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

E.L. Shishkina

doi:10.1515/math-2018-0005

Article Open Access

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

General Euler-Poisson-Darboux equation

E.L. Shishkina

Published/Copyright: February 8, 2018

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$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

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.

Keywords: Bessel operator; Euler-Poisson-Darboux equation; Singular Cauchy problem

MSC 2010: 26A33; 44A15

1 Introduction

The classical Euler-Poisson-Darboux equation has the form

∑i=1n∂2u∂xi2=∂2u∂t2+kt∂u∂t,u=u(x,t),x∈Rn,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=∂2∂t2+kt∂∂t.

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=1n∂2u∂xi2+γixi∂u∂xi=∂2u∂t2+kt∂u∂t,0<γi,i=1,...,n,k>0(2)

with the initials conditions

u(x,0)=f(x),∂u∂t|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,limt→0tk∂u∂t=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 ℝ_n which is symmetric correspondingly to each hyperplane x_i=0, i=1, …, n, Ω₊ = Ω ∩ R+n and Ω₊ = Ω ∩ R¯+n where

R¯+n={x=(x1,…,xn)∈Rn,x1≥0,…,xn≥0}.

We have Ω₊ ⊆ R+n and Ω₊ ⊆ R¯+n. Consider the set C^m(Ω₊), m ≥ 1, consisting of differentiable functions on Ω₊ by order m. Let C^m(Ω₊) be the set of functions from C^m(Ω₊) such that all their derivatives by x_i for all i = 1, …, n are continuous up to the x_i=0. Class Cevm(Ω¯+) consists of functions from C^m(Ω₊) such that ∂2k+1f∂xi2k+1|x=0=0 for all non-negative integers k≤m−12 and all x_i, i = 1, …, n (see [1], p. 21). A multi-index γ=(γ₁, …, γ_n) consists of fixed positive numbers γ_i > 0, i=1, …, n and |γ|=γ₁+…+γ_n.

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),x∈R+n,t>0,(5)

where

(△γ)x=△γ=∑i=1n(Bγ1)xi=∑i=1n∂2∂xi2+γixi∂∂x,(6)

(Bk)t=∂2∂t2+kt∂∂t,k∈R.

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

Statement 2.1

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

uk=t1−ku2−k,(7)

utk=tuk+2.(8)

Proof

Following [23] we prove (7). Putting w = t^k−1v, v = u^k we have

wt=(k−1)tk−2v+tk−1vt=k−1tw+tk−1vt,wtt=(k−1)(k−2)tk−3v+(k−1)tk−2vt+(k−1)tk−2vt+tk−1vtt==(k−1)(k−2)t2w+2(k−1)tk−2vt+tk−1vtt,2−ktwt=−(k−1)(k−2)t2w+(2−k)tk−2vt,wtt+2−ktwt=2(k−1)tk−2vt+tk−1vtt+(2−k)tk−2vt=tk−1vtt+ktvt

wtt+2−ktwt=tk−1vtt+ktvt.(9)

If w = t^k−1v satisfies the equation

Δγw=wtt+2−ktwt,

then using (9) we get

tk−1Δγv=tk−1vtt+ktvt

which means that v satisfies the equation

Δγv=vtt+ktvt.

Denoting w = u^2−k we obtain (7).

Now we prove the (8). Let tw = v_t, v = u^k. We obtain

wt=−1t2vt+1tvtt,wtt=2t3vt−2t2vtt+1tvttt.

We find now k+2twt:

k+2twt=−k+2t3vt+k+2t2vtt.

Then we get

wtt+k+2twt=2t3vt−2t2vtt+1tvttt−k+2t3vt+k+2t2vtt==1tvttt−kt3vt+kt2vtt=1tvttt−kt2vt+ktvtt=1t∂∂tvtt+ktvt

wtt+k+2twt=1t∂∂tvtt+ktvt.(10)

□

Recursion formulas (7) and (8) allow us to obtain, from a solution u_k 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 ∂2u∂t2+kt∂u∂t−△u=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+12Γγi2Γ12∫0πf(x1,...,xi−1,xi2+τi2−2xiτicos⁡αi,xi+1,...,xn)sinγi−1⁡αidαi.

