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On the structure and lifespan of smooth solutions for the two-dimensional hyperbolic geometric flow equation

  • Yinxia Wang , Yuzhu Wang EMAIL logo and Feiyan Zhao
Published/Copyright: August 29, 2025

Abstract

In this article, we investigate the initial value problem for the hyperbolic geometric flow equation, which is derived from hyperbolic geometric flow for Riemannian metric. We find that some of the quadratic nonlinear terms satisfy the null condition by rearranging the equation. Based on the ghost weight and standard energy estimate, the L 1 L estimate of solutions, and the bootstrap argument, lifespan of smooth solutions is established under the assumption of small data.

MSC 2010: 35L72; 35L15

1 Introduction

The hyperbolic geometric flow equation takes the following form:

(1.1) v t t Δ ln v = 0 ,

which comes from hyperbolic geometric flow introduced by Kong and Liu [18]. Let M be an n -dimensional complete Riemannian manifold with Riemannian metric g i j . The hyperbolic geometric flow is characterized by the following evolutionary equation for the metric g i j :

(1.2) 2 g i j t 2 = 2 R i j ,

where R i j stands for the Ricci curvature tensor of g i j . For the study on the hyperbolic geometric flow, we refer to the recent articles [5,6,1723].

On Riemann surface, all of the information about curvature is contained in the scalar curvature function R ; then, the hyperbolic geometric flow equation (1.2) can be simplified. Let R = 2 K and K be the Gauss curvature. The Ricci curvature is given by

(1.3) R i j = 1 2 R g i j ,

and then, the hyperbolic geometric flow equation (1.2) simplifies the following equation for the special metric:

(1.4) 2 g i j t 2 = R g i j .

The metric for a surface can always be written (at least locally) in the following form:

(1.5) g i j = v ( t , x , y ) δ i j ,

where v ( t , x , y ) > 0 . Therefore, we have

(1.6) R = Δ ln v v .

(1.1) is immediately derived by combining (1.4)–(1.6).

Equation (1.1) is a two-dimensional quasilinear wave equation if there exists C 0 > 0 such that C 0 1 v C 0 . Since the 1990s, global existence and lifespan of smooth solutions to the Cauchy problem for two-dimensional quasilinear wave equations with small initial data have been attracted many researcher’s interests, we refer to [1,3,9,2427,30]. If the nonlinear term on the right depends on the unknown function, lifespan of smooth solutions was established in [2426]. The null conditions play very important roles in studying global existence and lifespan of smooth solutions to the Cauchy problem for two-dimensional quasilinear wave equations. Global smooth solutions were obtained by Li and Zhou [27] if the equation at least has cubic nonlinearity, which satisfies the null condition. Zha [30] proved global existence of smooth solutions if both quadratic and cubic nonlinearities satisfy the null condition. The compact support condition is necessary to obtain sharp lifespan in all the aforementioned results because the nonlinear term depends on the unknown function itself.

If the nonlinear term on the right is dependent on the unknown function, Alinhac [1] proved global existence of smooth solutions with compact support small initial data under the assumption that both quadratic and cubic nonlinearities satisfy the null condition. Cai et al. [3] made full use of the nonlinear structure to obtain global smooth solutions with non-compactly supported small initial data. Hou and Yin [9] proved that the global existence of smooth solutions to the general 2-D null-form wave equations with non-compactly supported initial data.

Let

u = ln v ,

then (1.1) can be transformed to

(1.7) u t t e u Δ u = u t 2 .

In this article, we are concerned on the initial value problem for two-dimensional hyperbolic geometric flow equation (1.7) with

(1.8) t = 0 : u = u 0 ( x ) , u t = u 1 ( x ) , x R 2 ,

where ( t , x ) = ( t , x 1 , x 2 ) , and u = u ( t , x ) is the unknown function. u 0 ( x ) and u 1 ( x ) are given smooth functions.

To investigate the intrinsic structure of (1.7), it can be rearranged as

(1.9) u t t Δ u = [ ( e u 1 ) u ] + e u ( u 2 u t 2 ) + ( e u 1 ) u t 2 .

We find that e u ( u 2 u t 2 ) satisfies the null condition, which was introduced in [4,15]. In fact, let

(1.10) U = U ( s ) = U ( χ 0 t + χ 1 x 1 + χ 2 x 2 )

be any plane wave solution to two-dimensional linear homogeneous wave equation

(1.11) u = 0 .

Here, χ = ( χ 0 , χ 1 , χ 2 ) is a constant vector, and

(1.12) U ( 0 ) = U ( 0 ) = 0 .

It follows from (1.10) and (1.11) that

( χ 0 2 χ 1 2 χ 2 2 ) U ( s ) = 0 .

Noting that (1.12), then U is a nontrivial plane wave solution to linear wave equation, it suffices that the vector χ = ( χ 0 , χ 1 , χ 2 ) satisfies

χ 0 2 χ 1 2 χ 2 2 = 0 .

It is easy to know that

e U ( χ 1 2 + χ 2 2 χ 0 2 ) U = 0 .

Therefore, the nonlinear term e u ( u 2 u t 2 ) in the quadratic nonlinearity term satisfies the null condition.

Some results on smooth solutions to two-dimensional hyperbolic geometric flow equation have been established in [20] and [21]. But the structure of (1.7) such as the null condition in partial quadratic nonlinearity not been fully excavated and used. Therefore, our main aim of this article is to excavate the structure of (1.7) and then obtain sharp estimate for lifespan of smooth solutions.

We have the following theorem concerning the lifespan of smooth solutions for problems (1.7) and (1.8).

