CC BY
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УДК 517.956.223
Asymptotic Behavior at Infinity of the Dirichlet Problem Solution of the 2k Order Equation in a Layer
Mikhail S.Kildyushov* Valery A. Nikishkin^
Institute of Computer Technology, Moscow State University of Economics, Statistics and Informatics,
Nezhinskaya, 7, Moscow, 119501 Russia
Received 01.02.2014, received in revised form 01.03.2014, accepted 20.04.2014
For the operator (-A)ku(x) + v2ku(x) with x G Rn(n ^ 2, k ^ 2) an explicit fundamental solution is obtained, and for the equation (—Д)кu(x) + v2ku(x) = f (x) (for f G Cx (Rn) with compact support) the leading term of an asymptotic expansion at infinity of a solution is computed. The same result is obtained for the solution of the Dirichlet problem in a layer in Rn+1.
Keywords: asymptotic behavior, elliptic equation, fundamental solution, estimation of solution, G-Meyer function.
A fundamental solution for the operator (—A)ku(x) + v2ku(x), x £ Rn(n ^ 2, k ^ 2), is obtained in [1, Section 3, 2.8.3]. The leading term of an asymptotic expansion at infinity exponentially decreases and does not contain a rapidly oscillating factor. In [2] the general form of all solutions of the equation
Ak U + aiAfc-1 U + ... + ak U = 0,
in a domain is deduced. Here a1, a2,..., ak are complex constants. In [3] a fundamental solution for the operator
(-A)ku(x) + ^u(x),
satisfying the radiation condition, is considered.
Let h(£) be the Fourier transform of the function h(x) £ L1(Rn)
f-2ni:
' Rn
If g £ L1(Rn), then the inverse Fourier transform is
h(0 = F [h](£) = / e-2nlx« h(x)dx. J Rn
ourier transform is
F -1[g](x) = / e2nlx«g(£)d£.
J Rn
I Rn
Lemma. The equation
(-A)kE(x) + v2kE(x) = ¿(x) x £ Rn (n > 2, v> 0) (1)
has a radially symmetric solution E(r) of the form
E(r) = (2k)(n-4)/2 xGk+1 , 0 ((-A2k N (2)
(r ) (n_\n/2..2k-2..n-2 * G0, 2k Ol.
n —2 n — 2 . 1 n—2 . k —1 1 k —2 1 / ' V '
n—2 n—2 I 1 n —2 I k —1 k— 1 k —2 1 0
2k ' 2k + fc ' •••' 2fc + fc ' fc ' fc '•••' fc ' 0
(2n)n/2v2k-2rn-2 0,2k ^ \2k here r = -у/x1 + x2 + ... + xn, G( ) is the G-Meyer function (in the notation of [4]).
*eridan111@gmail.com tvnikishkin@mesi.ru © Siberian Federal University. All rights reserved
Proof. The Fourier transform E(£) of the fundamental solution E(x) of the equation (1) is a solution of the equation
((2np)2k + v2k )E(0 = 1.
Hence,
where p = ^£2 + £2 + ••• + &• So
E(£) = E(p)
E(x)) = i J R
1
(2np)2k + v2k'
e'
lRn (2np)2k + v2k
Using the formulas for the inverse Fourier transform of the radially symmetric function ^(p) (see [5, (2.113), (2.114)]), we have: for n = 2p
for n
E(x)) = 2p + 1
(—2n)-(p-1) lim f- ^ £—+o \ r dr /
c+tt
2n
e-ep^(p)pJ0(2nrp)dp
/1 AN P-1 E(x)) = (—2n)-(p-1) lim - — e—>+0 \ r dr /
/*+tt
e ep^(p)psin(2nrp)dp
(3)
(4)
Notice that in our case for e = 0 the integrals in square brackets in formulas (3), (4) converge. Therefore, we can consider formally the following expressions, although it is not yet proved that they define a solution of (1): for n = 2p
E(x) = (—2n)-(p-1)( -1
p-1
f + tt
2n
p
(2np)2k + v
,2k
Jo(2nrp)dp
for n = 2p +1
= ^2n)-(p-1w-
E(x) = (—2n)
p-1
dr 7
(2np)2k + v'
2k
sin(2nrp)dp
(5)
(6)
Consider the even dimension case. We calculate the integral on the right hand side of (5) at k = 1, 2, 3,4,..., using Mathematica (the licence L 3298-0846), and construct for it the following expression for any k:
f+tt p 1 / Jo (2np)2k + v2k Jo(2nrp)dp = 8kn2v2k-2 Gk+1,0 (
(2k)
2k
k-1 k-2 k > k ,
• (7)
Substituting it in (5) and using the formula for differentiation of the G-Meyer function (see [6, 8.2.2.32]) we obtain the function on the right hand side of the expression (2). Let us prove that this is the expression for the fundamental solution.
