Heat Flow and Perimeter in R m

Transcription

1 Potential Anal (2013) 39: DOI /s z Heat Flow and Perimeter in R m M. van den Berg Received: 25 September 2012 / Accepted: 6 February 2013 / Published online: 1 March 2013 Springer Science+Business Media Dordrecht 2013 Abstract Let be an open set in Euclidean space R m with finite perimeter P( ), and with m-dimensional Lebesgue measure. It was shown by M. Preunkert that if T(t) is the heat semigroup on L 2 (R m ) then H (t) := T(t)1 (x)dx = π 1/2 P( )t 1/2 + o(t 1/2 ), t 0. H (t) represents the amount of heat in if is at initial temperature 1 and if R m \ is at initial temperature 0. Inthispaperwewill compare the quantitative behaviour of H (t) with the usual heat content Q (t) associated to the Dirichlet heat semigroup on. We analyse the heat content for horn-shaped open sets of the form (α, ) ={(x, x ) R m : x (1 + x) α, x > 0}, where α>0, andwhere is an open set in R m 1 with finite perimeter in R m 1, which is star-shaped with respect to 0. For m 3 we find that there are four regimes with very different behaviour depending on α, and a further two limiting cases where logarithmic corrections appear. Keywords Heat flow Perimeter Euclidean space Mathematics Subject Classification (2010) 35K05 1 Introduction In this paper we will obtain some results for the heat flow from sets in Euclidean space R m into their complement. Let p m (x, y; t) = (4πt) m/2 e x y 2 /(4t), M. van den Berg (B) School of Mathematics, University of Bristol, University Walk, Bristol BS8 1TW, UK M.vandenBerg@bristol.ac.uk

2 370 M. van den Berg and let be a non-empty open set in R m.letu be the unique weak solution of u = u t, x Rm, t > 0, with initial condition u(x; 0) = 1 (x), x R m, where 1 : R m {0, 1} is the characteristic function of. We have that u (x; t) = dyp m (x, y; t). (1) We define the heat content of in R m at time t by H (t) = dxu (x; t) = dxdyp m (x, y; t). It was shown [8 10] thatif is an open set in R m with finite Lebesgue measure and with finite perimeter P( ) then ( π ) 1/2 P( ) = lim dxdyp m (x, y; t). t 0 t Since we conclude the following. (R m \ ) (R m \ ) dxdyp m (x, y; t) = H (t), Theorem 1 If is an open set in R m with f inite Lebesgue measure and with f inite perimeter then H (t) = π 1/2 P( )t 1/2 + o(t 1/2 ), t 0. Let v be the unique weak solution of with initial condition and with Dirichlet boundary condition v = v, x, t > 0, t v(x; 0) = 1, x, v(x; t) = 0, x, t > 0. The heat content of at time t is defined by Q (t) = dxv (x; t). The latter quantity has been studied extensively in the general setting of open bounded sets with smooth boundaries in complete Riemannian manifolds. See for example [4, 6]. We recall the following definition in [3].

