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−1 ′−1 for solutions of the Cauchy problem (1.3) is obtained, where p + p =1, (x, t) is an arbi- Rn+1 trary point in the layer T and ℓ stands for a unit n-dimensional vector. As a consequence of (1.4), the sharp coefficient p(t) = max p,ℓ(t) (1.6) K |ℓ|=1 K in the inequality u(x, t) (t) ϕ (1.7) |∇x |≤Kp k kp is found. In particular, |||A−1/2||| ect (t)= , (1.8) K∞ √π t1/2 where |||B||| = max|ℓ|=1 Bℓ is the spectral norm of the real-valued matrix B. It is known (e.g. | | 1/2 ′ [8], sect. 6.3) that |||B||| = λB , where λB is the spectral radius of the matrix B B. Here the symbol ′ denotes passage to the transposed matrix. As a special case of (1.8) one has 1 ect (t)= K∞ √aπ t1/2 for A = aI, where I is the unit matrix of order n. In Section 4 we consider a solution of the Cauchy problem ∂u n ∂2u n ∂u Rn+1 = ajk + bj + cu + f(x, t) in T , ∂t ∂xj ∂xk ∂xj  j,k=1 j=1 (1.9)  X X   u , t=0 =0  where A = ((a )) is a symmetric positive definite matrix of order n and f Lp(Rn+1)  jk ∈ T ∩ Cα Rn+1 with p > n + 2 and α (0, 1). By Cα Rn+1 we denote the space of functions T ∈ T f x, t Rn+1 ( ) which are continuous and bounded in T and locally H¨older continuous with expo- Rn p Rn+1 nent α in x , uniformly with respect to t [0, T ]. The space L T is endowed with the norm ∈ ∈ T 1/p  p f p,T = f(x, τ) dxdτ k k Rn Z0 Z  for 1 p< , and ≤ ∞ f = ess sup f(x, τ) : x Rn, τ (0, T ) . k k∞,T {| | ∈ ∈ } The explicit formula for the sharp coefficient 1/p′ p′+1 −1/2 t p′cτ A ℓ Γ 2 e (t)= dτ (1.10) p,ℓ 1/p ′(n+p′)/2 (n(p′−1)+p′)/2 C 2nπ(n+p −1)/2 det A1/2  p 0 τ   Z    3  in the inequality ∂u (x, t) (t) f (1.11) ∂ℓ ≤ Cp,ℓ k kp,t

for solutions of the Cauchy problem (1.9) is found, where (x, t) Rn+1. As a consequence T of (1.10), we arrive at the formula for the sharp coefficient ∈

p(t) = max p,ℓ(t) (1.12) C |ℓ|=1 C in the inequality u(x, t) (t) f . (1.13) |∇x | ≤ Cp k kp,t For instance, |||A−1/2||| t ecτ (t)= dτ . (1.14) C∞ √π √τ Z0 In the particular case A = aI, formula (1.14) takes the form

1 t ecτ (t)= dτ . C∞ √aπ √τ Z0 It can be of interest that the sharp coefficients in (1.4) and (1.10) do not depend on the coefficient vector b =(b1,...,bn).

2 Explicit formulas for solutions

By (x, y) we mean the inner product of the vectors x and y in Rn. The Schwartz class of rapidly decreasing C∞-functions on Rn will be denoted by (Rn). Since the matrices A and A−1 are positive definite, thereS exist positive definite matrices 2 2 A1/2 and A−1/2 of order n such that A1/2 = A and A−1/2 = A−1 (e.g. [8], sect. 2.14). We start with the Cauchy problem (1.3) for the homogeneous equation. The assertion below can be proved analogously to the correspondin g statement for the heat equation (e.g.[3], Th. 5.4; [9], Sect. 4.8; [11], Sect. 7.4). In the proof of this assertion we preserve only the formal arguments leading to the representation for a solution of problem (1.3). Lemma 1. Let ϕ (Rn). A solution u of problem (1.3) is given by ∈ S u(x, t)= G(x y, t)ϕ(y)dy , (2.1) Rn − Z where ct 2 e − 1 A−1/2(x+tb) G(x, t)= e 4t . (2.2) 2√πt n det A1/2 | | 

