4

If fCev1kis B–polyharmonic of order1k2then one of the solutions of the Cauchy problem(20)(21)for the k=−1,−3,−5,… is given by

u1(x,t)=f(x),(22)
uk(x,t)=f(x)+h=1k+12Δγhf(k+1)...(k+2h1)t2h24....2h,k=3,5,...(23)

The solution of(15)(16)is not unique for negative k.

The theorem 3.5 contains the explicit form of the transmutation operator for the solution. Definition, methods of construction and applications of the transmutation operators can be found in [27, 29, 30].

Theorem 3.5

Let k > 0. The twice continuously differentiable onR+n+1solution u=uk(x,t) of the Cauchy problem

(Δγ)xu=(Bk)tu,u=uk(x,t),xR+n,t>0,(24)
uk(x,0)=f(x),utk(x,0)=0(25)

such thatuxik(x1,...,xi1,0,xi+1,...,xn,t)=0,i=1,...,nis connected with the twice continuously differentiable onR+nsolution w=w(x,t) of the Cauchy problem

(Δγ)xw=wtt,w=w(x,t),xR+n,tR,(26)
w(x,0)=f(x),wt(x,0)=0(27)

such that wxi(x1, …, xi−1,0,xi+1, …, xn,t) = 0, i = 1, …, n by formula

uk(x,t)=(P1k12)αw(x,αt),(28)

where(Pτλ)αis transmutation Poisson operator (see [24]) acting by α

(Pτλ)αg(α)=2Γ(λ+1)πΓλ+121τ2λ0τg(α)[τ2α2]λ12dα.

Proof

The fact that the function uk defined by the equality (28) satisfies the conditions (31) is obvious. Let us show that uk defined by (28) satisfies (24)

(Δγ)xu=(P1k12)α(Δγ)xw(x,αt)=(P1k12)αwξξ(x,αt)==2Γk+12πΓk201(Δγ)xw(x,αt)[1α2]k21dα,

where ξ = α t. Further integrating by parts we obtain

ukt=2Γk+12πΓk201αwξ(x,αt)[1α2]k21dα==u=wξ(x,αt),dv=α[1α2]k21dα,du=twξξ(x,αt)dα,v=1k[1α2]k2==2Γk+12πΓk2tk01wξξ(x,αt)[1α2]k2dα==2Γk+12πΓk2tk01wξξ(x,αt)[1α2]k2dα.

For 2ukt2 we have

2ukt2=2Γk+12πΓk201α2wξξ(x,αt)[1α2]k21dα==2Γk+12πΓk201(Δγ)xw(x,αt)α2[1α2]k21dα.

Finally,

2ukt2+ktukt=2Γk+12πΓk201(Δγ)xw(x,αt)α2[1α2]k21dα+01(Δγ)xw(x,αt)[1α2]k2dα==2Γk+12πΓk201(Δγ)xw(x,αt)[1α2]k21dα=(Δγ)xuk.

Thus the function uk defined by equality (28) satisfies the problem (24)(31).

Let us prove that from the relation (28) we can uniquely obtain a solution of the problem (26)(27). By introducing new variables αt=τ,t=y, we get

yk12uk(x,y)=Γk+12πΓk20yw(x,τ)τ(yτ)k21dτ.

Let k > 0 then yk12uk(x,y) is the Riemann-Liouville left-sided fractional integral of the order k2 (see [31], p. 33):

yk12uk(x,y)=Γk+12πI0+k2w(x,τ)τ(y).

Thus we have unique representation of w(x,τ) (see [31], p. 44, theorem 24)

w(x,τ)=τπΓk+12D0+k2yk12uk(x,y)(τ)

or

w(x,t)=2Γnk2d2tdtn0tuk(x,z)zk(t2z2)k2n+1dz.

4 The second Cauchy problem for the general Euler-Poisson-Darboux equation

In this section we obtain solution of (2)-(4).