3 Heat Flow and Perimeter in R m 371 Definition 2 Let D be an open set in R m with boundary D. We say that D is R-smooth if for any point x 0 D there exist two open balls B 1 and B 2 with radius R such that B 1 D, B 2 (R m \ (D D)), B 1 B 2 ={x 0 }. In [3]itwasshownthatif is open, bounded and connected in R m with R-smooth boundary then Q (t) = 2π 1/2 H m 1 ( )t 1/2 + O(t), t 0, where H m 1 denotes (m 1)-dimensional Hausdorff measure. It is not hard to show that an R-smooth boundary is Lipschitz, and so by p.183 in [5], P( ) = H m 1 ( ). We note that in the R-smooth case and for t small the heat loss H (t) is comparable with the heat loss Q (t). A heuristic explanation for this is that at time 0, R m \ is initially at temperature 0 and so its neighbourhood in will loose heat at a comparable amount. In general P( ) H m 1 ( ), andthe following illustrates that very different behaviour may occur in the case of strict inequality. Example 3 Let K be the infinite triadic von Koch curve in the plane R 2 containing the origin 0. Let B(x, r) ={y R m : y x < r}, and let = B(0, 1) \ K. Since K =0, P( \ K) = P( ) = P(B(0, 1)) = 2π. Hence H (t) = 2π 1/2 t 1/2 + o(t 1/2 ). On the other hand the boundary of satisfies a capacitary density condition, and the interior Minkowski dimension of equals (log 4)/ log 3. It follows by the results in [1] that there exists c > 1, t 0 > 0 such that c 1 t 1 (log 2)/ log 3 Q (t) ct 1 (log 2)/ log 3, 0 t t 0. This paper is organized as follows. In Section 2 we will compare v with u and Q with H respectively. In Section 3 we obtain upper bounds for the heat content in R m for arbitrary open sets. These bounds are in general not sharp. However, in the case of one-sided horn-shaped open sets in R m a detailed analysis is possible. In Section 3 we shall prove the following. Theorem 4 Let be an open, bounded set in R m 1 with f inite perimeter in R m 1 which is star-shaped with respect to 0.Let (α, ) ={(x, x ) R m : x (1 + x) α, x > 0}. (i) If 0 <α (2m 1) 1 then H (α, ) (t) =+ for all t > 0. (ii) If (2m 1) 1 <α<(m 1) 1 then for t 0 H (α, ) (t) = (2α) 1 dθθ (1 (m+1)α)/(2α) H (θ)t (α(m 1) 1)/(2α) (1 α(m 1)) 1 H m 1 ( ) + O(t 1/2 ).

4 372 M. van den Berg (iii) If α = (m 1) 1,andift > 0 then for t 0 ( t H ((m 1) 1, )(t) = 2 1 (m 1)H m 1 ) ( ) log t (m 1) dθθ 1 (H (θ) H m 1 ( )) (m 1) [0,t ] [t, ) dθθ 1 H (θ) + O(t 1/2 ). (iv) If m = 2 and (m 1) 1 <α< or if m = 3, 4,... and (m 1) 1 <α< (m 2) 1 then for t 0 H (α, ) (t) = (α, ) (2α) 1 dθθ (1 (m 1)α)/(2α) [0, ) ( H m 1 ( ) H (θ) ) t ((m 1)α 1)/(2α) + O(t 1/2 ). (v) If m = 3, 4,...,and if α = (m 2) 1 then for t 0 H ((m 2) 1, )(t) = (α, ) +2 1 π 1/2 (m 2)P( )t 1/2 log t + o(t 1/2 log t). (vi) If m = 3, 4,...,and if α>(m 2) 1 then for t 0 H (α, ) (t) = (α, ) π 1/2 P( (α, ))t 1/2 + o(t 1/2 ). We see that in regime (i) the region cools down too slowly to give a finite heat content for any positive t. In regime (ii) the Lebesgue measure term of Theorem 1 is replaced by a term which diverges as t 0 with an α- dependent exponent of t. In regime (iv) the region has finite Lebesgue measure but infinite perimeter. In that case the exponentof t in the perimeter contribution is strictly less than 1/2. In regime (vi) both the Lebesgue measure and perimeter of (α, ) are finite and Theorem 1 applies. Note that regime (vi) does not show up in the planar case since (α, ) always has infinite perimeter. 2 Comparison of v and u and of Q and H In order to compare u and v it is useful to recall their probabilistic representations. Let (B(s), s 0; P x, x R m ) be Brownian motion with generator, andlet be the first exit time from. Then and T = inf{s 0 : B(s) / } v (x; t) = P x [T > t], (2) u (x; t) = P x [B(t) ]. (3)

