Well-Posedness of the Cauchy Problem for Hyperbolic Equations with Non-Lipschitz Coefficients

Abstract

We consider hyperbolic equations with anisotropic elliptic part and some non-Lipschitz coefficients. We prove well-posedness of the corresponding Cauchy problem in some functional spaces. These functional spaces have finite smoothness with respect to variables corresponding to regular coefficients and infinite smoothness with respect to variables corresponding to singular coefficients.

1. Introduction

Let us consider the Cauchy problem for a second-order hyperbolic equation: 𝑢̈−^𝑛_𝑖,𝑗=1𝑎_𝑖𝑗(𝑡)𝑢_{𝑥𝑖𝑥𝑗}+^𝑛_𝑗=1𝑏_𝑗(𝑡)𝑢_𝑥𝑗+𝑐(𝑡)𝑢=0,(𝑡,𝑥)∈[0,𝑇]×𝑅^𝑛,(1.1)𝑢(0,𝑥)=𝑢₀(𝑥),𝑢_𝑡(0,𝑥)=𝑢₁(𝑥),𝑥∈𝑅^𝑛,(1.2) where the matrix (𝑎_𝑖𝑗(𝑡)) is real and symmetric for all 𝑡∈(0,𝑇], 𝑢̈=𝑢_𝑡𝑡.

Suppose that (1.1) is strictly hyperbolic, that is, there exists 𝜆₀>0 such that 𝑎(𝑡,𝜉)≡^𝑛_𝑖,𝑗=1𝑎_𝑖𝑗𝜉(𝑡)^𝑖𝜉_𝑗|𝜉|²≥𝜆₀>0,(1.3) for all (𝑡,𝜉)∈(0,𝑇]×𝑅^𝑛⧵{0}.

It is known that if 𝑎(𝑡,𝜉) satisfies the Lipschitz condition and 𝑏_𝑗(𝑡),𝑐(𝑡)∈𝐿_∞(0,𝑇),𝑗=1,2,…,𝑛, then for any 𝑢₀∈𝐻^𝑠(𝑅^𝑛),𝑢₁∈𝐻^𝑠−1(𝑅^𝑛) the problem (1.1), (1.2) has a unique solution 𝑢(⋅)∈𝐶([0,𝑇];𝐻^𝑠(𝑅^𝑛))∩𝐶¹([0,𝑇],𝐻^𝑠−1(𝑅^𝑛)),(1.4) where 𝑠≥1 (see [1, Chapter 5] and [2, Chapter 3]).

If we reject the Lipschitz condition, this result, generally speaking, stops to be valid (see [3]).

In the paper [4] it is proved that if 𝑎(𝑡,𝜉)∈𝐿𝐿_𝜔(0,𝑇), that is, if 𝑎(𝑡,𝜉) satisfies the logarithmic Lipschitz condition: |𝑎(𝑡+𝜏,𝜉)−𝑎(𝑡,𝜉)|≤𝑐|𝜏|⋅|log|𝜏||⋅𝜔(|𝜏|),(1.5) where 𝜔(|𝜏|) monotonically decreasing tends to zero, and log|𝜏|⋅𝜔(|𝜏|) tends to infinity, then there exists 𝛿>0 such that, for all 𝑢₀∈𝐻^𝑠(𝑅^𝑛), 𝑢₁∈𝐻^𝑠−1(𝑅^𝑛) the problem (1.1), (1.2) has a unique solution 𝑢∈𝐶([0,𝑇];𝐻^𝑠−𝛿(𝑅^𝑛))∩𝐶¹([0,𝑇],𝐻^{𝑠−1−𝛿}(𝑅^𝑛)) (this behavior goes under the name of loss of derivatives).

In the paper [5] it is considered the case when 𝑎_𝑖,𝑗(𝑡)=0,𝑖≠𝑗, a part of coefficients belongs to the class 𝐿𝐿_𝜔(0,𝑇), and another part of coefficients satisfies the Lipschitz condition. It is proved that the loss of derivatives occurs in those variables 𝑥_𝑘 for which appropriate coefficient 𝑎_𝑘𝑘(𝑡) belongs to the class 𝐿𝐿_𝜔(0,𝑇).

It is interesting to investigate the Cauchy problem for (1.1), with singular coefficients. Many interesting results have been obtained in this direction. For example, in the paper [6] it is supposed that for each 𝜉∈𝑅^𝑛⧵{0}𝑎(𝑡,𝜉)∈𝐶¹(0,𝑇] and 𝑡^𝑞|𝑎(̇𝑡,𝜉)|≤𝑐,(𝑡,𝜉)∈(0,𝑇]×𝑅^𝑛⧵{0},(1.6) where 𝑞≥1, 𝑐>0. It is proved that if 𝑞=1, the problem (1.1), (1.2) is well-posed in 𝐶^∞(𝑅^𝑛). If 𝑞>1 and 𝑡^𝑝|𝑎(𝑡,𝜉)|≤𝑐,(𝑡,𝜉)∈(0,𝑇]×𝑅^𝑛⧵{0},(1.7) where 𝑝∈[0,1)∩[0,𝑞−1), then the problem (1.1), (1.2) is well-posed in the Geverey class 𝛾^(𝑠)(𝑅^𝑛), 𝑠<(𝑞−𝑝)/(𝑞−1) (see [6]). If the coefficients 𝑎_𝑖𝑗(𝑡) satisfy only Holder conditions of order 𝛼<1 then in [3] it is established that the problem (1.1), (1.2) is 𝛾^(𝑠) well-posed for all 𝑠<1/(1−𝛼). In this direction see also the results obtained in the papers [6–13].

In this paper we consider the Cauchy problem for a higher-order hyperbolic equation with anisotropic elliptic part: 𝑢̈+^𝑛_𝑘=1(−1)^𝑙𝑘𝑎_𝑘(𝑡)𝐷^2𝑙𝑘_𝑥𝑘𝑢+_{|𝛼∶𝑙|≤1}𝑏_𝛼(𝑡)𝐷^𝛼_𝑥𝑢=0,(𝑡,𝑥)∈[0,𝑇]×𝑅^𝑛,𝑢(0,𝑥)=𝑢₀(𝑥),𝑢_𝑡(0,𝑥)=𝑢₁(𝑥),𝑥∈𝑅^𝑛,(1.8) where 𝑙_𝑘∈𝑁,{1,2,…,},𝛼_𝑘∈𝑁∪{0},𝑘=1,2,…,𝑛,|𝛼∶𝑙|=𝛼₁/𝑙₁+⋯+𝛼_𝑛/𝑙_𝑛.

