Journal of Mathematical Sciences, Vol. 176, No. 6, August, 2011

CONDITIONS FOR STABILIZATION OF SOLUTIONS TO THE FIRST BOUNDARY VALUE PROBLEM FOR PARABOLIC EQUATIONS V. N. Denisov Moscow State University, 119991 Moscow, Russia [email protected]

UDC 517.9

Dedicated to V. V. Zhikov with admiration We study the stabilization to zero of a solution to the first boundary value problem for a parabolic equation. We obtain necessary and sufficient stabilization conditions in the case of the heat equation and sufficient conditions in the case of a parabolic equation with variable coefficients. Bibliography: 14 titles.

1

Statement of the Results

Let Q be an arbitrary (possibly, unbounded) domain in the N -dimensional Euclidean space RN , N  3. Introduce the right cylinder Z = Q × (0, ∞) with base Q and consider the divergence second order operator   N  ∂ ∂ aik (x, t) , (1.1) L(x, t) = ∂xi ∂xk i,k=1

where x = (x1 , . . . , xN ) ∈ Q, t > 0, and aik (x, t) are bounded measurable functions in Z. Throughout the paper, we assume that the coefficients of the operator (1.1) satisfy the uniform parabolicity condition (1.2) λ20 |ξ|2  (a(x, t)ξ, ξ)  λ21 |ξ|2 , where λ0 , λ1 > 0 for all x ∈ Q, t > 0, and ξ ∈ Q. We study the first boundary value problem L(x, t)u − ut = 0 in Z,  uS = 0, S = ∂Q × [0, ∞),  ut=0 = u0 (x), x ∈ Q,

(1.3) (1.4) (1.5)

in the cylinder Z, where L(x, t) is the operator (1.1) with bounded and measurable coefficients aik (x, t) in Q satisfying the symmetry condition aik = aki and the uniform parabolicity condition

Translated from Problems in Mathematical Analysis 58, June 2011, pp. 143–159. c 2011 Springer Science+Business Media, Inc. 1072-3374/11/1766-0870 

870

(1.2), whereas u0 (x) is a continuous bounded function in Q such that |u0 (x)|  M,

x ∈ Q.

(1.6)

Our goal is to find characteristics of the domain Q ⊂ RN that are sufficient for the stabilization of the solution u to the problem (1.3)–(1.5), i.e., lim u(x, t) = 0

(1.7)

t→∞

at each point x ∈ Q (x-uniformly in every compact set K in Q). We briefly discuss the existence of a weak solution to the problem (1.3)–(1.5). For this purpose, we recall definitions of some spaces of differentiable functions (cf., for example, [1]–[3]). Denote by Qba the direct product Q × (a, b), where Q is a domain in RN and a < t < b. Let Qba (R) = {Qba ∩ |x| < R}. o 1,0 b 2 (Qa (R))

The space W

is the completion of C0∞ (Qb+1 a−1 ) in the norm 

u

o

1,0 W 2 (Qba (R))

o 1,0 b 2 (Qa (R))

The space V

V 2 (Qba (R))



1/2 2

2

[u (x, t) + |∇u(x, t)| ]dxdt

=

.

(1.9)

Qba (R)

is the completion of C0∞ (Qb+1 a−1 ) in the norm  

u o 1,0

(1.8)

= vrai max atb

1/2 u2 (x, t)dx

BR



1/2



|∇u(x, t)|2 dxdt

+

.

(1.10)

Qba (R)

Definition 1.1. By a weak solution to the problem (1.3)–(1.5) in QT0 = Q × (0, T ) we mean o 1,0 T 2 (Q0 (R))

a function u(x, t) of class V  

− uηt + QT 0

N  i,k=1

satisfying the integral identity

  ∂u ∂η aik (x, t) dxdt = η(x, 0)ϕ(x)dx ∂xk ∂xi

(1.11)

Q

for all test functions η(x, t) ∈ C0∞ (QT−1 ) such that η(x, T ) ≡ 0. A function u(x, t) is a solution to the problem (1.3)–(1.5) in Q × (0, ∞) if it is a solution to the problem (1.3)–(1.5) in QT0 for any T > 0. The following assertions hold for solutions to the problem (1.3)–(1.5) (cf. [3, 4]). 1. Maximum principle. If u0 (x)  M for almost all x ∈ D, then u(x, t)  M for almost all x ∈ Q and all t > 0. 2. If u0 (x) is continuous and bounded in Q, then a bounded solution to the problem (1.3)– (1.5) exists and is unique. For a given u0 (x), together with the problem (1.3)–(1.5) in the cylinder Z = Q × (0, ∞), we consider the first boundary value problem for the heat equation 871

Δu − ut = 0 in Z,  uS = 0, S = ∂Q × [0, ∞),  ut=0 = u0 (x), x ∈ Q.

(1.12) (1.13) (1.14)

Without loss of generality we can assume that the origin O lies on the boundary ∂Q of the domain Q and cap (RN \Q) > 0, where cap (E) is the Wiener capacity [14] of the set E = RN \Q. Theorem 1.1. The solution to the problem (1.12)–(1.14) stabilizes, i.e., the limit (1.7) exists at a point x ∈ D (x-uniformly in every compact subset K of the domain Q) if and only if for some q > 1 the following series diverges: lim

n→∞

n 

cap(Bqm \ Q)q m(2−N ) = +∞,

(1.15)

m=1

where BR is the closed ball {|x|  R} in RN and cap (E) is the Wiener capacity of a compact set E in RN . We formulate the stabilization criterion for the problem (1.12)–(1.14) in the integral form. Theorem 1.2. The solution to the problem (1.12)–(1.14) stabilizes if and only if the following integral diverges: R

dτ cap Bτ \ Q N −1 = +∞, r0 > 0, (1.16) lim R→∞ τ r0

where BR is the closed ball {|x|  R} in RN and cap (E) is the Wiener capacity of a compact set E in RN . We formulate the main result of the paper. Theorem 1.3. If a domain Q ⊂ RN is such that the series (1.15) (the integral (1.16)) diverges, then the solution to the boundary value problem (1.3)–(1.5) stabilizes, i.e., the limit (1.7) exists. Theorem 1.3 can be specified as follows. Theorem 1.4. If a domain Q ⊂ RN is such that the series (1.15) (the integral (1.16)) diverges, then the solution to the boundary value problem (1.3)–(1.5) satisfies the inequality ⎧ ⎫ √ t

