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is considered in Section 3. Here f Lp Rn−1 (0 , + ) , 1 p , and the solution u is ∈ × ∞ ≤ ≤∞ represented by the heat double layer potential. The norm in the space Lp Rn−1 (0, t) is defined by × t 1/p f(x′, τ)) pdx′dτ for 1 p< , f = 0 Rn−1 | | ≤ ∞ (1.3) k kp,t Z Z ess sup f(x′, τ) : x′ Rn−1, τ (0, t) for p = . {| | ∈ ∈ } ∞ The main result obtained in Section 3 is the inequality u(x, t) (x, t) f ∇x x ≤ Wp || ||p,t n
2 with the best coefficient p−1 p cn,p n+p+2 p p(x, t)= max ωκ,λ (eσ, en) (eσ, en) p−1 (eσ, z) p−1 dσ , (1.4) 2+ n+1 W p |z|=1 Sn−1 | | | | xn Z where (x, t) is an arbitrary point in Rn (0, + ), + × ∞ ∞ λ −ξ ωκ,λ(u)= ξ e dξ; , (1.5) 2 Zκ/u and 1 1 p 2 1+ p 2 2 (4a ) qxn (n + 4)q cn,p = n , κ = , λ = 2 n −1 +1+ 1 2 π 2 q 2 p 4a t 2 − with p−1 + q−1 = 1. The extremal problem in (1.4) is solved for the case 2 p and the explicit formula ≤ ≤∞ p−1 π/2 ∞ p cn,p np+4 n+2(p+1) 2(p−1) −ξ p n−2 p(x, t)= 2ωn−1 ξ e dξ cos −1 ϑ sin ϑdϑ 2+ n+1 2 W p 0 qxn xn ( Z (Z 4a2t cos2 ϑ ) ) is obtained. In particular,
2 π/2 ∞ 16a √π n/2 −ξ 2 n−2 ∞(x, t)= ξ e dξ cos ϑ sin ϑdϑ . n−1 2 x2 W Γ xn 0 n 2 Z (Z 4a2t cos2 ϑ ) In Section 4 we obtain an analog of (1.2) for solutions of the Neumann problem
∂u 2 Rn ∂u ′ = a ∆u in + (0, + ), u t=0 =0, = g(x , t) ∂t × ∞ ∂xn xn=0
p Rn−1 with g L (0, + ) , represented by the heat single layer potential, 1 p . ∈ × ∞ Rn ≤ ≤∞ It is shown that for an arbitrary point (x, t) + (0, + ), the sharp coefficient p(x, t) in the inequality ∈ × ∞ N u(x, t) (x, t) g |∇x |≤Np || ||p,t is given by
p−1 p kn,p n−p+2 p e e e e p−1 e z p−1 p(x, t)= n+1 max ωκ,λ ( σ, n) ( σ, n) ( σ, ) dσ , (1.6) N p |z|=1 Sn−1 | | | | xn Z where ωκ,λ(u) is the same as in (1.5), and 2(3−p)/pa2/p qx2 (n + 2)q k = , κ = n , λ = 2 . n,p n + 1 2 πn/2q 2 p 4a t 2 − The extremal problem in (1.6) is solved for the case 2 p (n + 4)/2 and the explicit formula ≤ ≤ p−1 p π/2 ∞ (n−2)p+4 kn,p −ξ n+2 n−2 (x, t)= 2ω ξ 2(p−1) e dξ cos p−1 ϑ sin ϑdϑ p n+1 n−1 2 N p 0 qxn xn ( Z (Z 4a2t cos2 ϑ ) )
3 is obtained. In particular, 1/2 b π/2 ∞ (x, t)= n ξne−ξdξ cosn+2 ϑ sinn−2 ϑdϑ , 2 n+1 2 N 2 0 xn xn (Z (Z 2a2t cos2 ϑ ) ) where a bn = . n−1 n+1 2 4 n−1 2 π Γ 2 q 2 Extremal problems for integrals with parameters
2.1 Extremal problem for integrals with parameter on the space with measure Let X is the space with σ-finite measure µ defined on the σ-algebra S of measurable sets, S parameters y and y0 are elements of a set Y , ρ(x; y) and f(x; y) are [0, + ]-valued - measurable functions on X for any fixed y Y . ∞ A particular case of the assertion below∈ with ρ 1 and somewhat weaker assumption was proved in [5]. ≡ Proposition 1. Let y be a fixed point of Y . Let γ (0, + ) and let the integral 0 ∈ ∞ γ ρ(x; y0 )f (x; y)dµ (2.1) ZX attains its supremum on y Y at the point y0 Y (the case of + is not excluded). Further on, let ∈ ∈ ∞ (y,y )= ρ(x; y )f α(x; y)f β(x; y )dµ , (2.2) I 0 0 0 ZX where α> 0, β 0. Then the equality≥ holds
sup (y,y )= (y ,y )= ρ(x; y )f γ(x; y )dµ (2.3) I 0 I 0 0 0 0 y∈Y ZX for any α and β such that α + β = γ.