5 Heat Flow and Perimeter in R m 373 Proposition 5 Let be an open set in R m.then (i) (ii) (iii) (iv) (v) v (x; t) u (x; t) 1, x, t > 0. (4) Q (t) H (t), t > 0. (5) v (x; t) v (x; s), x, t s > 0. (6) Suppose H (s) < for some s 0. Then Q (t) Q (s), t s > 0. (7) H (t) H (s), t s > 0. Proof Since {T > t} {B(t) }, we have that P x [T > t] P x [B(t) ] 1. This implies (i) by (2) and(3). Since {T > t} {T > s} for 0 < s t we have that P x [T > t] P x [T > s] which implies (iii) by Eq. 2. Inequalities (5) and(7) follow by integrating both inequalities (4)and(6)withrespecttoxover. To prove (v) we suppose that H (s) < for some s 0.Then H (s) = dxu (x; s) = dxdydzp(x, y; s/2)p(x, z; s/2) R m = dxu (x; s/2) 2, R m implies that u ( ; s/2) L 2 (R m ). Since the heat semigroup T(t) acting in L 2 (R m ) is contractive we have for t > s that H (t) = u ( ; t/2) 2 L 2 (R m ) = T((t s)/2)u ( ; s/2) 2 L 2 (R m ) u ( ; s/2) 2 L 2 (R m ) = H (s). In the example below we show that t u (x; t) is not necessarily monotone. Example 6 Let ɛ (0, 1/2) be arbitrary. We construct an open set 0 in R 2 and 0 < t 1 < t 2 depending on ɛ and on such that u (0; t 1 )<ɛ,u (0; t 2 )>1 ɛ. Let 0 <δ<a < b <, and put ={x R 2 : x <δ} {x R 2 : a < x < b}.

6 374 M. van den Berg By formula (1) Hence We now choose u (0; t) = 1 e δ2 /(4t) + e a2 /(4t) e b 2 /(4t). 1 a 2 /(4t) e b 2 /(4t) u (0; t) δ 2 /(4t) + e a2 /(4t). and a such that e a2 ɛ/(4δ 2) = ɛ/2. So t 1 = δ 2 /ɛ, a = 2δ ɛ 1/2 (log(2/ɛ))1/2. Then u (0; t 1 )<ɛ. We subsequently choose so that t 2 > t 1, and t 2 = a 2 /(2ɛ), e a2 /(4t 2) 1 a2 4t 2 = 1 ɛ 2, e b 2 /(4t 2) = e b 2 ɛ/(2a 2). Finally we choose b such that e b 2 ɛ/(2a 2) = ɛ/2. Hence and u (0; t 2 )>1 ɛ. b = 23/2 δ ɛ log(2/ɛ), It is straightforward to extend this example to R m, and to finite sequences 0 < t 1 < t 2 < < t 2n with u (0; t 2 j )>1 ɛ, u (0; t 2 j 1 )<ɛ,j = 1, n, by adding larger and larger annuli. The following implies that the failure of monotonicity in Example 6 holds in a neighbourhood of 0. Proposition 7 Let be an open bounded set with diam( ) = sup{ z z :z, z }. Then for all x, x,t > 0 u (x; t) u (x ; t) min{1, 2 (m 2)/2 t 1 diam( ) x x }. (8) Proof First note that for y e x y 2 /(8t) e x y 2 /(8t) (8t) 1 x y 2 x y 2 (4t) 1 diam( ) x x.

7 Heat Flow and Perimeter in R m 375 Next note that u (x; t) u (x ; t) (4πt) m/2 dy e x y 2 /(8t) e x y 2 /(8t) (e x y 2 /(8t) + e x y 2 /(8t) ) (4πt) m/2 (4t) 1 diam( ) x x dy(e x y 2 /(8t) + e x y 2 /(8t) ) (4πt) m/2 (4t) 1 diam( ) x x dy(e x y 2 /(8t) + e x y 2 /(8t) ) R m = 2 (m 2)/2 t 1 diam( ) x x. This implies (8)since0 < u (x; t) <1. The following guarantees monotonicity of t u (0; t) with respect to a starshaped point 0. Proposition 8 (i) (ii) If 0 and if is open and star-shaped with respect to 0 then t u (0; t) is decreasing. If is convex then t u (x; t) is decreasing for all x. Proof (i) By the change of variable y = ηt 1/2, we have by Eq. 1 that u (0; t) = (4π) m/2 dηe η2 /4, t 1/2 where t 1/2 is the dilation of with respect to 0 by a factor t 1/2.Since is star-shaped with respect to 0 we have the inclusion t 1/2 s 1/2, 0 < s < t. This implies monotonicity since the integrand is non-negative. (ii) Convex sets are starshaped with respect to any interior point. The following shows that the principle of not feeling the boundary holds whether or not we impose Dirichlet 0 boundary conditions on. Proposition 9 Let be an open set in R m, and denote by δ(x) = min{ x y :y R m \ }. Then for t > 0 (i) (ii) (iii) u (x; t) 1 2 m/2 e δ(x)2 /(8t). (9) v (x; t) 1 2 (2+m)/2 e δ(x)2 /(8t). (10) H (t) 2 1 {x : δ(x) (m + 2)2 1/2 (log 2)t 1/2 }.