⎪ ⎪  ⎨ ⎬ cap Bτ \ Q dτ , (1.17) |u(x, t)|  C1 exp −C2 ⎪ ⎪ τ N −1 ⎩ ⎭ a0

where x0 is an arbitrary point in Q, C1 > 0 is a constant depending on λ1 , λ0 , and N , whereas a0 > 0 and C2 > 0 are constants depending on λ1 , λ0 , N , and |x0 |. In Section 9, we show that the compact set K in the above theorems cannot be replaced with the entire domain Q, i.e., the uniform stabilization of the solution on any compact set K in Q cannot be replaced with the uniform stabilization in x ∈ Q. 872

2

Growth Lemma

We recall the definition of the parabolic capacity (cf. [6]). On a B-set E in RN consider all possible measures μ such that  F (x − y, t − τ )dμ(y, τ )  1, (x, y) ∈ E,

+1

, we

(2.1)

E

where

⎧ 2 ⎨ 1 e− |x| 4t , t > 0, F (x, t) = tN/2 ⎩0, t  0 (except for the point x = 0, t = 0).

(2.2)

We introduce the heat (parabolic) capacity γ(E) of E by the formula γ(E) = sup μ(E),

(2.3)

where the supremum is taken over all possible measures μ satisfying (2.1). For the basic properties of heat capacity we refer the reader to [6, 7]. We consider the cylinder t1 ,t2 ≡ {x, t ∈ RN Z0,R

+1

: |x| < R, 0 < t1 < t < t2 }

with height t2 − t1 > 0 and base |x| < R lying in the hyperplane t = t1 . We consider three coaxial cylinders 0,2 Z1 = Z0,R

2 R2

,

5/42 R2 ,22 R2

Z2 = Z0,R

,

2

2

0, R Z3 = Z0,R ,

0 < < 1.

(2.4)

1,0 (Z) to the problem In Z = Q×[0, ∞), we consider a nonnegative solution u(x, t) of class V2,loc

Δu − ut = 0 in Z,  uS = 0, S = ∂Q × (0, ∞),  ut=0 = u1 (x), x ∈ Q, where u1 (x) is continuous in Q and

(2.5) (2.6) (2.7)

0  u1  M.

(2.8)

The problem (1.12)–(1.14) is reduced to the problem (2.5)–(2.7). Indeed, in the problem (1.12)– − + − (1.14), we can write u0 (x) = u+ 0 − u0 , where u0 = max(u0 , 0) and u = min(u0 , 0), and, using the linearity of the problem (1.12)–(1.14), obtain two problems of type (2.5)–(2.7). +1

We assume that the origin O lies on the boundary ∂Q of Q; moreover, CZ = RN + \Z is the +1 complement of Z in RN + and E = CZ ∩ Z3 , where Z3 is the cylinder defined in (2.4). The following assertion is similar to the growth lemma due to Landis [6]. Lemma 2.1. If u(x, t) is a nonnegative solution to the problem (2.5)–(2.7), then there exists 0 < 1 < 1 such that for 0 < < 1   η sup u  1 + N N γ(E) sup u, (2.9) R Z∩Z1 Z∩Z2 where the constant η > 0 depends only on N and γ(E) is the heat capacity of the cylinder E. 873

1,0 Remark 2.1. Since we deal with a weak solution u(x, t)  0 of class V2,loc (Z) to the problem (2.5)–(2.7), the classical proof due to Landis [6, 8] should be modified.

Proof of Lemma 2.1. Denote by S1 the lateral surface of the cylinder Z 1 . We fix an arbitrary point (y, τ ) ∈ Z3 and consider the function v(x, t) = F (x − y, t − τ ),

(2.10)

where F (x, t) is defined in (2.2). Let us estimate sup v(x, t) over (x, t) ∈ S1 . For this purpose we fix a point x such that |x| = R and find t > τ at which the function (2.10) attains the maximum:       2 2 2 |x − y| r r N ∂ (t − τ )−N/2 exp − = exp − − = 0, N N ∂t 4(t − τ ) 4(t − τ ) 4(t − τ ) 2 +2 2(t − τ ) 2 +1 where t − τ = r 2 /(2N ). For |x| = R and |y|  R, 0 < < 1, we have |x − y|  R(1 − ). Consequently, R2 (1 − )2 . t−τ  2N Since v(x, t) is monotone with respect to t up to the first maximum point, we have  sup v(x, t) 

(x,t)∈S1

 =

2N 2 R (1 − )2 2N R2 (1 − )2

N/2 N/2



R2 (1 − )2 2N exp − 4R2 (1 − )2



  N . exp − 2

(2.11)

Let us estimate inf v(x, t) over (x, t) ∈ Z2 . Since (y, τ ) ∈ Z3 and (x, t) ∈ Z2 , we have |x − y|2  (|x| + |y|)2  4 2 R2 ,

R 2 2  t − τ  2 2 R2 . 4

(2.12)

Therefore,   1 4 4 2 R2 1 = exp(−4) exp − . inf v(x, t)  2 2 R 4 (x,t)∈Z2 (2 2 R2 )N/2 2N/2 N RN

(2.13)

Since (y, τ ) is an arbitrary point in Z 3 , from (2.11) and (2.13) we find  sup (y,τ )∈Z3 , (x,t)∈S1

inf

(y,τ )∈Z3 , (x,t)∈Z2

F (x − y, t − τ ) 

2N 2 R (1 − )2

F (x − y, t − τ )  exp(−4)

N/2



N exp − 2

1 2N/2 N RN

.

