Singular Cauchy problem for generalized homogeneous Euler-Poisson-Darboux equation Текст научной статьи по специальности «Математика»

Математические заметки СВФУ Апрель—июнь, 2018. Том 25, № 2

UDC 517.9

SINGULAR CAUCHY PROBLEM FOR GENERALIZED HOMOGENEOUS EULER—POISSON—DARBOUX EQUATION E. L. Shishkina, M. Karabacak

Abstract. In this paper, we solve singular Cauchy problem for a generalised form of an homogeneous Euler—Poisson—Darboux equation with constant potential, where Bessel operator acts instead of the each second derivative. In the classical formulation, the Cauchy problem for this equation is not correct. However, S. A. Tersenov observed that, considering the form of a general solution of the classical Euler—Poisson—Darboux equation, the derivative in the second initial condition must be multiplied by a power function whose degree is equal to the index of the Bessel operator acting on the time variable. The first initial condition remains in the usual formulation. With the chosen form of the initial conditions, the considering equation has a solution. Obtained solution is represented as the sum of two terms. The first tern is an integral containing the normalized Bessel function and the weighted spherical mean. The second term is expressed in terms of the derivative of the square of the time variable from the integral, which is similar in structure to the first term.

DOI: 10.25587/SVFU.2018.98.14233 Keywords: Bessel operator, Euler—Poisson—Darboux equation, singular Cauchy problem.

1. Introduction

We study the initial value problem

Lu =

E

д2 ъ д

dx2 xi dxi

dt^ + ldt

u(x, 0; k) = ^(x),

lim t ut(x,t; k) = ^(x), t^+0

= u(x, t; k),

(1) (2)

where y > 0, x > 0, i = 1,... ,n, k G R, t > 0. The equation (1) is called generalized Euler—Poisson-Darboux equation (generalized EPD equation).

Using the terminology from the book [1], a problem for the equation of the type

A(t)—+B(t)—+C(t)u = Gu,

= u(t, x), x = (xi, . .. ,xn),

where G is a linear operator, acting only by variables ^C i, » » » , ^Cn, iS called singular if at least one of the operator coefficients tends to infinity in some sense as t ^ 0.

In [1] was given five general techniques for the solution of the singular Cauchy problem

E

dx2 dt2 + t дt

= u(x, t; k),

(3)

2

u = c u

u

© 2018 E. L. Shishkina, M. Karabacak

u(x, 0; k) = ^(x), ut(x, 0; k) = 0. (4)

These methods are

(1) Fourier transforms method in distribution space,

(2) spectral technique in a Hilbert space,

(3) transmutation method,

(4) studying related simpler differential equations,

(5) energy methods.

Some of these methods were successfully applied to generalized EPD equation (1) and to Lu = 0 in other pepers. Namely, using the Hankel transform instead of Fourier solutions to Lu = 0 and (1) with conditions (4) were obtained in [2,3], accordingly. The third and closely connected the forth methods were used to solution to Lu = 0 in [2, 4]. In [5] transmutation method was used for obtaining new integral initial conditions for EPD equation (3). Abstract differential equations with Bessel operator of EPD type were studied in the [1], cf. also recent papers [6-8]. In [9] the problem (3), (2) was solved using 'descent' operators which are special cases of Buschman-Erdelyi transmutations operators (see [10,11]). Here as a main result we obtain a solution to the problem (1), (2).

2. Basic definitions

In this section we need to introduce some of the basic definitions.

We deal with the subset of the Euclidean space

R++1 = {(t,x) = (t,xi,... ,xn) e Rn+1, t> 0, xi > 0,... ,xn > 0}.

Let x = (x1,..., xn), |x| = , x| and Q be a finite or infinite open set in

V i=i

symmetric with respect to each hyperplane t = 0, x^ = 0, i = 1,..., n, Q+ = QnR++1 and f2+ = 0 fl M™+1, where

R++1 = {(t, x) = (t,x1,... ,x„) e Rn+1, t> 0, x1 >0,... ,x„>0}.