Using the well-known asymptotic expansion of the G-Meyer function, we put
Gmq0(z) = Gmq0(Z\bl,b2,...,bJ .
From Theorem 2 (see [7, 5.9.2]) it follows that (for m ^ q — 1, arg z = 0, z ^
Gm'°(z) - Am'0HPjq(zeln(q-m)) + A"'0^(ze-ln(q-m)).
0
2
r
0
0
2
p
r
0
v2k r2k
k-1 k-2
0
0
kk
k
In the case under consideration m = k + 1,p = 0, q = 2k.
2k —1
, . (2n) 2 1 x 2k —1 x—-
Ho , 2k (Z) = .0 .1 exp ( - 2kZZ^ ]T MjZ = (2kj 2 j=0
where Mo = 1 ( [7, (5.9.1.13)]),
2k
• 0 = (-1)k-1(2ni)-(k-1) exp ( - in £ bj
j=k+1
( [7, 5.8.2.2]). Finally, for the G-Meyer function in (7) we obtain
N 2^2^ , n x , 3n(k - 1)
~ —==- exp ( — vr sin — cos--vr cos ,.
, ^, ... , 0 , , ^, k, ^ a/vT v 2k7 v 4k 2k
^k+1 , o( v2kr2k N 2v/2n ( . n , ( 3n(k - 1)
f ' 'I __I ^ / _ ûvn I _ 7/^T oin _ Ï r>r\<2 I _1
G°,2k \ (2k)2k
The exponential decrease ensures the convergence of the integrals and the applicability to the right hand side of (2) of the formula ( [5, (2.108)]) for the Fourier transform of the radially symmetric function:
2n f
*(P) = pn-2)T2 J0 *(r)rn/2J(„-2)/2(2npr)dr. (8)
In our case it looks like
(2k) n-2 r+— = (2np) 2-iv2k-r1-2 Jn-2 (2nPr)x
k+1,0 / fvr^ 2k
xG°.2k I V2k
dr.
_____.____; 1 n /
2k , 2k + k 2k + k , k , k
This integral (see [6, 2.24.4.1]) is expressed via the G-Meyer function and simplified to the form
2k fc-i\
1,1 ' (jlV
= (2np)2 v2k-2 GM I
k
k— 1 k
Using the integral representation of the G-Meyer function (( [6, Definition 8.2.1]), where the
k 2 k 2
integration contour L is the straight line (—---«to, —---+ «to)) and formula ( [8, 3.981(3)]),
2k 2k
we find
k-1 \ 2fc-2
G1,1 ( z
^ \ Z 2k
k—-1 y 1 + z
From here it follows that
2k k—1'
rG
1,1
k ' ^ (9)
(2np)2v2k-2 k-1 y (2np)2k + v2k'
Finally, the statement is proved for even n.
Let us prove the lemma for odd n. First, for n = 3 we substitute (9) into the inverse Fourier transform formula (formula (8), where p and r are interchanged). We obtain the integral (see [6, 2.24.4.1]) and substitute the result in (6). Now we proceed similarly to the case of even n. The lemma is proved. □
Let us specify the leading term of the asymptotic expansion of u(x) at infinity.
Theorem 1. Let x £ Rn (n ^ 2, v > 0), f (x) be a smooth function with compact support. Let the solution u(x) of the equation
(—A)k u(x) + v2k u(x) = f (x) (10)
exponentially decrease at infinity. Then the following representation holds
u(x) = r(1-n)/2 sin (vr cos n)e-vr sin 2k$1 (01,..., 0n-i)+
2k
+r(1-n)/2 cos (vr cos n)e-vrsin 2k$2(^1,..., 0n-i) + O(r-(n+1)/2e-vrsin 2k), (11) 2k
where $1(01,..., 0n-1), $2(02,..., 0n-1) are differentiable functions on the unit sphere.