In particular, the supremum of (y,y0 ) over y Y is independent of y0 if the value of integral I ∈ ρ(x; y)f γ(x; y)dµ ZX does not depend on y. Proof. Let α > 0 and β 0 are arbitrary numbers, α + β = γ. The case β = 0 is obvious. Now, let β > 0. By H¨older’s≥ inequality, the integral
(y,y ) = ρ(x; y )f α(x; y)f β(x; y )dµ I 0 0 0 ZX α β γ α γ β = ρ (x; y0 )f (x; y) ρ (x; y0 )f (x; y0 ) dµ X Z 4 does not exceed the product
α β γ γ α γ γ β γ β γ γ α α α γ β β ρ (x; y0 )f (x; y)dµ ρ (x; y0 )f (x; y0 )dµ . ZX ZX Since integral (2.1) attains its supremum on y Y at y , it follows that ∈ 0 sup (y,y ) ρ(x; y )f γ(x; y )dµ . (2.4) I 0 ≤ 0 0 y∈Y ZX On the other hand, by (2.2) we have
sup (y,y ) (y ,y )= ρ(x; y )f γ(x; y )dµ , I 0 ≥ I 0 0 0 0 y∈Y ZX which together with (2.4) completes the proof.
2.2 Extremal problem for integral over Sn−1
n−1 Let eσ be the n-dimensional unit vector joining the origin to a point σ S . We denote by e and z the n-dimensional unit vectors and assume that e is a fixed vector.∈ Let ρ and f be non-negative Lebesgue measurable functions in [ 1, 1]. The next assertion is an immediate consequence− of Proposition 1. Corollary 1. Let γ > 0 and let the integral
γ ρ (eσ, e) f (eσ, z) dσ (2.5) Sn−1 Z attains its supremum on z Rn, z = 1 at the vector e. Further, let α 0,β > 0 and α + β = γ. Then ∈ | | ≥
α β sup ρ (eσ, e) f (eσ, e) f (eσ, z) dσ z Sn−1 | |=1 Z γ = ρ (eσ, e) f (eσ, e) dσ. (2.6) Sn−1 Z Remark By the equality π n−2 F (eσ, e) dσ = ωn−1 F cos ϑ sin ϑdϑ, Sn−1 Z Z0 we conclude that value of the integral in the right-hand side of (2.6) is independent of e. In the case of the even function F , the last equality can be written as π/2 n−2 F (eσ, e) dσ =2ωn−1 F cos ϑ sin ϑdϑ. (2.7) Sn−1 Z Z0 Further, we consider a special case of Corollary 1 with γ = 2, ρ (u)= ω (u) u µ, f(u)= u , (2.8) κ,λ,µ κ,λ | | | | 5 where κ,λ,µ 0 and ≥ ∞ λ −ξ κ ωκ,λ(u)= ξ e dξ =Γ λ +1, 2 . (2.9) κ/u2 u Z Here by ∞ Γ(α, x)= ξα−1e−ξdξ (2.10) Zx is denoted the additional incomplete Gamma-function. Lemma 1. Let
µ+ν 2−ν Fκ,λ,µ,ν(z)= ωκ,λ (eσ, e) (eσ, e) (eσ, z) dσ . (2.11) Sn−1 | Z Then for any κ,λ,µ 0, 0 ν < 2, the equality ≥ ≤ µ+2 max Fκ,λ,µ,ν(z)= Fκ,λ,µ,ν(e)= ωκ,λ (eσ, e) (eσ, e) dσ (2.12) |z|=1 Sn−1 | Z holds. Proof. (i) The case ν = 0. By (2.11),