Theorem 1.1

Suppose that u 0 and u 1 are the smooth functions such that u 0 ( x ) = u 1 ( x ) = 0 , when x 1 , and let K 8 . Then, there exist constants ε 0 > 0 and C > 0 , such that if

a K ( a u 0 L 2 + a u 1 L 2 ) = ε ε 0 ,

then problems (1.7) and (1.8) admit a unique smooth solution u with the lifespan

(1.13) T ε C ε 2 .

Remark 1.2

There are also twofolds in this article. First, on the one hand, we find that the quadratic nonlinearity e u ( u 2 u t 2 ) satisfy the null condition by rearranging the equation; on the other hand, the quadratic nonlinearity [ ( e u 1 ) u ] has divergence form, which is useful for the energy estimate and L estimate. Second, we may directly use Alinhac’s ghost weight energy method to obtain the L 2 estimate for u since the appearance of the null form.

Remark 1.3

To obtain the sharp estimate for smooth solutions, the compact support condition on the initial data that is precisely caused by the nonlinear term that depends on the unknown function u is necessary.

Remark 1.4

The reason the lifespan of the smooth solution obtained in Theorem 1.1 is sharp lies in that the time decay of solutions to the two-dimensional linear wave equation is t 1 2 .

Key points We first observe that Alinhac’s ghost weight method can be applied to the hyperbolic geometric flow equation since one of the quadratic nonlinearities satisfies the null condition. Second, the L 2 -type norm of u itself cannot be bounded by the natural wave energy, and we make estimate for the corresponding linear equation using the energy method in the Fourier space. Third, in view of the appearance of u in the nonlinear term and its L 2 norm is growth with respect to time, we need to resort to the Hardy-type inequality to overcome the difficulty.

Main ideas In view of the null condition and divergence structure in the quadratic nonlinearity, we obtain the L 2 estimate for u = ( t u , u ) by combining Klainerman’s vector field and Alinhac’s ghost weight energy method. But the nonlinear term of the equation depends on the unknown function u itself, and we have to make estimate for the unknown function u itself. Unfortunately, we cannot obtain the L 2 estimate for u by applying the energy method to the equation itself, then we consider the corresponding linearized equation of the nonlinear equation by the energy method in the Fourier space and establish the L 2 estimate for u . The result is the logarithmic growth with respect to time, and we refer to Lemma 3.1 for the linear estimate and Lemma 4.2 for the nonlinear estimate. In additional, to close the energy, the L estimate for u and u is necessary. The L estimate for u is simple by the L 2 estimate for u and Klainerman Sobolev inequality. The same procedure for L estimate for u fails since the L 2 estimate for u is growth with respect to time. If we do, we cannot obtain the sharp lifespan. So we have to build the L estimate for u by making analysis for the corresponding linearized equation of the nonlinear equation. Based on the L estimate of solutions to the linearized equation, the L estimate for u follows. Then, the lifespan of smooth solutions to problems (1.7) and (1.8) is achieved by combining L 2 and L estimates for u and u , the local well-posedness, and the bootstrap argument.

This article is organized as follows. In Section 2, we introduce the vector fields and give some results. The L 2 and L 1 L estimates of solutions to the corresponding linear wave equation are established in Section 3, which play very important roles in investigating hyperbolic geometric flow equation. In Section 4, we shall build the L 2 and L estimates of smooth solutions using Alinhac’s ghost weight energy method, and L 2 and L 1 L estimates of solutions to the corresponding linear wave equation are obtained in Section 3. Then, lifespan of smooth solutions follows by combining L 2 and L estimate and the bootstrap argument in Section 5.

2 Vector fields and some useful lemmas

In this section, we give Klainerman’s vector field and list some useful lemmas. In order to apply Klainerman’s vector field method, we first introduce the vector fields:

Translations:

= ( i ) i = 1 2 , = ( t , 1 , 2 ) = ( t , ) .

Lorentz boosts:

L i = x i t + t i , i = 1 , 2 ,

Scaling vector field:

S = t t + r r = t t + i = 1 2 x i i ,

Rotations:

Ω x 1 2 x 2 1 .

We shall use Γ to denote a general vector field in { , L 1 , L 2 , S , Ω } . For the multi-index a = ( a 1 , , a 7 ) , we set

Γ a = Γ 1 a 1 Γ 7 a 7 , Γ { , L 1 , L 2 , S , Ω } .

From [27], it holds that

(2.1) ϕ C x t 1 Γ ϕ ,

with x t = 1 + x t .

The good derivatives are

T i = i + ω i t , i = 1 , 2 ,

where ω i = x i x with i = 1 , 2. The good derivatives will appear in Alinhac’s ghost weight method. Note that

T i = ω i ( t + r ) + ( ω ) i x Ω ,

t + r = 1 t + x i = 1 2 ω i L i + S ,

and

1 x Ω = 1 t + x ( ω Λ L + Ω ) ,

where ( ω ) i denotes the ith component of ω = { ω 2 , ω 1 } .

Then, we have

(2.2) T ϕ C x + t 1 Γ ϕ .

Klainerman’s vector field Γ has the following commutation relation between the wave operator and , which is critical in dealing with nonlinear wave equation. We may refer to [27] for the proof.

Lemma 2.1

For any given multi-index k = ( k 1 , , k 7 ) , we achieve

(2.3) [ α , Γ k ] = l k 1 C ˜ k l Γ l = l k 1 C ¯ k l Γ l , α = 0 , 1 , 2 ,

where [ , ] stands for the Poissons bracket, i.e., [ A , B ] = A B B A , and = ( t , 1 , 2 ) = ( 0 , 1 , 2 ) , and C ˜ k l and C ¯ k l are the constants.

