Mathematical Methods in the Applied Sciences, Vol. 12, 275-291 (1990) 
MOS subject classification: 35 J 30, 35 B40, 35 E 15 
A Uniqueness Condition for the Polyharmonic 
Equation in Free Space 
P. Lesky Jr 
Institut fur Angewandte Mathematik, Uniuersitut Bonn, Wegelerstr. 10, D 5300 Bonn I ,  West 
Germany 
Communicated by P. Werner 
Consider the polyharmonic wave equation a:u + ( - A)% =f in [w" x [0, m ) with time-independent 
tight-hand side. We study the asymptotic behaviour of u(x, t )  as t -t m and show that u(x, t )  either 
converges or increases with order t' or In f as t -t 03. In the first case we study the limit uo(x):= lim u(x, t) 
and give a uniqueness condition that characterizes uo among the solutions of the polyharmonic equation 
( - A)% =f in  58". Furthermore we prove in the case 2m 3 n that the polyharmonic equation has a solution 
satisfying the uniqueness condition if and only iff is orthogonal to certain solutions of the homogeneous 
polyharmonic equation. 
1-m 
1. Introduction 
Consider the problem 
a:u + ( -  A)% = Ciwff in R" x [O, a), 
u(x, 0) = a,u(x, 0) = 0 in R", 
where j - ~  CF (R"), o 2 0 and A : = 8; + . . . a % .  We are interested in the asymptotic 
behaviour of u(x, t) as t -+ co . In the case o > 0 it has been shown by Eidus [2] that 
1. If m < n, then the principle of limiting amplitude holds: 
u(x, t) = e-iW'u,(x) + o(1) (1.2) 
( - A)'%, - o'u, = f in R"; (1.3) 
as t -+ 00, 
where 
u, can be uniquely characterized by (1.3) and a suitable radiation condition. 
2. If m = n, then 
r 
u(x, t) = e-iW'u,(x) + c1 J f(x')dx' + o(1) as t --f 00, 
R" 
with a suitable constant c1 # 0, where u, is a solution of (1.3). 
01 704214/90/040275-17$08.50 
0 1990 by B. G. Teubner Stuttgart-John Wiley & Sons, Ltd. 
Received 19 April 1989 
276 P. Lesky Jr 
3. If m > n, then 
u(x, t )  = t' -Wrn c2 jiwnf(xf)dx' + o(tl-"/"') as t -+ co , (1.5) 
with c2 # 0. 
This shows that u(x, t )  is unbounded as t +. co if o > 0 and m =- n. As pointed out in 
[6], similar resonance effects can be observed in the case o = 0, m = 1, n = 1 or n = 2. 
In section 2 we study (1.1) in the case o = 0 for arbitrary m, n E  N. We discuss the 
asymptotic behaviour of the solution u as t + a3 and show: 
1. If 2m < n, then 
u(x, t) = uo(x) + o(1) as t -+ 03 (1.6) 
uniformly in every compact subset of [w", where uo satisfies the corresponding 
static equation 
( - A)%, =f in R". (1.7) 
2. If 2m >, n, then for odd n 
m - ( n  + l ) / Z  
u(x, t )  = 1 D s t 2 - y  jiwnf(x')lx - x'(2sdx' + uo(x) + o(1) 
s = o  
as t -+ co, (1.8) 
and for even n 
m- 1 - n / 2  
u(x, t) = 1 Dst2-* jiwnf(x')]x - x'Izsdx' 
s = o  
+ D*lntJ f(xf)lx - xfI2'"-"dx' + u,*(x) + o(1) as t -+ 00 (1.9) 
iw" 
uniformly in every compact subset of R", where uo and u: are solutions of (1.7) 
and D, and D* are specified in (2.25) below. 
Sections 3 and 4 deal with the polyharmonic equation (1.7) and with the solution uo 
determined by (1.6). Note that (1.6) holds also in the case 2m 2 n iff satisfies the 
condition 
f(x')Ix - x'(2sdx' = 0 for s = 0, 1, . . . , and every XER"  (1.10) 
([I] : = max(n E No: n < I}), or, equivalently, 
(compare (4.21), (4.22) in [ 3 ] ;  la[:= a1 + * . * + a, for every multi-index U E  N;, 
xfa: - xtal . . . x?). We study the question of how uo can be characterized uniquely 
among the solutions of (1.7) by imposing a suitable asymptotic condition as 1x1 -+ 00 . 