Here the coefficients 𝑎_𝑘(𝑡) satisfy different conditions of type (1.6) and (1.7), so that 𝑞_𝑘 and 𝑝_𝑘 corresponding to different 𝑘 are different. The smoothness of the solution depending on smoothness on initial data with respect to each variable 𝑥_𝑘 depends not only on 𝑙_𝑘 but also on 𝑞_𝑘 and 𝑝_𝑘.

2. Statement of the Problem and Results

We considered the Cauchy problem (1.8). Suppose that 𝑎_𝑘(𝑡) and 𝑏_𝛼(𝑡) satisfy the following conditions: 𝑎_𝑘𝑡(𝑡)≥𝑎>0,𝑡∈[0,𝑇],𝑘=1,2,…,𝑛,(2.1)_𝑞𝑘|𝑎̇_𝑘𝑏(𝑡)|≤𝑐,𝑡∈(0,𝑇],𝑘=1,2,…,𝑛,(2.2)_𝛼(𝑡)∈𝐿_∞(0,𝑇),|𝛼∶𝑙|≤1.(2.3)

In order to formulate the basic results we introduce some denotation. Let 𝐻 be some Hilbert space. By 𝑊₂^𝜆,𝐿(𝑅^𝑚,𝐻) we will denote a functional space with the norm ‖𝑢‖_𝑊2^𝜆,𝐿_{(𝑅𝑚,𝐻)}=[_𝑅𝑚(1+^𝑚_𝑘=1𝜂^2𝐿𝑘_𝑘)^𝜆‖𝑢(𝜂)‖²_𝐻𝑑𝜂]^1/2,(2.4) where 𝐿=(𝐿₁,…,𝐿_𝑚),𝐿_𝑗∈𝑁,𝑗=1,2,…,𝑚,𝜆≥0, and 𝑢(𝜂)=𝐹_𝑥[𝑢](𝜂);𝐹_𝑥 is a Fourier transformation with respect to variable 𝑥∈𝑅^𝑛.

For 𝑠≥1 by 𝛾_𝛽^𝑠,𝐿(𝑅^𝑚,𝐻) we will denote a functional space with the norm ‖𝑢‖_𝛾𝛽^𝑠,𝐿_{(𝑅𝑚,𝐻)}=[exp{𝛽|^𝑚_𝑘=1𝜂²_𝑘|^1/𝑠}‖𝑢(𝜂)‖²_𝐻𝑑𝜂]^1/2.(2.5)

Denote 𝑊₂^𝜆,𝐿(𝑅^𝑚,𝑅)=𝑊₂^𝜆,𝐿(𝑅^𝑚),𝛾_𝛽^𝑠,𝐿(𝑅^𝑚,𝑅)=𝛾_𝛽^𝑠,𝐿(𝑅^𝑚), 𝐶^∞(𝑅^𝑚;𝐻)=_𝜆≥0𝑊₂^𝜆,𝐿(𝑅^𝑚;𝐻),𝛾^(𝑠)(𝑅^𝑚;𝐻)=_𝛽≥0𝛾_𝛽^(𝑠)(𝑅^𝑚;𝐻).(2.6)

If 𝐿=(1,…,1) then 𝑊₂^𝜆,𝐿(𝑅^𝑚,𝐻)=𝐻^𝜆(𝑅^𝑚;𝐻), 𝛾_𝛽^𝑠,𝐿(𝑅^𝑚,𝐻)=𝛾^𝑠_𝛽(𝑅^𝑚,𝐻), and 𝛾_𝛽^𝑠,𝐿(𝑅^𝑚,𝑅)=𝛾_𝛽^(𝑠), where 𝛾_𝛽^(𝑠) is the Geverey space of order 𝑠 (see [12, 13]). If 𝜆∈𝐻 then 𝑊₂^𝜆,𝐿(𝑅^𝑚,𝐻) is Hilbert-valued anisotropic Sobolev space 𝑊^{(𝜆𝐿1,…,𝜆𝐿𝑚)}₂(𝑅^𝑚;𝐻). For the read valued functions the anisotropic Sobolev spaces are stated in [14]. The basic results led in [14] are also valid for abstract-valued functions.

We introduce also the following denotation: 𝑥^′=(𝑥₁,…,𝑥_𝑛1),𝑥^′′=(𝑥_𝑛1+1,…,𝑥_𝑛𝜉),_′=(𝜉₁,…,𝜉_𝑛1),𝜉^′′=(𝜉_𝑛1+1,…,𝜉_𝑛𝑙),_′=(𝑙₁,…,𝑙_𝑛1),𝑙^′′=(𝑙_𝑛1+1,…,𝑙_𝑛),|𝜉|=^𝑛_𝑘=1𝜉^2𝑙𝑘_𝑘,|𝜉^′|_𝑙′=^𝑛1_𝑘=1𝜉^2𝑙𝑘_𝑘,|𝜉^′′|_𝑙′′=^𝑛_𝑘=𝑛1+1𝜉^2𝑙𝑘_𝑘,𝑛₂=𝑛−𝑛₁.(2.7)

The main results are the following theorems.