Proof. Let f (x) have its support in a ball QR of radius R . Then for the solution u(x) of the equation (10) we have the following representation
i(x) = / E(\x - y|)f(y)dy. ■J Rn
Introduce the following notation x = (xi,x2, ...,xn). Suppose that x runs along a ray, we can turn the coordinate system so that this ray coincides with x1 > 0, x2 = ... = xn = 0. Then
u(xi, 0,..., 0) = ^ E (^(xi - yi)2 + y2 + ... + / (y)dy =
= / E (|(xi - yi|) /(y)dy + / (e( 7(xi - yi)2 + y2 + ... + yn) - E(|(xi - yi |)) / (y)dy. •/Qr JQr V /
Denote the integrals on the right hand side of the this equality by Ji and J2. For E(r) we employ the asymptotic expansion
n+1 —4 k , . \/j \ \
tv \ 2 -vr sin -n / n(n + 1)(k - 1) n\
E(r)--тe vr sin 2k cos —---- - vr cos— r—> +00, (12)
^ ' (2nr)1 V 4k ^ ' y >
and obtain, as xi ^
Ji = / (|xi - yi|(i-n)/2 sin (v|xi - yi| cos П)e-vlxi-y1sin2kCi + ./Qr 2k
+ |x1 — y1|(1-n)/2 cos (v|x1 — y1| cos n)e-v 1 Xl-yi1 sin 2kC2 +
2k
+O(|x1 — y1|-(n+1)/2e-v 1 xi-yi1 sin 2k))f (y)dy = = x11-n)/2 sin (vx1 cos 2k)e-vxi sin 2kC3 + x11-n)/2 cos (vx1 cos 2k)e-vxi sin 2kC4+
+O(x-(n+1)/2e-vxisin 2k),
where c1, c2, c3, c4 are constants. Turn now to the estimation of J2.
Using (12) and the mean value theorem we arrive at the following inequalities for some 0 < © < 1: _
^(x1 — y1)2 + y2 + ... + yn) — E (|x1 — y11) = E (V1 — y11 + © ^(x1 — y1)2 + y2 + ... + yn — |x1 — y1|^
(V(x1— y1)2 + y2 +... + yn— |x1 — y1^ <
n + i " 2
I I • n n+i . _ - n+i
< C°e-v|xi-yi1 sin 2k |x1 — y1| ^ < C1e-vxi sin 2kx- 2
if |y| < R, x1 > 2R. Therefore
|J2| < Ce-vxi sin 2k x1
where C°, C1, C are constants.
Going back to initial coordinates, we get the representation (11). We shall apply the obtained results to the Dirichlet problem in a layer. Denote
n = {(x,xn+1) £ Rn+1|x £ Rn, xn+1 £ (a, b)}, —to < a < b < +to, n > 2. Consider the problem
(—1)k
^ $2k ( n 02 \ \
--+ y] v + av = h, (x, xn+1) £ n,
dxn+1 dx22M (13)
d j
v
dxn+1
v
= 0, j = 0, ...,k — 1.
Xn+i=b
Let 0 < A1 < A2 < ... be the eigenvalues and , (l = 1, 2,...) be the eigenfunctions of the problem
f y(2k)(t) + ( —1)k+1A2ky(t) = 0, t £ (a, b),
\ y(j)(a) = y(j)(b) = 0, j = 0, ...,k — 1. (14)
Put w = 2ka + A2k (l = 1, 2,...).
In[9] (Theorem 6) the solvability of the problem (13) and uniqueness of the solution v was proved for h(x, xn+1) £ CTO(n) with compact support and a + Aj^ > 0, as well as the estimate
|v(x,xn+1)| < Ce-(Mi sin 2k-£)|x|, (x,xn+1) £ n, (15)
(here e > 0 is sufficiently small).
Study the proof of the estimate (15) more closely. Denote by v = 5(£1,..., £n, xn+1) the Fourier transform with respect to x of the function v(x,xn+1). Then v is a solution of the one-dimensional boundary value problem in xn+1 on [a, b] with the parameters £1,..., £n :
i ( —1)kv2k + av + (2n)2k(e2 + ... + en)kv = F[h], xn+1 £ (a,b),
\ v(j)(a) = v(j)(b) = 0, j = 0, ...,k — 1.
The singular sets in this case are given by the conditions
—M2k = (2n)2k(e2 +... + en)k, i = 1,2,3,.... (16)
Put
Zj = Re ej, Tj = Im e (j = 1,..., n), Z = (C1, ...,Zn), T = (T1, ...,Tn).
To apply Theorem 6 of [9], the intersection of the cylinder (2n)2 |t|2 = y2 with the singular sets (16) must be empty. We shall find 7, for which this condition is fulfilled, that is the following system has no solutions:
j <2n>V!2 = 7»,
I (2n)2k(e2 +... + en)k = —w' v = 1,2,3,...).