 ,

(2.14) (2.15)

Let E1 ⊂ Z3 , where E1 = K × [0, 2 R2 ] and K is a compact set in the ball BR . We consider the heat potential  F (x − y, t − τ )dμ(y, τ ),

w(x, t) =

(2.16)

E1

where μ is the measure realizing the heat capacity of the cylinder E1 . Without loss of generality we assume that γ(E1 ) > 0. The existence of a measure realizing the heat capacity of the 874

compact set E1 is proved in [9]–[12], where various properties of the heat potential (2.16) are also established. In particular, it is proved in [11, 12] that w(x, t) is a solution of class V21,0 (ΠT \E1 ) to +1 +1 the heat equation, where ΠT = {x, t ∈ RN + , x ∈ RN , 0 < t  T } is a strip in RN + , T > 2 2 R2 , w(x, t) = 1 in the pointwise sense on the upper base t = 2 R2 of the cylinder E1 , whereas w = 1 o 1,0 2 (ΠT \E1 )).

on the lateral surface of E1 is understood in the sense of V21,0 (i.e., w − 1 ∈V Furthermore, w(x, t)  1 in ΠT \E1 pointwise. From (2.14) and (2.15) we find   2N N/2 1 γ(E1 ), sup w(x, t)  N e R (1 − )N S1 inf w(x, t)  Z2

1 e4 2N/2 RN N

γ(E1 ),

(2.17) (2.18)

where γ(E1 ) is the heat capacity of the cylinder E1 . In Z ∩ Z1 , we consider the function U (x, t) = M [1 − w(x, t) + λ],

(2.19)

where M = sup u(x, t), Z∩Z1

λ=

sup (x,t)∈Z1 , (y,τ )∈Z3

F (x − y, t − τ )γ(Z 3 \Z).

(2.20)

We compare u(x, t) and U (x, t) on the parabolic boundary of the cylinder Z1 ∩Z. By [11, 12], the function (2.19) is a solution of class V21,0 to the heat equation ΔU − Ut = 0

in ΠT \E1 .

 On the lateral surface S1 in Z 1 ∩ Z, we have U |S > M since w(x, t)S1  λ and u(x, t)  M , which implies   U S1  us1 . Furthermore, w(x, t)  1 in the sense of V21,0 on the lateral surface of the cylinder Z, contained in Z1 , and, consequently, U (x, t)  0 in the sense of V21,0 on the lateral surface of Z, contained in Z1 . Furthermore, u(x, t) = 0 on the same part of the lateral surface of Z ∩ Z1 . Finally, for t = 0 we have 0  u(x, 0)  M pointwise (by that, in the sense of V21,0 ) on the lower base of the cylinder Z 1 ∩ Z, whereas U (x, t) > M on the lower base of Z1 ∩ Z since w(x, 0) = 0, x ∈ K. By the maximum principle [2], we see that u(x, t) < U (x, t) everywhere in Z1 ∩ Z. Therefore, sup u < sup U (x, t)  M (1 − inf w(x, t) + λ).

Z2 ∩Z

Z2 ∩Z

Z2 ∩Z

Taking into account (2.18) in (2.21), we find      γ(Z 3 \Z) 2N N/2 1 1 − . sup u  M 1 − RN e (1 − )N e4 2N/2 N Z∩Z2

(2.21)

(2.22)

Now, we consider the function f ( ) =

A2 A1 − , N (1 − )N

where A1 =

1 , e4 2N/2

 A2 =

2N e

N/2 , 875

and note that lim f ( ) = −∞,

lim f ( ) = +∞,

→+0

→1−0

and f  ( ) < 0 for 0 < < 1. Therefore, there is 0 < 1 < 1 such that for 0 <  1 f ( ) >

A1 . 2 N

(2.23)

We choose 1 such that the inequality (2.23) holds for 0 <  1 . Then (2.22) implies  sup u  M

Z∩Z2

1−η

γ(Z 3 \Z) R N N

 ,

(2.24)

where η = 1/(e4 2N/2+1 ) and 0 <  1 < 1. From (2.24) we obtain the required inequality (2.9). Indeed, since Z2 ⊂ Z1 , (2.24) implies sup u  sup u + sup u · η

Z1 ∩Z

3

Z∩Z2

Z∩Z1

  γ(Z 3 ∩ Z) η  sup γ(Z \Z . 1 + 3 R N N R N N Z∩Z2

Iteration Inequality and Its Consequences

We set q = 1/ > 1, where 0 < < 1 is taken from Lemma 2.1. Suppose that t0 > 0, m ∈ N, and t0 − 2q 2m  0. Consider the system of coaxial cylinders 2m ,t 0

t0 −2q Z1m = Z0,q m+1

,

t −3/4q 2m ,t0

0 Z2m = Z0,q m

,

t0 −2q Z3m = Z0,q m

2m , t −q 2m 0

(3.1)

(i.e., in the notation of Lemma 2.1, we set q m = R and = 1/q < 1). We assume that q 2 > 8/3, which guarantees the inclusion

m−1

Z2m ⊃ Z 1

(3.2)

.

(3.3)

Since q = 1/ , > 0, is bounded from below by zero, the inequality (3.2) is valid for

2 < min 3/8, 21 .