4 Proof. Applying the Fourier transform 1 u ξ, t e−i(x,ξ)u x, t dx ˆ( )= n/2 ( ) (2.3) (2π) Rn Z to the Cauchy problem (1.3), we obtain duˆ = (Aξ,ξ)+ i(b, ξ)+ c uˆ , uˆ(ξ, 0)=ϕ ˆ(ξ) . (2.4) dt −  The solution of problem (2.4) is

uˆ(ξ, t)=ϕ ˆ(ξ)e{−(Aξ,ξ))+i(b,ξ)+c}t . (2.5)

By the inverse Fourier transform 1 u x, t ei(x,ξ)u ξ, t dξ, ( )= n/2 ˆ( ) (2π) Rn Z we deduce from (2.5) 1 u x, t ei(x,ξ)ϕ ξ e{−(Aξ,ξ))+i(b,ξ)+c}tdξ ( )= n/2 ˆ( ) (2π) Rn Z ect ei(x,ξ)e{−(Aξ,ξ))+i(b,ξ)}t e−i(y,ξ)ϕ y dy dξ = n ( ) (2π) Rn Rn Z Z  ect ei(x−y+tb,ξ)e−(tAξ,ξ)dξ ϕ y dy. = n ( ) (2.6) (2π) Rn Rn Z Z  Let us denote ect G x, t e−(tAξ,ξ)+i(x+tb,ξ)dξ. ( )= n (2.7) (2π) Rn Z The known formula (e.g.[13], Ch. 2, Sect. 9.7)

n/2 −(Mξ,ξ)+i(ζ,ξ) π − 1 (ζ,M −1ζ) e dξ = e 4 , Rn √det M Z where M is a symmetric positive definite matrix of order n, together with (2.7) leads to

ct e − 1 A−1(x+tb),x+tb G(x, t)= e 4t ( ) . (2.8) (2√πt)n√det A

Since (A−1ζ,ζ)=(A−1/2A−1/2ζ,ζ)=(A−1/2ζ, A−1/2ζ) = A−1/2ζ 2 for any ζ Rn and 2 | | ∈ det A = det A1/2 , we can write (2.8) as (2.2). It follows from (2.6) and (2.7) that the solution of problem (1.3) can be represented as (2.1), where G(x, t) is given by (2.2). 

5 Remark 1. Changing the variable ξ = A−1/2(x + tb) in the integral

2 − 1 A−1/2(x+tb) e 4t | | dx , Rn Z we arrive at the equality

ct 2 e − 1 A−1/2(x+tb) ct G( , t) 1 = G(x, t)dx = e 4t dx = e , (2.9) n 1/2 | | k · k Rn 2√πt det A Rn Z Z which generalizes the analogous fact for the heat equation. In particular, (1.2) follows from (2.9). The next assertion can be proved on the base of Lemma 1 similarly to the analogous statement for the heat equation (e.g.[3], Th. 5.5; [11], Sect. 7.4). p Rn Rn+1 R Proposition 1. Suppose that 1 p and ϕ L ( ). Define u : T by (2.1), where G is given by (2.2). Then ≤u(x,≤ t) ∞is solution∈ of the equation →