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

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

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

|S1+(n)|γ=∫S1+(n)∏i=1nxiγidS(y)=∏i=1nΓγi+122n−1Γn+|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 f ∈ Cev2satisfies 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),x∈R+n,t>0,(15)

uk(x,0)=f(x),utk(x,0)=0.(16)

Theorem 3.2

Let f ∈ Cev2. Then for the case k > n+|γ | − 1 the solution of(15)–(16)is unique and given by

uk(x,t)=2nΓk+12Γk−n−|γ|+12∏i=1nΓγi+12∫B1+(n)[γTtyf(x)](1−|y|2)k−n−|γ|−12yγdy.(17)

Using weighted spherical mean we can write

uk(x,t)=2t1−kΓk+12Γk−n−|γ|+12Γn+|γ|2∫0t(t2−r2)k−n−|γ|−12rn+|γ|−1Mfγ(x;r)dr.(18)

Theorem 3.3

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

uk(x,t)=t1−k∂t∂tm(tk+2m−1uk+2m(x,t)),(19)

where m is a minimum integer such thatm≥n+|γ|−k−12anduk+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+2m−1),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 f ∈ Cev1−kis B–polyharmonic of order1−k2then one of the solutions of the Cauchy problem(20)–(21)for the k=−1,−3,−5,… is given by

u−1(x,t)=f(x),(22)

uk(x,t)=f(x)+∑h=1−k+12Δγhf(k+1)...(k+2h−1)t2h2⋅4⋅....⋅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=u^k(x,t) of the Cauchy problem

(Δγ)xu=(Bk)tu,u=uk(x,t),x∈R+n,t>0,(24)

uk(x,0)=f(x),utk(x,0)=0(25)

such thatuxik(x1,...,xi−1,0,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),x∈R+n,t∈R,(26)

w(x,0)=f(x),wt(x,0)=0(27)

such that w_{x_i}(x₁, …, x_i−1,0,x_i+1, …, x_n,t) = 0, i = 1, …, n by formula

uk(x,t)=(P1k−12)αw(x,αt),(28)

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

(Pτλ)αg(α)=2Γ(λ+1)πΓλ+121τ2λ∫0τg(α)[τ2−α2]λ−12dα.

Proof

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

(Δγ)xu=(P1k−12)α(Δγ)xw(x,αt)=(P1k−12)αwξξ(x,αt)==2Γk+12πΓk2∫01(Δγ)xw(x,αt)[1−α2]k2−1dα,

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

∂uk∂t=2Γk+12πΓk2∫01αwξ(x,αt)[1−α2]k2−1dα==u=wξ(x,αt),dv=α[1−α2]k2−1dα,du=twξξ(x,αt)dα,v=−1k[1−α2]k2==2Γk+12πΓk2tk∫01wξξ(x,αt)[1−α2]k2dα==2Γk+12πΓk2tk∫01wξξ(x,αt)[1−α2]k2dα.

For ∂2uk∂t2 we have

∂2uk∂t2=2Γk+12πΓk2∫01α2wξξ(x,αt)[1−α2]k2−1dα==2Γk+12πΓk2∫01(Δγ)xw(x,αt)α2[1−α2]k2−1dα.

Finally,

∂2uk∂t2+kt∂uk∂t=2Γk+12πΓk2∫01(Δγ)xw(x,αt)α2[1−α2]k2−1dα+∫01(Δγ)xw(x,αt)[1−α2]k2dα==2Γk+12πΓk2∫01(Δγ)xw(x,αt)[1−α2]k2−1dα=(Δγ)xuk.

Thus the function u^k 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

yk−12uk(x,y)=Γk+12πΓk2∫0yw(x,τ)τ(y−τ)k2−1dτ.