8 376 M. van den Berg (iv) Proof We have that u (x; t) = 1 Q (t) 2 1 {x : δ(x) (m + 4)2 1/2 (log 2)t 1/2 }. 1 R m \ dyp m (x, y; t) R m \B(x,δ(x)) 1 e δ(x)2 /(8t) 1 e δ(x)2 /(8t) = 1 2 m/2 e δ(x)2 /(8t). dyp m (x, y; t) R m \B(x,δ(x)) dy(4πt) m/2 e x y 2 /(8t) R m dy(4πt) m/2 e x y 2 /(8t) For the proof of assertion (ii) we refer to Lemma 6.3 and Corollary 6.4 in [3]. Assertions (iii) and (iv) follow immediately by inequalities (9)and(10) respectively. The bounds in Proposition 9 do not give information for t large compared with δ(x) 2. We have the following. Proposition 10 Let be an open set in R m. Then for all x,t > 0, u (x; t) 2 1 min{4 m Ɣ((2 + m)/2) 1 t m/2 δ(x) m, 1}. Proof u (x; t) B(x,δ(x)) dyp m (x, y; t) (4πt) m/2 e δ(x)2 /(4t) 1dy B(x,δ(x)) = 2 m Ɣ((2 + m)/2) 1 t m/2 δ(x) m e δ(x)2 /(4t). (11) For e δ(x)2 /(8t) 2 (2+m)/2 we have that the right hand side of inequality (9) is bounded from below by 1/2. For e δ(x)2 /(8t) 2 (2+m)/2 we have that the right hand side of inequality (11) is bounded from below by 2 1 2m Ɣ((2 + m)/2) 1 t m/2 δ(x) m. The combination of these two lower bounds completes the proof. Additivity for the heat content in R m of disjoint open sets fails, and we have the following.

9 Heat Flow and Perimeter in R m 377 Proposition 11 Let be an open set in R m,andlet = i I i,where{ i, i I} is the collection of components of. Then (i) (ii) H (t) = i I H i (t) + {(i, j) I 2 :i = j} i j dxdyp m (x, y; t) Q (t) = i I Q i (t) Proof The proof of (i) is immediate from the definition of H (t). The proof of (ii) follows from the fact that the Dirichlet heat kernel p (x, y; t) = 0 for x and y in disjoint components. Both H (t) and Q (t) satisfy similar scaling relations. Proposition 12 Let be an open set in R m and suppose that both H (t) and Q (t) are f inite for all t > 0. For λ>0 we have that (i) (ii) H λ (t) = λ m H (λ 2 t). Q λ (t) = λ m Q (λ 2 t). Proof H λ (t) = dxdyp m (x, y; t) λ λ = λ 2m dxdyp m (λx,λy; t) = λ m H (λ 2 t). (12) The proof of assertion (ii) follows by replacing p m (x, y; t) in Eq. 12 by the Dirichlet heat kernel p (x, y; t) for. 3 Unbounded and Horn-shaped Regions For an arbitrary open set in R m and x R we define the cross-section of with the plane {(x, x ) : x R m 1 } by x ={x R m 1 : (x, x ) }.