The answer is easy in the case 2m < n. Then 
(1.12) 
A Uniqueness Condition for the Polyharmonic Equation 277 
for a~ NE with la1 < 2m - 1, and there exists only one solution of (1.7) with this 
property. The same statement holds in the case 2m = n iff satisfies (1.11) (compare 
(2.32) and (4.4) below). If 2m > n, then uo(x) may be unbounded as 1x1 -, 00, even if 
(1.11) is valid, as we shall see in an example at the beginning of Section 3. We shall 
show that uo is uniquely determined by (1.7) and the property 
uo(x)dS, = o(R"-') as R -+ co for every xo E R". (1.13) 
Note that this condition is weaker than (1.12), so that uo satisfies (1.13) also in the case 
2m < n. 
The verification of (1.13) yields as a further result the following statement on the 
solvability of (1.7) in the case 2m 2 n: (1.7) has a solution satisfying (1.13) if and only if 
(1.11) holds. Condition (1.1 1) says that fis orthogonal to certain polynomial solutions 
of ( - A)"k = 0 in R". 
The analysis in Section 2 is based on the spectral theory for unbounded self-adjoint 
operators. Most conclusions are analogous to some in [S] and [3]. Here we give only 
a short description of the main steps. In Section 3 we use Green's formula to derive an 
expansion of the form 
s /x - x,,( = R 
m -  1 
u(x)dS, = C c ~ A ~ v ( ~ ~ ) R ~ - ~ + Z ~  (1.14) s IX - x0( = R j = O  
for every solution V E  Czm(Rn) of the homogenous equation ( - A)% = 0, where cj  # 0 
are suitable real constants. This shows that (1.7) has at most one solution with the 
property (1.13). A Taylor expansion yields that uo satisfies (1.13) if and only if (1.11) 
holds. This, together with (1.14), implies the above statement on the solvability of (1.7) 
for 2m 2 n. 
2. The time-dependent problem 
We study the problem 
a:u + ( -  = f  in (w" x [O, a), 
u(x, 0) = atu(x, 0) = 0 in R", 
with givenfeC,"(LW"). We require U E  CZm([W" x [0, 00)) and 
u(., t ) ~  Hm(R") for every t 2 0, (2.2) 
where Hm(R") denotes the mth Sobolev space. Then u is uniquely determined (com- 
pare the discussion in [3] in a related situation). We extend the operator ( - A)m to 
a positive self-adjoint operator in L2(R") by setting 
D(A) : = { U E H,(R"): AmU E L ~ ( R " ) } ,  
AU :=(-A)mU for U E D ( A ) .  (2.3) 
Let { P A }  denote the (left continuous) spectral family of A.  The functional calculus for 
unbounded self-adjoint operators and the elliptic regularity theory yield 
U(X, t )  = - (1 - cos &t)d(P,f(x)). jom : 
278 P. Lesky Jr 
In order to obtain the asymptotic behaviour of u(x, t )  as t + 00 we proceed as in [3], 
to which we refer for a more detailed presentation of the argument. 
A modification of (3.1 1) in [3] yields the following representation of the resolvent 
R, = ( A  - z q - 1  of A: 
where c = (42) - 1 and 
H?'(i) = J,(O + iN,(O (5 E ~ \ ( O } )  (2.6) 
denotes Hankel's function. By means of Stone's formula it follows that P,f is 
continuous with respect to 1 E R and differentiable for 1 # 0. In particular, we have 
Note that P, f = 0 for 1 < 0, since A is positive. Using 
m 
J,(O = c Cs12s+a, 
s = o  
with 
( -  1)" 
20+zw-(0 + s + 1) c, = 
(compare [4]), we obtain for 110 
if 2m < n, 
Jb.f(d)[x - x'12sdx' + O(R1/2") if 2m 2 n, ' 
uniformly in every compact subset of R". Note that 
dP,f (x) d l  
d1  
and set 
(2.11) 
(2.12) 
(2.13) 
(2.14) 
(6 > 0). Let K be an arbitrary compact subset of R". At first we study the case 2m 3 n. 