(3.4)

Mm = sup u,

(3.5)

Consider the sequence Z1m ∩Z

where u(x, t) is a solution to the problem (2.5)–(2.7). Lemma 3.1. The sequence (3.5) satisfies the inequality   η m Mm  Mm−1 1 + mN γ(CZ ∩ Z 3 ) , q m

m

(3.6) m

where γ(CZ ∩ Z 3 ) ≡ γ(Z 3 \Z) is the heat capacity of the compact set Z 3 \Z and η > 0. 876

Proof. Applying Lemma 2.1 to the solution u(x, t)  0 to the problem (2.5)–(2.7) in the cylinders (3.1), we obtain the inequality   η m

. sup u  sup u 1 + mN γ CZ ∩ Z 3 q Z∩Z1m Z∩Z2m Since (3.4) implies the inclusion (3.3), from (2.9) and the inequality sup u  sup u we find Z2m

m−1

Z1

Mm = sup u  sup u m

Z 1 ∩Z



Z2m ∩Z

 sup u 1 +

m−1

Z1

∩Z

m

γ CZ ∩ Z 3 mN η

q



 = Mm−1 1 +

 m

γ CZ ∩ Z . 3 mN η

q

Lemma 3.2. The sequence (3.5) satisfies the inequality m   b i γ(CZ ∩ Z 3 ) , Mm  M1 exp qN i

(3.7)

i=2

where b > 0 depends on η > 0. Proof. Iterating the inequality (3.6) and using the logarithm property, we obtain      η η m−1 m

1 + mN γ CZ ∩ Z 3 Mm  Mm−2 1 + (m−1)N γ CZ ∩ Z 3 q q m        η η η 1 + iN γ  M1 1 + 2N γ . . . 1 + mN γ = M1 exp ln q q q = M1 exp

m  i=2

 ln 1 +

η q iN

i=2

 γ .

Since ln(1 + ηx) > bx, x > 0, with a constant b depending on η, we conclude that    m i  bγ CZ ∩ Z  3 . Mm  M1 exp iN q

(3.8)

i=2

The lemma is proved.

4

Lower Estimate for the Heat Capacity of a Cylinder via the Wiener Capacity of Its Base

Lemma 4.1. Let γ(K) be the heat capacity of the cylinder K = F × [0, a] of height a, where F is a compact set in RN . Then γ(K)  a cap (F ), (4.1) where cap (E) is the Wiener capacity of a compact set E. 877

Proof. Denote ΠT ≡ {x, t : x ∈ RN , 0 < t  T, T > 0} and K = F × [0, a] ⊂ ΠT . Let γ(K) be the heat capacity of the cylinder K. We use the following property of heat capacity [7]. If ∞  Ki , then Ki is a sequence of compact sets such that Ki ⊃ Ki+1 , i = 1, . . ., and K = i=1

γ(K) = lim γ(Ki ).

(4.2)

i→∞

By (4.2), for estimating the heat capacity γ(K) we can assume that the base F of the cylinder K is a domain with smooth boundary ∂F . The property (4.2) of heat capacity is also valid for the Wiener capacity cap (F ) = lim cap (Fi ),

(4.3)

i→∞

where Fi ⊃ Fi+1 and F =

∞  i=1

Fi .

As is known [9]–[11], there exists a measure μ realizing the heat capacity of the compact set K = F × [0, a] and the heat potential  μ (4.4) w (x, t) = F (x − y, t − τ )dμ(y, τ ) K

satisfying the problem Δwμ − wtμ = 0 in ΠT \K,   wμ Γ(Π ) = 1, wμ t=0 = 0.

(4.5) (4.6)

T

Since the boundary ∂F of the compact set F is smooth, we have wμ (x, t) ∈ C ∞ (RN to the lateral surface and upper base of the cylinder K. The function wμ (x, t) has finite V21,0 (ΠT \K)-norm, i.e., ⎡ ⎢ wμ V 1,0 = vrai sup ⎣ 2

τ

⎤1/2



⎥ (wμ (x, τ ))2 dx⎦

⎡ ⎢ +⎣

ΠT \K∩{t=T }



+1

\K) up

⎤1/2 ⎥ (∇wμ (x, t))2 dxdt⎦

,

(4.7)

ΠT \K

where 1 γ(K)  2



μ



2

|w (x, τ )| dx + RN

|∇wμ (x, t)|2 dxdt.

(4.8)

ΠT \K

The inequality (4.8) is proved in [11, 12]. We need some properties (the proof can be found in [13]) of the Wiener capacity of a compact set F ⊂ RN , N  3.  1. cap (F ) =

inf

ϕ∈C ∞ (RN )

|∇ϕ|2 dx, ϕ = 1 in a neighborhood of F .

2. There exists a unique function U (x) such that 1 (RN ), (a) U (x) ∈ W2,loc

878

(b) ΔU (x) = 0 in RN \F ,  (c) U ∂F = 1,  |∇U (x)|2 dx. (d) cap (F ) = RN

Since ∂F is smooth, we can assume that U (x) ∈ C ∞ (RN \F ) up to the boundary ∂F . We fix t ∈ (0, a). Taking into account that ΔU (x) = 0 in the sense of W21 (RN \F ), we find  ΔU (x)[U (x) − wμ (x, t)]dx = 0. (4.9) RN \F

 Integrating by parts and noting that (U (x) − wμ (x, t))∂F = 0, we have  ∇U (x)∇[U (x) − wμ (x, t)]dx = 0, RN \F

i.e.,





2

∇U (x)∇wμ (x, t)dx.

|∇U (x)| dx = RN \F

RN \F

Applying the inequality ab  (a2 + b2 )/2 to the right-hand side of the last equality, we get   2 |∇U (x)| dx  |∇wμ (x, t)|2 x, RN \F

RN \F

which implies



|∇wμ (x, t)|2 dx.

cap (F ) 

(4.10)

RN \F

Integrating (4.10) with respect to t from 0 to a and using (4.8), we obtain a 

μ

2



|∇w (x, t)| dxdt 

a cap (F )  0 RN \F

ΠT \K

1 |∇w (x, t)| dxdt + 2 μ

2



(wμ (x, T ))2 dx  γ(K),

RN

which completes the proof. Lemma 4.2. The sequence (3.5) satisfies the inequalities m   b cap (B qi \Q) , Mm  M1 exp q i(N −2) i=1