Lemma 2.2

For any given multi-index k = ( k 1 , , k 7 ) , it holds that

(2.4) [ , Γ k ] u = l k 1 C k l Γ l u ,

where [ , ] stands for the Poissons bracket, i.e., [ A , B ] = A B B A , and C k l are the constants.

Let

Q ( ϕ , ψ ) = Q α β α ϕ β ψ = ϕ ψ t ϕ t ψ .

Obviously, Q ( ϕ , ψ ) = Q α β α ϕ β ψ satisfies the null condition, and we call Q is a null form. For null form Q , we have the following estimate.

Lemma 2.3

Assume that Q ( ϕ , ψ ) is a null form, then we have

(2.5) Q ( ϕ , ψ ) = Q α β α ϕ β ψ C ( T ϕ ψ + ϕ T ψ ) .

The Klainerman Sobolev inequality that is proved in [11] is to use the L 2 norms of products Γ a of Lorentz vector fields Γ to ϕ and obtain a weighted control of ϕ ( t , x ) .

Lemma 2.4

There exists a constant C > 0 such that

(2.6) x + t n 1 2 x t 1 2 ϕ ( t , x ) C a n + 2 2 Γ a ϕ ( t ) L 2 .

Since the appearance of u ( x , t ) in the nonlinear term and the L 2 norm of u ( x , t ) is growth with respect to t , to obtain sharp lifespan, we need the following lemma, which is called as Hardy-type inequality and comes from [28].

Lemma 2.5

Assume that Supp ϕ = { x x 1 + t } , then it holds that

(2.7) x t 1 ϕ ( t ) L 2 x ϕ ( t ) L 2 .

To deal with some nonlinear terms, we need the following lemma, which is found in [27].

Lemma 2.6

Suppose that G = G ( ϕ ) is a sufficiently smooth function of ϕ with

G ( 0 ) = 0 .

For any given integer N 0 , if a vector function ϕ = ϕ ( x , t ) satisfies

a [ N 2 ] Γ a ϕ ( , t ) L ν 0 , t [ 0 , T ] ,

where [ ] stands for the integer part of a real number and ν 0 is a positive constant, then it holds that

(2.8) a N Γ a G ( ϕ ( , t ) ) L P C ( ν 0 ) a N Γ a ϕ ( , t ) L p , t [ 0 , T ] ,

provided that all norms appearing on the right-hand side of (2.8) are bounded, where C ( ν 0 ) is a positive constant depending on ν 0 , and p is a real number with 1 p .

3 Linear wave equation

In this section, our main aim is to build and give the estimate of solutions to two-dimensional linear wave, which will play very important roles in dealing with hyperbolic geometric flow equation.

Lemma 3.1

Assume that ϕ = ϕ ( t , x ) is the solution to the initial value problem for two-dimensional wave equation

(3.1) ϕ = Φ + j = 0 2 j Ψ ,

with the initial value

(3.2) t = 0 : ϕ = ϕ 0 ( x ) , t ϕ = ϕ 1 ( x ) ,

where Φ = Φ ( t , x ) and Ψ = Ψ ( t , x ) : R 1 + 2 R are two sufficiently nice functions, then we have

(3.3) ϕ ( t ) L 2 C ( ϕ 0 L 2 + ϕ 1 L 2 + Ψ ( ξ , 0 ) L 2 ) + C log ( 2 + t ) 1 2 ( ϕ 1 L 1 + Ψ ( ξ , 0 ) L 1 ) + C 0 t ( Φ ( τ ) L 2 + Ψ ( τ ) L 2 ) d τ + C log ( 2 + t ) 1 2 0 t Φ ( τ ) L 1 d τ .

Proof

Fourier transform and integration by parts entail that the solution to problems (3.1) and (3.2) in Fourier space is given by

ϕ ( ξ , t ) = cos ( ξ t ) ϕ ^ 0 ( ξ ) + sin ( ξ t ) ξ ϕ ^ 1 ( ξ ) + 0 t sin ( ξ ( t τ ) ) ξ ( Φ ^ + j = 0 2 i ξ j Ψ ^ + t Ψ ^ ) ( ξ , τ ) d τ = cos ( ξ t ) ϕ ^ 0 ( ξ ) + sin ( ξ t ) ξ ( ϕ ^ 1 ( ξ ) Ψ ^ ( ξ , 0 ) ) + 0 t sin ( ξ ( t τ ) ) ξ Φ ^ + j = 1 2 i ξ j Ψ ^ ( ξ , τ ) d τ + 0 t cos ξ ( t τ ) Ψ ^ ( ξ , τ ) d τ .

Then, we have from Plancherel theorem

(3.4) ϕ ( t ) L 2 cos ( ξ t ) ϕ ^ 0 ( ξ ) L 2 + sin ( ξ t ) ξ ( ϕ ^ 1 ( ξ ) Ψ ^ ( ξ , 0 ) ) L 2 + 0 t sin ( ξ ( t τ ) ) ξ Φ ^ ( τ ) L 2 d τ + j = 1 2 0 t sin ( ξ ( t τ ) ) ξ i ξ j Ψ ^ ( τ ) L 2 d τ + 0 t cos ξ ( t τ ) Ψ ^ ( τ ) L 2 d τ : A 1 + A 2 + A 3 + A 4 + A 5 .

In what follows, we estimate A j ( j = 1 , , 4 ) . Obviously,

A 1 ϕ 0 L 2 .