A Uniqueness Condition for the Polyharmonic Equation 279 
We insert (2.10) into (2.12) and obtain 
(2.15) 
where 
n + 2s 
2m 
&:= 1 - ___ 
(2.16) 
(2.17) 
and wl(x, t; 6) - 0 as 6 10 uniformly with respect to (x, ~ ) E K  x [O, 00). In order to 
compute I;, we substitute p:= $t. This yields 
f i t  1 - cosp 
Ip*(t; 6) = 2t28 jo p1 + 20 dP 
If f l  > 0, it follows that 
(compare integral l lc ,  section 1.1.3.4 in [l]) with 
(2.18) 
(2.19) 
(2.20) 
as an integration by parts shows. If fl = 0, we obtain 
f i l  I 1  - cosp OD cosp 
I,*(t; 6 )  = 2 j1  idp + 210 
P dp + W,(t; 6; 0) 
= 21nt + ln6 + 2C, + Wl(t; 6; 0) (2.21) 
(C, denotes the Euler-Mascheroni constant; compare (3.67) in [7]). Setting 
p,(x) : = jR:(x’)lx - x’1 2s dx’, 
we obtain from (2.15), (2.19) and (2.21) for odd n 
m - ( n  + 1)/2 
I1(X, t; 6) = 1 Dst”~p,(x) - 
s = o  
+ WI@, t; 4 + WZ(X, t; 61, 
(2.22) 
(2.23) 
280 P. Lesky Jr 
and for even n 
where 
(2.24) 
and 
Now consider I, defined by (2.13). Note that 
In order to study Ri, as t 10 we use (2.5), (2.6), (2.8) and 
r m  
(2.26) 
(2.27) 
( -  l)s+l s + a  
cy = n2a+%!(O. + s)! ( il: + } 
J ct! = - (0 - S - I)! n S! 22s-a 
(compare [4]). This, together with (2.10) and (2.27), implies that for odd n 
1 m - b + l ) / Z  c 
2rn(27~)’”~ s = o  pS6 c + P S ( X )  + W A X ;  6) 1 2 h  6) = uo(x) + 
(2.29) 
(2.30) 
A Uniqueness Condition for the Polyharmonic Equation 281 
(2.3 1 )  
with w,(x; 6) + 0 as 6 10 uniformly in K ,  where 
if 2m 2 n and n odd, 
if 2m > n and n even 
- 
(2.32) 
f(x')lx - x'12m-"lnlx - x'ldx' Cm - n / Z  -~ 
uo(x): = 
f(x')lx - x'12m-ndx' 
 i 
and 
ln2 Cm-n,2 + Ci-n/2 . 
(2.33) 
13(x, t; 6 )  = o(1) as t -+ co (2.34) 
uniformly with respect to x E K ,  as a slight modification of the proof of Lemma 5.2 in 
[3] shows, we conclude from (2.23), (2.24), (2.30) and (2.31) that (1.8) and (1.9) hold 
uniformly in K.  By (2.22), we have ug = uo iff satisfies (1.1 1). In this case (1.8) and (1.9) 
reduce to (1.6). 
Now we study the case 2m < n. Let K be an arbitrary compact subset of R". By 
(2.10) and (2.12) we obtain 
ll(x; t; 6)+0 as 6 1 0  (2.35) 
uniformly with respect to (x, t)E K x [0, co). Taking into account that 
R,, f(x) --+ uo(x) as z 10 for 2m < n with 
{+e!s m ) I u t  (x) : = ~ O ( X )  - ~m - n / 2  (x) 4 ( 2 q / 2  - 1 
Note that u,, and uz are solutions of (1.7). Since 
[ T(n/2 - m) 7~"/~4'"(m - I ) !  iw" J X  - '(") dx' (2m < n), uo(x) : = (2.36) 
we conclude from (2.27) that 
I2(x;  6) = uo(x) + o(1) as 6 1 0  (2.37) 
uniformly in K. Thus it follows by (2.34) that (1.6) holds uniformly in K ,  where uo is 
given by (2.36). In particular, uo is a solution of (1.7). Thus we have verified the 
following Theorem: 
Theorem 2.1. Let UE C2"(R" x [0, a)) be the unique solution of (2.1), (2.2). Then the 
following statements hold: 
1. If 2m < n, then (1.6) holds uniformly in every compact subset of R", and uo is given 
by (2.36). 
