Based on Herring’s physical formula for anisotropic surface free energy, Taylor, Cahn, and Handwerker [1] reformulated crystal-interface dynamics into a sharp-interface geometric evolution model. They introduced the concept of crystalline curvature κ γ , a generalization of mean curvature adapted to orientation-dependent surface energy γ ( n ) . The interface evolves according to the crystalline curvature flow V = − κ γ , where V denotes the interface normal velocity. In the isotropic limit, κ γ reduced to ordinary mean curvature κ . In 2007, Lou [2] investigated the following curvature flow

and showed the existence and uniqueness of the traveling waves. In this paper, we consider a generalized curvature flow

in heterogeneous media with periodic oblique striations in the x y -plane, where α ( 0 < α < π / 2 ) is the inclination angle of the oblique striations, for any time t and any simple plane curve Γ t and for any P ∈ Γ t , κ denotes the curvature of Γ t at P , n denotes the unit normal vector at P and V denotes the velocity along n , H ∈ C 1 ( ℝ × S 1 × ℝ ) and satisfying

(A) H κ ( κ , n , η ) > 0 in ℝ × S 1 × ℝ ; H ( κ , n , η ) is 1-periodic in η ; 0 < H m ≤ H ( 0 , n , η ) ≤ H M .

Clearly, the typical example

satisfies the condition (A) if a and b are continuous, positive and 1-periodic functions in x .

The local-in-time existence and the asymptotic behavior of solutions of (1) have been investigated by many authors(see [3][4] and references therein). References [5] and [6] studied traveling wave solutions of the special case of (2): V = a ( n ) κ + b ( n ) .

In 1993, Altschuler and Wu [7] considered the problem with a ( x , n ) ≡ A ( n ) and b ( x , n ) ≡ 0 in (2) and proved that any global solution with some initial data converges to the unique (up to a shift) traveling wave solution. In 2006 and 2013, references [8] and [9] considered the problem (2) with a ( x , n ) ≡ A ( n ) and b ( x , n ) ≡ 0 in a band domain with undulating boundaries. They constructed the periodic traveling wave solution and studied its uniqueness and asymptotic stability. Later, Cai and Lou [10] studied the analogue problem with the almost periodic boundaries and obtained the analogue results. In 1994, Altschuler and Wu [11] extended their result in [7] to the case of two dimensions. Recently, the results in [11]-[13] were further improved by Ma, Wang and Wei [14] to N -dimensional space ( N > 1 ).

Since the 1980s, various (mean) curvature flows, including the curvature flow V = a ( n ) κ β with β > 0 or even β = − 1 (which corresponds to an expanding flow), have been extensively studied by geometers from a mathematical point of view (see [3][15]). In 2023, Lou and Yuan [16] studied the generalized mean curvature flow V = κ β + b with β > 0 , b ∈ ℝ and proved the existence, uniqueness, and the asymptotic behavior of radial flows in a cylinder with constant boundary angles.

Our main purpose in this paper is to consider the generalized curvature flow (1) in the case that the curve Γ t is the smooth graph of a function y = u ( x , t ) for each t . To avoid sign confusion, the normal to the curve will always be chosen upward (toward y -direction), and the V and κ will be understood in accordance with this choice of the direction of the normal. Consequently,

and (2) is equivalent to

In the following, a solution u ( x , t ) of (3) is called a periodic traveling wave if

where φ ( x , t ) is X α -periodic in x and T α -periodic in t and X α = 1 / sin α , T α satisfies

Note that the speed of the wave is time-periodic so that c = 1 / ( T α cos α ) represents the average speed over one period.

In Section 2, we transform Problems (3) and (4) into the boundary value problem for a first-order ordinary differential equation by means of several changes of variables. In Section 3, we first prove that the boundary value problem for a first-order ordinary differential equation has a unique solution. Then, we verify the existence and uniqueness of periodic traveling waves of (3) and derive the upper and lower bounds for the average speed of the wave. ψ ( 0 ) = 0 in (7) is used to give the uniqueness of the periodic traveling waves of (3). Our main result is the following Theorem.

Theorem 1.For any α ∈ ( 0 , π / 2 ) satisfying ( H M 2 − H m 2 ) tan 2 α ≤ H m 2 . Equation(3)has a uniqueperiodic traveling wave u * ( x , t ) = φ * ( x , t ) + c * t , x ∈ ℝ , t ∈ ℝ , and

where the uniqueness holds under our chosen phase normalization condition ψ ( 0 ) = 0 in(7).

For fixed α ∈ ( 0 , π / 2 ) satisfying ( H M 2 − H m 2 ) tan 2 α ≤ H m 2 , in order to show the existence and uniqueness of periodic traveling wave solutions of (3) with inclination angle α , the method we adopt is to reduce (3) to a first-order ordinary differential equation.