By the detailed calculation, we arrive at

A 2 = R 2 sin ( ξ t ) ξ ( ϕ ^ 1 Ψ ^ ( ξ , 0 ) ) ( ξ ) 2 d ξ 1 2 ξ 1 sin ( ξ t ) ξ ϕ ^ 1 ( ξ ) Ψ ^ ( ξ , 0 ) 2 d ξ 1 2 + ξ 1 sin ( ξ t ) ξ ϕ ^ 1 ( ξ ) Ψ ^ ( ξ , 0 ) 2 d ξ 1 2 ( ϕ ^ 1 L + Ψ ^ ( ξ , 0 ) L ) 0 t sin 2 ρ ρ d ρ 1 2 + ϕ 1 ( t ) L 2 + Ψ ^ ( ξ , 0 ) L 2 ( ϕ 1 L 1 + Ψ ( ξ , 0 ) L 1 ) ( log ( 2 + t ) ) 1 2 + ϕ 1 ( t ) L 2 + Ψ ( ξ , 0 ) L 2 .

The same procedure leads to

A 3 0 t ( Φ ( τ ) L 1 ( log ( 2 + t τ ) ) 1 2 + Φ ( τ ) L 2 ) d τ 0 t ( ( log ( 2 + t ) ) 1 2 Φ ( τ ) L 1 + Φ ( τ ) L 2 ) d τ .

Plancherel theorem implies that

A 4 + A 5 C 0 t Ψ ^ ( τ ) L 2 d τ C 0 t Ψ ( τ ) L 2 d τ .

Plugging the estimates for A i ( i = 1 , , 5 ) into (3.4) immediately yields (3.3).□

Lemma 3.2

Assume that ϕ = ϕ ( t , x ) is the solution to the initial value problem for two-dimensional wave equation

ϕ = 0 ,

with the initial value

t = 0 : ϕ = ϕ 0 ( x ) , t ϕ = ϕ 1 ( x ) .

Assume that ϕ 0 W 3,1 , ϕ 1 W 2,1 , then we have

(3.5) ϕ ( x , t ) C ( 1 + t ) 1 2 ( ϕ 0 W 3,1 + ϕ 1 W 2,1 ) .

Proof

The proof of (3.5) has been given in [27], and the proof process is complex. Hou and Yin [9] obtained weighted decay estimate under some strong condition, and since we only need the time-decay rate, the analysis given in this article is simpler. The result holds for non-compactly supported initial data.

Poisson formula gives

(3.6) u ( x , t ) = 1 2 π y x t ϕ 1 ( y ) t 2 y x 2 d y + 1 2 π t y x t ϕ 0 ( y ) t 2 y x 2 d y .

We first prove that

(3.7) y x t ϕ 1 ( y ) t 2 y x 2 d y C ( 1 + t ) 1 2 ϕ 1 W 2,1 .

In fact, when 0 t 1 , it holds that

(3.8) y x t ϕ 1 ( y ) t 2 y x 2 d y ϕ 1 L σ y x t ( t 2 y x 2 ) σ 2 d y 1 σ C ϕ 1 L σ 0 t r ( t 2 r 2 ) σ 2 d r 1 σ C ϕ 1 W 2,1 0 t r ( t 2 r 2 ) σ 2 d r 1 σ C t 1 2 + 1 σ ϕ 1 W 2,1 C ( 1 + t ) 1 2 ϕ 1 W 2,1 ,

where σ satisfies 2 < σ .

When t 1 , it is suffice to prove

(3.9) y x t ϕ 1 ( y ) t 2 y x 2 d y t 1 2 ϕ 1 W 1,1 .

We rewrite y x t ϕ 1 ( y ) t 2 y x 2 d y as

y x t ϕ 1 ( y ) t 2 y x 2 d y = y t ϕ 1 ( x y ) t 2 y 2 d y = y t 1 2 ϕ 1 ( x y ) t 2 y 2 d y + t 1 2 y t ϕ 1 ( x y ) t 2 y 2 d y .

Obviously, we have

y t 1 2 ϕ 1 ( x y ) t 2 y 2 d y C t 1 2 ϕ 1 L 1 .

Let ϖ be a smooth cutoff function satisfying

0 ϖ ( r ) 1 , ϖ ( r ) = 1 , t 1 2 r t , 0 , 0 r t 2 3 .

Then

t 1 2 y t ϕ 1 ( x y ) t 2 y 2 d y C t 1 2 t 1 2 y t ϕ 1 ( x y ) t y d y = C t 1 2 t 1 2 y t ϖ ( y ) ϕ 1 ( x y ) y y t y d y C t 1 2 0 y t ϖ ( y ) ϕ 1 ( x y ) y y t y d y C t 1 2 0 y t ϖ ( y ) ϕ 1 ( x y ) t y d y + C t 1 2 0 y t ϕ 1 ( x y ) + ϕ 1 ( x y ) y ϖ ( y ) t y d y C t 1 2 R 2 ( ϕ 1 ( y ) + ϕ 1 ( y ) ) d y C t 1 2 ϕ 1 W 1,1 .

The aforementioned two estimates entail that (3.9) holds.

Noting that

(3.10) t y x t ϕ 0 ( y ) t 2 y x 2 d y = 1 t y x t ϕ 0 ( y ) t 2 y x 2 d y + y x t y x t ϕ 0 ( y ) t 2 y x 2 d y .

Applying (3.7) to the second term on the right, we immediately obtain

(3.11) y x t y x t ϕ 0 ( y ) t 2 y x 2 d y C ( 1 + t ) 1 2 ϕ 0 W 3,1 .

In what follows, we estimate 1 t y x t ϕ 0 ( y ) t 2 y x 2 d y .

When t 1 , the aforementioned same procedure implies that

(3.12) 1 t y x t ϕ 0 ( y ) t 2 y x 2 d y C ( 1 + t ) 1 2 ϕ 0 W 3,1 .