2. If 2m 2 n, then the asymptotic behaviour of u as t -+ 03 is given by the estimates 
(1.8) and (1.9), which hold uniformly in every compact subset of R". If, in addition, 
f satisjes (1.1 l ) ,  then (1.6) holds uniformly in every compact subset of R"; in this case 
uo is given by (2.32). 
282 P. Lesky Jr 
3. The polyharmonic equation 
3.1. An example 
Assume that 2m 2 n and thatfeCo"(R") satisfies (1.11). Consider the solution uo of 
(1.7) given by (2.32). In order to find a condition that singles out uo among the 
solutions of (1.7), we study first the special case m = n = 3. Since 
we obtain by (1.1 1) and (2.32) 
This formula shows that uo is unbounded as 1x1 --i 00 if for example one of the 
integrals fRnf(x')xt2 dx'(i = 1,2,3) does not vanish. In particular, we have 
This asymptotic condition does not suffice for the unique characterization of uo, since 
also u(x) = uo(x) + c-x + d with C E  R3, d e  R is a further solution of (1.7) with the 
property (3.3). In order to characterize uo uniquely, note that 
uo(x) dS, = O ( R )  as R + a, 
(XI = R 
since 
by (1.11). Moreover, it can be shown in the same way that 
uo(x) dS, = O(R) as R -+ 00 for every xo E R3. (3.4) s IX - X,, = R 
Note that u(x) = uo(x) + c - x  satisfies the asymptotic estimate in (3.4) for xo = 0. On 
the other hand, uo is the only function of the form u(x) = uo(x) f c - x  + d, that 
satisfies (3.4) for every x0sR3. In the following we prove that uo is uniquely 
characterized by (1.7) and (3.4) in the general case. 
3.2. The uniqueness proof 
We prove: 
Lemma 3.1. I f  u E CZm(Rn) satisfies 
( - A)mu = 0 in R" (3.5) 
A Uniqueness Condition for the Polyharmonic Equation 283 
and 
v(x)dS, = o(R"-') as R -+ co for every xo E R", (3.6) 
l x  -xoI  = R 
then u = 0 in R". 
Remark. In the case n = 1 the integral in (3.6) has to be understood in the sense 
V(X) dS, : = V ( X ~  + R) + U(XO - R). s (X - x O I  = R (3.7) 
ProoJ: Let xo E R" be fixed and assume that R > 0. First we derive a representation of 
Jlx-xor = g(x)dS, for gECZ(Rn). We set B E : =  {XE R": E < Ix - xo( < R} for 
0 < E < R. In the case n 2 3 we use Ax(l/(x - X ~ I " - ~ )  = 0 for x # xo and conclude 
from Green's formula that 
where n denotes the normal unit vector on aB, pointing into the exterior of B,. Letting 
E 10, we obtain by the theorem of Gauss 
(rn:= surface measure of the unit sphere in R"), and hence 
Ag(x)dS, dr. (3.10) 
This formula holds also in the case n = 1 with rl := 2. In fact, integrating by parts 
twice, we obtain 
I 
jBe Ix - x0lg'Wdx 
= R(g'(xo + R) - g'(x0 - R)} - E(g'(Xo + E )  - g'(xo - E ) }  
and from this and (3.7), (3.10) follows. 
284 P. Lesky Jr 
In the case n = 2 we use Axln (x - xo( = 0 for x # xo. As above Green's formula 
and the theorem of Gauss yield 
n 
In (x - xo 1. Ag(x) dx 
IX - x O I  Q R 
= (In R ) s  Ag(x)dx - - s g(x)dSx + r,g(Xo), (3.12) J 
IX - x0I = R r = O  I x - x o I = r  1 s 
r = O  Ix - yo/ = r I 
IX - x0I < R Ix-x , l=R 
and therefore 
g(x) dSx = T,Rg(x,) + R In RJR { s Ag(x) dSx dr 
(3.13) 
u(x)dS,. Taking into account 
- R s R  (lnr) { [ Ag(x)dSx dr. 
