1. PRELIMINARIES 9
Then the estimate (1.2.3) with C a and the estimate (1.2.4) with m =
(2, ... , 2) for solutions of the Cauchy problem (1.2.1) are valid.
PROOF.
By simple calculations we obtain that the left-hand side in (1.2.2)
equals Hx + H2 where
HX =
4^ac,|2
+ + i g
2
-
dc%+l\2
-
(i/2)ecxn+lDn+lc\zn+l\2
+ {\/2){{x , Vc) - bDnc)
ReC„2+1
+ exn+£H+y(Vc, £»,
H2 =
4a2p\xlCl
+ ••• + xnCn - bCn +
6cxn+lCn+l\2.
In view of homogeneity reasons we may assume |C| = 1. In the case under
consideration £ = * + ixo(p{x", xn -b, -6xn+x). If T = 0 and A(x; Q = 0
/ 2 2
we have | C | = c£n+\ = c/(l + c). So using the Cauchy inequality we get
Hx op{\-9c- (\/2)dxn+lDn+lc + (l/(2c))((x , Vc) - bDnc)
-d\xn+yc\cl/2)c/(i+c)
eo(p(x)
for certain e , by condition (1.2.11). Considering the imaginary part of H2
we get
H2 =
4o2(po2(p2T2{x]
+ + x\_x + (xn - bf -
02cx2n+l)2
^
A
2 2 2 2
4a (pa cp x e.
Fix x e CI. Via continuity, eacp H on the compactum
{C:|C|=l,^
m
(-; O = 0, xope}
for some e . From the estimate above we have H2 a pe if xacp e . On
the other hand, it is clear that H{ -Cacp. So choosing a large we get
strict positivity of H{ + H2. Using the choice of parameters b , s, ... we
obtain cp 0 on 9(? \ T, so we deduce the conclusion of Corollary 1.2.5
from Theorem 1.2.1.
We remark that in case 1) one can obtain the exact description of the
dependence domain. In case 2) the set Q n {xn+l = 0} contains the set
Qn{xn -hx, xn+x =0} where
hx = b((l +
(d2
+
e)/b2)l/2
- 1)
(d2
+ e)/b,
Finally, a large observation time is required in order to satisfy the condition
h(h +
2b)6T2.
Now we give some interesting and important examples of nonuniqueness
in the Cauchy problem.
EXAMPLE
1.2.6
(PLISH
[116]). There is an elliptic operator
1 1 2 2 2 1 2
a (x3)Dxu +D2u +D3u +a (x)Dxu + a (x)D2u + a(x)u = 0
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