(4.11)

where cap (E) is the Wiener capacity of a compact set E. Proof. Applying (3.8) to the right-hand side, using the inequality (4.1), and taking into m account that the height of the cylinder Z 3 is equal to q 2m , we obtain (4.11). 879

5

Proof of Theorems 1.1 and 1.2

Proof of Theorem 1.1. Sufficiency. We fix a compact set K and choose q > 1 such that K ⊂ B q = {|x|  q}. Let the series (1.15) diverge. By (4.11), for any natural m > 2     m m   cap (B qi \Q) cap (B qi \Q) M1  Mm exp −b  M exp −b . (5.1) q i(N −2) q i(N −2) i=2 i=2 We fix ε > 0 and choose m(ε) ∈ N such that the right-hand side of (5.1) is less than ε/2. Then M1 =

sup

u(x, τ ) < ε/2.

m

x, τ ∈Z∩Z 1

(5.2)

Such a choice of m(ε) is possible since the solution u(x, t) to the problem (2.5)–(2.7) is bounded by the number M for all x ∈ Q, t > 0 and the series (1.15) converges. For m(ε) we find t(ε) from the conditions (5.3) t(ε) − 2q 2m(ε)  0. Let t  t(ε). Then the requirement t − 2q 2m  0 leads to the inequality sup u(x, τ ) < ε/2 m

Z∩Z 1

since for m  m(ε) the height of the cylinder Z ∩ Z1m is not less than the height of the cylinder m(ε) Z ∩ Z1 , i.e., for any ε > 0 there exists t(ε)  2q m(ε) such that for t  t(ε) and all x ∈ K u(x, t) < ε.

(5.4)

Thus, sufficiency is proved. Necessity. Let the series (1.15) converge, and let cap (RN \Q) > 0. By [14, Theorem 5.1], there exists an equilibrium measure μ of the closed set (RN \Q) and the corresponding potential  dμ(y) μ (5.5) U (x) = |x − y|N −2 RN \Q

such that U μ (x) = 1 μ-a.e. on RN \Q, μ

lim U (x) = 0,

|x|→∞

ΔU μ (x) = 0

in Q.

Lemma 5.1. The potential (5.5) possesses the following properties: 1 (RN ), (1) U μ (x) ∈ W2,loc 1 (RN \Q), (2) U μ (x) = 1 on RN \Q in the sense of W2,loc  ∇U μ (x)∇ϕ(x)dx = 0 for all ϕ(x) ∈ C0∞ (RN ). (3) Q

880

(5.6) (5.7) (5.8)

1 Remark 5.1. The equality U μ (x) = 1 on RN \Q in the sense of W2,loc (RN \Q) means that for 1 (RN \Q)-norm any ball BR = {|x| < R} the function U μ (x)−1 can be approximated in the W2,loc by smooth functions vanishing near the boundary ∂Q ∩ BR .

Proof of Lemma 5.1. Let {Kn }∞ n=1 be an increasing sequence of compact sets, Kn+1 ⊃ Kn such that Kn → RN \Q as n → ∞. We consider the family of problems Δun (x) = 0 in RN \Kn ,  lim un (x) = 0, un ∂(RN \Kn ) = 1, |x|→∞

(5.9) (5.10)

1 (RN \K ) with unit whose solutions are understood in a distributional sense in the space W2,loc n trace on ∂(RN \Kn ). As is known [13], any n-solution to the problem (5.9), (5.10) has the form  dμn (y) μn , (5.11) un (x) = U (x) = |x − y|N −2 Kn

where μn is the measure realizing the Wiener capacity of the compact set Kn . As is known [14], if the series (1.15) converges, then the sequence of measures {μn } weakly converges to a measure μ, called an equilibrium measure of the set RN \Q. We note that the functions (5.11) possess the following properties: 1 (RN \Kn ), (1) U μn (x) ∈ W2,loc

(2) un = 1 in the sense of W21 on Kn , (3) the following relation holds: 

N

μ(R \Q)  μn (Kn ) = C(N )

|∇U μn (x)|2 dx

(5.12)

RN \Kn

 (4)

∇un (x)∇ϕ(x)dx = 0 for all ϕ(x) ∈ C0∞ (RN \Kn ).

RN \Kn

Since the left-hand side of (5.12) is bounded, the weak convergence of μn to the measure μ implies the weak compactness of ∇U μn in L2 (RN ). Therefore, in (4), we can pass to the limit along a weakly converging subsequence ∇unk (x). On the other hand (cf. [14]), lim un (x) = U μ (x) pointwise.

n→∞

(5.13)

Thus, the potential U μ (x) possesses properties (1)–(3) indicated in Lemma 5.1. Remark 5.2. From (5.12), passing to the limit, we find  μ(RN \Q) = C(N ) |∇U μ (x)|2 dx > 0.

(5.14)

Q

From (5.14) it follows that U μ (x) ≡ 1 in Q; otherwise, we have μ(RN \Q) = 0. 881

To prove the necessity in Theorem 1.1, we consider the problem ΔU − Ut = 0 in Z = Q × (0, ∞),  U S = 0, S = ∂Q × (0, ∞),  U t=0 = w(x), x ∈ Q.