A m - k  Now we set g := A m - k ~  and compute SIX - xol = 
that u satisfies (3.5), we have for k = 1 by (3.10) and (3.13), respectively, 
Am-l~(x)dSx = rnRn-lAm-lu(~o) .  s IX - x0I = R 
If n # 2, then we obtain by (3.10) and induction with respect to k 
k -  1 
Am-ku(x)dS, = rnR"-lAm-ku(xo) + C ~ ~ ~ ( l t ) A ~ - ~ + j ~ ( ~ ~ ) R " - l + ~ j  (3.14) 
with suitable constants ckj(n)~R\{O). If n = 2, then (3.13) and induction yield also 
(3.14), since 
{x -xoI = R j =  1 
Thus we have in R" (with arbitrary n E N) 
m -  1 
v(x)dS, = T,R"-'~(x,) + C , ~ ( I ~ ) A ~ U ( X ~ ) R " - ~ + ~ ~  (3.15) 
for every solution u E CZm(Rn) of (3.5). This and (3.6) imply u(xo) = 0, which proves 
Lemma 3.1. 
J1. - xol = R j =  1 
3.3. The existence of the solution 
Lemma 3.1 implies that the problem 
( - A ) m ~  =f in R", 
u(x)dSx = o(R"-') as R -+ co for every x ~ E R "  s Ix - xoI = R 
admits at most one solution UEC'~((W~). If 2m < n, then the function uo defined by 
(2.36) is the solution of (3.16). In fact, under the assumption fECA(R") we have 
uo E C y R " ) ,  
(3.17) 
A Uniqueness Condition for the Polyharmonic Equation 285 
and ( - A)'%, = f i n  R", which follows from 
T(n/2 - 1) 
( -  A ) m - ' ~ o ( ~ )  = 
In the following we suppose that 2m 2 n and that feC:(R") satisfies (1.11) and 
therefore (1.10). It is our aim to prove that the function uo given by (2.32) is the 
solution of (3.16). As above we have u o ~ C Z m ( R n )  and ( - A)"u0 =f in  R". Hence it 
suffices to verify the infinity condition in (3.16). For the sake of simplicity, we set 
if n is odd, 
if n is even. 
First we study the case n = 1. Let xo E R be fixed. We choose an a > 0 such that 
f(x) = 0 for 1x1 > a. For R > max{a - xo, a + xo} we obtain from (2.32) 
C m -  1 - 
4(2n)"" - 
(3.18) 
-- C m - n / Z  i (2n)"'Z D(n, m) : = 
and 
= 20(1, m ) m i l  ( 2m- 2j 1 ) R z m - 1 - 2 j  f(x')(x, - x')'jdx' 
.i = 0 - a  
(3.19) 
This implies by (1.10) 
uo(x)dS, = 0 for R > max {a - xo, a + xo>. (3.20) s IX - x0I = R 
Hence uo is the solution of (3.16). 
Now we study the case n 2 3, n odd. By (2.32) and (3.18) we have with z:= x - xo 
u,(x)dS, = D(n, m) fb') IZ - (x' - Xo)12m-ndS, s IX - x0I = R s R n  { 1 2 1  = R 
We use the expansion 
as 1zJ -+ co . Substituting 1:  = j + k in the inner sum, we obtain 
286 P. Lesky Jr 
as (z( + 03 with dj ,  = dj ,  (n, m) := (" 5 " I 2 )  ( ! j). Inserting (3.22) into (3.21), we have 
as R + m . Note that the inner integral does not depend on x': 
dS, = a ( j ,  I, n)Rn-' ,  
J z I = R (  I4 Ix' - XOI 
with 
( -  2~,)~j- 'dS, .  s 121 = 1 a( j ,  I ,  n) : = 
Thus we obtain 
(3,23) 
(3.24) 
(3.25) 
uo(x) dS, 
/X - x0J = R 
2 m - n  2 j  
j = o  , = j  
J 
= D(n, m) c 1 dj,a( j, 1, n)R2"-l- '  
+ O(R"-') as R +  m.  (3.26) 
Note that it suffices to restrict the inner summation in (3.26) to even indices 1 with 
1 < 2m - n, since a( j, 1, n) = 0 for odd 1 by (3.25) and R2"'-'-' = O(R"-*)  as R -+ m if 
I >, 2m - n + 1. We substitute k:= tj2 in (3.26) and change the order of the 
summations. Taking into account that n is odd, we conclude that 
m - ( n  + 1)/2 
uo(x)dS, = D(n, rn) pk(n, m)R2m-' -2k  
i x  - x0I = R k = O  
P 
f(x')lx' - x0lZkdx' + O ( R n - 2 )  as R -+ co, (3.27) 
x JRn 
with 
Since we have assumed that fsatisfies (1.10) it follows from (3.27) that 
uo(x)dSx = O ( R n - 2 )  as R 3 00. .r IX - x0I = R 
This shows that uo is the solution of (3.16) if n >, 3, n odd. 