µ 2 Fκ,λ,µ,0(z)= ωκ,λ (eσ, e) (eσ, e) (eσ, z) dσ . (2.13) Sn−1 | | Z ′ Let z = z (z, e)e. We choose the Cartesian coordinates with origin at the center of −n−1 ′ O the sphere S such that e1 = e and en is collinear to z . Then z = αe1 + βen, where α2 + β2 =1. (2.14) Now, we rewrite (2.13) in the form
µ 2 Fκ,λ,µ,0(z)= ωκ,λ (eσ, e1) (eσ, e1) eσ,αe1 + βen dσ Sn−1 | Z µ 2 2 2 2 = ωκ,λ (eσ, e1) (eσ, e1) α (eσ, e1) +2αβ(eσ, e1)(eσ, en)+β (eσ, en) dσ. (2.15) Sn−1 | Z Let us show that µ ωκ,λ (eσ, e1) (eσ, e1) (eσ, e1)(eσ, en)dσ =0 . (2.16) Sn−1 | | Z The last equality is obvious for the case n = 2. We suppose that n 3. We denote ≥ by ϑ1, ϑ2,...,ϑn−1 the spherical coordinates with the center at , where ϑi [0, π] for 1 i n 2, and ϑ [0, 2π]. Then for any σ =(σ ,...,σ ) OSn−1 we have∈ ≤ ≤ − n−1 ∈ 1 n ∈ σ1 = cos ϑ1,
σ2 = sin ϑ1 cos ϑ2, ......
σn−1 = sin ϑ1 ... sin ϑn−2 cos ϑn−1,
σn = sin ϑ1 ... sin ϑn−2 sin ϑn−1.
6 Using the equalities
(eσ, e1)= σ1 = cos ϑ1, (eσ, en)= σn = sin ϑ1 ... sin ϑn−2 sin ϑn−1 and n−2 n−3 dσ = sin ϑ1 sin ϑ2 ... sin ϑn−2 dϑ1dϑ2 ...dϑn−1, we calculate the integral on the left-hand side of (2.16):
µ ωκ,λ (eσ, e1) (eσ, e1) (eσ, e1)(eσ, en)dσ Sn−1 | | Z π π 2π n−2 = ... ω cos ϑ cos ϑ µ cos ϑ sinn−i ϑ sin ϑ dϑ ...dϑ dϑ κ,λ 1 | 1| 1 i n−1 1 n−2 n−1 0 0 0 i=1 ! Z Z Z Y π π n −2 2π n−i = Iκ,λ ... sin ϑi dϑ2...dϑn−2 sin ϑn−1dϑn−1 , (2.17) 0 0 i=2 ! 0 Z Z Y Z
where π I = ω cos ϑ cos ϑ µ cos ϑ sinn−1 ϑ dϑ . κ,λ κ,λ 1 | 1| 1 1 1 Z0 Since the inner integral in (2.17) is equal to zero, we arrive at (2.16). So, by (2.14), (2.15) and (2.16), we have
µ 2 2 2 2 Fκ,λ,µ,0(z)= ωκ,λ (eσ, e1) (eσ, e1) α (eσ, e1) +β (eσ, en) dσ max U, V , (2.18) Sn−1 | | ≤ { } Z where µ+2 U = ωκ,λ (eσ, e1) (eσ, e1) dσ (2.19) Sn−1 | | Z and µ 2 V = ωκ,λ (eσ, e1) (eσ, e1) (eσ, en) dσ. (2.20) Sn−1 | | Z In view of (2.7) and the evenness of ωκ,λ(u) in u, we can write (2.19) as
π/2 µ+2 n−2 U =2ωn−1 ωκ,λ(cos ϑ1) cos ϑ1 sin ϑ1dϑ1. Z0 π By the change of variable ϑ1 = 2 ϕ in the integral on the right-hand side of the last equality, we obtain −