10 378 M. van den Berg Proposition 13 For open in R m we have that H (t) (4πt) (m 1)/2 Proof then H (t) = = R R 2 1 = R R dxdydx dy p m ((x, x ), (y, y ); t) dxdydx dy (4πt) m/2 e x y 2 /(4t) R dx(h m 1 ( x )) 2. dxdy(4πt) m/2 e x y 2 /(4t) H m 1 ( x )H m 1 ( y ) R R = (4πt) (m 1)/2 dxdy(4πt) m/2 e x y 2 /(4t) ((H m 1 ( x )) 2 + (H m 1 ( y )) 2 ) dxdy(4πt) m/2 e x y 2 /(4t) (H m 1 ( x )) 2 In a very similar fashion we show that if R x dx(h m 1 ( x )) 2. (13) ={x R : (x, x ) } H (t) (4πt) 1/2 R m 1 dx (H 1 ( x )) 2. Since H (t) is invariant under rotations of the above estimates are also invariant under rotations. An estimate involving the L p norm of H m 1 (. ) for 1 < p < 2 is the following. Proposition 14 If p (1, 2) is such that dx(h m 1 ( x )) p < R then H (t) C p,m ( R dx(h m 1 ( x )) p ) 2/p t (2p mp 2)/(2p), (14)

11 Heat Flow and Perimeter in R m 379 where C p,m = ((p 1)/(pe)) (p 1)/p 2 (2p 2 mp)/p π (3p mp 4)/(2p) Ɣ((2 p)/(2p)). Ɣ(1/p) Proof Let 1 < p < 2 and λ = 2(p 1)/p then 0 <λ<1. Since e x y 2 /(4t) (λ/(2e)) λ/2 (4t) λ/2 x y λ we have by the third line of inequality (13)that H (t) (2λ/e) λ/2 (4πt) m/2 t λ/2 dxdyh m 1 ( x ) x y λ H m 1 ( y ). R R Inequality (14) follows immediately from the Hardy-Littlewood-Sobolev inequality (Theorem 4.3 in [7]). We recall the following definition from [3]. Definition 15 Aset R m is (one-sided) horn-shaped if (i) is open and connected, (ii) x1 x2 for all x 1 x 2 > 0, (iii) x = for x 0. (iv) (0,δ) dxhm 1 ( x )< for all δ>0. We define 0 + = x>0 x. Note that 0 + is an open set in R m 1. The heat content Q is well understood in the planar case (Theorem 5.4 in [3], and [2]). Our main result for the heat content of a horn-shaped open set in R m is the following. Theorem 16 Let be a horn-shaped set in R m, and let H x (t) denote the heat content of the cross-section x in R m 1 at time t. If dxh x (t) < for all t > 0, then (i) (ii) (iii) H (t) lim t 0 H (t) dxh x (t), (15) dxh x (t) dxe x2 /(4t) H x (t), (16) dxe x2 /(4t) H x (t) dxh x (t) = 0. (17)

12 380 M. van den Berg Proof To prove (i) we have that H (t) = = 2 2 = = dxdy dx dy p m ((x, x ), (y, y ); t) x y dxdy dx dy p m ((x, x ), (y, y ); t) x y dy dx dy p m ((x, x ), (y, y ); t) x x {(x,y) :x<y} dx {y>x} dx dx dy p m 1 (x, y ; t) x x dxh x (t). (18) To prove (ii) we have the following from the second identity in inequality (18). H (t) 2 dxdy dx dy p m ((x, x ), (y, y ); t) {(x,y) :x<y} y y = 2 dy dx dx dy p m ((x, x ), (y, y ); t) [0, ) (,y] y y 2 dx dy dx dy p m ((x, x ), (y, y ); t) (,0] [0, ) y y = dyh y (t) 2(4πt) 1/2 dxdye (x+y)2 /(4t) H y (t) = To prove (iii) we note that dyh y (t) 2(4πt) 1/2 dxh x (t) dxe x2 /(4t) H x (t). dxe x2 /(4t) H x (t) (0,δ) (0,δ) (0,δ) On the other hand for a > 0 and fixed dxh x (t) dxh x (t) + e δ2 /(4t) dxh x (t) + e δ2 /(4t) dxdye (x2 +y 2 )/(4t) H y (t) [δ, ) dxh m 1 ( x ) + e δ2 /(4t) (0,a) dxh x (t). dxh x (t) dxh x (t) dxh x (t).