Let k > 0 then yk−12uk(x,y) is the Riemann-Liouville left-sided fractional integral of the order k2 (see [31], p. 33):

yk−12uk(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+k2yk−12uk(x,y)(τ)

w(x,t)=2Γn−k2d2tdtn∫0tuk(x,z)zk(t2−z2)k2−n+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

Ifφ∈Cevn+|γ|+k−12then the solution v = v^k(x,t) of

(△γ)xv=(Bk)tv,0<γi,i=1,...,n,k<1,x∈R+n,t>0,(29)

vk(x,0)=0,limt→0tk∂v∂t=φ(x)(30)

is given by

vk(x,t)=Γ3−k2∏i=1nΓγi+12Γ2−k+2q−n−|γ|+122n+q(1−k)Γ3−k+2q2Γ2−k+2q21t∂∂tq××t1−k+2q∫B1+(n)[γTtyφ(x)](1−|y|2)2−k+2q−n−|γ|−12yγdy

if n+|γ|+k is not an odd integer and

vk(x,t)=2−qΓ3−k2(1−k)Γ3−k+2q21t∂∂tqtn+|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+2q ≥ n+|γ| − 1.

Proof

Let q ≥ 0 be the smallest positive integer number such that 2−k+2q ≥ n+|γ| − 1 i.e. q=n+|γ|+k−12 and let v^2−k+2q(x,t) be a solution of (29) when we take 2 − k+2q instead of k such that

v2−k+2q(x,0)=φ(x),vt2−k+2q(x,0)=0.(31)

Then by property (7) we obtain that

vk−2q=t1−k+2qv2−k+2q

is a solution of the equation

(△γ)xv=∂2v∂t2+k−2qt∂v∂t.

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

1t∂∂tqvk−2q=1t∂∂tq(t1−k+2qv2−k+2q)

is a solution of the (29).

Let’s consider

vk(x,t)=2−qΓ3−k2(1−k)Γ3−k+2q21t∂∂tq(t1−k+2qv2−k+2q).(32)

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

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

1t∂∂tq(t1−k+2qv2−k+2q)=∑s=0q2q−sCqsΓ1−k2+q+1Γ1−k2+s+1t1−k+2s1t∂∂tsv2−k+2q.(33)

Taking into account formula (33) we obtain v^k(x,0) = 0 and

limt→0tkvtk(x,t)=2−qΓ3−k2(1−k)Γ3−k+2q2limt→0tk∂∂t1t∂∂tq(t1−k+2qv2−k+2q)==2−qΓ3−k2(1−k)Γ3−k+2q2limt→0tk∂∂t∑s=0q2q−sCqsΓ1−k2+q+1Γ1−k2+s+1t1−k+2s1t∂∂tsv2−k+2q==11−klimt→0tk∂∂tt1−kv2−k+2q=11−klimt→0tk(1−k)t−kv2−k+2q+t1−kvt2−k+2q==11−klimt→0(1−k)v2−k+2q+tvt2−k+2q=φ(x).

Now we obtain the representation of v^k through the integral. Using formula (18) we get

v2−k+2q=2Γ3−k+2q2Γ3−k+2q−n−|γ|2Γn+|γ|2∫01(1−r2)1−k+2q−n−|γ|2rn+|γ|−1Mφγ(x;rt)dr.

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

vk=2−qΓ3−k2(1−k)Γ3−k+2q2∑s=0q2q−sCqsΓ1−k2+q+1Γ3−k2+st1−k+2s1t∂∂tsv2−k+2q==Γ3−k21−k∑s=0qCqst1−k+2s2sΓ3−k2+s1t∂∂tsv2−k+2q==Γ3−k+2q2Γ1−k2Γ3−k+2q−n−|γ|2Γn+|γ|2∑s=0qCqst1−k+2s2sΓ3−k2+s××∫01(1−r2)1−k+2q−n−|γ|2rn+|γ|−11t∂∂tsMφγ(x;rt)dr.

If 2−k+2q = n+|γ|−1 then v2−k+2q=Mφγ(x;t) and

vk=2−qΓ3−k2(1−k)Γ3−k+2q21t∂∂tqtn+|γ|−2Mfγ(x;t)==2−1−qΓ1−k2Γ3−k+2q2∑s=0q2q−sCqsΓ3−k2+qΓ3−k2+st1−k+2s1t∂∂tsMfγ(x;t)==∑s=0qCqsΓ1−k22s+1Γ3−k2+st1−k+2s1t∂∂tsMfγ(x;t).

□

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Received: 2016-10-31

Accepted: 2017-12-22

Published Online: 2018-02-08

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Keywords for this article

Bessel operator; Euler-Poisson-Darboux equation; Singular Cauchy problem

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