π/2 µ+2 n−2 U =2ωn−1 ωκ,λ(sin ϕ) sin ϕ cos ϕdϕ . (2.21) Z0
7 Now, we calculate the integral on the right-hand side of (2.20):
µ 2 V = ωκ,λ (eσ, e1) (eσ, e1) (eσ, en) dσ Sn−1 | | Z π π 2π n−1 = ... ω (cos ϑ ) cos ϑ µ sinn+1−i ϑ dϑ ...dϑ dϑ κ,λ 1 | 1| i 1 n−2 n−1 0 0 0 i=1 ! Z Z Z Y π π π n−1 = ω (cos ϑ ) cos ϑ µ sinn ϑ dϑ 2 ... sinn+1−i ϑ dϑ ...dϑ .(2.22) κ,λ 1 | 1| 1 1 i 2 n−1 0 ( 0 0 i=2 ! ) Z Z Z Y π Putting ϑ1 = ϕ + 2 in the first integral on the right-hand side of (2.22), we arrive at equality π π/2 ω (cos ϑ ) cos ϑ µ sinn ϑ dϑ =2 ω (sin ϕ) sinµ ϕ cosn ϕdϕ . (2.23) κ,λ 1 | 1| 1 1 κ,λ Z0 Z0 Evaluating the multiple integral on the right-hand side of (2.22), we obtain
π π n−1 n−1 π/2 2 ... sinn+1−i ϑ dϑ ...dϑ =2 2n−2 sink ϑdϑ i 2 n−1 · 0 0 i=2 ! 0 Z Z Y kY=2 Z 2n−1 n−1 Γ k+1 Γ 1 2π(n−1)/2 ω = 2 2 = = n−1 , 2n−2 Γ k+2 (n 1)Γ n−1 n 1 k=2 2 2 Y − − which together with (2.22) and (2.23) leads to 2ω π/2 V = n−1 ω (sin ϕ) sinµ ϕ cosn ϕdϕ . (2.24) n 1 κ,λ − Z0 Let us show that U>V . Integrating by parts in (2.21), we have U 1 π/2 = ω (sin ϕ) sinµ+1 ϕd cosn−1 ϕ 2ω −n 1 κ,λ n−1 − Z0 1 π/2 = cosn−1 ϕd ω (sin ϕ) sinµ+1 ϕ n 1 κ,λ − Z0 1 π/2 d = cosn−1 ϕ (µ + 1) sinµ ϕ cos ϕω (sin ϕ) + sinµ+1 ϕ ω (sin ϕ) dϕ. n 1 κ,λ dϕ κ,λ − Z0 In view of (2.24), we can rewrite the last equality as U V 1 π/2 d = + cosn−1 ϕ sinµ+1 ϕ ω (sin ϕ) dϕ . (2.25) 2ω 2ω n 1 dϕ κ,λ n−1 n−1 − Z0 By definition (2.9) of the function ωκ,λ, we arrive at λ d κ 2 2κ cos ϕ π ω (sin ϕ)= e−κ/ sin ϕ > 0 for ϕ 0, , dϕ κ,λ sin2 ϕ sin3 ϕ ∈ 2 8 which together with (2.25) implies U > V. This, by (2.18) and (2.19), leads to the inequality
µ+2 max Fκ,λ,µ,0(z) ωκ,λ (eσ, e1) (eσ, e1) dσ . (2.26) |z|=1 ≤ Sn−1 | | Z By (2.7), the value of the integral
µ+2 ωκ,λ (eσ, e) (eσ, e) dσ Sn−1 | | Z is independent of e. Hence, by (2.26),
µ+2 max Fκ,λ,µ,0(z) ωκ,λ (eσ, e) (eσ, e) dσ . (2.27) |z|=1 ≤ Sn−1 | | Z The obvious lower estimate
µ+2 max Fκ,λ,µ,0(z) Fκ,λ,µ,0(e)= ωκ,λ (eσ, e) (eσ, e) dσ |z|=1 ≥ Sn−1 | | Z together with (2.27), leads to (2.12) for the case ν = 0. (ii) The case ν (0, 2). By (2.8), we rewrite (2.11) as ∈ ν 2−ν Fκ,λ,µ,ν(z)= ρκ,λ,µ (eσ, e) f (eσ, e) f (eσ, z) dσ . Sn−1 Z By part (i) of the proof,