Theorem 2.1. Let the conditions (2.1)–(2.3) be satisfied, where 𝑞_𝑘∈[0,1),for𝑘=1,2,…,𝑛₁,𝑞(2.8)_𝑘=1,for𝑘=𝑛₁+1,…,𝑛.(2.9)
Then for any 𝜆^′≥0,𝜆^′′≥0 the energy estimates 𝐸(𝑡,𝜆^′,𝜆^′′)≤𝑀𝐸(0,𝜆^′,𝜆^′′+𝜆₀),(2.10) hold, where 𝑀 and 𝜆₀ are some constants indepent of 𝑡∈[0,𝑇], 𝐸(𝑡,𝜆^′,𝜆^′′+𝜆)=_𝑅𝑛(1+|𝜉^′|_𝑙′)^𝜆′(1+|𝜉^′′|_𝑙′′)^{𝜆′′+𝜆}[|𝑣(̇𝑡,𝜉)|²+(1+|𝜉|_𝑙)|𝑣(𝑡,𝜉)|²̇]𝑑𝜉,𝜆≥0,𝑣(𝑡,𝜉)=𝜕𝑣(𝑡,𝜉).𝜕𝑡(2.11)

Theorem 2.2. Let the conditions (2.1)–(2.3) be satisfied, where 𝑞_𝑘∈[0,1),for𝑘=1,2,…,𝑛₁,𝑞(2.12)_𝑘=𝑞>1,for𝑘=𝑛₁+1,…,𝑛.(2.13) Additionally, let the conditions 𝑡^𝑝|𝑎_𝑘(𝑡)|≤𝑐,𝑡∈[0,𝑇],for𝑘=𝑛₁+1,…,𝑛.(2.14) be satisfied, where 𝑝∈[0,1)∩[0,𝑞−1). Then for any 𝛽>0,𝜆^′≥0, and 1≤𝑠<(𝑞−𝑝)/(𝑞−1) the energy estimates, ℰ(𝑡,𝛽,𝑠,𝜆^′)≤𝑀ℰ(0,𝛽+𝛿,𝑠,𝜆^′),(2.15) hold, where 𝑀 and 𝛿 are some constants independent of 𝑡∈[0,𝑇], ℰ(𝑡,𝛽,𝑠,𝜆^′)=_𝑅𝑛exp{𝛽|𝜉^′′|_𝑙^1/𝑠′′}(1+|𝜉^′|_𝑙′)^𝜆′[|𝑣(̇𝑡,𝜉)|²+(1+|𝜉|_𝑙)|𝑣(𝑡,𝜉)|²]𝑑𝜉.(2.16)

Remark 2.3. It is clear by our notation that 𝐸(𝑡,𝜆^′,𝜆^′′)≤‖𝑢̇(𝑡,⋅)‖_{𝑊𝜆2′′,𝑙′′(𝑅𝑛2𝑥′′;𝑊𝜆′′2+1,𝑙(𝑅𝑛1𝑥′))}+‖𝑢(𝑡,⋅)‖_{𝑊𝜆2′′,𝑙′′(𝑅𝑛2𝑥′′;𝑊𝜆′′2+1,𝑙(𝑅𝑛1𝑥′))}+‖𝑢(𝑡,⋅)‖_{𝑊𝜆2′′+1,𝑙′′(𝑅𝑛2𝑥′′;𝑊𝜆′′2,𝑙(𝑅𝑛1𝑥′))}≤2𝐸(𝜆^′,𝜆^′′,𝑡),(2.17) and we can write ℰ(𝑡,𝛽,𝑠,𝜆^′)=‖𝑢(𝑡,⋅)‖_{𝛾𝛽𝑠,𝑙′′(𝑅𝑛2𝑥′′;𝑊𝜆′′2,𝑙(𝑅𝑛1𝑥′))}.(2.18)

Remark 2.4. It is possible to replace the conditions 𝑎₁(𝑡),…,𝑎_𝑛1(𝑡)∈𝐶¹(0,𝑇] and (2.8) or (2.12) by Lipschitz conditions.

The following theorems are obtained from Theorems 2.1 and 2.2.

Theorem 2.5. Let condition (2.1)–(2.9) be satisfied. Then for any 𝑠≥0,𝑢₀∈𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^{𝑠+1,𝑙′}₂(𝑅^𝑛1_𝑥′)), 𝑢₁∈𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^𝑠,𝑙′₂(𝑅^𝑛1_𝑥′)) the problem (1.1), (1.2) admits a unique solution 𝑢∈𝐶([0,𝑇];𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^{𝑠+1,𝑙′}₂(𝑅^𝑛1_𝑥′)))∩𝐶¹([0,𝑇];𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^𝑠,𝑙′₂(𝑅^𝑛1_𝑥′))).(2.19)

Theorem 2.6. Let conditions (2.1)–(2.3) and (2.12)–(2.14) be satisfied. Then for any 𝑠^′≥0,1≤𝑠^′′<(𝑞−𝑝)/(𝑞−1),𝑢₀∈𝛾^𝑠′′(𝑅^𝑛2_𝑥′′;𝑊^{𝑠′+1,𝑙′}₂(𝑅^𝑛1_𝑥′)), 𝑢₁∈𝛾^𝑠′′(𝑅^𝑛2_𝑥′′;𝑊^{𝑠′,𝑙′}₂(𝑅^𝑛1_𝑥′)) the problem (1.1), (1.2) admits a unique solution 𝑢∈𝐶([0,𝑇];𝛾^𝑠′′(𝑅^𝑛2_𝑥′′;𝑊^{𝑠′+1,𝑙′}₂(𝑅^𝑛1_𝑥′)))∩𝐶¹([0,𝑇];𝛾^𝑠′′(𝑅^𝑛2_𝑥′′;𝑊^{𝑠′,𝑙′}₂(𝑅^𝑛1_𝑥′))).(2.20)

In particular it follows from Theorem 2.1 that if the conditions (2.1)–(2.3) are satisfied, then the problem (1.1), (1.2) is well-posed in 𝐶^∞(𝑅^𝑛), and if the conditions (2.1)–(2.3) and (2.12)–(2.14) are satisfied then the problem (1.1), (1.2) is well-posed in the Geverey class 𝛾^(𝑠).

3. Proof of Theorems

At first we reduce some auxiliary statements.