Consider the class Cm (Q+) consisting of m times differentiable on Q+ functions and denote by Cm(il+) the subset of functions from Cm(il+) such that all derivatives of these functions with respect to t and x^ for any i = 1,..., n are continuous up to t = 0 and Xi = 0. Function / £ Cm(i!+) we will call even with respect to t and Xj, i = 1,...,n if

d 2k+1 f

dt2k+1

d 2k+1 f 0, J

t=0,x=0 dx2k+

=0

t=0,x=0

for all nonnegative integer k (see [12, p. 21]. Class C™(f2+) consists of functions from Cm(il+) even with respect to each variable t and Xi, i = 1,... ,n. In the

_ri+l

following we will denote Cm (R+ ) by Cm. We set

C(o+) = flC(n+)

_n-j- \

with intersection taken for all finite m. Let CV (R+ ) = CV.

We will deal with the singular Bessel differential operator Bv (see, for example, [12, p. 5]:

d2 v d 1 d d and the elliptical singular operator or the Laplace-Bessel operator AY :

A, = «M. = ¿(B„).„ «

i=1 i=1 v 1 ' i=1 '

The operator (5) belongs to the class of B-elliptic operators by I. A. Kipriyanovs' classification (see [12,13]). Operator

(Dk,7 )t,x = (Bk )t — (A7 )x

is a B-hyperbolic by the same classifications.

The symbol jv is used for the normalized Bessel function:

. , 2Vr(v + 1) . ,

where Jv (t) is the Bessel function of the first kind of order v (see [14]). The function jv (t) is even by t and

>(0) = 1, =0. (6) dt t=0 Using formulas 9.1.27 from [15] we obtain

(B„)tJ^i(Tt) = -T2j^1(Tt). (7)

We deal with a multi-index 7 = (71,..., 7„) which consists of positive fixed reals Yi > 0, i = 1,... ,n, |y| = 71+.. .+7„.

The operator kTtT for k > 0 is a generalized translation acts by a variable t defined by the next formula (see [16, p. 122, formula (5.19)])

Wkbl) I

kTt\f(t,x) = J /( 02 + T2 _ 2tT cos <p, x) sin*-1 v ép (8)

and YTy =Y1 TX1 .. TXyn* is multidimensional generalized translation, where each of the one-dimensional generalized translations Ty acts by a variable x for i = 1,..., n according to the formula (8). It is known that

(at) = {at)jk=i (ar), (9)

where a G R.

Based on the multidimensional generalized translation YTy the weighted spherical mean MY[f (x)] of a suitable function is constructed by the formula

m№)]=k+7-TT f ^/(x^dS, (10)

|S1 (n)|Y J

S+(n)

where

nr№)

07 = rK4. S+(n) = {0:\0\ = l,emrl}, Ist(r

»n1 =

2n-lr(™)

It is easy to see that

d

limC[/(x)] = /(a;), ]im—M?\f(x)\ = 0. (11)

r^o r^o dr

3. Singular initial value hyperbolic problems

We will be concerned with the solutions of the following singular initial value hyperbolic problem

Lu =

+ ilJL) -(I- + t<L

\ dxj Xi dxi J V dt2 t dt

u = c2u,

(12)

u(x, 0; k) = ^(x), lim t ut(x, t; k) = ^(x), u = u(x, t; k). t^+0

We will call (12) the generalized Euler—Poisson-Darboux equation. In [3] the next theorem was proven.

Theorem 1. The solution u e Ce2v (R++1) to the

[(A7)x - (Bk)t]u = c2u, c > 0, u = u(x,t; k), (13)

u(x, 0; k) = ^(x), ut(x, 0; k) = 0, (14)

for k > n + |y| - 1 is unique and defined by the formula

k — n — | y | — 1

J k — n — |-y| —1 2

u(a:, t; k) = A(n, 7, J(t2 - r2)'

0

x (cVi2 - r^+M^Miïvix)] dr, (15)

where

A(n, y, k) =

2r(i±i)

r(«^L)r(

2

It is known (see [1]) that if u(x, t; k) is a solution to the (13) when the next two fundamental recursion formulas hold

u(x,t; k) = t1-ku(x, t; 2 - k), (16)

t

u(x, t; k)t = tu(x, t; 2 + k).