When t 1 , we obtain from direct calculation and Sobolev embedding theorem that

(3.13) 1 t y x t ϕ 0 ( y ) t 2 y x 2 d y = z 1 ϕ 0 ( x + t z ) 1 z 2 d z C ϕ 0 L C ( 1 + t ) 1 2 ϕ 0 W 2,1 .

Collecting (3.6) and (3.10)–(3.13) immediately yields (3.5). We complete the proof of Lemma 3.2.□

To estimate the L norm of u , we need the following two lemmas on L 1 L estimate, which have been established in [10].

Lemma 3.3

Assume that ϕ = ϕ ( t , x ) is the solution to the initial value problem

ϕ = Φ ,

with the initial value

t = 0 : ϕ = 0 , t ϕ = 0 ,

where Φ = Φ ( t , x ) : R 1 + 2 R is a sufficiently nice function, then we have

(3.14) ϕ ( x , t ) C ( 1 + t ) 1 2 0 t α 1 ( 1 + τ ) 1 2 Γ α Φ ( τ ) L 1 d τ .

Lemma 3.4

Assume that ϕ = ϕ ( t , x ) is the solution to the initial value problem

ϕ = j = 0 2 j Ψ

with the initial value

t = 0 : ϕ = 0 , t ϕ = 0 ,

where Ψ = Ψ ( t , x ) : R 1 + 2 R is a sufficiently nice function, then we have

(3.15) ϕ ( x , t ) C ( 1 + t ) 1 2 0 t ( 1 + τ ) 1 2 Ψ ( τ ) L + ( 1 + τ ) 3 2 α n + 1 Γ α Ψ ( τ ) L 1 d τ .

4 A priori estimate

In this section, we make a priori estimate. To this end, for K 8 and 0 < σ < 1 24 , we define

(4.1) X ( t ) = sup 0 τ t a K Γ a u ( τ ) L 2 2 + sup 0 τ t a K ( 1 + τ ) 2 σ Γ a u ( τ ) L 2 2 + sup 0 τ t ( 1 + τ ) a K 2 + 1 Γ a u ( τ ) L 2 .

By Lemmas 2.1 and 2.2, it follows from (1.7) that

(4.2) Γ a u = [ ( e u 1 ) Γ a u ] + b a C 1 b a Γ b [ ( e u 1 ) u ] [ ( e u 1 ) Γ a u ] + b a Γ b e u ( u 2 u t 2 ) + b a Γ b [ ( e u 1 ) u t 2 ] = [ ( e u 1 ) Γ a u ] + c + d < a , 0 α , β 2 C 1 b a α β α [ Γ c ( e u 1 ) β Γ d u ] + b + c a C 2 b a Γ b e u Γ c Q ( u , u ) + b a C 3 b a Γ b [ ( e u 1 ) u t 2 ] = [ ( e u 1 ) Γ a u ] + c + d < a , 0 α , β 2 C 1 b a α β α [ Γ c ( e u 1 ) β Γ d u ] + b + c 1 + c 2 a C 2 b c 1 c 2 a Γ b e u Q ( Γ c 1 u , Γ c 2 u ) + b a C 3 b c d a Γ b ( e u 1 ) Γ c u Γ c u I 1 + I 2 + I 3 + I 4 .

Let

q ( τ ) = arctan τ , τ = x t ,

then

q ( τ ) = 1 ( 1 + x t 2 )

and

e q ( τ ) ( e π 2 , e π 2 ) .

Lemma 4.1

Under the assumption of Theorem 1.1, it holds that

(4.3) sup 0 τ t a K Γ a u ( τ ) L 2 2 C ε 2 + C ( 1 + t ) 1 2 ( X 3 2 ( t ) + X 2 ( t ) + X 3 ( t ) ) .

Proof

Multiplying (4.2) by e q Γ a u and integrating it with respect to x and t , we have

(4.4) 0 t R 2 e q Γ a u t Γ a u d x d τ = 0 t R 2 e q ( I 1 a + I 2 a + I 3 a + I 4 a ) t Γ a u d x d τ .

Using integration by parts and Stokes formula, we have

R 2 e q Γ a u t Γ a u d x = 1 2 R 2 t [ e q ( t Γ a u 2 + Γ a u 2 ) ] d x R 2 ( e q t Γ a u Γ a u ) d x R 2 1 2 ( t Γ a u 2 + Γ a u 2 ) t e q d x + R 2 t Γ a u Γ a u e q d x = 1 2 d d t R 2 [ e q ( t Γ a u 2 + Γ a u 2 ) ] d x + 1 2 i = 1 2 R 2 e q T i Γ a u 2 1 + x t 2 d x .

Then, we have

(4.5) 0 t R 2 e q Γ a u t Γ a u d x d τ = 1 2 e q 2 t Γ a u L 2 2 + e q 2 Γ a u L 2 2 1 2 e q 2 t Γ a u ( x , 0 ) L 2 2 + e q 2 Γ a u ( x , 0 ) L 2 2 + 1 2 i = 1 2 0 t R 2 e q T i Γ a u 2 1 + x τ 2 d x d τ .