X
This system splits into k systems of the form ( (2n)2 |t |2 = y 2,
\ (2n)2(e2 + ... + en) = m2 (cos ^ + isin (l = 1, 2, 3,...),
where s = 0,1,..., k — 1.
For each s the system consists of the real equations
(2n)2 |t |2 = y2,
(2n)2(|Z |2 - |t |2) = M2 cos (l = 1, 2, 3,...),
(2n)22(Z, t) = M2 sin i1±|Sn (l =1, 2, 3,...).
The Cauchy-Schwarz inequality implies that the contradiction is achieved if the following condition is fulfilled:
2 A 2 (1 + 2s)n , 2\ 4 • 2 (1 + 2s)n
4Y ( Ml cos-k-+ 1 ) < sin k '
(1 + 2sW
Solving this inequality for y, we obtain y < Mi sin-—-.
2k
n
Thus, if y < M1 sin —, then the system (17) is inconsistent. 2k
Theorem 2. Let h(x,xn+1) £ CTO(n) in the problem (13) have compact support, the constant a satisfy the condition a + A2k > 0, where A1 is the first eigenvalue, and y be the corresponding eigenfunction of the problem (14). Let the solution v(x, xn+1) of the problem (13) exponentially decrease at infinity. Then
v(x,xn+1) = (sin ( a + A1kr cos 2k)$1(01,..., #„-1)+
+ cos (a + A1kr cos 2k)$2№,..., ^-1)) x r(1-n)/2e- 2Va+^Fr sin ife y (x„+1)+
+ 0(r-(„+1)/2e- V^r sin 2k ),
where $1(01,..., #n-1), $2(#2,..., #n-1) are differentiable functions on the unit sphere. Proof. Put
h-1(x) = h(x, x„+1)y1(x„+1)dx„+1,
a
V1(x) = / v(x, x„+1)^1(x„+1 )dxn+1.
a
Note that h1(x) has compact support and h1(x) £ CTO(R"). The function v1(x) is a solution of
(-A)kv1(x) + M1kv1(x) = h1(x) x £ R„.
Then for the solution v(x, xn+1) of (13) we have the representation
v(x,x„+1) = V1(x)y1(x„+1) + v (x, x„+1), (18)
and for £(x,xn+1), by Theorem 6 of [9], we have the estimate
|v(x,xn+1)| < Ce-(M2 sin 2k-e)H,
(e > 0 is sufficiently small).
The asymptotic expansion for v(x, xn+1) follows from (18) and Theorem 1. □
References
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[3] A.V.Filinovsky, About asymptotic behavior of solutions of one non-stationary mixed problem. Dif. Uravn, 21(1985), no. 3, 443-454 (in Russian).
[4] G.S.Meijer, On the G-function, Nederl. Akad. Wetensch. Proc. Ser. A, 49(1946), 227-237, 344-357, 457-469, 632-641, 765-772, 936-943, 1063-1072, 1162-1175.
[5] S.Mizohata, The theory of partial differential equations, Cambridge University Press, New York, 1973.
[6] A.P.Prudnikov, J.A.Brychkov, O.I.Marichev, Integrals and series. Supplementary chapters, Moscow, Nauka, 1986 (in Russian).
[7] Y.L.Luke, The Special Functions and Their Approximations, New York, Academic Press, vol. I-II, 1969.
[8] S.Gradshteyn, I.M.Ryzhik, Table of integrals, series, and products, Fourth edition prepared by Ju. V. Geronimus and M. Ju. Ceitlin. Translated from the Russian by Scripta Technica, Inc. Translation edited by Alan Jeffrey, Academic Press, New York, 1965. MR 0197789 (33 #5952)
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Об асимптотике решения задачи Дирихле для уравнения порядка 2k в слое
Михаил С.Кильдюшов Валерий А.Никишкин
Для оператора (-Л)кu(x) + v2кu(x) в Rn (n ^ 2, k ^ 2) получен явный вид фундаментального решения, а для уравнения (-A)ku(x) + v2ku(x) = f (x) (с финитной бесконечно дифференцируемой функцией f) — первый член асимптотики решения на бесконечности. Изучается также задача Дирихле в слое из Rn+1.
Ключевые слова: асимптотика, эллиптическое уравнение, фундаментальное решение, оценки решений, G-функция Мейера, слой.
