Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 76, pp. 1–30. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS ON R2 BUI DUC NAM, BUI DAI NGHIA, NGUYEN ANH TUAN Abstract. In this article, we study a non-local Love problem on unbounded domains where the non-locality in the main equation is interpreted as a fractional Laplacian operator. With various assumptions on the initial conditions, we derive several estimates for mild solutions for the homogeneous source scenario. For the nonlinear problem, we show the existence and uniqueness of a global mild solution. In two cases, we obtain convergence results. The first one states that the solution to the fractional Love equation converges to the mild solution of the fractional wave equation according to a cross-section radius parameter. The second result shows that solutions of the fractional Love equation incorporating the fractional Laplacian operator converge to those of the classical problem, involving the usual Laplacian, as the fractional orders approach 1. This work is the first that we are aware of that deals with mild solutions of Love equations on unbounded domains. 1. Introduction In this article, we study the solution u(x, y, t) : R× R× [0,∞) → R to the equation utt (I + k(−∆)s) + (−∆)θu = G(u), (1.1) associated with conditions u(x, y, 0) = a(x, y), ut(x, y, 0) = b(x, y). (1.2) Here k is a positive constant. a, b are initial state functions, and G is the source term that describes the external forces. For s, θ ∈ (0, 1), (−∆)s, and (−∆)θ are the nonlocal (or fractional) Laplace operators which are defined in [7, Theorem 1.1a]. 1.1. State of the art and main contributions. When θ = s = 1, Equation (1.1) becomes the classical Love’s equation utt −∆u− k∆utt = 0, (1.3) which was derived by Love in [8]. This equation models how solid bodies deform under stress, which is essential to the theory of elasticity. It clarifies how elastic materials respond to outside forces, particularly with regard to stress and strain analysis. Equation (1.3) seems to be similar to the wave models considered in [3, 4], but requires a different approach because the presence of the fractional Laplacian operator of different orders. In this work, we investigate the properties of solution to a different version of Equation (1.3), obtained by additional considering of external forces (via G(u)) and nonlocal effects (via fractional Laplacians). In the literature, there are many works on modified forms of (1.3). However, it appears that no study has taken the same approach as our study. Let us briefly review some of the top general references to clarify the motivation behind this work. The forms of tension and variational motion (used to construct (1.3)) were adjusted by Radochová [15] in 1978 to obtain the equation utt − E ρ uxx − 2µ2k2uxxtt = 0, 2020 Mathematics Subject Classification. 35L05, 35Q74, 35B40. Key words and phrases. Love type equation; global solution; regularity; behavior of solutions; convergence. ©2025. This work is licensed under a CC BY 4.0 license. Submitted March 19, 2025. Published July 22, 2025. 1 2 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 where µ, E, and ρ represent, respectively, the displacement, Young’s modulus of the material, and the mass density. This model describes rod vibrations caused by extension. In [12], Ngoc et al. studied a nonlinear Love equation in the one dimensional domain utt(x, t)− uxx(x, t)− uxxtt = G(x, t, u, ut), 0 < x < 1, 0 < t < T, u(0, t) = u(1, t) = 0, 0 < t < T, u(x, 0) = ũ0(x), ut(x, 0) = ũ1(x), 0 < x < 1, (1.4) where ũ0 and ũ1 are functions that represents the initial state and G is a nonlinear source term. They applied the Faedo-Galerkin method to show the existence of a local weak solution to Problem (1.4). Ngoc, Duy and Long [10] focused on the one-dimensional nonlinear Love equation utt − uxx − εuxxtt + λ |ut|q−2 ut +K|u|p−2u = G(x, t), 0 < x < 1, 0 < t < T, εuxtt(0, t) + ux(0, t) = hu(0, t) + g(t), u(1, t) = 0 0 < t < T, u(x, 0) = ũ0(x), ut(x, 0) = ũ1(x), 0 < x < 1, where p > 1, q > 1, ε > 0, λ > 0, K > 0, h ≥ 0 are real numbers and ũ0, ũ1, G, g are given functions, which satisfy some appropriate assumptions. The aforementioned problem has primarily been studied using the Faedo-Galerkin approach, the compactness method, and the monotone method. The authors obtained the existence, uniqueness, regularity and asymptotic behavior of the weak solution. Following the work [10], Ngoc and Long [11] investigated the solution u(x, t) : (0, 1)× [0, T ) → R to the nonlinear Love equation utt − uxx − uxxtt − λ1uxxt + λut = F (x, t, u, ux, ut, uxt)− ∂ ∂x [G (x, t, u, ux, ut, uxt)] +G(x, t), u(x, 0) = ũ0(x), ut(x, 0) = ũ1(x), where λ, λ1 > 0 are constants and ũ0, ũ1 ∈ H1 and F,G have been supposed satisfying some necessary requirements. In their study, the authors proved the existence of a weak solution by using the Faedo-Galerkin method. They also obtained the results of blow-up and decay of the weak solutions. Zennir et al. [16, 2, 17] studied the nonlinear Love-equation associated with infinite memory. They proved the local existence and uniqueness of weak solutions by combining the linearization method, the Faedo-Galerkin approximation, and the theory of weak compactness. They also obtained the existence of a global weak solution under some appropriate assumptions on the initial datum and the kernel function. Furthermore, in certain instances, the finite time blow up of weak solutions was also investigated. Another version with biharmonic and polynomial nonlinearities was considered in [9]. Precisely, Xu and Liu have studied the multidimensional double dispersion equations utt −∆u−∆utt +∆2u = ∆G(u), x ∈ Rn, where G(u) = a|u|p. Using potential well method, they showed the existence and nonexistence of global weak solutions without establishing the local existence theory. They also provided some sharp c onditions for global wellposedness using the Galerkin method. While numerous intriguing articles have explored the Love type equation, as previously men- tioned, the analysis of Equation (1.3) in an unbounded domain has not yet been studied extensively. Our paper appears to be the first to study mild solutions to the Love equation on R2. The funda- mental difference between our work and many previous studies on the Love type equation is that we do not focus on weak solutions. Instead, we delve into the topic of mild solutions in unbounded domains. One of the most challenging problems we encounter is the presence of singular compo- nents in R2 integrals. From a technical standpoint, this makes the situation more difficult than problems in bounded domains. We refer the reader to (1.1) and the interesting papers [1, 6] for problems on Rn. These authors studied the initial-value problem for a general class of nonlinear nonlocal wave equations arising in one-dimensional nonlocal elasticity. They established the global existence of solutions and also investigated the conditions for finite-time blow-up. Our principal contributions in this paper are described in the following. EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 3 • Firstly, we focus on the regularity of mild solutions to Problem (1.1)-(1.2) when G ≡ 0. To accomplish this, we introduce specific techniques to handle integrals in R2. • Secondly, we examine the global well-posedness of Problem (1.1)-(1.2). We establish an upper bound on the solution in various function spaces, assuming certain conditions on the initial data. The main challenge arises when addressing the existence of solutions to the nonlinear problem. To obtain global results, we employ a delicate norm in a weighted space, using methods distinct from those in [1, 6]. • Thirdly, we explore the convergence of the mild solution as θ, s → 1−. This interest is inspired by a recent article by Oscar and Loachaman [13], where they demonstrate that solutions to the fractional Navier-Stokes equations, involving the fractional Laplacian operator (−∆)s with 1 2 < s < 1, converge to a solution of the classical case with −∆ as s → 1−. This motivates us to investigate whether a similar phenomenon occurs in Problem (1.1)-(1.2). • Lastly, we show that the solution to the fractional Love equation converges to the solution of the fractional wave equation as k → 0. A recent paper by Nam et al. [14] analyzed this convergence for the homogeneous Love equation as k → 0. Motivated by their work, we demonstrate the convergence of the mild solution for the nonlinear Love Problem (1.1)- (1.2) as k → 0, and we also provide an error estimate for this convergence. 