Let η ( x , t ) : = x sin α + u cos α , then (3) is rewritten as

Let z = x + c t cos α , ψ ( z ) = η ( x , t ) = x sin α + u cos α , then (3) and (4) are rewritten as

A pair ( c , ψ ( z ) ) solving (7) is called a solution of (7). Next, we prove the existence and uniqueness of a solution ( c , ψ ( z ) ) of (7); this implies the existence and uniqueness of a periodic traveling wave of (3). ψ ( 0 ) = 0 in (7) is used to give the uniqueness of ψ ( z ) , without this condition, ψ ( z ) is unique up to a translation.

We first prove the following result.

Lemma1.Let ( c , ψ ( z ) ) be a solution of (7); then the following properties hold.

1) The function ψ ( z ) − z sin α isperiodic with period X α ,namely,

2) ψ ′ ( z ) isperiodic with period X α .

3) c > 0 and ψ ′ ( z ) > 0 , z ∈ ℝ .

Proof. 1) Since H ( κ , n , η ) is 1-periodic in η , the function ϕ ( z ) = ψ ( z + X α ) − 1 satisfies the same differential equation in ℝ as ψ ( z ) does. Since ϕ ( 0 ) = ψ ( 0 ) and ϕ ′ ( 0 ) = ψ ′ ( 0 ) , we easily obtain, by the uniqueness of the initial value problem for second-order ODEs, that ϕ ( z ) = ψ ( z ) . This implies that (8) holds.

2) From (1), we have ψ ( z ) = ψ ( z + X α ) − 1 . So, ψ ′ ( z ) = ψ ′ ( z + X α ) in ℝ .

3) From (2), there exists z * ∈ [ 0 , X α ) such that ψ ′ ( z * ) = max ℝ ψ ′ ( z ) > 0 . As ψ ″ ( z * ) = 0 , we can get that c > 0 by (7).

Now, let z * ∈ [ 0 , X α ) such that ψ ′ ( z * ) = min ℝ ψ ′ ( z ) . Then, ψ ″ ( z * ) = 0 so that the differential equation in (7) yield c ψ ′ ( z * ) > 0 . Then, ψ ′ ( z ) > 0 , z ∈ ℝ .

Lemma2.Let s = ψ ( z ) , v ( s ) = sin α ψ ′ ( z ) , then ( c , ψ ( z ) ) is a solution of(7) if and only if ( c , v ( s ) ) be a solution of the following problem

where the transformation from ( c , ψ ( z ) ) to ( c , v ( s ) ) is given by ψ ′ ( z ) = sin α v ( s ) , s = ψ ( z ) , andthe transformation from ( c , v ( s ) ) to ( c , ψ ( z ) ) is given by s = ψ ( z ) , v ( s ) = sin α ψ ′ ( z ) .

Proof.This Lemma is easily shown by direct verification and is omitted.

For fixed α ∈ ( 0 , π / 2 ) satisfying ( H M 2 − H m 2 ) tan 2 α ≤ H m 2 , we prove the existence and uniqueness of periodic traveling wave solutions of (3) with inclination angle α in this section. For this purpose, we first show the following Theorem.

Theorem 2.For any α ∈ ( 0 , π / 2 ) satisfying ( H M 2 − H m 2 ) tan 2 α ≤ H m 2 . Problem(9)has a unique solution ( c * , v * ( s ) ) , and

The proof of Theorem 2 easily follows from the following three lemmas.

Lemma3.For any given α ∈ ( 0 , π / 2 ) satisfying ( H M 2 − H m 2 ) tan 2 α ≤ H m 2 and for any given p ≥ 0 , the following problem

has a unique solution ( c ( p , α ) , v ( s , p , α ) ) .Moreover

Proof. We first consider the initial value problem for the following ODE

Since H κ > 0 , for any given c ≥ 0 , by the implicit function theorem, the equation can be rewritten in (13) as V ′ = G ( V , s , α , c ) , where G is a C 1 function. Thus (13) is equivalent to

By the theory of ODEs, (14) has a unique solution V ( s , p , α , c ) on intervals. Denote by [ 0 , s ( c ) ) the maximum existence interval.

1) When c = c 1 : = H M ( p + 1 ) 2 + ( p + 2 ) 2 tan 2 α , we consider the existence interval of V ( s , p , α , c 1 ) . If V ( s , p , α , c 1 ) ∈ ( − p − 1 , p + 1 ) , then we have

Since H κ > 0 , we have V ′ ( s , p , α , c 1 ) < 0 , as long as V ( s , p , α , c 1 ) ∈ ( − p − 1 , p + 1 ) .

On the other hand, if V ( s , p , α , c 1 ) < − V 0 : = − c 1 H m , then

H κ > 0 implies that V ′ ( s , p , α , c 1 ) > 0 for V < − V 0 .