It follows from integration by parts, Stokes formula, and (4.1) that

(4.6) 0 t R 2 e q I 1 a t Γ a u d x d τ = 0 t R 2 e q [ ( e u 1 ) Γ a u ] t Γ a u d x d τ = 0 t R 2 [ e q ( e u 1 ) Γ a u t Γ a u ] d x 0 t R 2 e q ( e u 1 ) Γ a u t Γ a u d x 0 t R 2 e q ( e u 1 ) Γ a u t Γ a u d x = 1 2 0 t d d t R 2 [ e q ( e u 1 ) Γ a u 2 ] d x + 0 t R 2 t ( e q ) ( e u 1 ) Γ a u 2 d x d τ 1 2 0 t R 2 e q t u Γ a u 2 d x d τ 0 t R 2 ( e q ) ( e u 1 ) Γ a u t Γ a u d x d τ 1 2 0 t R 2 [ e q ( e u 1 ) Γ a u 2 ] d x + 1 2 0 t R 2 [ e q ( e u ( x , 0 ) 1 ) Γ a u ( x , 0 ) 2 ] d x + C X 3 2 ( t ) 0 t ( 1 + τ ) 1 2 d τ 1 2 0 t R 2 [ e q ( e u 1 ) Γ a u 2 ] d x + 1 2 0 t R 2 [ e q ( e u ( x , 0 ) 1 ) Γ a u ( x , 0 ) 2 ] d x + C X 3 2 ( t ) ( 1 + t ) 1 2 .

Thanks to Hölder’s inequality, Lemmas 2.5 and 2.6, and (4.1), we arrive at

(4.7) 0 t R 2 e q I 2 a t Γ a u d x = c + d < a , 0 α , β 2 R 2 e q α [ Γ c ( e u 1 ) β Γ c u ] t Γ a u d x d τ = c + d < a , 0 α , β 2 , c a 2 0 t R 2 e q α [ Γ c ( e u 1 ) β Γ d u ] t Γ a u d x d τ + c + d < a , 0 α , β 2 , c a 2 0 t R 2 e q α [ Γ c ( e u 1 ) β Γ d u ] t Γ a u d x d τ C c + d < a , c a 2 0 t Γ c ( e u 1 ) L Γ d u L 2 t Γ a u L 2 d τ + C c + d < a , c a 2 0 t Γ c ( e u 1 ) L 2 Γ d u L 2 t Γ a u L 2 d τ + C c + d < a , c a 2 0 t Γ c ( e u 1 ) L 2 Γ d u L t Γ a u L 2 d τ + C c + d < a , c a 2 0 t Γ c ( e u 1 ) x t L 2 Γ Γ d u L t Γ a u L 2 d τ C X 3 2 ( t ) 0 t ( 1 + τ ) 1 2 d τ C ( 1 + t ) 1 2 X 3 2 ( t ) .

The same procedure leads to

0 t R 2 e q I 3 a t Γ a u d x d τ = c 1 + c 2 a 0 t R 2 e q e u Q ( Γ c 1 u , Γ c 2 u ) t Γ a u d x d τ + b + c 1 + c 2 a , 0 < b a 2 0 t R 2 e q Γ b e u Q ( Γ c 1 u , Γ c 2 u ) t Γ a u d x d τ + b + c 1 + c 2 a , b a 2 0 t R 2 e q Γ b e u Q ( Γ c 1 u , Γ c 2 u ) t Γ a u d x d τ C c 1 + c 2 a , c 1 a 2 0 t R 2 e q T Γ c 1 u Γ c 2 u t Γ a u d x d τ + C c 1 + c 2 a , c 1 a 2 0 t R 2 e q T Γ c 1 u Γ c 2 u t Γ a u d x d τ + b + c 1 + c 2 a , 0 < b a 2 0 t R 2 e q Γ b ( e u 1 ) ( T Γ c 1 u Γ c 2 u + Γ c 1 u T Γ c 2 u ) t Γ a u d x d τ

(4.8) + b + c 1 + c 2 a , b a 2 0 t R 2 e q Γ b ( e u 1 ) ( T Γ c 1 u Γ c 2 u + Γ c 1 u T Γ c 2 u ) t Γ a u d x d τ C c 1 + c 2 a , c 1 a 2 0 t T Γ c 1 u L Γ c 2 u L 2 t Γ a u L 2 d τ + 1 24 c 1 + c 2 a , c 1 a 2 0 t R 2 e q T Γ c 1 u 2 1 + x τ 2 d x + x τ Γ c 2 u t Γ a u L 2 2 d τ + 1 24 b + c 1 + c 2 a , b a 2 , c 1 c 2 0 t R 2 e q T Γ c 2 u 2 1 + x τ 2 d x + x τ Γ b ( e u 1 ) Γ c 1 u t Γ a u L 2 2 d τ + 1 24 b + c 1 + c 2 a , b a 2 , c 1 c 2 0 t R 2 e q T Γ c 1 u 2 1 + x τ 2 d x + x τ Γ b ( e u 1 ) Γ c 2 u t Γ a u L 2 2 d τ + b + c 1 + c 2 a , b a 2 0 t T Γ c 1 u L x τ Γ c 2 u L Γ b ( e u 1 ) x τ L 2 t Γ a u L 2 d τ C ( X 3 2 ( t ) + X 2 ( t ) + X 3 ( t ) ) ( 1 + t ) 1 2 + 1 8 c 1 a 0 t R 2 e q T Γ c 1 u 2 1 + x τ 2 d x d τ

and

(4.9) 0 t R 2 e q I 4 a t Γ a u d x d τ = 0 t R 2 e q Γ a [ ( e u 1 ) ( t u ) 2 ] t Γ α u d x d t = b + c + d = a , b a 2 0 t R 2 e q Γ b ( e u 1 ) Γ c u Γ d u t Γ a u d x d τ + b + c + d = a , b a 2 0 t R 2 e q Γ b ( e u 1 ) Γ c u Γ d u t Γ a u d x d τ C b + c + d = a , b a 2 , c d 0 t Γ b ( e u 1 ) L Γ c u L Γ d u L 2 t Γ a u L 2 d τ + C b + c + d = a , b a 2 , c d 0 t Γ b ( e u 1 ) L Γ c u L 2 Γ d u L t Γ a u L 2 d τ

+ C b + c + d = a , b a 2 , c d 0 t Γ b ( e u 1 ) x τ L 2 Γ Γ c u L Γ d u L t Γ a u L 2 d τ + C b + c + d = a , b a 2 , c d 0 t Γ b ( e u 1 ) x τ L 2 Γ c u L Γ Γ d u L t Γ a u L 2 d τ C X 2 ( t ) 0 t ( 1 + τ ) 1 d τ C X 2 ( t ) log ( 2 + t ) .