We denote 𝑣(𝑡,𝜉)=𝐹_𝑥[𝑢](𝑡,𝜉) and define the weighted energetic function in the following way: Φ(𝑡)=Φ(𝑡,𝜉,𝜆^′,𝜆^′′,𝛽,𝑟)=[|𝑣(̇𝑡,𝜉)|²+(1+|𝜉^′|_𝑙′+𝑑(𝑡,𝜉^′′))|𝑣(𝑡,𝜉)|²]⋅𝐻(𝑡,𝜉),(3.1) where 𝐻(𝑡,𝜉)=𝐻(𝑡,𝜉,𝜆^′,𝜆^′′,𝛽,𝑟)=(1+|𝜉^′|_𝑙′)^𝜆′(1+|𝜉^′′|_𝑙′′)^𝜆′′×exp[−^𝑡₀𝛼(𝜏,𝜉^′′)𝑑𝜏+𝛽|𝜉^′′|_𝑙^{(𝑞−1)/𝑟}_′′],𝜆^′≥0,𝜆^′′≥0,𝛽>0,𝑟=𝑠(𝑞−1),for𝑞>1,1,for𝑞=1,𝑑(𝑡,𝜉^′′)^{𝑛𝑘=𝑛1+1}𝑎_𝑘(𝑇)𝜉^2𝑙𝑘_𝑘,for𝑇^𝑟|𝜉^′′|_𝑙′′≤1,_{𝑛𝑘=𝑛1+1}𝑎_𝑘(|𝜉^′′|_𝑙^{−1/𝑟′′})𝜉^2𝑙𝑘_𝑘,for𝑇^𝑟|𝜉^′′|_𝑙′′>1,𝑡^𝑟|𝜉^′′|_𝑙′′≤1,_{𝑛𝑘=𝑛1+1}𝑎_𝑘(𝑡)𝜉^2𝑙𝑘_𝑘,for𝑡^𝑟|𝜉^′′|_𝑙′′>1,𝛼(𝑡,𝜉^′′)|𝑑(𝑡,𝜉^′′)−^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2𝑙𝑘_𝑘|,for𝑡^𝑟|𝜉^′′|_𝑙′′|∑≤1,_{𝑛𝑘=𝑛1+1}𝑎̇_𝑘(𝑡)𝜉^2𝑙𝑘_𝑘|∑^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2𝑙𝑘_𝑘,for𝑡^𝑟|𝜉^′′|_𝑙′′>1.(3.2)

The following auxiliary lemmas are proved similar to the paper [6]. The proofs of the lemmas are in appendix.

Lemma 3.1. If 𝑞_𝑘=1,𝑘=𝑛₁+1,…,𝑛, then there exits such 𝑐₁>0,𝑐₂>0, that 𝑎|𝜉^′′|_𝑙′′≤𝑑(𝑡,𝜉^′′)≤[𝑐₁+𝑐₂ln(1+|𝜉^′′|_𝑙′′)]|𝜉^′′|_𝑙′′.(3.3)
If 𝑞_𝑘>1,𝑘=𝑛₁+1,…,𝑛, then there exits such 𝑐₁>0,𝑐₂>0, that 𝑎|𝜉^′′|_𝑙′′≤𝑑(𝑡,𝜉^′′)≤[𝑐₁+𝑐₂|𝜉^′′|_𝑙^{𝑝/𝑟′′}]|𝜉^′′|_𝑙′′.(3.4)

Lemma 3.2. If 𝑞_𝑘=1,𝑘=1,2,…,𝑛₁, then there exits such constant 𝑐₃>0,𝛾>0, that ∫^𝑡₀𝛼(𝜏,𝜉)𝑑𝜏≤𝑐₃+𝑐₄ln(1+|𝜉^′′|_𝑙′′).
If 𝑞_𝑘>1,𝑘=1,2,…,𝑛₁ then there exits such 𝑐₃>0,𝑐₄>0, that ^𝑡₀𝛼(𝜏,𝜉)𝑑𝜉≤𝑐₃+𝑐₄|𝜉^′′|_𝑙^{(𝑞−1)/𝑟′′}.(3.5)

By the definition of Φ(𝑡)=Φ(𝑡,𝜉,𝜆^′,𝜆^′′,𝛽,𝑟) we have 𝑑Φ(𝑡)̈𝑑𝑡=2Re[𝑣(𝑡,𝜉)𝑣(̇𝑡,𝜉)+(1+|𝜉^′|_ℓ′+𝑑(𝑡,𝜉^′′))𝑣(𝑡,𝜉)̇+̇𝑣(𝑡,𝜉)]𝐻(𝑡,𝜉)𝑑(𝑡,𝜉^′′)|𝑣(𝑡,𝜉)|²𝐻(𝑡,𝜉)−𝛼(𝑡,𝜉)Φ(𝑡).(3.6)

On the other hand from (1.8) we have 𝑣(̈𝑡,𝜉)+^𝑛_𝑘=1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘𝑣(𝑡,𝜉)+_{|𝛼∶ℓ|≤1}𝑏_𝛼(𝑡)(𝑖𝜉)^𝛼𝑣(𝑡,𝜉)=0,(3.7)𝑣(0,𝜉)=𝑣₀(𝜉),𝑣(̇0,𝜉)=𝑣₁(𝜉),(3.8) where 𝑣₀(𝜉)=𝐹[𝑢₀](𝜉),𝑣₁(𝜉)=𝐹[𝑢₁](𝜉),𝑣(̈𝑡,𝜉)=𝜕²𝑣(𝑡,𝜉)/𝜕𝑡².

From (3.6) and (3.7) we obtain 𝑑Φ(𝑡)𝑑𝑡=2Re[−^𝑛1_𝑘=1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘+(1+|𝜉^′|_ℓ′)+(𝑑(𝑡,𝜉^′′)−^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘)]×𝑣(𝑡,𝜉)𝑣(̇𝑡,𝜉)𝐻(𝑡,𝜉)−2Re_{|𝛼∶ℓ|≤1}𝑏_𝛼(𝑡)(𝑖𝜉)^𝛼𝑣(𝑡,𝜉)̇+̇𝑣(𝑡,𝜉)𝐻(𝑡,𝜉)𝑑(𝑡,𝜉^′′)|𝑣(𝑡,𝜉)|²𝐻(𝑡,𝜉)−𝛼(𝑡,𝜉)Φ(𝑡).(3.9)

If 𝑡^𝑟|𝜉^′′|<1, then by definition of 𝑑(𝑡,𝜉) and 𝛼(𝑡,𝜉^′′) we have 𝑑Φ(𝑡)𝑑𝑡=2Re[−^𝑛1_𝑘=1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘+(1+|𝜉^′|_ℓ′)+𝛼(𝑡,𝜉)]𝑣(𝑡,𝜉)𝑣(̇𝑡,𝜉)𝐻(𝑡,𝜉)−2Re_{|𝛼∶ℓ|≤1}𝑏_𝛼(𝑡)(𝑖𝜉)^𝛼𝑣(𝑡,𝜉)𝑣(̇𝑡,𝜉)𝐻(𝑡,𝜉)−𝛼(𝑡,𝜉)Φ(𝑡).(3.10)

By our supposition 𝑞_𝑘<1 for𝑘=1,2,…,𝑛₁. Therefore we can easily see that 𝑎≤𝑎_𝑘(𝑡)≤𝑎_𝑇,𝑘=1,2,…,𝑛₁(3.11) with some constant 𝑎_𝑇>𝑎.