13 Heat Flow and Perimeter in R m 381 By Definition 15 (iv) the measure of the set x is finite, and so lim dxh x (t) = dxh m 1 ( x ). t 0 (0,a) (0,a) By monotonicity there exists t 0 such that for 0 t t 0, (0,a) dxh x (t) 2 1 (0,a) dxhm 1 ( x ). We conclude that for 0 t t 0 the quotient in the left hand side of formula (17) is bounded from above by (0,δ) 2 dxhm 1 ( x ) (0,a) dxhm 1 ( x ) + /(4t). e δ2 Next choose δ = (at) 1/3 so that lim t 0 (0,(at) 1/3 ) dxhm 1 ( x ) (0,a) dxhm 1 ( x ) = 0, by Lebesgue s dominated convergence theorem. This completes the proof of (iii). The following lemmas, needed in the proof of Theorem 4, give uniform bounds for the heat content. Lemma 17 If is an open bounded set in R m with f inite perimeter then H (t) π 1/2 P( )t 1/2, t > 0. (19) Proof Proposition 8 in [10] shows that u (x; t)dx π 1/2 P( )t 1/2. R m \ By formula (1) R m dxu (x; t) =. This proves inequality 19 since R m \ u (x; t)dx = R m u (x; t)dx H (t). Lemma 18 (i) If is an open set in R m with <. Then H (t) min{(4πt) m/2 2, }. (20) (ii) If is an open set in R m with diam( ) < then H (t) (4πt) m/2 e diam( )2 /(4t) 2. (21)

14 382 M. van den Berg (iii) If is an open bounded set in R m then Proof The proof is immediate since H (t) = lim t tm/2 H (t) = (4π) m/2 2. (22) dxdyp m (x, y; t) dxdy(4πt) m/2 = (4πt) m/2 2, (23) and inequality (20) follows by the second inequality in Eq. 5. Furthermore if diam( ) < then for x, y p m (x, y; t) (4πt) m/2 e diam( )2 /(4t). (24) Integrating (24) withrespecttox and y over yields (21). The assertion under (iii) follows from the assertions under (i) and (ii). Note that the behaviour of H (t) for large t given by inequality (24) is very different from the corresponding behaviour of Q (t). Proposition 19 If is an open set in R m and if Q (t 0 )< for some t 0 > 0 then lim t t 1 log Q (t) = infspec( ), where is the Dirichlet Laplacian acting in L 2 (R m ). Proof It follows by Lemma 2.6 in [3] thatifq (t 0 )< for some t 0 > 0 then the Dirichlet heat semigroup is trace class, and the spectrum of the Dirichlet Laplacian acting in L 2 (R m ) is discrete. We have that p (x, y; t) = e tλ j φ j (x)φ j (y), j=1 where λ 1 λ 2 is the spectrum of,and{φ 1,φ 2, } is a corresponding orthonormal set of eigenfunctions in L 2 ( ).Hence Q (t) = ( e tλ j j=1 φ j ) 2 e tλ1 ( Note that ( φ 1) 2 > 0 since φ1 does not change sign. Hence φ 1 ) 2. lim inf t t 1 log Q (t) infspec( ).