(17)

Theorem 2. Let ip = ip(x), p € Clev 2 . Then the solution of (13), (14) for k < n + |y| - 1, k = -1, -3, -5,... is

( d \m

u(x,t;k) = t1-k i—j (tk+2m-1u(x,t;k + 2mj), (18)

where m is a minimum integer such that m > k—- and u(x, t; k + 2to) is the

solution of the Cauchy problem

[(Ay)x - (Bfe+2m)t]u = c2u, c > 0, (19)

U{^k + 2m)= (k + mJ^k + 'lrn-iy ^>0;ifc + 2m)=0. (20)

Proof. In order to proof that (18) is a solution of (13), (14) when k < «.+ |y|- 1, k= — 1, -3, -5,..., we will use the recursion formulas (16) and (17). Let choose minimum integer m such that k + 2m > n + |y | - 1. Now we can write the solution of the Cauchy problem

[(Ay )x - (Bk+2m )t]u = c2u, c > 0, u(x, 0; k + 2m) = g(x), ut(x, 0; k + 2m) = 0, g G Cfv, by (15). We have

t

-> j / k + 2m — n — | yI — 1

u(x, t;k + 2to) = A(n, 7, A: + 2TO)i1-fe-2m / (i2 - r2)-s-

0

X Jk+2m-n-hl-i (cVt2 - r2)r"+lTl-1MJ[q(x)l dr, 2

where

A(n, 7, A; + 2to) - 1 2 '

) P ( fc+2m—re—17| + 1 ^ '

Considering (16), it is easy to see that

tk+2m-1u(x, t; k + 2m) = u(x, t; 2 - k - 2m). Applying (17) to the last formula m times we get d \m

— J (¿^"-^(x, t; k + 2 to)) = m(x, t;2-k).

Applying again (16) we can write

( d \m

u(x,t;k) = t1-k i—j (tk+2m-1u(x,t;k + 2mj), (21)

which gives the solution of the (19). Now we obtain the function g such that the (20) is true. From (21) it follows that

u(x, t; k) = (k + 1)(k + 3) ... (k + 2m - 1)u(x, t; k + 2m) + Ctu(x, t; k + 2m) + O(t2),

when t ^ 0, where C is a constant. Evidently, if

g(x)

(k + 1)(k + 3) »»» (k + 2m - 1)

then u(x,t; k) defined by (18) satisfies the initial conditions (14).

Let us recall that for u(x, t; k + 2m) to be a solution of (19), (20) it is sufficient that f G . In order to be able to carry out the construction (21), it is sufficient

to require that / G Cev 2

rn+H+fc-l]

Theorem 3. Let if) G Cev 2 • The solution u = u(x,t; k) to the

[(Ay)x - (B)t]u = c2u, c > 0, (22)

u(x, 0; k) = 0, lim t ut(x, t; k) = ^(x), t^+o

for k < 1 is defined by the formula

( 1 d V

u(t,x;k) = B(n,7, k,ç) I -— I

(t2 - r2)

1 —fc + 2q —n — |-y|

(23)

x .7i-fc+2g-n-i-r (c\/i2 — r2)r

2)rn+|7|-l

MY[^(x)] dr I , (24)

where

B(n, y, k, q)

Proof. Let q > 0 is the smallest positive integer number such that 2 — k + 2q > ■ + |y| — 1 i-e. q > 3 and let u(x, t; 2 — k + 2q) is a solution to (22) when

we take 2 — k + 2q instead of k such that

u(x, 0; 2 - k + 2q) = ^(x), ut(x, 0; 2 - k + 2q) = 0»

(25)

When by property (16) we obtain

u(t, x; k - 2q) = t1-k+2qu(t, x; 2 - k + 2q)

is a solution to the equation

(A7)xu -

<9i2

k - 2q dv

= c2u»

Further, applying q-times the formula (17) we obtain

Q 9 u(t, x; k - 2q) = Q^y (i1-^2^^, x; 2 - k + 2q))

is a solution to the (23). In order to get a solution to (22) satisfying to the conditions (23) we use the multiplier ,,)r(3-2fc+4?r- Let

u(t, x; k)