Using the Cauchy inequality, definition of 𝛼(𝑡,𝜉), 𝐻(𝑡,𝜉), and 𝜑(𝑡) we have

2Re𝛼(𝑡,𝜉)𝑣(𝑡,𝜉)𝑣(̇𝑡,𝜉)𝐻(𝑡,𝜉)−𝛼(𝑡,𝜉)Φ(𝑡)≤0,(3.12)2Re_{|𝛼∶ℓ|≤1}𝑏_𝛼(𝑡)(𝑖𝜉)^𝛼𝑣(𝑡,𝜉)𝑣(̇𝑡,𝜉)𝐻(𝑡,𝜉)≤2𝑏_{𝑇|𝛼∶𝑙|≤1}|𝜉^𝛼||𝑣(𝑡,𝜉)|⋅|𝑣(̇𝑡,𝜉)|⋅𝐻(𝑡,𝜉)≤2𝑏_𝑇𝑐₅[(1+^𝑛_𝑘=1|𝜉_𝑘|^2𝑙𝑘)|𝑣(𝑡,𝜉)|²+|𝑣(̇𝑡,𝜉)|²]𝐻(𝑡,𝜉),(3.13) where 𝑏_𝑇=sup_{|𝛼∶𝑙|≤1}‖𝑏_𝛼(𝑡)‖_{𝐿∞(0,𝑇)},𝑐₅=sup_{𝜉∈𝑅𝑛}(∑_{|𝛼∶ℓ|≤1}|𝜉^𝛼|)²/(∑^𝑛_𝑘=1|𝜉_𝑘|^2𝑙𝑘+1).

From (3.10)–(3.13) we get that when 𝑡^𝑟|𝜉^′′|_𝑙′′<1, then there exists such a constant 𝑀₁>0, that 𝑑Φ(𝑡)𝑑𝑡≤𝑀₁Φ(𝑡).(3.14) If 𝑡^𝑟|𝜉^′′|_𝑙′′≥1 then by definition of 𝑑(𝑡,𝜉) and 𝛼(𝑡,𝜉^′′) from (3.9) we have that 𝑑Φ(𝑡)𝑑𝑡=2Re[−^𝑛1_𝑘=1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘−_{|𝛼∶ℓ|≤1}𝑏_𝛼(𝑡)(𝑖𝜉)^𝛼𝑣(𝑡,𝜉)̇+𝑣(𝑡,𝜉)]𝐻(𝑡,𝜉)_{𝑛𝑘=𝑛1+1}𝑎̇_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘|𝑣(𝑡,𝜉)|²𝐻(𝑡,𝜉)−|∑^𝑛_𝑘=𝑛1+1𝑎̇_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘|∑^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘Φ(𝑡).(3.15) On the other hand 𝑎̇^𝑛_{𝑘=𝑛1+1𝑘}(𝑡)𝜉^2ℓ𝑘_𝑘|𝑣(𝑡,𝜉)|²|∑𝐻(𝑡,𝜉)−^{𝑛𝑘=𝑛1+1}𝑎̇_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘|∑^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘=Φ(𝑡)_{𝑛𝑘=𝑛1+1}𝑎̇_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘|𝑣(𝑡,𝜉)|²𝐻(𝑡,𝜉)−|∑^𝑛_𝑘=𝑛1+1𝑎̇_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘|∑^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘×[|𝑣(̇𝑡,𝜉)|²+(1+|𝜉^′|²_𝑙^′+^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2ℓ𝑘_𝑘)|𝑣(𝑡,𝜉)|²]𝐻(𝑡,𝜉)≤0.(3.16) From (3.13), (3.15), and (3.16) we again get inequality (3.14).

It follows from (3.14) that Φ(𝑡)≤𝑀Φ(0),𝑡∈[0,𝑇],(3.17) where 𝑀=𝑀₁𝑒^𝑇.

Proof of Theorem 2.1. Let 𝑞_𝑘=𝑞=1,𝑘=𝑛₁+1,…,𝑛, then 𝑟=1. From Lemmas 3.1 and 3.2 we have _𝑅𝑛Φ(𝑡,𝜉,𝜆^′,𝜆^′′,0,1)𝑑𝜉≤_𝑅𝑛(1+|𝜉^′|_ℓ′)^𝜆′(1+|𝜉^′′|_ℓ′′)^𝜆′′×[|𝑣(̇𝑡,𝜉)|²+(1+|𝜉^′|²_𝑙^′+|𝜉^′′|_𝑙′′(𝑐₁+𝑐₂ln(1+|𝜉^′′|_𝑙′′)))]|𝑣(𝑡,𝜉)|²×exp(𝑐₃+𝑐₄ln(1+|𝜉^′′|_𝑙′′))𝑑𝜉≤𝑐_6𝑅𝑛(1+|𝜉^′|_ℓ′)^𝜆′(1+|𝜉^′′|_ℓ′′)^{𝜆′′+𝑐2}[|𝑣(̇𝑡,𝜉)|²+(1+|𝜉|_ℓ)|𝑣(𝑡,𝜉)|²]𝑑𝜉=𝑐₆𝐸(𝑡,𝜆^′,𝜆^′′+𝜆₀),(3.18) where 𝜆₀=𝑐₄+1, 𝑐₆=max{1,𝑐₁𝑒^𝑐3,𝑐₂𝑒^𝑐3}.
Thus, _𝑅𝑛Φ(𝑡,𝜉,𝜆^′,𝜆^′′,0,1)𝑑𝜉≤𝑐₆𝐸(𝑡,𝜆^′,𝜆^′′+𝜆₀).(3.19)
On the other hand from the definition of Φ and 𝐸 we have _𝑅𝑛Φ(𝑡,𝜉,𝜆^′,𝜆^′′,0,1)𝑑𝜉≥𝑐₇𝐸(𝑡,𝜆^′,𝜆^′′).(3.20) It follows from (3.17)–(3.20) that 𝐸(𝑡,𝜆^′,𝜆^′′)≤𝑐₈𝐸(0,𝜆^′,𝜆^′′+𝑑).(3.21)