(5.15)

w(x) = 1 − U μ (x),

(5.18)

We set

(5.16) (5.17)

where U μ (x) is the equilibrium potential of the set RN \Q. According to Remark 5.2, w(x) ≡ 1,0 (Z) to the problem (5.15)–(5.17) has the form 0 in Q and a bounded solution of class V2,loc u(x, t) = w(x) for all t > 0; moreover, lim u(x, t) = w(x) = 0. Necessity is proved. t→∞

To prove Theorem 1.2, we need the following result. Lemma 5.2. For any k ∈ N, k  2, q > 1 qk

 k k   cap (B qm+1 \Q) cap (B τ \Q) cap (B qm \Q)  dτ  B , B1 2 τ N −1 q (m−1)(N −2) q m(N −2) m=2 m=2

(5.19)

q

where B1 and B2 are positive constants. Proof. As is known [14], the capacity of a compact set possesses the monotonicity property in the sense that cap (E1 )  cap (E)  cap (E2 ) for compact sets E1 , E2 , E such that E1 ⊂ E ⊂ E2 . We consider the integral q

m

Im = q m−1

cap (B τ \Q) dτ, τ N −1

m  2.

(5.20)

For q m−1  τ  q m , using the inequalities 1 1 1  N −1  (m−1)(N −1) τ q m(N −1) q and the monotonicity property of capacity, from (5.20) we find cap (B q( m−1) \Q) q m(N −1)

(q m − q m−1 )  Im 

cap (B qm \Q) m (q − q m−1 ). (m−1)(N −1) q

(5.21)

The following identities hold: cap (B q( m−1) \Q) q m(N −1)

cap (B q( m−1) \Q) q (m−1)(N −2) m · (q − q m−1 ) q (m−1)(N −2) q m(N −1)   cap (B qm−1 \Q) 1 1 . 1− = q q (m−1)(N −2) q N −2

(q m − q m−1 ) =

(5.22)

cap (B qm \Q) q m(N −2) cap (B qm \Q) m m−1 (q − q ) = · (q m − q m−1 ) q (m−1)(N −1) q m(N −2) q (m−1)(N −1) =

882

cap (B qm \Q) N −2 q (q − 1). q m(N −2)

(5.23)

Using the relations (5.22) and (5.23) for (5.20) and taking the sum with respect to m from 2 to k, we obtain the required inequalities (5.19). Proof of Theorem 1.2. By Lemma 5.2, the series (1.15) converges or diverges depending on the divergence of the integral (1.16). It is clear that q > 0 in (5.14), (5.17) can be any number > 1 since the choice of q > 1 does not affect the convergence of the integral (1.16). The divergence of the series (1.15) is equivalent to the stabilization to zero of the solution to the problem (1.12)–(1.14).

6

Properties of Heat Potential and Parabolic Capacities

To prove Theorem 1.3 by methods similar to the methods used in Theorems 1.1 and 1.2 for proving sufficiency, we should study properties of parabolic capacities and potentials corresponding to Equation (1.3). +1

We extend the coefficients aik (x, t), i, k = 1, . . . , N , of the operator in (1.3) to RN \Z by +1 setting aik (x, t) = δik for x ∈ RN \Z and preserve the notation Lu for the extended operator. Let Γ(x, y, t, τ ) be the fundamental solution to the extended parabolic operator L(x, t)u − ut = 0

in RN

+1

.

The fundamental solution Γ(x, y, t, τ ) satisfies the two-sided Aronson estimates (cf. [5])     r2 r2  Γ(x, y, t, τ )  C1 (t − τ )−N/2 exp − , C2 (t − τ )−N/2 exp − 4β(t − τ ) 4α(t − τ ) where C1 , C2 , α, and β are positive constants depending only on N , λ0 , and λ1 . Let  Γ(x, y, t, τ ), t > τ, F (x, y, t, τ ) = 0, t  τ.

(6.1)

(6.2)

+1

Following [6], we define the parabolic capacity of a B-set E ⊂ RN . For this purpose, we consider all possible measures μ such that  μ (6.3) w (x, t) = F (x, y, t, τ )dμ(y, τ )  1, (x, t) ∈ E, E

and introduce the parabolic capacity γ(E) of E by the formula γ(E) = sup μ(E),

(6.4)

where the supremum is taken over all possible measures for which the potentials wμ satisfy (6.3). The existence of the parabolic capacity corresponding to the kernel (6.2) is proved in [9, 11, 12]. As is shown in [11, 12], the function (6.3) is a solution of class V21,0 (RN \E) to the problem Lwμ − wtμ = 0 in RN  wμ ∂(RN \E) = 1,  wt=0 = 0.

+1

\E,

(6.5) (6.6) (6.7) 883

To prove an analogue of Lemma 2.1, even a simpler fact is sufficient: the existence of a measure μ that is equivalent to the heat measure and  F (x, y, t, τ )dμ(y, τ )  1. E

Such a measure can be obtained as follows. Let a measure μ realize the heat capacity of a set E with the kernel   r2 −N/2 F1 (x − y, t − τ ) = (t − τ ) . (6.8) exp − 4α(t − τ ) Then from the upper estimate in (6.1) it follows that the function  1 μ F1 (x, y, t, τ )dμ(y, τ ) w1 (x, t) = C1

(6.9)

E

1) satisfies the inequality w1μ (x, t)  1,

(x, t) ∈ RN

+1

\E,

(6.10) +1

μ = 0 in the sense of V21,0 on RN \E, 2) is a solution to the equation Lw1μ − w1t 3) satisfies the inequality w1μ  1 in the sense of V21,0 on the lateral surface of E. o 1,0 2 (Z)

Let u(x, t) be a nonnegative bounded solution of class V

to the problem

Lu − ut = 0 in Z,  uS = 0, S = ∂Q × (0, ∞),  ut=0 = u1 (x), x ∈ Q,

(6.11) (6.12) (6.13)

in Z, where u1 (x) is a continuous function in Q such that 0 < u1 (x)  M.

