Singular Cauchy problem for the general Euler ...

Open Math. 2018; 16: 23–31

Open Mathematics Research Article E.L. Shishkina*

Singular Cauchy problem for the general Euler-Poisson-Darboux equation General Euler-Poisson-Darboux equation https://doi.org/10.1515/math-2018-0005 Received October 31, 2016; accepted December 22, 2017.

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: 26A33, 44A15

1 Introduction The classical Euler-Poisson-Darboux equation has the form n

∑ i=1

∂2 u ∂2 u k ∂u = 2 + , ∂t t ∂t ∂x2i

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) ∂2 k ∂ (B k )t = 2 + . ∂t t ∂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 [11–15]). 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 n

∑ i=1

∂2 u γi ∂u ∂2 u k ∂u = 2 + , + x i ∂x i ∂t t ∂t ∂x2i

0 < γi ,

i = 1, ..., n,

k>0

(2)

with the initials conditions u(x, 0) = f (x),

∂u ∣ = 0. ∂t t=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

*Corresponding Author: E.L. Shishkina: Voronezh State University, Faculty of Applied Mathematics, Informatics and Mechanics, Universitetskaya square, 1, Voronezh 394006, Russia, E-mail: [email protected] Open Access. © 2018 Shishkina, published by De Gruyter Open. NonCommercial-NoDerivs 4.0 License.

This work is licensed under the Creative Commons Attribution-

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24 | E.L. Shishkina 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 [20–22]. 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 lim t k

u(x, 0) = 0,

t→0

∂u = ϕ(x). ∂t

(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 , . . . , x n ) ∈ Rn , x1 >0, . . . , x n >0} and Ω is open set in Rn 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 , . . . , x n ) ∈ Rn , x1 ≥0, . . . , x n ≥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 m all i = 1, ..., n are continuous up to the x i =0. Class C m ev (Ω + ) consists of functions from C (Ω + ) such that ∂2k+1 f ∣ ∂x2k+1 i

= 0 for all non-negative integers k ≤

m−1 2

and all x i , i = 1, ..., n (see [1], p. 21). A multi-index

x=0

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 )x u = (B k )t u,

−∞ < k < ∞,

where

u = u k (x, t),

n

n

i=1

i=1

(△γ )x = △γ = ∑(Bγ1 )x i = ∑

x ∈ R+n ,

t > 0,

γi ∂ ∂2 , + x i ∂x ∂x2i

(5)

(6)

∂2 k ∂ + , k ∈ R. ∂t2 t ∂t Equation (5) we will call the general Euler-Poisson-Darboux equation. (B k )t =

Statement 2.1. Let u k = u k (x, t) denote the solution of (5) when the next two fundamental recursion formulas hold u k = t1−k u2−k , (7) u kt = tu k+2 .

(8)

Proof. Following [23] we prove (7). Putting w = t k−1 v, v = u k we have w t = (k − 1)t k−2 v + t k−1 v t =

k−1 w + t k−1 v t , t

w tt = (k − 1)(k − 2)t k−3 v + (k − 1)t k−2 v t + (k − 1)t k−2 v t + t k−1 v tt = =

(k − 1)(k − 2) w + 2(k − 1)t k−2 v t + t k−1 v tt , t2 Authenticated | [email protected] author's copy Download Date | 2/10/18 7:01 AM

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

25

(k − 1)(k − 2) 2−k wt = − w + (2 − k)t k−2 v t , t t2 w tt +

k 2−k w t = 2(k − 1)t k−2 v t + t k−1 v tt + (2 − k)t k−2 v t = t k−1 (v tt + v t ) t t

or w tt +

2−k k w t = t k−1 (v tt + v t ) . t t

If w = t k−1 v satisfies the equation ∆γ w = w tt +

(9)

2−k wt , t

then using (9) we get t k−1 ∆γ v = t k−1 (v tt +

k vt ) t

which means that v satisfies the equation ∆γ v = v tt +

k vt . t

Denoting w = u2−k we obtain (7). Now we prove the (8). Let tw = v t , v = u k . We obtain wt = − w tt = We find now

Then we get

k+2 t

wt :

1 1 v t + v tt , t2 t

2 2 1 v t − 2 v tt + v ttt . t3 t t

k+2 k+2 k+2 w t = − 3 v t + 2 v tt . t t t k+2 2 2 1 k+2 k+2 w t = 3 v t − 2 v tt + v ttt − 3 v t + 2 v tt = t t t t t t k k 1 ∂ k 1 k 1 k (v tt + v t ) = v ttt − 3 v t + 2 v tt = (v ttt − 2 v t + v tt ) = t t t t t t t ∂t t w tt +

or w tt +

k+2 1 ∂ k wt = (v tt + v t ) . t t ∂t t

(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 2 − △u = 0. equation ∂∂t2u + kt ∂u ∂t

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 : γ

Tt =

γ1

T xt11 ...γn T xt nn ,

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26 | E.L. Shishkina where each γi T xτii is defined by the formula (see [24]) γi

Γ ( γi2+1 )

T xτii f (x) =

Γ

( γ2i ) Γ

π

∫ ( 12 ) 0

f (x1 , ..., x i−1 ,

√

x2i + τi2 − 2x i τi cos αi , x i+1 , ..., x n ) sinγi −1 αi dαi .