Proof of Theorem 2.2. Let 𝑞_𝑘=𝑞>1,𝑘=𝑛₁+1,…,𝑛, then 𝑟=(𝑞−1)𝑠. Taking into account Lemmas 3.1 and 3.2 and Theorem 2.5 we have _𝑅𝑛Φ(𝑡,𝜉,𝜆^′≤,0,𝛽,𝑟)𝑑𝜉_𝑅𝑛(1+|𝜉^′|_ℓ)^𝜆′[|𝑣(𝑡,𝜉)|²+(1+|𝜉^′|_𝑙′+𝑐₁|𝜉^′′|_𝑙′+𝑐₂|𝜉^′′|^𝑝/2+1)]×|𝑣(𝑡,𝜉)|²exp(𝑐₃+(𝛽+𝑐₄)|𝜉|^(𝑞−1)/2)𝑑𝜉.(3.22)
Further using the inequality 𝜂^𝑝/2+1≤𝑐₉exp(𝑐𝜂^1/𝑠) we obtain _𝑅𝑛Φ(𝑡,𝜉,𝜆^′,0,𝛽,𝑟)𝑑𝜉≤𝑐₁₀ℰ(𝑡,𝜆^′,𝑠,𝛽+𝛿),(3.23) where 𝛿=𝑐₄+𝑐.
On the other hand from the definition of 𝜙 and ℰ we have _𝑅𝑛Φ(𝑡,𝜉,𝜆^′,0,𝛽,𝑟)𝑑𝜉≥𝑐₁₁ℰ(𝑡,𝜆^′,𝑠,𝛽).(3.24)
From inequalities (3.17), (3.23), and (3.24) it follows that ℰ(𝑡,𝜆^′,𝑠,𝛽)≤𝑐₁₂ℰ(0,𝜆^′,𝑠,𝛽+𝑑).(3.25)

Proof of Theorem 2.5. For any 𝜉∈𝑅^𝑛 the problem (3.7), (3.8) has a unique solution 𝑣(𝑡,𝜉)∈𝐶¹[0,𝑇] (see [15, Chapter I]).
Let 𝑢₀∈𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^{𝜆′+1,𝑙′}₂(𝑅^𝑛1_𝑥′)), 𝑢₁∈𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^{𝜆′,𝑙′}₂(𝑅^𝑛1_𝑥′)), then for any 𝑠≥0,𝜆^′≥0, 𝐸(0,𝜆^′,𝑠+𝜆₀)≤𝑐_𝑠,𝜆′,(3.26) where 𝑐_𝑠,𝜆>0 is some constant dependent on 𝑠≥0 and 𝜆^′≥0.
Taking into account Theorem 2.1 it follows from (3.20) that 𝐸(𝑡,𝜆^′,𝑠)≤𝑀𝑐_𝜆′,𝑠,𝑡∈[0,𝑇],(3.27) that is, ‖𝑢̇(𝑡,⋅)‖_{𝑊2𝑠,𝑙′′(𝑅𝑛2𝑥′′;𝑊𝜆′′2,𝑙(𝑅𝑛1𝑥′))}+‖𝑢(𝑡,⋅)‖_{𝑊2𝑠+1,𝑙′′(𝑅𝑛2𝑥′′;𝑊𝜆′′2,𝑙(𝑅𝑛1𝑥′))}+‖𝑢(𝑡,⋅)‖_{𝑊2𝑠,𝑙′′(𝑅𝑛2𝑥′′;𝑊𝜆′′2+1,𝑙(𝑅𝑛1𝑥′))}≤𝑀𝑐_𝜆′,𝑠,𝑡∈[0,𝑇].(3.28) It follows from (3.28) that 𝑢∈𝐶([0,𝑇];𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^{𝜆′+1,𝑙′}₂(𝑅^𝑛1_𝑥′̇))),𝑢∈𝐶([0,𝑇];𝐶^∞(𝑅^𝑛2_𝑥′′;𝑊^{𝜆′,𝑙′}₂(𝑅^𝑛1_𝑥′))).(3.29)
By the expression of 𝑢(𝑡,𝑥) it follows that the function 𝑢(𝑡,𝑥) is the solution of problem (1.8).
The uniqueness of the solution is proved by standard method.

The proof of Theorem 2.6 is carried out in the similar way.