Theorem 4.1

IfφCevn+|γ|+k12then the solution v = vk(x,t) of

(γ)xv=(Bk)tv,0<γi,i=1,...,n,k<1,xR+n,t>0,(29)
vk(x,0)=0,limt0tkvt=φ(x)(30)

is given by

vk(x,t)=Γ3k2i=1nΓγi+12Γ2k+2qn|γ|+122n+q(1k)Γ3k+2q2Γ2k+2q21ttq××t1k+2qB1+(n)[γTtyφ(x)](1|y|2)2k+2qn|γ|12yγdy

if n+|γ|+k is not an odd integer and

vk(x,t)=2qΓ3k2(1k)Γ3k+2q21ttqtn+|γ|2Mφγ(x;t).

if n+|γ|+k is an odd integer, where q ≥ 0 is the smallest positive integer number such that 2−k+2qn+|γ| − 1.

Proof

Let q ≥ 0 be the smallest positive integer number such that 2−k+2qn+|γ| − 1 i.e. q=n+|γ|+k12 and let v2−k+2q(x,t) be a solution of (29) when we take 2 − k+2q instead of k such that

v2k+2q(x,0)=φ(x),vt2k+2q(x,0)=0.(31)

Then by property (7) we obtain that

vk2q=t1k+2qv2k+2q

is a solution of the equation

(γ)xv=2vt2+k2qtvt.

Further, applying q-times the formula (8) we obtain that

1ttqvk2q=1ttq(t1k+2qv2k+2q)

is a solution of the (29).

Let’s consider

vk(x,t)=2qΓ3k2(1k)Γ3k+2q21ttq(t1k+2qv2k+2q).(32)

We have shown that (32) satisfies the equation (29).

Now we will prove that vk satisfies the conditions (31). For vkCevq(Ω+) we have the formula (see [19], p.9)

1ttq(t1k+2qv2k+2q)=s=0q2qsCqsΓ1k2+q+1Γ1k2+s+1t1k+2s1ttsv2k+2q.(33)

Taking into account formula (33) we obtain vk(x,0) = 0 and

limt0tkvtk(x,t)=2qΓ3k2(1k)Γ3k+2q2limt0tkt1ttq(t1k+2qv2k+2q)==2qΓ3k2(1k)Γ3k+2q2limt0tkts=0q2qsCqsΓ1k2+q+1Γ1k2+s+1t1k+2s1ttsv2k+2q==11klimt0tktt1kv2k+2q=11klimt0tk(1k)tkv2k+2q+t1kvt2k+2q==11klimt0(1k)v2k+2q+tvt2k+2q=φ(x).

Now we obtain the representation of vk through the integral. Using formula (18) we get

v2k+2q=2Γ3k+2q2Γ3k+2qn|γ|2Γn+|γ|201(1r2)1k+2qn|γ|2rn+|γ|1Mφγ(x;rt)dr.

If 2 − k+2q > n+|γ| − 1 then by applying (32) and (33) we write

vk=2qΓ3k2(1k)Γ3k+2q2s=0q2qsCqsΓ1k2+q+1Γ3k2+st1k+2s1ttsv2k+2q==Γ3k21ks=0qCqst1k+2s2sΓ3k2+s1ttsv2k+2q==Γ3k+2q2Γ1k2Γ3k+2qn|γ|2Γn+|γ|2s=0qCqst1k+2s2sΓ3k2+s××01(1r2)1k+2qn|γ|2rn+|γ|11ttsMφγ(x;rt)dr.

If 2−k+2q = n+|γ|−1 then v2k+2q=Mφγ(x;t) and

vk=2qΓ3k2(1k)Γ3k+2q21ttqtn+|γ|2Mfγ(x;t)==21qΓ1k2Γ3k+2q2s=0q2qsCqsΓ3k2+qΓ3k2+st1k+2s1ttsMfγ(x;t)==s=0qCqsΓ1k22s+1Γ3k2+st1k+2s1ttsMfγ(x;t).

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Received: 2016-10-31
Accepted: 2017-12-22
Published Online: 2018-02-08

© 2018 Shishkina, published by De Gruyter

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