We institute (4.5)–(4.9) into (4.4) and obtain

sup 0 τ t a K Γ a u ( τ ) L 2 2 C ε 2 + C ( 1 + t ) 1 2 ( X 3 2 ( t ) + X 2 ( t ) + X 3 ( t ) ) .

Then, Lemma 3.1 follows.□

Lemma 4.2

Under the assumption of Theorem 1.1, it holds that

(4.10) sup 0 τ t a K ( 1 + τ ) 2 σ Γ a u ( τ ) L 2 2 C ε 2 + C ( 1 + t ) ( X 2 ( t ) + X 3 ( t ) ) .

Proof

We apply Lemma 3.1 to (4.2) and obtain

(4.11) Γ α u L 2 C ( Γ α u ( 0 ) L 2 + t Γ α u ( 0 ) L 2 + I 2 ( 0 ) L 2 ) + C log 1 2 ( 2 + t ) ( t Γ α u ( 0 ) L 1 + I 2 ( 0 ) L 1 ) + C log 1 2 ( 2 + t ) 0 t ( I 3 L 1 + I 4 L 1 ) d τ + C 0 t ( I 1 + I 2 L 2 + I 3 L 2 + I 4 L 2 ) d τ .

Noting that Γ α u ( 0 ) , t Γ α u ( 0 ) , and I 2 ( 0 ) are the initial data u 0 and u 1 , and Supp u 0 , Supp u 1 = { x x 1 } , then we have

(4.12) Γ α u ( 0 ) L 2 + t Γ α u ( 0 ) L 2 + I 2 ( 0 ) L 2 C ε

and

(4.13) t Γ α u ( 0 ) L 1 + I 2 ( 0 ) L 1 C ε .

We conclude from Lemma 2.3, Hölder’s inequality, Lemma 2.5, and (4.1) that

(4.14) 0 t I 3 L 1 d τ C b + c + d a 0 t Γ b e u ( T Γ c 1 u Γ c 2 u + Γ c 1 u T Γ c 2 u ) L 1 d τ C c 1 + c 2 a 0 t T Γ c 1 u L 2 Γ c 2 u L 2 + Γ c 1 u L 2 T Γ c 2 u L 2 d τ + C b + c + d a , b a 2 , b 0 0 t Γ b ( e u 1 ) L T Γ c 1 u L 2 Γ c 2 u L 2 d τ + C b + c + d a , b a 2 0 t Γ b ( e u 1 ) x τ L 2 T Γ c 1 u L 2 x τ Γ c 2 u L d τ C X ( t ) 0 t ( 1 + τ ) 1 + σ d τ + X 3 2 ( t ) 0 t ( 1 + τ ) 1 2 d τ C X ( t ) + X 3 2 ( t ) ( 1 + t ) 1 2 .

Using Hölder’s inequality, Lemmas 2.5, 2.6, and (2.1), (4.1), we achieve

(4.15) 0 t I 4 L 1 d τ C b + c + d a , b a 2 0 t Γ b ( e u 1 ) L Γ c u L 2 Γ d u L 2 d τ + C b + c + d a , b a 2 0 t Γ b ( e u 1 ) x τ L 2 x τ Γ c u L Γ d u L 2 d τ C X 3 2 ( t ) 0 t ( 1 + τ ) 1 2 d τ C X 3 2 ( t ) ( 1 + t ) 1 2 .

Hölder’s inequality, Lemmas 2.5, 2.6, and (2.1) and (4.1) entail that

(4.16) 0 t I 1 + I 2 L 2 d τ C c + d a , c a 2 0 t Γ c ( e u 1 ) L Γ d u L 2 d τ + C c + d a , c a 2 0 t Γ c ( e u 1 ) x τ L 2 x τ Γ d u L d τ C X ( t ) 0 t ( 1 + τ ) 1 2 d τ C X ( t ) ( 1 + t ) 1 2 .

Thanks to Lemma 2.3, Hölder’s inequality, Lemmas 2.5 and 2.6, and (2.1) and (4.1), we arrive at

(4.17) 0 t I 3 L 2 d τ C b + c 1 + c 2 a 0 t Γ b e u ( T Γ c 1 u Γ c 2 u + Γ c 1 u T Γ c 2 u ) L 2 d τ C c 1 + c 2 a 0 t ( T Γ c 1 u L Γ c 2 u L 2 + Γ c 1 u L T Γ c 2 u L 2 ) d τ + C b + c 1 + c 2 a , b a 2 0 t Γ b ( e u 1 ) L ( T Γ c 1 u L Γ c 2 u L 2 + Γ c 1 u L T Γ c 2 u L 2 ) d τ + C b + c 1 + c 2 a , b a 2 0 t Γ b ( e u 1 ) x τ L 2 T Γ c 1 u L x τ Γ c 2 u L d τ C X ( t ) 0 t ( 1 + τ ) 1 d τ + X 3 2 ( t ) 0 t ( 1 + τ ) 3 2 d τ C X ( t ) log ( 2 + t ) + X 3 2 ( t ) .