2 max Fκ,λ,µ,0(z)= Fκ,λ,µ,0(e)= ρκ,λ,µ (eσ, e) f (eσ, e) dσ , |z|=1 Sn−1 Z which, by Corollary 1 with γ = 2, implies
2 max Fκ,λ,µ,ν(z)= ρκ,λ,µ (eσ, e) f (eσ, e) dσ . |z|=1 Sn−1 Z Last inequality combined with (2.8), proves (2.12) for any ν (0, 2). ∈ 3 Weighted estimate for solutions of the Dirichlet problem
Here we deal with a solution of the first boundary value problem for the heat equation: ∂u = a2∆u , (x, t) Rn (0, + ), ∂t ∈ + × ∞ u =0 , (3.1) t=0
u = f(x′, t) . xn=0 9 Here f Lp Rn−1 (0, + ) , 1 p , and u is represented by the heat double layer potential∈ × ∞ ≤ ≤ ∞ 2 − |x−y| t 4a2(t−τ) xn e ′ ′ u(x, t)= n/2 (n+2)/2 f(y , τ)dy dτ (3.2) 2 Rn−1 (t τ) 4a π Z0 Z − with y =(y′, 0),y′ Rn−1. The norm f was introduced in (1.3). ∈ k kp,t Rn Proposition 2. Let (x, t) be an arbitrary point in + (0, + ). The sharp coefficient (x, t) in the inequality × ∞ Wp u(x, t) (x, t) f (3.3) ∇x x ≤ Wp || ||p,t n
is given by
p−1 p cn,p n+p+2 p p p p(x, t)= max ωκ,λ (eσ, en) (eσ, en) −1 (eσ, z) −1 dσ , (3.4) 2+ n+1 W p |z|=1 Sn−1 | | | | xn Z where 1 1+ 1 2 p (4a2) p cn,p = n , (3.5) n −1 +1+ 1 π 2 q 2 p −1 −1 p + q =1, ωκ,λ(x) is defined by (2.9) and qx2 px2 (n + 4)q np +4 κ = n = n , λ = 2= . (3.6) 4a2t 4a2(p 1)t 2 − 2(p 1) − − In particular,
p−1 π/2 ∞ p c np+4 n+2(p+1) n,p 2(p−1) −ξ n−2 p(x, t)= 2ωn−1 ξ e dξ cos p−1 ϑ sin ϑdϑ (3.7) 2+ n+1 2 W p 0 qxn xn ( Z (Z 4a2t cos2 ϑ ) ) for 2 p . As≤ a special≤∞ case of (3.7) one has
2 π/2 ∞ 16a √π n/2 −ξ 2 n−2 ∞(x, t)= ξ e dξ cos ϑ sin ϑdϑ . (3.8) n−1 2 x2 W Γ xn 0 n 2 Z (Z 4a2t cos2 ϕ ) Proof. (i) General case. By (3.2),
2 − |x−y| t 4a2(t−τ) u(x, t) 1 e ′ ′ = n/2 (n+2)/2 f(y , τ)dy dτ . x 2 Rn−1 (t τ) n 4a π Z0 Z −
Differentiating with respect to xj, j =1,...,n, we obtain
t 2 u(x, t) 2π y x − |x−y| 4a2(t−τ) ′ ′ x = (n+2)/2 −(n+4)/2 e f(y , τ)dy dτ. ∇ x 2 Rn−1 (t τ) n 4a π Z0 Z −
10 Hence,
t 2 u(x, t) 2π y x exy, z − |x−y| z | − | 4a2(t−τ) ′ ′ x , = (n+2)/2 (n+4)/2 e f(y , τ)dy dτ, (3.9) ∇ x 2 Rn−1 (t τ) n 4a π Z0 Z −