2-«r(V) fldYtf-k+^^-k + iq)). (26)

(1 — k)r(——\tdt

2

2

We have shown that (26) satisfies the equation (22). Now we will prove that u(t, x; k) satisfies the conditions (25). Using formula 1.13 from [17, p. 9] we obtain

(i^Mi,^ - k + 2q))

and u(0, x; k) = 0 for k < 1. For the second condition (25) we get

lim tkut(t, x; k) t^o

-_2 gr(V0_limtk— (- —y (t^+^uft x-2-k + 2a))

~ (i-k)r(3-fc2+2") t™ dt\tdt) [ {t' ' + q))

f.—rH 2-^t Wl-fc I „ I

s

x (jJ^J x; 2 — A; + 2ç)

1 d

lim (i1-fcu(i, x; 2 -k + 2q))

1 - k t^o dt

1 lim tk ((1 - k)rku{t, x; 2 - A; + 2q) + t^u^t, x;2-k + 2q))

1 - k t^o

1 lim ((1 - k)u(t, x]2-k + 2q)+ tut(t, x; 2 -k + 2q))

1 - k t^o

lim u(t, x; 2 - k + 2q) = ^(x)» t^o

Now we write the representation of u(t, x; k) through the integral. Using formula (15) we get

t

u(x, t;2-k + 2q) = A{n, 7, 2 - k + 2q)tk-1-2q J (t2 - r2) 1~*+2,

o

x n-^-n-Ki (cy^2 - r2)r"+l7l~1Mj[^(x)1 dr.

2

Considering (26) we write

u(t,x;k) = A(n,~f,2 - k+ 2q)- V 2 7

t

(jlJ J(t2 -ry-'+y-H 7-1_t+ag_n_K| dr.

Simplifying we get (24) and this completes the proof.

The union of the theorems 2 and 3 gives the following statement.

r T1+ |-y I — k 1 . 0 r |-r I 1 1

Theorem 4. Let <p = <p(x), tp G Clev 2 1 , ip = ip(x), ip G Cev 2 • Then the solution of

[(Ay)x - (Bk)t]u = c2u, c > 0, (27)

u(x, 0; k) = ^(x), lim tkut(x, t; k) = ^(x), (28)

t^+0

for k < min{n + |y| - 1,1}, k = -1, -3, -5,... is given by formula

u(x, t; k) = u1(x, t; k) + u2(x, t; k), where u1(x, t; k) is found by Theorem 2 and u1(x, t; k) is found by Theorem 3.

4. Examples

1. Let's start with an example

[(BY)x - (Bk)t]u = c2u, c > 0, u = u(t, x; k), (t,x) G R+, (29) u(x, 0; k) = jY(ax), ut(x, 0; k) = 0, (30)

where n =1, 7 - 2 < k < 7, k = -1, 0<y, a G R. When m =1 and considering that (see (9))

1Txh(ax) = {ax)j^i (ar) (31)

we obtain

u(x, t;k + 2) = A(1',7'fc + 2) J^i (ax)^1

k + 1 2

where

A(l,7,*: + 2) - V 2

Passing to the functions Jv in the integral we get

i; A + 2) =-LlIl- ^^-^^ (ax)t

(k + lja 2 c 2 t

x / (¿2 - r2)^1 Jk-i (c\Jt2 - r2)r^ hl J-,-1 (ar) dr.

J 2 2

0

Applying the formula 2.12.35.2 from [18] of the form t

/«" - = w

0

t> 0, Re v> -l - 1, Re -m - 1, (32)

t

we obtain t

(t2 - r2) k41 Jk-1 (c\/t2 - r2)r"r21+1 (ar) dr 2 2

= ife21a",21cfe2"'(v/a2 ''t1 jfc+1 (¿y/a2 c2~)

and

u(x,t;k + 2) =- K 2 )i^-i(ax)t-^(y/a2+c2)-i:%1Jk:+i(tv/a2 + c2)

k + 1 2 2

(ivWc2).

k + 1 2 2 Then the solution of (29), (30) is

r) f) /-

ii(ar, i; fe) =i1~'c — (ife+1w(x,i;A; + 2)) = --J-r-i(ax) — (tk+1jk+i (ty/a2+c2))

=- 2 , —i (ax)—-(t^ Jk+1 (t\/a2 + c2))

i-t - -

— 2~rrj-'-1 (a'x)t 2 Jfc-i (¿v a2 + c2) = 7-r-i (ax)jk-i (ty a2 + c2).