(6.14)

Consider the cylinder Z1 , Z2 , and Z3 defined in (2.4). Assume that the origin O lies on the +1 boundary ∂Q of Q; moreover, CZ = RN + \Z and E = CZ ∩ Z3 . Lemma 6.1. If u(x, t) is a nonnegative solution to the problem (6.11)–(6.13), then there exists 1 ∈ (0, 1) such that for 0 <  1   ηγ(E) sup u  sup u 1 + N N , (6.15) R Z1 ∩Z Z2 ∩Z where γ(E) is the heat capacity of the cylinder E corresponding to the heat kernel (6.8). Proof. Denote by S1 the lateral surface of the cylinder Z1 . We fix an arbitrary point (y, τ ) ∈ Z3 and estimate the function (6.2) by using the upper estimate for Γ(x, y, t, τ ) in (6.1):   r2 −N/2 . (6.16) sup C1 (t − τ ) exp − sup F < 4α(t − τ ) Z1 ∩|x|=R (x,t)∈S1 ∩Z 884

We fix an arbitrary point x on S1 , |x| = R, and find t > τ for which the function on the right-hand side of (6.16) attains the maximum: " #  r2 ∂ (t − τ )−N/2 exp − = 0. (6.17) ∂t 4α(t − τ ) Hence t − τ = r 2 /(4N α). For |x| = R and |y| < R , 0 < < 1, we have r = |x − y|  R(1 − ). Therefore, R2 (1 − )2 . t−τ  2N α By (6.16) and the monotonicity of v1 (x, y) = F1 (x − y, t − τ ) up to the first maximum (F1 is defined in (6.8)), we have  sup (x,t)∈S1 ∩Z

F  C1

2N α 2 R (1 − )2

N/2

  N . exp − 2

(6.18)

Let us estimate inf F over (x, t) ∈ Z2 . Since (y, τ ) ∈ Z3 and (x, t) ∈ Z2 , we have |x − y|2  (|x| + |y|)2  4 2 R2 ,

R2 2 /4  t − τ  2 2 R2 .

Using (6.1) and the above estimates, we find   C2 4 4R2 2 C2 exp(−4/β) = exp − 2 2 . inf F  2 2 N/2 R 4β (x,t)∈Z2 (2 R ) 2N/2 RN N

(6.19)

Since (y, τ ) is an arbitrary point on Z 3 , from (6.18) and (6.19) we obtain the estimates sup (x,t)∈S1 , (t,τ )∈Z3

F  C1

inf

(x,t)∈S1 , (t,τ )∈Z3

(2N α)N/2 − N e 2, RN (1 − )N

(6.20)

exp(−4/β) . 2N/2 RN N

(6.21)

F  C2

We consider the potential (6.9), where μ is the equilibrium heat measure of the cylinder E1 contained in Z 3 . Then (6.20) and (6.21) imply C2 sup w1 (x, t)  C1 S1 inf

(x,t)∈Z2 , (y,τ )∈Z3



2N α e

N/2

w1 (x, t) 

In the cylinder Z1 ∩ Z, we consider the function $ U (x, t) = M 1 − w1 (x, t) +

1 γ(E1 ), − )N

RN (1

C2 exp(−4/β) γ(E1 ). C1 2N/2 N RN

sup

(x,t)∈Z1 , (y,τ )∈Z3

% F1 γ(Z 3 \Z) . C1

(6.22) (6.23)

(6.24)

Compare the function (6.24) and the solution u(x, t) to the problem (6.11)–(6.13) on the parabolic boundary of Z1 ∩ Z. By the results in [11], the function (6.24) is a solution of class V21,0 to the equation LU − Ut = 0 in ΠT \E1 . 885

On the lateral surface S1 = {|x| = R} in Z1 ∩ Z, we have U (x, t)  0 in the sense of V21,0 (Z1 ∩ Z) since w1  1 in the sense of V21,0 (Z1 ∩ Z), and u(x, t)  M on the lower base in the pointwise sense (moreover, in the sense of V21,0 ), the function w1 (x, t) vanishes at t = 0 in the pointwise sense. Consequently, by the maximum principle, u(x, t) < U (x, t) in Z ∩ Z1 . Therefore,   F1 γ(Z 3 \Z) , (6.25) sup u  sup U  M 1 − inf w1 (x, t) + sup Z1 ∩Z Z2 ∩Z Z2 ∩Z Z1 ∩Z C1  1 F1 (x, y, t, τ ) F1 (x, y, t, τ )dμ  inf γ(Z 3 \Z). (6.26) inf w1 = inf Z1 ∩Z Z1 ∩Z C1 C1 (x,t)∈Z2 , (y,τ )∈Z3 Z 3 \Z

Taking into account (6.26), (6.22), and (6.23), from (6.25) we find    (2N α)N/2 γ(Z3 \Z) C2 e−4/β − N/2 . + sup u < M 1 − RN C1 2N/2 N e (1 − )N Z2 ∩Z Further, consider the function f ( ) =

A1 A2 − , N (1 − )N

where

C2 e−4/β (2N α)N/2 , A = . 2 C1 2N/2 eN/2 By the same argument as in the proof of Lemma 2.1, we obtain the required assertion. A1 =

7

Proof of Theorem 1.3

Let 0 <  1 < 1 be the same as in Lemma 6.1. We set q = 1/ > 1. Assume that t0 > 0, m ∈ N, and t0 − 2q m  0. We consider the system of coaxial cylinders (3.1) under the condition

(7.1) 2  min 3/8, 21 . We introduce the sequence Mm = sup u(x, t), m

(7.2)

Z 1 ∩Z

where u(x, t) is a solution to the problem (6.11)–(6.13). Using the inequality (6.15), we conclude that the sequence (7.2) satisfies the following inequality. Lemma 7.1. The following inequality holds:   η m (7.3) Mm  Mm−1 1 + N m γ(Z 3 \Z) . q As in Section 3, from (7.3) we obtain the inequality m   bγ(Z i \Z) 3 . Mm  M1 exp q iN

(7.4)

i=2

Applying to (7.4) the estimate (4.1), we obtain m   b cap (B qi \Q) . Mm  M1 exp i(N −2) q i=2 Arguing in the same way as in Theorem 1.1, we establish sufficiency in Theorem 1.3. 886