∂u n ∂2u n ∂u = a + b + cu ∂t jk ∂x ∂x j ∂x j k j=1 j j,kX=1 X in Rn+1. If 1 p< , then u( , t) ϕ in Lp as t 0+. T ≤ ∞ · → → Remark 2. It is known (e.g. [1], Ch.1, Sect. 6, 7 and 9) that (2.1), where G is given by (2.2), represents a unique bounded solution of the Cauchy problem (1.3) and u(x, t) ϕ(x) as t 0+ at any x Rn under assumption that ϕ belongs to C(Rn) L∞(Rn). This→ fact together→ with the Lusin’s∈ theorem (see [14], Ch. VI, Sect. 6) implies that∩ u( , t) ϕ almost everywhere in Rn as t 0+, where u is a solution of problem (1.3) with ϕ · L∞→(Rn). → ∈ Further, let us consider the Cauchy problem (1.9) for the nonhomogeneous equation. The next statement can be proved analogously to the similar assertion for the heat equation (e.g. [9], Sect. 4.8). In the proof of this assertion we preserve only the formal arguments leading to the representation for a solution of problem (1.9). Lemma 2. Let f( , t) (Rn) for any t [0, T ] and let the quantities C in the estimates · ∈ S ∈ α,m (1 + x m) ∂αf(x, t) C | | | x | ≤ α,m are independent of t for any integer m 0 and multiindex α. The solution of problem (1.9) is given≥ by

t u(x, t)= G(x y, t τ)f(y, τ)dydτ , (2.10) Rn − − Z0 Z where G is defined by (2.2).

6 Proof. Applying the Fourier transform to the Cauchy problem (1.9), we obtain duˆ = (Aξ,ξ)+ i(b, ξ)+ c uˆ + fˆ(ξ, t) , uˆ(ξ, 0)=0 . (2.11) dt −  The solution of problem (2.11) is

t uˆ(ξ, t) = e{−(Aξ,ξ))+i(b,ξ)+c}t fˆ(ξ, τ)e{(Aξ,ξ))−i(b,ξ)−c}τ dτ 0 t Z = fˆ(ξ, τ)e{−(Aξ,ξ))+i(b,ξ)+c}(t−τ)dτ . (2.12) Z0 By the inverse Fourier transform in (2.12), we have

1 t u x, t fˆ ξ, τ e{−(Aξ,ξ))+i(b,ξ)+c}(t−τ)dτ ei(x,ξ)dξ ( ) = n/2 ( ) (2π) Rn Z Z0  1 t ei(x,ξ)fˆ ξ, τ e{−(Aξ,ξ))+i(b,ξ)+c}(t−τ)dξ dτ = n/2 ( ) (2π) Rn Z0 Z  1 t ei(x,ξ) e−i(y,ξ)f y, τ dy e{−(Aξ,ξ))+i(b,ξ)+c}(t−τ)dξ dτ = n ( ) (2π) Rn Rn Z0 Z Z   t ec(t−τ) ei(x−y+ib(t−τ),ξ)e−(Aξ,ξ)(t−τ)dξ f y, τ dydτ. = n ( ) (2.13) Rn (2π) Rn Z0 Z  Z  By (2.7), expression inside of the braces in the right-hand side of (2.13) is equal to G(x y, t τ). So, equality (2.13) can be written as (2.10), where G is given by (2.2). − − Remark 3. It is known (e.g. [1], Ch.1, Sect. 7 and 9) that formula (2.10), where G is defined by (2.2), solves problem (1.9) with f Cα Rn+1 . ∈ T u x, t f Cα Rn+1 x, t Rn+1 Let ( ) be a solution of problem (1.9) with  T and ( ) T . Then, by (2.9) and (2.10), we arrive at the sharp pointwise∈ estimates ∈  ect 1 u(x, t) − f , | | ≤ c k k∞,t where c = 0, and 6 u(x, t) t f , | | ≤ k k∞,t where c = 0. The last estimate is well known for solutions of the Cauchy problem with zero initial data for the nonhomogeneous heat equation (e.g. [13], Ch. III, Sect. 16).