The below-considered weighted spherical mean generated by a multidimensional generalized translation γ T t has the form (see [25]) 1 γ rθ (11) T x f (x)θγ dS, M γf (x; r) = + ∣S1 (n)∣γ +∫ S1 (n)

n

where θγ = ∏ θiγi , S+1 (n)={θ∶∣θ∣=1, θ∈R+n } and the coefficient ∣S+1 (n)∣γ is computed by the formula i=1

n

∣S+1 (n)∣γ

n

= ∫

S+ (n) 1

γ ∏ x i i dS(y) i=1

=

∏Γ(

i=1

γi +1 ) 2

(12)

2n−1 Γ ( n+∣2γ ∣ )

(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 ∈ C2ev satisfies the general equation Euler–Poisson–Darboux equation (∆γ )x M γf (x; t) = (B k )t M γf (x; t) , k = n + ∣γ ∣ − 1 (13) and the conditions M γf (x; 0) = f ,

(M γf )′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. u = u k (x, t),

(∆γ )x u = (B k )t u,

u k (x, 0) = f (x),

x ∈ R+n ,

t > 0,

(15)

u kt (x, 0) = 0.

(16)

Theorem 3.2. Let f ∈ C2ev . Then for the case k > n + ∣γ ∣ − 1 the solution of (15)–(16) is unique and given by ) 2n Γ ( k+1 2

u k (x, t)= Γ

( k−n−∣2γ ∣+1 )

n

∏Γ

i=1

γ ty 2 ∫ [ T f (x)](1−∣y∣ )

( γi2+1 )B+1 (n)

k−n−∣γ ∣−1 2

yγ dy.

(17)

Using weighted spherical mean we can write u (x, t)= [

Γ

( k−n−∣2γ ∣+1 ) Γ

n+∣γ ∣−k ]+2 2

Theorem 3.3. If f ∈ C ev

t

) 2t1−k Γ ( k+1 2

k

∫ ( n+∣2γ ∣ ) 0

(t2 − r2 )

k−n−∣γ ∣−1 2

r n+∣γ ∣−1 M γf (x; r)dr.

(18)

then the solution of (15)–(16) for k < n + ∣γ ∣ − 1, k ≠ −1, −3, −5, ... u k (x, t) = t1−k (

where m is a minimum integer such that m ≥

∂ m k+2m−1 k+2m ) (t u (x, t)), t∂t

n+∣γ ∣−k−1 2

(19)

and u k+2m (x, t) is the solution of the Cauchy problem

(B k+2m )t u k+2m (x, t) = (∆γ )x u k+2m (x, t),

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(20)

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

u k+2m (x, 0) =

f (x) , (k + 1)(k + 3)...(k + 2m − 1)

(x, 0) = 0. u k+2m t

27

(21)

The solution of (15)–(16) is unique for k ≥ 0 and not unique for negative k. Theorem 3.4. If f ∈ C1−k ev is B–polyharmonic of order (20)–(21) for the k=−1, −3, −5, ... is given by

1−k 2

then one of the solutions of the Cauchy problem

u−1 (x, t) = f (x),

k

− k+1 2

u (x, t) = f (x) + ∑

h=1

(22)

∆γh f

t2h , (k + 1)...(k + 2h − 1) 2 ⋅ 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 on R+n+1 solution u=u k (x, t) of the Cauchy problem (∆γ )x u = (B k )t u, u = u k (x, t), x ∈ R+n , t > 0, (24) u k (x, 0) = f (x),

u kt (x, 0) = 0

(25)

such that u kxi (x1 , ..., x i−1 , 0, x i+1 , ..., x n , t) = 0, i = 1, ..., n is connected with the twice continuously differentiable on R+n × R solution w=w(x, t) of the Cauchy problem (∆γ )x w = w tt ,

x ∈ R+n ,

w = w(x, t),

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

t ∈ R,

w t (x, 0) = 0

(26)

(27)

such that w x i (x1 , ..., x i−1 , 0, x i+1 , ..., x n , t) = 0, i = 1, ..., n by formula k−1

u k (x, t) = (P1 2 )α w(x, αt),

(28)

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

(Pτλ )α g(α)