(Va2 + c2)— 2

2 2 2

As might be seen from (7) and (6)

ji^l (ax) lim jfc^i (iv7 a2 + c2) = ji^i (ax),

jf-i (ax) lim —jk-i (t\/a2 + c2) = 1 (ax) = 0,

(B^-rii-i (ax)jk-i (is/a2 + c2) = —a2?-,-! (ax)jk-i (is/a2 + c2),

(Bh)tii-i (ax)jk-1 (¿\/a2 + c2) = —(a2 + c2)?-,-i (ax)jk-1 (¿\/a2 + c2) which showss that the function

ii(i, x; A;) — 7-r-i (ax)jk-i (t\J a? + c2) (33)

satisfies (29), (30).

2. Consider the problem

[(BY)x - (Bk)t]u = c2u, c > 0, u = u(t, x; k), (t, x) G R+, (34) u(x, 0; k) = 0, linio tkut(x,t; k) = jY(bx), (35)

where n =1, k < 1,0 < 7 < 3, b G R. When q =1 and considering (31) we obtain u(t, x; k) = ¿3(1, 7, k, l) ?-,-! (bx)

1 d

I I (t2 - r2)1 ^ j1 k+'jicy't2 - r2)?->-! (br)r"1 dr

j?(l,7,M)r(2±l)r(2-*±2) . ih^d(}2

-—- . ■ ,-—h-i ® "T \t ~ r 4

Vo

2

x Jt )=+-, (cyt2 — r2) J-,-1 (br)r 2 + 1dr

where

Applying the formula 2.12.35.2 from [18] of the form

i

x J^+^+m+i+liiV7C2 + h2)], t> 0, Re v> —l — 1, Re — m — 1,

we obtain t

#2 „2

J{t2-r2)2 4 fc+-r (c\A2 - r2)J-,-i (br)r',21+1 dr

and

u(t, x; k) = 2^5(1,7, A, l)r ( -^J T ( 2 - ^ ) (62 + c2)

tdV —v "

i1"* /-

(bx)jl^k (t\/b2 + c2).

1 - r 2

Taking into account (7) and (6) it is easy to check that

fi-k - fi-k -

(B7)x---j^iibxl^itVV + c2) = -b2--^(tVb2+c2),

1 — K 2 2 1 — K 2 2

fl-fc - fl-fc -

(Bfc)t—■¿J^lMji^dVb2 +c2) = ~(b2 +c2)T-^j2_1(bx)j1_!,(ty/b2 +c2),

fi-k -

lim-— 7-r-i (bx)ji^k (ty b2 + c2) = 0

t—>o 1 — K 2 2

t

and

8 ( t1^k /-\

which confirms that the function

t1-k -

u(t, x; k) = --fcj2^1 (t v b2 + c2) (36)

satisfies (34), (35).

3. From examples 1 and 2 it is plain to see that the solutions of

[(B7)x - (Bk)t]u = c2u, c > 0, u = u(t, x; k), (t, x) G R+, (37)

u(x, 0; k) = jY(ax), lim tkut(x, t; k) = jY(bx), (38)

t^+0

where n =1,0 < 7 < 1 7 - 2 < k < 7, k = -1, a, b G R is

- -

w(i, x; k) = j't^i (ax)jfc^i_ (iy a2 + c2) + -—(tyb2 + c2). (39)

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Submitted February 28, 2018

Elina Leonidovna Shishkina Voronezh State University,

Faculty of Applied Mathematics, Informatics and Mechanics,

1 Universitetskaya Square, Voronezh 394063, Russia ilina_dico@mail. com

Mesut Karabacak Atatürk University,

Department of Mathematics, Science Faculty 25240 Yakutiye, Erzurum, Turkey [email protected]