3 Estimates for solutions of the homogeneous equation

In this section we consider the Cauchy problem (1.3). Here we suppose that ϕ Lp(Rn), where p [1, ]. ∈ ∈ ∞ 7 Rn+1 Theorem 1. Let (x, t) be an arbitrary point in T and u be solution of problem (1.3). The sharp coefficient p,ℓ(t) in inequality (1.5) is given by (1.4). As a consequence,K the sharp coefficient (t) in inequality (1.7) is given by Kp 1/p′ p′+1 −1/2 ct |||A ||| Γ 2 e t . p( )= 1/p ′ (3.1) K 2nπ(n+p−1)/2 det A1/2  p′(n+p )/2  t(n+p)/(2p)   As a special case of(3.1) one has (1.8).   Proof. By (2.1), (2.2) and (2.8), we have ∂u ∂ = G(x y, t)ϕ(y)dy , (3.2) ∂x Rn ∂x − j Z j where

ct 2 ct e − 1 A−1/2(x+tb) e − 1 A−1(x+tb),x+tb G(x, t)= e 4t = e 4t ( ) . (3.3) 2√πt n det A1/2 | | (2√πt)n det A1/2

Differentiating in (3.3) with respect to xj, j =1,...,n, we obtain

ct 2 ∂ e −1 − 1 A−1/2(x−y+tb) G(x y, t)= A (x y+tb) e 4t , (3.4) n 1/2 j | | ∂xj − −2t 2√πt det A −  which together with (3.2), leads to

ct 2 ∂u e −1 − 1 A−1/2(x−y+tb) = A (x y+tb),ℓ e 4t ϕ(y)dy. (3.5) n 1/2 | | ∂ℓ −2t 2√πt det A Rn − Z  Applying the H¨older inequality to the right-hand side of (3.5), we conclude that the sharp coefficient in estimate (1.5) is given by

1 ct ′ ′ 2 p′ e −1 p − p A−1/2(x−y+tb) p,ℓ(t)= A (x y+tb),ℓ e 4t dy . (3.6) n 1/2 | | K 2t 2√πt det A Rn − Z   Further, we introduce the new variable

ξ = A−1/2(x y + tb) . − Since y = A1/2ξ + x + tb, we have dy = det A1/2dξ, which together with (3.6) leads to the following representation−

′ 1 ct 1/2 1/p ′ ′ 2 p′ e (det A ) −1/2 p − p |ξ| p,ℓ(t)= A ξ,ℓ e 4t dξ . n 1/2 K 2t 2√πt det A Rn Z   

8 By the symmetricity of A−1/2, we have

1 ct ′ ′ 2 p′ e −1/2 p − p |ξ| p,ℓ(t)= ξ, A ℓ e 4t dξ . (3.7) n 1/2 1/p K 2t 2√πt (det A ) Rn Z   Passing to the spherical coordinates in (3.7), we obtain

1/p′ ct ∞ ′ 2 ′ e p′+n−1 − p ρ −1/2 p p,ℓ(t)= ρ e 4t dρ eσ, A ℓ dσ , (3.8) n 1/2 1/p K 2t 2√πt (det A ) Sn−1 Z0 Z   where eσ is the n-dimensional unit vector joining the origin to a point σ of the unit sphere Sn−1 in Rn. −1/2 Let ϑ be the angle between eσ and A ℓ. We have

′ ′ π/2 −1/2 p −1/2 p p′ n−2 eσ, A ℓ dσ =2ωn−1 A ℓ cos ϑ sin ϑdϑ Sn−1 Z Z0  (n−1)/2 p′+1 ′ ′ ′ 2π Γ −1/2 p p +1 n 1 −1/2 p 2 = ωn−1 A ℓ B , − = A ℓ ′ . (3.9) 2 2 Γ n+p    2

Further, making the change of variable ρ = √u in the integral 

∞ ′ 2 p′+n−1 − p ρ ρ e 4t dρ Z0 and using the formula (e.g. [2], 3.381, item 4)

∞ xα−1e−βxdx = β−αΓ(α) (3.10) Z0 with positive α and β, we obtain

p′+n ∞ ′ 2 ∞ ′ ′ 2 ′ p′+n−1 − p ρ 1 p +n −1 − p u 1 4t n + p ρ e 4t dρ = u 2 e 4t du = Γ . (3.11) 2 2 p′ 2 Z0 Z0     Combining (3.9) and (3.11) with (3.8), we arrive at (1.4). Formula (3.1) follows from (1.4) and (1.6). As a particular case of (3.1) with p = , we obtain (1.8). ∞