1 2Γ (λ + 1) 1 =√ g(α)[τ 2 − α2 ]λ− 2 dα. 1 τ 2λ ∫ πΓ (λ + 2 )

0

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) 1

k−1 k−1 ) 2Γ ( k+1 2 k −1 2 (∆γ )x u = (P1 2 )α (∆γ )x w(x, αt) = (P1 2 )α wξξ (x, αt) == √ ∫ (∆γ )x w(x, αt)[1 − α ] 2 dα, k πΓ ( 2 ) 0

where ξ = αt. Further integrating by parts we obtain 1

k+1 ∂u k 2Γ ( 2 ) 2 k −1 =√ ∫ α wξ (x, αt)[1 − α ] 2 dα = k ∂t πΓ ( 2 ) 0 k k 1 = {u = wξ (x, αt), dv = α[1 − α2 ] 2 −1 dα, du = twξξ (x, αt)dα, v = − [1 − α2 ] 2 } = k

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28 | E.L. Shishkina 1

) t 2Γ ( k+1 2 k 2 =√ ∫ wξξ (x, αt)[1 − α ] 2 dα = πΓ ( 2k ) k 0 1

) t 2Γ ( k+1 k 2 =√ wξξ (x, αt)[1 − α2 ] 2 dα. ∫ k πΓ ( 2 ) k 0 For

∂2 u k ∂t2

we have 1

k+1 ∂ 2 u k 2Γ ( 2 ) 2 2 k −1 = √ ∫ α wξξ (x, αt)[1 − α ] 2 dα = k ∂t2 πΓ ( 2 ) 0 1

) 2Γ ( k+1 2 2 k −1 2 =√ ∫ (∆γ )x w(x, αt)α [1 − α ] 2 dα. k πΓ ( 2 ) 0 Finally, 1 ⎡ 1 ⎤ ⎢ ⎥ ⎢∫ (∆γ )x w(x, αt)α2 [1 − α2 ] 2k −1 dα + ∫ (∆γ )x w(x, αt)[1 − α2 ] 2k dα⎥ = ⎢ ⎥ ⎢ ⎥ ⎣0 ⎦ 0

k+1 ∂2 u k k ∂u k 2Γ ( 2 ) + = √ ∂t2 t ∂t πΓ ( 2k )

1

) 2Γ ( k+1 2 k −1 k 2 =√ ∫ (∆γ )x w(x, αt)[1 − α ] 2 dα = (∆γ )x u . k πΓ ( 2 ) 0 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 y Let k > 0 then y 33):

k−1 2

u k (x,

√

k−1 2

u k (x,

√

√ k w(x, τ ) (y − τ ) 2 −1 dτ . √ ∫ y

) Γ ( k+1 2

y) = √

πΓ ( 2k )

τ

0

y) is the Riemann-Liouville left-sided fractional integral of the order y

k−1 2

u k (x,

√

y) =

) Γ ( k+1 2 √

π

k

2 (I0+

k 2

(see [31], p.

√ w(x, τ ) ) (y). √ τ

√

Thus we have unique representation of w(x, τ ) (see [31], p. 44, theorem 24) √ √ k k−1 √ √ τ π 2 w(x, τ ) = (D0+ y 2 u k (x, y)) (τ ) k+1 Γ( 2 ) or

t

w(x, t) =

u k (x, z)z k d n 2 ( ) ∫ dz. k k 2 2 2 −n+1 Γ (n − 2 ) 2tdt 0 (t − z )

4 The second Cauchy problem for the general Euler-Poisson-Darboux equation In this section we obtain solution of (2)-(4). [

n+∣γ ∣+k−1 ] 2

Theorem 4.1. If ϕ ∈ C ev

then the solution v = v k (x, t) of

(△γ )x v = (B k )t v,

0 < γi ,

i = 1, ..., n,

k < 1,

x ∈ R+n ,

t > 0,

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(29)

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

v k (x, 0) = 0,

lim t k t→0

∂v = ϕ(x) ∂t

29

(30)

is given by n

) ∏ Γ ( γi2+1 )Γ ( Γ ( 3−k 2 i=1

k

v (x, t) =

2n+q (1 − k)Γ

2−k+2q−n−∣γ ∣+1 ) 2

( 3−k+2q )Γ 2

( 2−k+2q ) 2

(

1 ∂ q ) × t ∂t

⎞ ⎛ 2−k+2q−n−∣γ ∣−1 γ 1−k+2q γ ty 2 ⎟ 2 y dy ×⎜ t [ T ϕ (x)](1 − ∣y∣ ) ∫ ⎟ ⎜ ⎠ ⎝ B+ (n) 1 if n + ∣γ ∣ + k is not an odd integer and v k (x, t) =