where z is a unit n-dimensional vector and exy =(y x)/ y x . By (3.9), we conclude that the sharp coefficient (x, t) in inequality (3.3) is given− by| − | Wp 1/q t q q 2 2π y x exy, z − q|y−x| | − | | | 4a2(t−τ) ′ p(x, t)= n/2 max (n+4)q/2 e dy dτ . W 2 |z|=1 Rn−1 (t τ) 4a π (Z0 Z − ) We write the last equality in the form
2 1/q − q|y−x| q+n e z q t 4a2(t−τ) 2π y x xy, xn ′ e p(x, t)= max | − | | | dy dτ . (3.10) n/2 z n (n+4)q/2 W 4a2π | |=1Rn−1 xn y x 0 (t τ) Z | − | Z − Setting q x y 2 s = | − | , 4a2(t τ) − we represent the inner integral on the right-hand side of (3.10) as
q|y−x|2 − (n+4)q −1 t 4a2(t−τ) 2 2 ∞ e 4a (n+4)q −2 −s dτ = s 2 e ds. (3.11) (n+4)q/2 2 2 0 (t τ) q y x q|y−x| Z − | − | Z 4a2t By (2.9) and the equality y x e , e = x , (3.12) | − || xy n | n we write (3.11) as
q|y−x|2 (n+4)q −1 t − 2 2 2 2 e 4a (t−τ) 4a exy, en dτ = ω (e , e ) , (3.13) (t τ)(n+4)q/2 qx2 κ,λ xy n Z0 − n ! where κ and λ are defined by (3.6). In view of (3.12), we have
q+n x y x q+n = n , (3.14) | − | e , e | xy n |! which, in combination with (3.10) and (3.13), leads to
p−1 2 1+ 1 p 2(4a ) p n+p+2 p e e e e p−1 e z p−1 p(x, t)= n+1 max ωκ,λ ( σ, n) ( σ, n) ( σ, ) dσ , (3.15) n n 1 2+ z W −1 +1+ p | |=1 Sn−1 | | | | π 2 q 2 p xn (Z − )
11 where Sn−1 = σ Sn−1 :(e , e ) < 0 . − { ∈ σ n } Using the evenness of the integrand in (3.15) with respect to eσ, we obtain
p−1 1 2 1+ 1 2 p (4a ) p n+p+2 p p e e e e p−1 e z p−1 p(x, t)= n+1 max ωκ,λ ( σ, n) ( σ, n) ( σ, ) dσ , (3.16) n n 1 2+ z W −1 +1+ p | |=1 Sn−1 | | | | π 2 q 2 p xn Z which proves (3.4). (ii) The case p [2, ]. Solving the system ∈ ∞ p n + p +2 2 ν = , µ + ν = − p 1 p 1 − − with respect to ν and µ, we arrive at p 2 n +4 ν = − , µ = . p 1 p 1 − − So, µ> 0 for any p> 1 and ν [0, 1) for p 2. Applying Lemma 1 to (3.16), we conclude ∈ ≥ p−1 p cn,p n+2(p+1) p(x, t)= ωκ,λ (eσ, en) (eσ, en) p−1 dσ , (3.17) 2+ n+1 W p Sn−1 | | xn Z where p [2, ] and the constant cn,p is defined by (3.5). By (2.7) and (2.9), we write (3.17) as (3.7).∈ ∞
4 Estimate for solutions of the Neumann problem