Therefore, V ( s , p , α , c 1 ) ∈ ( − V 0 , p ) and s ( c 1 ) = + ∞ . Especially, V ( 1 , p , α , c 1 ) < p .

2) V ( s , p , α , c ) is decreasing in c. In fact, using (14), V c : = ∂ V ∂ c and V p : = ∂ V ∂ p satisfy

So V p = exp ( ∫ 0 s G V d τ ) > 0 . By using the variation of constants technique, we have for μ : = V c V p ,

Substituting V ′ = G ( V , s , α , c ) into the equation of (13), we have

Differentiating it with respect to c, we get that

Hence, G c < 0 and we have μ ′ < 0 . This implies that μ < 0 for s > 0 . Therefore, V c < 0 for any s ∈ ( 0 , s ( c ) ) .

3) Set Z : = { c ∈ [ 0 , c 1 ] : s ( c ) > 1   and   V ( 1 , p , α , c ) < p } . Since for V ( s , p , α , 0 ) , we have

so H κ > 0 implies that V ′ ( s , p , α , 0 ) > 0 . This implies that 0 ∉ Z . The above (1) means that c 1 ∈ Z and (2) tells us that if c 2 ∈ ( 0 , c 1 ) ∩ Z , then [ c 2 , c 1 ] ⊂ Z .

4) For any fixed c ∈ Z . On the one hand, V ( s , p , α , c ) ≥ V ( s , p , α , c 1 ) > − V 0 . On the other hand, if υ > p is a value of V ( s , p , α , c ) at some s ∈ ( 0 , 1 ) , then there exists s 1 ∈ ( 0 , 1 ) such that V ( s , p , α , c ) = υ and V ′ ( s 1 , p , α , c ) ≤ 0 , so at this point s 1 , we have

i.e., c 1 ≥ c ≥ H m V 2 + ( 1 − V ) 2 tan 2 α . Since υ can be chosen arbitrarily, as long as it is a value of V greater than p , we have V 2 ≤ c 1 2 / H m 2 when V > p . Hence, V ( s , p , α , c ) ∈ [ − c 1 / H m , p ] or [ − c 1 / H m , c 1 / H m ] for s ∈ [ 0 , 1 ] .

From (14), there exists M 1 = M 1 ( H , c 1 , V 0 , p ) = M 1 ( H , p ) such that, for any c ∈ Z , V ( s , p , α , c ) satisfies

5) Denote c 3 : = inf Z c , then (15) and (2) deduce that there exists a function v ( s , p , α ) ∈ C 1 ( [ 0 , 1 ] ) such that

Taking the limit c → c 3 + 0 in (14), we have

Therefore, v ( s , p , α ) ≡ V ( s , p , α , c 3 ) on [ 0 , 1 ] . Clearly, V ( s , p , α , c 3 ) satisfies (15) and V ( 1 , p , α , c 3 ) ≤ p .

6) Next, we show that V ( 1 , p , α , c 3 ) = p . Otherwise, V ( 1 , p , α , c 3 ) < p − 2 δ 1 for some δ 1 ∈ ( 0 , 1 ) . Denote M 2 : = max { | G V | + | G c | : s ∈ [ 0 , 1 ] , V ∈ [ − M 1 − 1 , M 1 + 1 ] , c ∈ [ 0 , c 1 ] } . Consider (14) with c = c 4 : = c 3 − δ 2 , where δ 2 < min { δ 1 M 2 − 1 exp ( − M 2 ) , c 3 } .

First, we prove that

In fact, the first inequality is true on [ 0 , s ( c 4 ) ) ∩ [ 0 , 1 ] . The second inequality holds on [ 0 , s 2 ] for some small s 2 . Set s 3 : = sup { s : the second inequality holds on   [ 0 , s ] } . If s 3 ≥ 1 , then (16) is proved. If s 3 < 1 , then

Set V ˜ ( s ) = V ( s , p , α , c 4 ) − V ( s , p , α , c 3 ) ≥ 0 . Then, on [ 0 , s 3 ] , there exists ζ ( s ) lies between V ( s , p , α , c 3 ) and V ( s , p , α , c 4 ) , and c 5 ∈ ( c 4 , c 3 ) ⊂ ( 0 , c 1 ) such that

Since s 3 < 1 , we have V ˜ ( s 3 ) ≤ δ 2 M 2 exp ( M 2 ) ≤ δ 1 < 1 , in contradiction with (17). Therefore, (16) holds.