(7.5)

8

Proof of Theorem 1.4

We prove the estimate (1.17). Let the assumptions of Theorem 1.4 be satisfied, and let the series (1.15) (or the integral (1.16)) be divergent. From the inequality (7.5) for t  2q 2m , q > 1, we obtain  m   b cap (B qi \Q) M1  M exp − , (8.1) q m(N −2) i=2 where M1 =

sup (x,τ )∈Z1 ∩Z

Z1 = {x, t : |x| < q, t − 2q 2 < τ < t}.

u(x, τ ),

Applying the lower estimate from (5.19) to the right-hand side of (8.1), we find ⎧ ⎫ m q ⎨ b cap (B τ \Q) ⎬ dτ , b1 = . M1  M exp −b1 N −1 ⎩ ⎭ τ B2

(8.2)

q

For any t > 2 there is m ∈ N such that

Hence q m  (8.2), we find

&

2q 2m  t  2q 2m+2 .

(8.3) & t/(2q 2 ). Replacing the upper integral limit with the larger limit t/(2q 2 ) in √

 sup

|x|q, t2q m

u(x, t)  M exp

q

√

 sup



− b1

Since K ⊂ Bq , the last inequality implies

x∈K, t2q m

t/(2q 2 )

u(x, t)  M exp

t/(2q 2 )



− b1 q

 cap (B τ \Q) dτ . τ N −1

 cap (B τ \Q) dτ . τ N −1

The estimate (1.17) and, consequently, Theorem 1.4 is proved. Proposition 8.1. Let Z be a given cylinder in RN , N  3, with the base radius ρ and height h. Then the following estimates hold: cap (Z)  α1 ρN −3 h, N −3

cap (Z)  α2 ρ

h,

(8.4) (8.5)

where α1 and α2 are positive constants depending on N . The inequality (8.4) is proved in [6, Section 2]. Let RN \Q be the cone $ ' % γ = RN \Q = x : x21 + x22 + . . . + x2N −1 < xN , 0 < xN < ∞

(8.6)

located in the half-space RN + = {x : xN  0}. We consider two cylinders: the cylinder Z1 of height 4τ with upper base of radius τ > 0 located in the hyperplane xN = τ and the cylinder Z2 of height τ with upper base of radius τ /4 located in the hyperplane xN = τ /4. It is clear that Z1 ⊃ Bτ \Q ⊃ Z2 .

(8.7) 887

By (8.7) and the monotonicity property of capacity [14], cap (Z1 )  cap (Bτ \Q)  cap (Z2 ).

(8.8)

Using (8.4) and (8.5) with ρ = τ /4, h = τ and ρ = τ , h = 4τ respectively, from (8.8) we find α2 4τ N −2  cap (Bτ \Q))  α1 τ N −2 By (8.9), α1 N 4 −4

√ t/C3 a0

dτ  τ

By (8.10),

√ t/C3 a0

√ t/C3 lim

t→∞

a0

1 4N −4

cap (Bτ \Q)) dτ  4α2 τ N −1

.

√ t/C3 a0

(8.9)

dτ . τ

(8.10)

cap (Bτ \Q)) dτ = ∞, τ N −1

i.e., by Theorem 1.1, the solution to the problem (1.3)–(1.5) stabilizes to zero as t → ∞. Furthermore, the estimate (1.17) and the lower estimate in (8.10) imply  |u(x, t)|  C1 exp

−

α1

4N −4

√ t/C3 a0

dτ τ

 = C4

1 , tC5 /2

(8.11)

√ N−1 . where C5 = α1 /4N −4 and C4 = C1 (a C3 )α/4 ' $ % N x21 + x22 + . . . + x2N −1 < xN , the solution to the By (8.11), in the case R \Q = x : problem (1.3)–(1.5) stabilizes to zero as t → ∞ at power rate.

9

Impossibility of Uniform Stabilization over Entire Domain Q

Let D = Q × (0, ∞) be a cylinder, where Q coincides with a cone. We consider the problem Δu − ut = 0 in D,   = 1, u = 0, u t=0

S

(9.1) (9.2)

where S = ∂Q×(0, ∞) and f (x, t) is a smooth function such that f (x, t) = 1 for t = 0, f (x, t) = 0 for t  1, and 0  f (x, t)  1. It is obvious that for t = 1 the solution to the problem (9.1), (9.2) is bounded and continuously vanishes on the lateral surface of the cylinder. Furthermore, for the solution to the problem Δu − ut = 0 in D,   ut=1 = u0 (x), u∂Q×(1,x) = 0, where u0 (x) = u(x, 1), we have lim u(x, t) = 0 x-uniformly in every compact set K in Q. t→∞

888

(9.3) (9.4)

We extend the solution u to the problem (9.1), (9.2) to the cylinder D1 = D × (−∞, ∞) by setting u = 1 on D1 . The extended function remains to be a positive solution to the heat yR = |x−yR | < equation. We choose a sufficiently large R > 0 and a point yR such that the ball BR R is contained in the cone Q. By the Harnack inequality (cf., for example, [1, 6]), inf u > C(N ) sup u, 1 ZR

2 ZR

1 = B yR × (R2 , 2R2 ) and Z 2 = B yR × (−R2 , 0). Since u = 1 for t  0, we have where ZR R R R

inf u  C(N ) > 0, 1 ZR

i.e., the solution to the problem (9.1), (9.2) is greater than the constant C(N ), where R2 is as large as desired. Consequently, the stabilization to zero of the solution to the problem (9.1), (9.2) is not uniform in G.

Acknowledgment The work is financially supported by the Russian Foundation for Basic Research (grant No. 09-01-00446). The author expresses his deep thanks to Professors Yu. A. Alkhutov and V. V. Zhikov for valuable advices and attention.

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2.

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Submitted on June 2, 2011

890