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BY-NC-ND 4.0 license 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


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


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


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


with the initials conditions


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


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).



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


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:




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



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




If w = tk−1v satisfies the equation


then using (9) we get


which means that v satisfies the equation


Denoting w = u2−k we obtain (7).

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


We find now k+2twt:


Then we get





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:


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


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


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


(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


and the conditions


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.


Theorem 3.2

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


Using weighted spherical mean we can write


Theorem 3.3

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


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


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


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


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


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


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



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)


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


For 2ukt2 we have




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


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


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




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


is given by


if n+|γ|+k is not an odd integer and


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


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


Then by property (7) we obtain that


is a solution of the equation


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


is a solution of the (29).

Let’s consider


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)


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


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


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


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



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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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