Repeating the above discussion on [0, 1], we have V ˜ ( 1 ) ≤ δ 1 , that is, V ( 1 , p , α , c 4 ) ≤ V ( 1 , p , α , c 3 ) + δ 1 < p − δ 1 . This implies that c 4 ∈ Z , which contradicts the definition of c 3 . Therefore, V ( 1 , p , α , c 3 ) = p , and so the pair c ( p , α ) : = c 3 and v ( s , p , α ) ≡ V ( s , p , α , c 3 ) is the unique solution of (11).

7) Now, we prove that (12) holds. Since v ( s , p , α ) is 1-periodic, for any υ ∈ [ min v ( ⋅ , p , α ) , max v ( ⋅ , p , α ) ] , there exists s 4 , s 5 ∈ [ 0 , 1 ) such that v ( s 4 , p , α ) = v ( s 5 , p , α ) = υ , v ′ ( s 4 , p , α ) ≥ 0 ≥ v ′ ( s 5 , p , α ) . Then, at this point s = s 4 , we have

that is, c ( p , α ) ≥ H m v 2 + ( 1 − v ) 2 tan 2 α . Since υ ∈ [ min v ( ⋅ , p , α ) , max v ( ⋅ , p , α ) ] can be chosen arbitrarily, we have c ( p , α ) ≥ H m max s ∈ [ 0 , 1 ] v 2 ( s , p , α ) + ( 1 − v ( s , p , α ) ) 2 tan 2 α .

Similarly, one can show that c ( p , α ) ≤ H M min s ∈ [ 0 , 1 ] v 2 ( s , p , α ) + ( 1 − v ( s , p , α ) ) 2 tan 2 α .

Lemma 4.For any given α ∈ ( 0 , π / 2 ) satisfying ( H M 2 − H m 2 ) tan 2 α ≤ H m 2 ,thereexists a unique p = p α such that the corresponding unique solution of(11),that is,the pair

satisfies v * ( s , α ) > 0 for s ∈ [ 0 , 1 ] , and ∫ 0 1 v * ( s , α ) d x = 1 .Hence, ( c * , v * ( s ) ) isthe unique solution of(9).

Proof. For any p ≥ 0 , recall the solution of (11) is v ( s , p , α ) ≡ V ( s , p , α , c ( p ) ) . Let I ( p , α ) = ∫ 0 1 v ( s , p , α ) d s , we need to prove that there exists a unique p = p α such that I ( p α , α ) = 1 .

1) By (12), we have

The second inequality is impossible when p > 0 . Hence, v ( s , p , α ) > 0 and min v ( ⋅ , p , α ) → + ∞ when p → + ∞ by the first inequality. This means that I ( p , α ) → + ∞ as p → + ∞ .

When p = 0 , by (12) we know that

Since ( H M 2 − H m 2 ) tan 2 α ≤ H m 2 , it follows that max s ∈ [ 0 , 1 ] v ( s , p , α ) ≤ 1 . Clearly, v ( s , α ) ≡ 1 , hence I ( 0 , α ) < 1 . In fact, if v ( s , α ) ≡ 1 , then ψ ( z ) = z sin α . So u ( x , t ) = c t sin α is the periodic traveling wave of (3), but u ( x , t ) = c t sin α does

not satisfy (4).

2) Denote V ¯ ( p , α , c ( p , α ) ) : = V ( 1 , p , α , c ( p , α ) ) , then by V ¯ ( p , α , c ( p , α ) ) = p , we have V ¯ p + V ¯ c ⋅ c p = 1 , and so c p = ( 1 − V ¯ p ) / V ¯ c . Using the notation μ in the proof of Lemma 2, we have

The above two assertions imply that there exists a unique p α > 0 such that I ( p α , α ) = 1 . Thus, ( c * ( α ) , v * ( s , α ) ) defined by (18) is the unique solution of (9) and v * ( s , α ) satisfies that ∫ 0 1 v * ( s , α ) d x = 1 and v * ( s , α ) > 0 for s ∈ [ 0 , 1 ] .

Lemma5.Let ( c * ( α ) , v * ( s , α ) ) be the unique solution obtained in Lemma4,then(10) holds.

Proof. v * ( s , α ) > 0 for s ∈ [ 0 , 1 ] and ∫ 0 1 v * ( s , α ) d x = 1 imply that 0 < min s ∈ [ 0 , 1 ] v * ( s , α ) ≤ 1 ≤ max s ∈ [ 0 , 1 ] v * ( s , α ) . Hence, (10) follows from (12).

Proof of Theorem1. From Theorem 2 and Lemma 2, we easily obtain that Problem (9) has a unique solution ( c * , ψ * ( z ) ) , and (10) holds. Let

By using Lemma 1, it is easy to check that

is X α -periodic in x and T α -periodic in t . From the discussion in Section 2, we follow that Equation (3) has a unique periodic traveling wave u * ( x , t ) = φ * ( x , t ) + c * t , x ∈ ℝ , t ∈ ℝ , and H m ≤ c * ≤ H M .