4 Estimates for solutions of the nonhomogeneous equation

In this section we consider the Cauchy problem (1.9) for the nonhomogeneous equation. Here we suppose that f Lp(Rn+1) Cα Rn+1 , where p > n + 2 and α (0, 1). ∈ T ∩ T ∈ 

9 Rn+1 Theorem 2. Let (x, t) be an arbitrary point in T and let u solve problem (1.9). The sharp coefficient p,ℓ(t) in inequality (1.11) is given by (1.10). As a consequenceC of (1.10), the sharp coefficient (t) in inequality (1.13) is given by Cp 1/p′ p′+1 −1/2 t p′cτ |||A ||| Γ 2 e (t)= dτ . (4.1) p 1/p ′(n+p′)/2 (n(p′−1)+p′)/2 C 2nπ(n+p−1)/2 det A1/2  p 0 τ   Z  As a special case of (4.1) one has (1. 14).   Proof. By (3.4) and (2.10), we have ∂u t = xG(x y, t τ),ℓ f(y, τ)dydτ, (4.2) ∂ℓ Rn ∇ − − Z0 Z where 

c(t−τ) 2 e − 1 A−1/2(x−y+(t−τ)b) G(x y, t τ),ℓ = c A−1(x y+(t τ)b),ℓ e 4(t−τ) . ∇x − − − n (t τ)(n+2)/2 − − | | −   Here 1 c = . (4.3) n 2n+1πn/2 det A1/2 Applying the H¨older inequality to the right-hand side of (4.2), we conclude that the sharp coefficient (t) in the estimate (1.11) is given by Cp,ℓ 1 2 ′ t p′c(t−τ) p′ A−1/2(x−y+(t−τ)b) p e p′ − t c A−1 x y t τ b ,ℓ e | 4(t−τ) | dydτ . p,ℓ( )= n (n+2)p′/2 ( +( ) ) C Rn − − (Z0 Z t τ ) −  Changing the variable τ = t η , we rewrite the previous representation for (t) as − Cp,ℓ 1 ′ ′ −1/2 2 p′ t p cη ′ p A (x−y+ηb) e −1 p − t c A x y ηb ,ℓ e | 4η | dydη . p,ℓ( )= n (n+2)p′/2 ( + ) (4.4) C Rn η − (Z0 Z )  Now, we introduce the new variable ξ = A−1/2(x y)+ ηb − in the inner integral in the right-hand side of (4.4). By the symmetricity of A−1/2 and by the relation dy = det A1/2dξ, we arrive at

1 ′ ′ ′ t cp η ′ p′|ξ|2 p 1/2 1/p e −1/2 p − ℓ t c A ξ, A ℓ e 4η dξdη . p, ( )= n det (n+2)p′/2 (4.5) C Rn η Z0 Z    Passing to the spherical coordinates in (4.5), we obtain

1 ′ ′ ′ t cp η ∞ ′ 2 ′ p 1/2 1/p e n+p′−1 − p ρ −1/2 p ℓ t c A ρ e 4η dρ dη e , A ℓ dσ . p, ( )= n det (n+2)p′/2 σ C η Sn−1 Z0 Z0  Z    10 The last equality together with (3.9), (3.11) and (4.3) leads to

1/p′ p′+1 −1/2 t p′cτ A ℓ Γ 2 e (t)= dτ , (4.6) p,ℓ 1/p ′(n+p′)/2 (n(p′−1)+p′)/2 C 2nπ(n+p −1)/2 det A1/2  p 0 τ   Z  where the integral converges for p > n + 2. Thus, formula (1.10) is proved.  Equality (4.1) follows from (4.6) and (1.12). Putting p = in (4.1), we arrive at (1.14). ∞

Acknowledgement. The publication has been prepared with the support of the ”RUDN University Program 5-100”.

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