Appendices

A. Proof of Lemmas

Proof of Lemma 3.1. Let 𝑞_𝑘=1,𝑘=𝑛₁+1,…,𝑛. Then from (2.2) we have 𝑎_𝑘(𝑡)≤𝑎_𝑘(𝑇)+|𝑎_𝑘(𝑡)−𝑎_𝑘(𝑇)|≤𝑎_𝑘(𝑇)+^𝑇_𝑡|𝑎̇_𝑘(𝑠)|𝑑𝑠≤𝑎_𝑘𝑇(𝑇)+𝑐ln𝑡≤𝑐₁+𝑐₂1ln(1+𝑡).(A.1) It follows from (2.1) and (2.14) that 𝑎|𝜉^′′|_𝑙′′≤𝑑(𝑡,𝜉^′′).(A.2)
By definition of 𝑑(𝑡,𝜉^′′) for 𝑇|𝜉^′′|_𝑙′′≤1 we have 𝑑(𝑡,𝜉^′′)=^𝑛_𝑘=𝑛1𝑎_𝑘(𝑇)𝜉^2𝑙𝑘_𝑘≤𝑐₁|𝜉^′′|_𝑙′′.(A.3) If 𝑇|𝜉^′′|_𝑙′′>1, and 𝑡|𝜉^′′|_𝑙′′<1, then from (A.1) we have 𝑑(𝑡,𝜉^′′)=^𝑛_𝑘=𝑛1𝑎_𝑘(|𝜉^′′|_𝑙^−1′′)𝜉^2𝑙𝑘_𝑘≤[𝑐₁+𝑐₂1ln(1+|𝜉^′′|_𝑙^−1′′)]^𝑛_𝑘=𝑛1𝜉^2𝑙𝑘_𝑘=(𝑐₁+𝑐₂ln(1+|𝜉^′′|_𝑙′′))|𝜉^′′|_𝑙′′.(A.4) If 𝑡|𝜉^′′|_𝑙′′>1, then using (A.1) we get 𝑑(𝑡,𝜉^′′)=^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2𝑙𝑘_𝑘≤[𝑐₁+𝑐₂1ln(1+𝑡)]|𝜉^′′|_𝑙′′≤[𝑐₁+𝑐₂ln(1+|𝜉^′′|_𝑙′′)]|𝜉^′′|_𝑙′′.(A.5) Consequently if 𝑞=1, the statement of the lemma follows from (A.2)–(A.5).
Let 𝑞_𝑘>1,𝑘=𝑛₁+1,…,𝑛. By definition of 𝑑(𝑡,𝜉^′′) for 𝑇^𝑟|𝜉^′′|_𝑙′′<1 we have 𝑑(𝑡,𝜉^′′)≤𝑐₁|𝜉^′′|_𝑙′′.(A.6)
If 𝑇^𝑟|𝜉^′′|_𝑙′′>1 and 𝑡^𝑟|𝜉^′′|<1 then 𝑑(𝑡,𝜉^′′)=^𝑛_𝑘=𝑛1+1𝑎_𝑘(|𝜉^′′|_𝑙^{−1/𝑟′′})𝜉^2𝑙𝑘_𝑘≤^𝑛_𝑘=𝑛1+1𝑀(|𝜉^′′|_𝑙^{−1/𝑟′′})^𝑝𝜉^2𝑙𝑘_𝑘=𝑀|𝜉^′′|^1+p_𝑙^/𝑟′′.(A.7)
If 𝑡^𝑟|𝜉^′′|>1 then 𝑑(𝑡,𝜉^′′)=^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝑡)𝜉^2𝑙𝑘_𝑘≤^𝑛_𝑘=𝑛1+1𝑀𝑡^𝑝𝜉^2𝑙𝑘_𝑘=𝑀|𝜉^′′|_𝑙′′⋅|𝜉^′′|^𝑝/𝑟=𝑀|𝜉^′′|^1+𝑝/𝑟.(A.8)
Thus if 𝑞_𝑘>1,𝑘=𝑛₁+1,…,𝑛 then the statement of the lemma follows from (A.2), (A.6), and (A.8).
The lemma is proved.