15 Heat Flow and Perimeter in R m 383 On the other hand we have for any ɛ (0, 1) and t such that ɛt > t 0 that It follows that Q (t) e (1 ɛ)tλ1 j=1 e (1 ɛ)tλ1 Q (t 0 ). e ɛtλ j ( φ j ) 2 lim sup t 1 log Q (t) (1 ɛ)infspec( ). t Letting ɛ 0 completes the proof. Proof of Theorem 4 (i) (ii) Let t > 0 and arbitrary, let ρ = min{ x :x / }, andlet n (α, ) ={(x, x ) R m : x (1 + n) α,n < x < n + 1} where n = 1, 2,. Since (α, ) n=1 n(α, ) we have that H (α, ) (t) n=1 H n(α, )(t). Each n (α, ) contains an open cylinder with height 1 over a (m 1)-dimensional disc with radius (1 + n) α ρ. Each such cylinder contains a sub-cylinder with height 1/3 over a (m 1)-dimensional disc with radius (2 + n) α ρ/2. Each point of this subcylinder has at least distance min{(1 + n) α ρ/2, 1/3} to the boundary of the larger cylinder, and hence to n (α, ). The measure of this sub-cylinder equals 3 1 ω m 1 ((1 + n) α ρ/2) m 1,whereω m 1 is the volume of the ball with radius 1 in R m 1.LetN(α, ρ, t) be such that for n N(α, ρ, t) both (1 + n) α ρ/2 1/3 and (1 + n) α ρ/2 t 1/2. It then follows by Proposition 10 that for all n N(α, ρ, t), H n(α, )(t) c m t m/2 ((1 + n) α ρ/2) 2m 1,wherec m depends on m only. Summing both sides of this inequality over all n N(α, ρ, t) yields (i) since 0 <α (2m 1) 1. First observe that the second term in the right hand side of inequality (16) is bounded in absolute value by dxe x2 /(4t) H (α, )x (t) dxe x2 /(4t) H m 1 ( ) = O(t 1/2 ). (25) The first term in the right hand side of inequality (16) equals dxh (α, )x (t) = dxx α(m 1) H (x 2α t) [0, ) = [1, ) [0, ) dxx α(m 1) H (x 2α t) dxx α(m 1) H (x 2α t) [0,1] :=A 1 A 2. (26) It is easily seen that A 2 is finite since H (x 2α t) H m 1 ( ), andα(m 1) <1. Lemma 18 shows that A 1 is finite as well. Hence applying Lemma 18 to the open set in R m 1 gives that A 1 dxx α(m 1) min{h m 1 ( ), (H m 1 ( )) 2 (4π x 2α t) (m 1)/2 }. [0, )

16 384 M. van den Berg The above is finite by the hypothesis under (ii). It follows that, using the change of variable x 2α t = θ A 1 = (2α) 1 dθθ (1 (m+1)α)/(2α) H (θ)t (α(m 1) 1)/(2α). By Lemma 18 we have that dxx α(m 1) H (x 2α t) [0,1] By Lemma 17 we conclude that dxx α(m 1) H (x 2α t) [0,1] [0,1] [0,1] dxx α(m 1) H m 1 ( ) = (1 α(m 1)) 1 H m 1 ( ). dxx α(m 1) (H m 1 ( ) C(x 2α t) 1/2 ) = (1 α(m 1)) 1 H m 1 ( ) + O(t 1/2 ). (iii) This completes the proof of case (ii). By the first equality in formula (26) and a change of variable we have for any t > 0 that dxh ((m 1) 1, ) x (t) = 2 1 (m 1) dθθ 1 H (θ) [0, ) [t, ) ( t = 2 1 (m 1)H m 1 ) ( ) log t (m 1) dθθ 1 (H (θ) H m 1 ( )) [0,t ] (m 1) dθθ 1 H (θ) 2 1 (m 1) [t, ) [0,t] dθθ 1 (H (θ) H m 1 ( )). (27) The fourth term in the right hand side of formula (27) is bounded in absolute value using Lemma 17 by 2 1 (m 1)P( ) dθθ 1/2 = O(t 1/2 ). [0,t] Lemma 17 also implies that the integral in the second term in the right hand side of formula (27) converges. Finally inequality (23) implies that for the open