(3.29) 
A Uniqueness Condition for the Polyharmonic Equation 287 
Finally we assume that n is even. By (2.32) and (3.18) we have with z:= x - xo 
c 
uo ( 4  dSx J jx - xoI = R 
(z - (x‘ - xo)(2m-nln (z - (x‘ - xo)l dS, 
Z/ = R 
It holds that 
1 (z - z‘l2 2m-n 
InT 2 IZI 
Iz - z’)2m-nlnlz - z’( = (1nlzl)lz - z ’ ( ~ ~ - ~  + -1z - z I 
(“ ,”’):= 0 for j >/ m - n/2 + 1. 
A Taylor expansion yields 
)z - 2‘12 
)z - z’J2m-nln--- 
1Zl2 
with suitable real constants cj. Thus we have 
Iz - z’(2m-nln)z - z’l 
as JzI + cc . By the argument leading to (3.22) it follows that 
/z - z’(2m-nlnlz - z’l 
as 121 + cc , where 
Inserting (3.32) into (3.30), we obtain by (3.24) 
-+ co. 
I 2 m - n  2 j  uo(x)dS, = D(n, m) c 2 (dj,ln R + d;.I a( j ,  I, n)R2”-’-’ /x - xo/  = R j - 0  [ = j  
x sRnf(xr)/x’ - x,(’dx’ + O(R”-’) as R 
s 
(3.31) 
(3.32) 
(3.33) 
288 P. Lesky Jr 
As in (3.26) it suffices to restrict the inner summation in (3.33) to even indices 1 with 
1 < 2m - n. Setting k : = 112 and changing the order of the summations we conclude 
that 
m - "12 
uo(x)dSx = D(n, m) c {Bk(n, m) 1nR + P;(n, m))R2m- ' -2k  
Jx - Yo/ = R k = O  
x f(x')/x' - xOIZkdx' + 0(Rn-') as R + co. (3.34) 
s 
(3.35) 
From (3.34) and (1.10) it follows (3.29). Therefore the function uo is the solution of 
(3.16) in the case of even n. Hence we have proved: 
Theorem 3.1 Let fECh( Iw" ) .  Furthermore assume that 2m < n or that 2m > n and 
f satisjies (1.11). Then problem (3.16) has a unique solution U E  CZm(Rn), which is given 
by (2.32). 
3.4. An alternative theorem 
In the case 2m < n, problem (3.16) has a solution for every fECh(IW") by 
Theorem 3.1. In the case 2m 2 n we prove the following alternative: 
Theorem 3.2. Assume that 2m 2 n and that f E  CA(R"). Then: 
1. Iff satisjies ( l . l l ) ,  then problem (3.16) has a uniquely determined solution 
2. If  (1.11) is not valid, then (3.16) has no solution UEC~'" (R") .  
u E c Z m ( R " ) .  
Proof. It suffices to prove part 2 of the theorem, since part 1 is contained in Theorem 
3.1. We suppose that 2m 2 n and that U E  CZm(Rn) is a solution of (3.16). We show that 
f satisfies (1.1 1). 
We set v : = u - uo, where uo is given by (2.32). Let xo E R" be fixed. Note that (3.15) 
holds, since u is a solution of the homogenous equation (3.5). We combine (3.15) with 
(3.19) in the case n = 1, with (3.27) in the case n 2 3, n odd and 
with (3.34) in the case of even n. Then we obtain for odd n 
and for even n 
r m -  1 m - n l 2  
u(x)dS,  = c y:Rn-1+2k + 2 y;(lnR)R2m-'-2k + O(R"-2) 
Jx - x0/ = R k = O  k = O  
(3.37) 
as R + co ; here the constants yk  E R depend on v and &' E Iw depend on v and f ,  since 
A Uniqueness Condition for the Polyharmonic Equation 289 
the first sum in (3.37) contains a part of the sum in (3.34). Furthermore, 
Y; = D(n, m)Pk(n, m) f(x')lx - x'J2kdx' for 0 d k d 
with Pk(n, m) defined by (3.28) if n 2 2 and by 
2m - 1 
Pk(Lm):=  2( 2k ) (0 < k < m - 1) (3.39) 
(compare (3.19)). 