) 2−q Γ ( 3−k 2

1 ∂ q n+∣γ ∣−2 γ ) (t Mϕ (x; t)) . (1 − k)Γ ( 3−k+2q ) t ∂t 2 (

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+∣γ2∣+k−1 ] and let v2−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),

v2−k+2q (x, 0) = 0. t

(31)

Then by property (7) we obtain that v k−2q = t1−k+2q v2−k+2q is a solution of the equation (△γ )x v =

∂2 v k − 2q ∂v + . ∂t2 t ∂t

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

1 ∂ q k−2q 1 ∂ q 1−k+2q 2−k+2q ) v =( ) (t v ) t ∂t t ∂t

is a solution of the (29). Let’s consider v k (x, t) =

) 2−q Γ ( 3−k 2

1 ∂ q 1−k+2q 2−k+2q ) (t v ). (1 − k)Γ ( 3−k+2q ) t ∂t 2 (

(32)

We have shown that (32) satisfies the equation (29). Now we will prove that v k satisfies the conditions (31). For v k ∈ C qev (Ω+ ) we have the formula (see [19], p.9) q 2q−s C s Γ ( 1−k + q + 1) 1 ∂ q 1−k+2q 2−k+2q 1 ∂ s 2−k+2q q 2 ( ) (t v )=∑ t1−k+2s ( ) v . (33) 1−k t ∂t t ∂t Γ ( 2 + s + 1) s=0 Taking into account formula (33) we obtain v k (x, 0) = 0 and lim t k v kt (x, t) = t→0

= =

) 2−q Γ ( 3−k 2 (1 − k)Γ ( 3−k+2q ) 2

) 2−q Γ ( 3−k 2 (1 − k)Γ ( 3−k+2q ) 2 lim t k t→0

lim t k t→0

∂ 1 ∂ q 1−k+2q 2−k+2q ( ) (t v )= ∂t t ∂t

q−s s 1−k ∂ q 2 C q Γ ( 2 + q + 1) 1−k+2s 1 ∂ s 2−k+2q t ( ) v = ∑ ∂t s=0 t ∂t Γ ( 1−k + s + 1) 2

1 ∂ 1−k 2−k+2q 1 lim t k (t v )= lim t k ((1 − k)t−k v2−k+2q + t1−k v2−k+2q )= t 1 − k t→0 ∂t 1 − k t→0 Authenticated | [email protected] author's copy Download Date | 2/10/18 7:01 AM

30 | E.L. Shishkina =

1 lim ((1 − k)v2−k+2q + tv2−k+2q ) = ϕ(x). t 1 − k t→0

Now we obtain the representation of v k through the integral. Using formula (18) we get v2−k+2q =

) 2Γ ( 3−k+2q 2 Γ

γ∣ ( 3−k+2q−n−∣ )Γ 2

1 2 ∫ (1 − r )

( n+∣2γ ∣ ) 0

1−k+2q−n−∣γ ∣ 2

γ (x; rt)dr. r n+∣γ ∣−1 Mϕ

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

) 2−q Γ ( 3−k 2

q

∑

2q−s C sq Γ ( 1−k + q + 1) 2 Γ ( 3−k + s) 2

(1 − k)Γ ( 3−k+2q ) s=0 2 =

=

) Γ ( 3−k 2

q

∑

1−k

s=0

C sq t1−k+2s 2s Γ ( 3−k + s) 2

) ) Γ ( 1−k Γ ( 3−k+2q 2 2 Γ(

3−k+2q−n−∣γ ∣ )Γ 2

1

× ∫ (1 − r2 )

1−k+2q−n−∣γ ∣ 2

t1−k+2s (

1 ∂ s 2−k+2q ) v = t ∂t

(

1 ∂ s 2−k+2q ) v = t ∂t

q

C sq t1−k+2s

∑

( n+∣2γ ∣ ) s=0 2s Γ ( r n+∣γ ∣−1 (

0

3−k 2

+ s)

×

1 ∂ s γ ) Mϕ (x; rt)dr. t ∂t

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

vk =

=

) 2−q Γ ( 3−k 2

1 ∂ q n+∣γ ∣−2 γ ) (t M f (x; t)) = (1 − k)Γ ( 3−k+2q ) t ∂t 2

) 2−1−q Γ ( 1−k 2 ) Γ ( 3−k+2q 2

q

∑ s=0

(

2q−s C sq Γ ( 3−k + q) 2 Γ ( 3−k + s) 2

q

) C sq Γ ( 1−k 2

s=0

2s+1 Γ ( 3−k + s) 2

=∑

t1−k+2s (

t1−k+2s (

1 ∂ s γ ) M f (x; t) = t ∂t

1 ∂ s γ ) M f (x; t). t ∂t

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

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