It follows from Hölder’s inequality, Lemmas 2.5 and 2.6 and (2.1) and (4.1) that

0 t I 4 L 2 d τ C b + c + d a , b a 2 0 t Γ b ( e u 1 ) L Γ c u L Γ d u L 2 d τ

(4.18) + C b + c + d a , b a 2 0 t Γ b ( e u 1 ) x τ L 2 x τ Γ c u L Γ d u L d τ C X 3 2 ( t ) 0 t ( 1 + τ ) 1 d τ C X 3 2 ( t ) log ( 2 + t ) .

We institute (4.12)–(4.18) into (4.11) immediately, which yields (4.10). Lemma 4.2 is proved.□

Lemma 4.3

Under the assumption of Theorem 1.1, it holds that

(4.19) sup 0 τ t a K 2 + 1 ( 1 + τ ) Γ a u ( τ ) L 2 C ε 2 + C ( ( 1 + t ) 2 σ X 2 ( t ) + ( 1 + t ) X 2 ( t ) + ( 1 + t ) X 3 ( t ) ) .

Proof

Let

(4.20) Γ a u = u + u ˜ + u ¯ ,

where u , u ˜ , and u ¯ satisfy

(4.21) u = 0 , u ( x , 0 ) = Γ a u ( x , 0 ) , t u ( x , 0 ) = t Γ a u ( x , 0 ) ,

(4.22) u ˜ = b + c a C 1 b c a ( Γ b ( e u 1 ) Γ c u ) , u ˜ ( x , 0 ) = 0 , t u ˜ ( x , 0 ) = 0 ,

and

(4.23) u ¯ = b + c 1 + c 2 a C 2 b c 1 c 2 a Γ b e u Q ( Γ c 1 u , Γ c 2 u ) + b + c + d a C 2 b c d a Γ b ( e u 1 ) Γ c u Γ d u , u ¯ ( x , 0 ) = 0 , t u ¯ ( x , 0 ) = 0 ,

respectively.

Applying Lemma 3.2 to (4.21) yields

(4.24) u ( x , t ) C ( 1 + t ) 1 2 ( Γ a u ( x , 0 ) W 3,1 + t Γ a u ( x , 0 ) W 2,1 ) C ( 1 + t ) 1 2 ε .

We conclude from Lemma 3.3 and (4.22) that

(4.25) u ˜ ( x , t ) C ( 1 + t ) 1 2 0 t b + c a ( 1 + τ ) 1 2 Γ b ( e u 1 ) Γ c u L + b + c a + 3 ( 1 + τ ) 3 2 Γ b ( e u 1 ) Γ c u L 1 d τ C ( 1 + t ) 1 2 X ( t ) 0 t ( 1 + τ ) 1 2 d τ + 0 t ( 1 + τ ) 3 2 + σ d τ C X ( t ) .

Lemma 3.4, Hölder’s inequality, and 4.23 entail that

(4.26) u ¯ ( x , t ) C ( 1 + t ) 1 2 0 t ( 1 + τ ) 1 2 b + c 1 + c 2 a + 1 Γ b Q ( Γ c 1 u , Γ c 2 u ) L 1 + b + c + d a + 1 Γ b ( e u 1 ) Γ c 1 u Γ c 2 u L 1 d τ C ( 1 + t ) 1 2 X ( t ) 0 t ( 1 + τ ) 1 + σ d τ + X 3 2 ( t ) 0 t ( 1 + τ ) 1 2 d τ

C ( 1 + t ) 1 2 ( 1 + τ ) σ X ( t ) + ( 1 + τ ) 1 2 X 3 2 ( t ) .

Substituting (4.24)–(4.26) into (4.20) immediately yields (4.19). Then, we complete the proof of Lemma 4.3.□

5 Lifespan of smooth solutions

In this section, we give the lifespan of smooth solutions to problems (1.7) and (1.8). In other words, we shall complete the proof of Theorem 1.1.

Proof

From (4.3), (4.10), (4.19), and (4.1), we conclude that

(5.1) X ( t ) C 0 ε 2 + C 1 ( 1 + t ) 1 2 X 3 2 ( t ) + C 2 ( 1 + t ) ( X 2 ( t ) + X 3 ( t ) ) .

Assume that

(5.2) X ( t ) M ε 2 ,

where M = 8 C 0 .

Then, (5.1) and (5.2) entail that

(5.3) X ( t ) C 0 ε 2 + C 1 ( 1 + t ) 1 2 M 3 2 ε 3 + C 2 ( 1 + t ) ( M 2 ε 4 + M 3 ε 6 ) .

Taking ε 1 M and t + 1 min { 1 64 C 1 2 M ε 2 , 1 8 C 2 M ε 2 } , then from (5.3), we arrive at

X ( t ) 1 2 M ε 2 .

Thus, by the local well-posedness of problems (1.7) and (1.8) and the continuous induction method, then problems (1.7) and (1.8) admit a global smooth solution u on [ 0 , T ε ] , with

T ε C ε 2 .

We complete the proof of Theorem 1.1.□

Acknowledgments

The authors wish to thank the referees for their very helpful comments and suggestions that improved the manuscript.

  1. Funding information: This work was supported in part by the Natural Science Foundation of Henan (Grant No. 252300421305) and Program for Innovative Research Team (in Science and Technology) in University of Henan Province (Grant No. 25IRTSTHN013).

  2. Author contributions: All authors contributed equally to the writing of this article. All authors read and approved the final manuscript.

  3. Conflict of interest: The authors declare that they have no competing interests.

  4. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2024-09-21
Revised: 2025-06-16
Accepted: 2025-07-14
Published Online: 2025-08-29

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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