Note that in (3.36) the exponents n - 1 + 2k in the first sum are even and that the 
exponents 2m - 1 - 2k in the second sum are odd. Since u is supposed to satisfy the 
asymptotic condition in (3.16) and since 2m 2 n, it follows from (3.36) and (3.37), 
respectively, that y k  = y ;  = 0 if n is odd and y i  =: y ;  = 0 if n is even. Since D(n, m) # 0 
for every n, mE N by (3.18), (2.9) and (2.29), we have to show that Pk(n, m) # 0 for 
0 d k d [m - (n/2)]. Then (3.38) and y ;  = 0 for 0 < k < [m - (n/2)] imply that 
fsatisfies (1.10) and therefore (1.1 1). 
If n = 1, we have Pk(n, m) # 0 by (3.39). In the case n 2 2 we consider (3.28). Note 
that by (3.25) 
( - 2z,)2j-2kdSZ = 2 2 j - 2 k + l  71 (n-1)12r(j - k + +) 
I-( j - k + 4)' (3.40) a( j ,  2k, n) = 
We set 
(with (" 3 ' I2) := 0 if n is even and j 2 m + 1 - n/2 . 
Then we have by (3.28) and (3.40) 
2k 
P k ( %  m, = 6j(k, n, m). 
j =  k 
It holds that 
(3.42) 
m S . l - 3  m - 5  
k + l  71 
aj+,(k + 1, n,m + 1) = { ~ ~ ( k ,  n, m) + __ 6,- l(k, n + 2, m) 
f o r k + l  d j d 2 k a n d  
Hence it follows that 
(3.43) 
290 P. Lesky Jr 
Taking into account that 
for n, m e  N, we obtain by induction 
(3.44) 
(3.45) 
for 0 < k < [m - (n/2)] (n 2 2). This concludes the proof of Theorem 3.2. 
4. Remarks 
1. Assume thatfc Cg(R”). Then the problems (2.1), (2.2) and (3.16) are related: the 
solution u(x, t )  of (2.1), (2.2) converges to a limit uo(x)  as t -+ 00 if and only if 
(3.16) has a solution. In this case the limit uo is the unique solution of (3.16). 
2. The alternative Theorem 3.2 says that (3.16) has a solution if and only if 
f~ Cb(W”) is orthogonal to the polynomial solutions of ( - A)% = f i n  R” given by 
(4.1) 
If 2m < n, then the set of the polynomials (4.1) is empty in agreement with the 
fact that problem (3.16) has a unique solution for everyfE CA(R”) in this case. If 
rn = 1 and n < 2, then p(x) = 1 is the only polynomial of the form (4.1). Thus the 
polynomials (4.1) can be considered as a generalization of the standing wave 1, 
introduced by Morgenrother and Werner [ 5 ]  in the special case rn = 1, to 
equations of arbitrary order 2rn. The polynomials (4.1) occur in the resonance 
terms in (1.8) and (1.9), since 
S! 
JRnf(x’)Ix - x’JZsdx’ = __ f(x’)lx12j( - ~ X . X ’ ) ~ ~ X ’ / ~ ~ ~ X ’  
j + k + l = s  J ’! k! I !  
( - A)% =f in R“, 
u(x) = o(1) as 1x1 -+ m 
(4.3) 
has at most one solution. In the case rn = 1 this result is a well known 
consequence of the maximum principle. Note that the maximum principle does 
not hold in the case m > 1, as the solution p ( x )  = - 1xI2 of ( - A)”p = 0 shows. 
If 2m < n, then problem (4.3) has a solution, which is given by (2.36). In the 
case 2rn = n, (4.3) has a solution if and only iff€ CA(R”) satisfies fRn f(x’)dx’ = 0. 
A Uniqueness Condition for the Polyharmonic Equation 291 
This follows from (2.32), the asymptotic estimate 
and the second part of Theorem 3.2. If 2m > n, then (4.3) may have no solution, 
even iff satisfies (Lll), as the example at the beginning of Section 3 shows. 
Acknowledgement 
This work has been supported by the Deutsche Forschungsgemeinschaft (SFB 256) 
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