Let us consider the Neumann problem for the heat equation: ∂u = a2∆u , (x, t) Rn (0, + ), ∂t ∈ + × ∞ u =0 , t=0 (4.1) ∂u = g(x′, t) ∂xn xn=0 p n−1 with g L R (0, + ) , 1 p . Here u is represented as the heat single layer potential∈ × ∞ ≤ ≤ ∞ 2 − |x−y| 2 t 4a2(t−τ) 2a e ′ ′ u(x, t)= n/2 n/2 g(y , τ)dy dτ , (4.2) − 2 Rn−1 (t τ) 4a π Z0 Z − where y =(y′, 0),y′ Rn−1. ∈ Proposition 3. Let (x, t) be an arbitrary point in Rn (0, + ). The sharp coefficient + × ∞ p(x, t) in the inequality N u(x, t) (x, t) g (4.3) |∇x |≤Np || ||p,t 12 is given by
p−1 p kn,p n−p+2 p e e e e p−1 e z p−1 p(x, t)= n+1 max ωκ,λ ( σ, n) ( σ, n) ( σ, ) dσ , (4.4) N p |z|=1 Sn−1 | | | | xn Z where 2(3−p)/pa2/p kn,p = , (4.5) n + 1 πn/2q 2 p −1 −1 p + q =1, ωκ,λ(x) is defined by (2.9) and qx2 px2 (n + 2)q (n 2)p +4 κ = n = n , λ = 2= − . (4.6) 4a2t 4a2(p 1)t 2 − 2(p 1) − − In particular,
p−1 p π/2 ∞ (n−2)p+4 kn,p −ξ n+2 n−2 (x, t)= 2ω ξ 2(p−1) e dξ cos p−1 ϑ sin ϑdϑ (4.7) p n+1 n−1 2 N p 0 qxn xn ( Z (Z 4a2t cos2 ϑ ) ) for 2 p (n + 4)/2. As≤ a special≤ case of (4.7) one has
1/2 b π/2 ∞ (x, t)= n ξne−ξdξ cosn+2 ϑ cosn−2 ϑdϑ , (4.8) 2 n+1 2 N 2 0 xn xn (Z (Z 2a2t cos2 ϑ ) ) where a bn = . n−1 n+1 2 4 n−1 2 π Γ 2 q Proof. (i) General case. Differentiating in (4.2) with respect to xj, j =1,...,n, we obtain
t 2 1 y x − |x−y| 4a2(t−τ) ′ ′ xu(x, t)= n/2 −(n+2)/2 e g(y , τ)dy dτ, ∇ − 2 Rn−1 (t τ) 4a π Z0 Z − which leads to
t 2 1 y x exy, z − |x−y| z | − | 4a2(t−τ) ′ ′ xu(x, t), = n/2 (n+2)/2 e g(y , τ)dy dτ, (4.9) ∇ − 2 Rn−1 (t τ) 4a π Z0 Z − where z is a unit n-dimensional vector and exy =(y x)/ y x . It follows from (4.9) that the sharp coefficient (x, t) in inequality (4.3) is given− by| − | Np 1/q t q q 2 1 y x exy, z − q|y−x| | − | | | 4a2(t−τ) ′ p(x, t)= n/2 max (n+2)q/2 e dy dτ . N 2 |z|=1 Rn−1 (t τ) 4a π (Z0 Z − )
13 Now, we write the last equality as
2 1/q − q|y−x| q+n e z q t 4a2(t−τ) 1 y x xy, xn ′ e p(x, t)= max | − | | | dy dτ . (4.10) n/2 z n (n+2)q/2 N 4a2π | |=1 Rn−1 xn y x 0 (t τ) Z | − | Z − Putting q x y 2 s = | − | , 4a2(t τ) − we represent the inner integral on the right-hand side of (4.10) in the form
q|y−x|2 − (n+2)q −1 t 4a2(t−τ) 2 2 ∞ e 4a (n+2)q −2 −s dτ = s 2 e ds. (4.11) (n+2)q/2 2 2 0 (t τ) q y x q|y−x| Z − | − | Z 4a2t By (2.9) and equality (3.12), we write (4.11) as follows