Proof of Lemma 3.2. At first we consider the case when 𝑞_𝑘=1,𝑘=𝑛₁+1,…,𝑛. If 𝑇|𝜉^′′|_ℓ′′≤1, then ^𝑡₀𝛼(𝜏,𝜉^′′)𝑑𝜏≤^𝑇₀𝛼(𝜏,𝜉^′′≤)𝑑𝜏_𝑇0|^𝑛_𝑘=𝑛1𝑎_𝑘(𝑇)𝜉^2ℓ𝑘_𝑘−^𝑛_𝑘=𝑛1𝑎_𝑘(𝜏)𝜉^2ℓ𝑘_𝑘≤|𝑑𝜏_{𝑛𝑘=𝑛1}𝜉^2ℓ𝑘_𝑘^𝑇₀|𝑎_𝑘(𝑇)−𝑎_𝑘(𝜏)|𝑑𝜏≤𝑇⋅max_{𝑘=𝑛1+1,…,𝑛}𝑎_𝑘(𝑇)|𝜉^′′|_ℓ′′+|𝜉_𝑘^′′|_ℓ′′^𝑇₀𝑎_𝑘(𝜏)𝑑𝜏≤𝑎_𝑇,(A.9) where 𝑎_𝑇=max_{𝑘=𝑛1+1,…,𝑛}𝑎_𝑘(𝜏)+(1/𝑇)max_{𝑘=𝑛1+1,…,𝑛}∫^𝑇₀𝑎_𝑘(𝜏)𝑑𝜏<+∞.
If 𝑇|𝜉^′′|>1, then ^𝑡₀𝛼(𝜏,𝜉^′′)𝑑𝑠≤^{|𝜉′′|𝑙−1′′}₀𝛼(𝑠,𝜉^′′)𝑑𝜏+^𝑇_{|𝜉′′|ℓ−1′′}𝛼(𝑠,𝜉^′′≤)𝑑𝑠_{|𝜉′′|𝑙−1′′0}|𝑑(𝜏,𝜉^′′)−^𝑛_𝑘=𝑛1𝑎_𝑘(𝜏)𝜉^2𝑙𝑘_𝑘|𝑑𝜏+^𝑇_{|𝜉′′|ℓ−1′′}|∑^𝑛_𝑘=𝑛1𝑎̇_𝑘(𝜏)𝜉^2ℓ𝑘_𝑘|∑^𝑛_𝑘=𝑛1𝑎_𝑘(𝜏)𝜉^2ℓ𝑘_𝑘≤𝑑𝜏_{|𝜉′′|𝑙−1′′0}𝑑(𝜏,𝜉^′′)𝑑𝜏+^𝑛_𝑘=𝑛1𝜉^2𝑙𝑘_𝑘⋅^{|𝜉′′|𝑙−1′′}₀𝛼_𝑘𝑐(𝜏)𝑑𝜏+𝑎^𝑛_𝑘=𝑛1𝜉^2ℓ𝑘_𝑘^𝑇_{|𝜉′′|ℓ−1′′}𝑑𝜏𝜏≤^{|𝜉′′|ℓ−1′′}₀[𝑐₁+𝑐₂ℓ𝑛(1+|𝜉^′′|_ℓ′′)]|𝜉^′′|_ℓ′′+𝑑𝜏_{𝑛𝑘=𝑛1}𝜉^2ℓ𝑘_𝑘⋅𝑐^{|𝜉′′|ℓ−1′′}₀𝑇ℓ𝑛𝜏𝑐𝑑𝜏+𝑎^𝑛_𝑘=𝑛1^𝑇_|𝜉^′′_|ℓ⁻¹_′′𝑑𝜏𝜏=𝑐₁+𝑐₂ℓ𝑛(1+|𝜉^′′|_ℓ^−1′′)+𝑐|𝜉^′′|_ℓ′′⋅^{|𝜉′′|ℓ−1′′}₀𝑇ℓ𝑛𝜏𝑐𝑑𝜏+𝑎^𝑇_|𝜉^′′_|ℓ⁻¹_′′𝑑𝜏𝜏≤𝑐₃+𝑐₄ℓ𝑛(1+|𝜉^′′|_ℓ′′).(A.10)
Now let us consider the case 𝑞_𝑘>1,𝑘=𝑛₁+1,…,𝑛. In this case 𝑟=(𝑞−1)𝑠. If 𝑇^𝑟|𝜉^′′|_ℓ′′≤1, then ^𝑡₀𝛼(𝜏,𝜉^′′)𝑑𝜏≤^𝑇₀𝛼(𝜏,𝜉^′′≤)𝑑𝜏_{𝑛𝑘=𝑛1+1𝑇0}|𝑎_𝑘(𝑇)−𝑎_𝑘(𝜏)|𝜉^2ℓ𝑘_𝑘𝑑𝜏≤max_{𝑘=𝑛1+1,…,𝑛}𝑎_𝑘(𝑇)𝑇|𝜉^′′|_ℓ′′+^𝑇₀𝑐𝜏^−𝑝𝑑𝜏|𝜉^′′|_ℓ′′≤𝑎_𝑇⋅𝑇^1−𝑟1+𝑐⋅𝑇1−𝑝^1−𝑝|𝜉^′′|_ℓ′′≤𝑎_𝑇𝑇^1−𝑟+𝑐𝑇1−𝑝^{1−𝑝−𝑟}.(A.11) If 𝑇^𝑟|𝜉′′|_ℓ′′>1, then ^𝑡₀𝛼(𝜏,𝜉^′′)𝑑𝜏≤^{|𝜉′′|−1/𝑟}₀𝛼(𝜏,𝜉)𝑑𝜏+^𝑇_{|𝜉′′|−1/rℓ′′}𝛼(𝜏,𝜉^′′)𝑑𝜏≤^{|𝜉′′|−1/rℓ′′}₀|𝑑(𝜏,𝜉)−^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝜏)𝜉^2𝑙𝑘_𝑘|𝑑𝜏+^𝑇_{|𝜉′′|−1/rℓ′′}𝛼(𝜏,𝜉^′′≤)𝑑𝜏_{𝑛𝑘=𝑛1+1}𝑎_𝑘(|𝜉^′′|^−1/r_ℓ^′′)𝜉^2ℓ𝑘_𝑘^{|𝜉′′|−1/r𝑙′′}₀𝑑𝜏+^𝑛_𝑘=𝑛1+1𝜉^2ℓ𝑘_𝑘^{|𝜉′′|−1/rℓ′′}₀𝑎_𝑘+(𝜏)𝑑𝜏_{𝑇|𝜉′′|−1/rℓ′′}|∑^𝑛_𝑘=𝑛1+1𝑎̇_𝑘(𝜏)𝜉^2ℓ𝑘_𝑘|∑^𝑛_𝑘=𝑛1+1𝑎_𝑘(𝜏)𝜉^2ℓ𝑘_𝑘≤𝑐𝑑𝜏(|𝜉^′′|^−1/r_ℓ^′′)^𝑝⋅|𝜉^′′|_ℓ′′⋅^{|𝜉′′|−1/rℓ′′}₀𝑑𝜏+𝑐|𝜉^′′|_ℓ′′⋅^{|𝜉′′|−1/𝑟}₀𝑑𝜏𝜏^𝑝𝑐𝑑𝜏+𝑎^𝑇_|𝜉^′′_|^−1/𝑟𝑑𝜏𝜏^𝑞≤𝑐|𝜉^′′|_ℓ^{𝑝/𝑟+1′′}⋅|𝜉^′′|^−1/𝑟+𝑐|𝜉^′′|_ℓ′′⋅11−𝑝(|𝜉^′′|^−1/𝑟)^1−𝑝+𝑐𝑎11−𝑞(𝑇^1−𝑞−(|𝜉^′′|^−1/𝑟)^1−𝑞)<𝑐|𝜉^′′|_ℓ^{1−((1−𝑝)/𝑟)′′}+𝑐1−𝑝|𝜉^′′|_ℓ^{1−((1−𝑝)/𝑟)′′}+𝑐𝑎(𝑞−1)|𝜉^′′|^{(𝑞−1)/𝑟}.(A.12)
As 𝑟=(𝑞−1)𝑠, and 𝑠<(𝑞−𝑝)/(𝑞−1), it follows that 1−(1−𝑝)/𝑠<1/𝑠 and (𝑞−1)/𝑟=1/𝑠. Then according to the Young inequality there exists such 𝛿>0 that |𝜉^′′|^{1−((1−𝑝)/𝑟)}≤𝑐₁𝛿+𝛿₁|𝜉^′′|^1/𝑠.(A.13) Thus, by (A.9)–(A.13) the following inequality is valid: ^𝑡₀𝛼(𝜏,𝜉^′′)𝑑𝜏≤𝛿|𝜉|^1/𝑠+𝑐_𝛿,(A.14) where 𝛿=𝛿₁𝑎(2+𝑝)/(1−𝑝)+(𝑐/𝑎(𝑞−1))𝑐_𝛿=𝑐_1𝛿𝑐(2+𝛿)/(1−𝑝).

B. Example

Let us consider the Cauchy problem in [0,𝑇)×𝑅²: 𝑢_𝑡𝑡−(1+𝑡²)𝑢_𝑥𝑥−(1+√³𝑡²)𝑢_𝑦𝑦=0,𝑢(0,𝑥,𝑦)=𝜑₁(𝑥)𝜓₁𝑢(𝑦),_𝑡(𝑜,𝑥,𝑦)=𝜑₂(𝑥)𝜓₂(𝑦),(B.1) where 𝜑₁(𝑥),𝜑₂(𝑥)∈𝐶^∞(𝑅)=∩_𝑠≥0𝑊^𝑠₂(𝑅),𝜓₁(𝑦)∈𝑊²₂(𝑅),𝜓₂(𝑦)∈𝑊¹₂(𝑅),𝑢=𝑢(𝑡,𝑥,𝑦).

It follows from Theorem 2.5 that the problem (B.1) has a unique solution 𝑢∈𝐶([0,𝑇];𝐶^∞(𝑅;𝑊²₂(𝑅)))∩𝐶¹([0,𝑇];𝐶^∞(𝑅;𝑊¹₂(𝑅))).(B.2)