17 Heat Flow and Perimeter in R m 385 (iv) set R m 1, H (θ) (4πθ) (m 1)/2 (H m 1 ( )) 2. Hence the integral in the thirdterminformula(27) converges. This completes the proof of case (iii). By the first equality in formula (26) and a change of variables we have that dxh (α, )x (t) = (α(m 1) 1) 1 H m 1 ( ) [0, ) + (2α) 1 dθθ (1 (m+1)α)/(2α) (H (θ) [0, ) H m 1 ( ))t ((m 1)α 1)/(2α) (2α) 1 dθθ (1 (m+1)α)/(2α) (H (θ) [0,t] H m 1 ( ))t ((m 1)α 1)/(2α). (28) Next note that (α, ) =(α(m 1) 1) 1 H m 1 ( ). (29) By Lemma 17 we have that dθθ (1 (m+1)α)/(2α) (H (θ) H m 1 ( )) = O(t (1 mα+2α)/(2α) ). [0,t] We conclude that the third term in the right hand side of formula (28)isO(t 1/2 ). Combining this with inequality (25) completes the proof of case (iv). (v) By the first equality in formula (26), (29), and the change of variable x 2/(m 2) t = θ we have that dxh ((m 2) 1, ) x (t) = ((m 2) 1, ) Let t be strictly positive and arbitrary. Then 2 1 (m 2) [t, ) 2 1 (m 2) (m 2) dθθ 3/2 (H (θ) H m 1 ( ))t 1/2. [t, ) dθθ 3/2 (H (θ) H m 1 ( ))t 1/2 [t, ) dθθ 3/2 (H (θ) + H m 1 ( ))t 1/2 (m 2) dθθ 3/2 H m 1 ( )t 1/2 [t, ) ( ) t 1/2 2(m 2)H m 1 ( ), (30) t

18 386 M. van den Berg and 2 1 (m 2) dθθ 3/2 (H (θ) H m 1 ( ))t 1/2 [t,t ) ( t = 2 1 π 1/2 (m 2)P( )t 1/2 ) log π 1/2 (m 2) t dθθ 3/2 (H (θ) H m 1 ( ) + π 1/2 P( )θ 1/2 )t 1/2. (31) [t,t ] The integral in the second term in the right hand side of identity (31) is bounded in absolute value by ( t t 1/2 log t We now choose t depending on t such that for t 0 (i) log t = o(log t), (ii) t = o(1), ( (iii) t ) 1/2 t = o(t 1/2 log t). ) sup{ (H (θ) H m 1 ( ))θ 1/2 + π 1/2 P( ) :θ [0, t ]}. (32) Requirement (i) guarantees that the leading term in the right hand side of Eq. 31 equals 2 1 π 1/2 (m 2)P( )t 1/2 log t. Requirement (ii) guarantees that the supremum in expression (32) is o(1). Finally requirement (iii) guarantees that the right hand side of inequality (30) iso(t 1/2 log t). These requirements can be met by choosing t = (log(1/t)) 1. This proves the assertion under (v). (vi) For α>(m 2) 1, (α, ) < and is given by formula (29). Let f α (x) = (1 + x) α, x 0. Since ( ) P( (α, )) = H m 1 ( ) + P( ) dxf α (x) m 2 (1 + ( f α (x))2 ) 1/2, [0, ) we also have that P( (α, )) < for α>(m 2) 1. The assertion under (vi) follows by Theorem 1. References 1. van den Berg, M.: Heat content and Brownian motion for some regions with a fractal boundary. Probab. Theory Relat. Fields 100, (1994) 2. van den Berg, M.: Heat content asymptotics for planar regions with cusps. J. Lond. Math. Soc. 57, (1998) 3. van den Berg, M., Davies, E.B.: Heat flow out of regions in R m.math.zeit.202, (1989) 4. van den Berg, M., Gilkey, P.: Heat content asymptotics of a Riemannian manifold with boundary. J. Funct. Anal. 120, (1994) 5. Evans, L. C., Gariepy, R.F.: Measure Theory and Fine Properties of Functions. Chapman & Hall / CRC, Boca Raton (1992) 6. Gilkey, P.B.: Asymptotic Formulae in Spectral Geometry. Chapman & Hall / CRC, Boca Raton (2004)

19 Heat Flow and Perimeter in R m Lieb, E.H., Loss, M.: Analysis. Graduate Studies in Mathematics, vol. 14. American Mathematical Society, Providence, Rhode Island (1997) 8. Miranda, M. Jr., Pallara, D., Paronetto, F., Preunkert, M.: On a characterisation of perimeters in R N via heat semigroup. Ric. Mat. 44, (2005) 9. Miranda, M. Jr., Pallara, D., Paronetto, F., Preunkert, M.: Short-time heat flow and functions of bounded variation in R N. Ann. Fac. Sci. Toulouse 16, (2007) 10. Preunkert, M.: A Semigroup version of the isoperimetric inequality. Semigroup Forum 68, (2004)