q|y−x|2 (n+2)q −1 t − 2 2 2 2 e 4a (t−τ) 4a exy, en dτ = ω (e , e ) , (4.12) (t τ)(n+2)q/2 qx2 κ,λ xy n Z0 − n ! where κ and λ are defined by (4.6). Using (3.14) and substituting (4.12) into (4.10), we arrive at the representation p−1 2 1/p p (4a ) n−p+2 p e e e e p−1 e z p−1 p(x, t)= n+1 max ωκ,λ ( σ, n) ( σ, n) ( σ, ) dσ , (4.13) n 1 z N n/2 + p | |=1 Sn−1 | | | | π q 2 p xn (Z − ) where Sn−1 = σ Sn−1 :(e , e ) < 0 . − { ∈ σ n } In view of the evenness of the integrand in (4.13) with respect to eσ, we obtain p−1 (3−p)/p 2/p 2 a n−p+2 p p e e e e p−1 e z p−1 p(x, t)= n max ωκ,λ ( σ, n) ( σ, n) ( σ, ) dσ , (4.14) n 1 +1 z N n/2 + p | |=1 Sn−1 | | | | π q 2 p xn Z which proves (4.4). (ii) The case p [2, (n + 4)/2]. Solving the system ∈ p n p +2 2 ν = , µ + ν = − − p 1 p 1 − − with respect to ν and µ, we obtain p 2 n 2p +4 ν = − , µ = − . p 1 p 1 − − Therefore, the conditions 0 ν < 2,µ 0 hold for p [2, (n +4)/2]. Applying Lemma 1 to (4.14), we arrive at ≤ ≥ ∈ p−1 p kn,p n+2 e e e e p−1 p(x, t)= n+1 ωκ,λ ( σ, n) ( σ, n) dσ , (4.15) N p Sn−1 | | xn Z where p [2, (n +4)/2] and the constant kn,p is defined by (4.5). In view of (2.7) and (2.9), we write∈ (4.15) as (4.7).
14 References
[1] G. Kresin and V. Maz’ya, Sharp Real-Part Theorems. A Unified Approach, Lect. Notes in Math., 1903, Springer, Berlin, 2007. [2] G. Kresin, V. Maz’ya. Sharp real-part theorems in the upper half-plane and similar estimates for harmonic functions. J. Math. Sci. (New York), 179:1 (2011), 144–163. [3] G. Kresin and V. Maz’ya, Maximum Principles and Sharp Constants for Solutions of Elliptic and Parabolic Systems, Math. Surveys and Monographs, 183, Amer. Math. Soc., Providence, Rhode Island, 2012. [4] G. Kresin and V. Maz’ya, Optimal estimates for derivatives of solutions to Laplace, Lam´eand Stokes equations, J. Math. Sci., New York, 196:3 (2014), 300–321. [5] G. Kresin, An extremal problem for integrals on a measure space with abstract parame- ters, Complex Analysis and Operator Theory, 11:7 (2017), 1477–1490. [6] G. Kresin and V. Maz’ya, Generalized Poisson integral and sharp estimates for har- monic and biharmonic functions in the half-space, Mathematical Modelling of Natural Phenomena, to appear.
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