1.2. Notation and outline. In our estimations we will denote the implied positive constant by C, whose value may vary from line to line. When the dependence of C on some parameters β need to be specified, we write Cβ . The symbol T always stands for a positive finite constant. We also use the notation ∫∫ A×B (·)dξ dη instead of ∫ A ∫ B (·)dξ dη. The Fourier transform of a function G(x, y) is defined by f̂(ξ, η) := ∫∫ R2 e−ixξ−iyηG(x, y) dx dy. We recall the notion of non-homogeneous and homogeneous Sobolev spaces Hm(R2), Ḣm(R2) of order m ≥ 0 as follows Hm(R2) := {tempered distribution f such that f̂ ∈ L2 loc(R2) and ∥f∥2Hm < ∞}, r Ḣm(R2) := {tempered distribution f such that f̂ ∈ L1 loc(R2) and ∥f∥2 Ḣm < ∞}, where ∥f∥Hm(R2) := (∫∫ R2 (1 + ξ2 + η2)m|f̂(ξ, η)|2 dξ dη )1/2 , ∥f∥Ḣm(R2) := (∫∫ R2 (ξ2 + η2)m|f̂(ξ, η)|2 dξ dη )1/2 . Remark 1.1. The family of Hm(R2) is decreasing with respect to m ≥ 0. The space Ḣm(R2) is a Hilbert space if and only if m < 1. This article is organized as follows. In section 2, we give some preliminaries. Section 3 provides the regularity of the mild solution. Theorem 2.1 shows an upper bound on the mild solution. Theorem 2.2 considers the convergence of solutions to the fractional Love equation when k → 0+. In Theorem 2.3, we show that the solution of the fractional Love equation converges to the solution of the classical Love equation. 2. Homogeneous case We devote this section to studying Equation (1.1) in the homogeneous case, i.e., G ≡ 0. We first introduce the definition of mild solutions to Problem (1.1). Obviously, if u(x, y, t) is a smooth 4 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 solution to Problem (1.1)-(1.2) with G ≡ 0, its Fourier representation û(ξ, η, t) in the frequency space will satisfy d2 dt2 û(ξ, η, t) + (ξ2 + η2)θû(ξ, η, t) + k(ξ2 + η2)s d2 dt2 û(ξ, η, t) = 0, û(ξ, η, 0) = â(ξ, η), d dt û(ξ, η, 0) = b̂(ξ, η). From this equation, we obtain û(ξ, η, t) = cos (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) â(ξ, η) + √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) b̂(ξ, η). The inverse Fourier transform yields the following relation u(x, y, t) = P(t)a(x, y) +Q(t)b(x, y), (2.1) where P(t)v = F−1 ( cos (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) , Q(t)v = F−1 (√1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) . We then define the mild solution to Problem (1.1)-(1.2) with G ≡ 0 as a function u(x, y, t) : R× R× [0,∞) → R, satisfying Equation (2.1). 2.1. Regularity of mild solution. Theorem 2.1. Let p ≥ 0. We have the following results. (1) Suppose that γ = max{p, (θ− s)β + p} for s, θ > 0, and β > 0 such that (θ− s)β + p > 0. Let a ∈ Hγ(R2),and b ∈ Hp(R2). Then, there exist a positive constant Cβ, which only depends on β, such that ∥u(t)∥Hp(R2) ≤ ∥a∥Hp(R2) + CβT β ( 1 min(1, k) )β/2 ∥a∥H(θ−s)β+p(R2) + T∥b∥Hp(R2). (2) Suppose that a ∈ L1(R2) ∩ L2(R2) and b ∈ L1(R2) ∩Hµ(R2) with 0 < s < θ and 0 < µ < s− θ. Then, ∥u(t)∥L2(R2) ≤ C ( ∥a∥L1(R2) + ∥a∥L2(R2) + ∥b∥L1(R2) + ∥b∥L2(R2) + ∥b∥Hµ(R2) ) . Proof. (1) We begin by estimating P(t)a. Thank to the inequality | cos(y)| ≤ 1 + Cβy β , we can find the mentioned constant Cβ such that∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣ ≤ 1 + Cβ ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s ) β 2 tβ . Thus, it holds ∥P(t)a∥2Hp(R2) = ∫∫ R2 (1 + ξ2 + η2)p ∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2∣∣â(ξ, η)∣∣2 dξ dη ≤ ∫∫ R2 (1 + ξ2 + η2)p ∣∣â(ξ, η)∣∣2 dξ dη + C2 βT 2β ∫∫ R2 (1 + ξ2 + η2)p ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s )β∣∣â(ξ, η)∣∣2 dξ dη. (2.2) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 5 The first term on the right-hand side (RHS) is bounded by∫∫ R2 (1 + ξ2 + η2)p ∣∣â(ξ, η)∣∣2 dξ dη = ∥a∥2Hp(R2). We treat the second term on the RHS of (2.2) as follows. Using the inequality (1 + d)s ≤ 1 + ds, we find that min(1, k)(1 + ξ2 + η2)s ≤ min(1, k) ( 1 + (ξ2 + η2)s ) ≤ min(1, k) + min(1, k)(ξ2 + η2)s ≤ 1 + k(ξ2 + η2)s. Accordingly, one has (1 + ξ2 + η2)p ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s )β ≤ ( 1 min(1, k) )β (1 + ξ2 + η2)(θ−s)β+p. Thus, we obtain immediately that C2 βT 2β ∫∫ R2 (1 + ξ2 + η2)p ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s )β∣∣â(ξ, η)∣∣2 dξ dη ≤ C2 βT 2β ( 1 min(1, k) )β ∫∫ R2 (1 + ξ2 + η2)(θ−s)β+p ∣∣â(ξ, η)∣∣2 dξ dη = C2 βT 2β ( 1 min(1, k) )β ∥a∥2H(θ−s)β+p(R2). From the assumption a ∈ Hp(R2), we obtain ∥P(t)a∥Hp(R2) ≤ ∥a∥Hp(R2) + CβT β ( 1 min(1, k) )β/2∥a∥H(θ−s)β+p(R2). (2.3) We turn to estimate the quantity Q(t)b. In view of the basic inequality | sin(y)| ≤ y, y ≥ 0 we find that 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2 ≤ 1 + k(ξ2 + η2)s (ξ2 + η2)θ tε ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s ) ≤ T. This implies that ∥Q(t)b∥2Hp(R2) = ∫∫ R2 (1 + ξ2 + η2)p 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2∣∣̂b(ξ, η)∣∣2 dξ dη ≤ T ∫∫ R2 (1 + ξ2 + η2)p ∣∣̂b(ξ, η)∣∣2 dξ dη. Therefore, by the assumption b ∈ Hp(R2), we obtain ∥Q(t)b∥Hp(R2) ≤ T∥b∥Hp(R2). (2.4) Combining (2.3) and (2.4), we have ∥u(t)∥Hp(R2) ≤ ∥P(t)a∥Hp(R2) + ∥Q(t)b∥Hp(R2) ≤ ∥a∥Hp(R2) + CβT β ( 1 min(1, k) )β/2 ∥a∥H(θ−s)β+p(R2) + T∥b∥Hp(R2). (2) Based on the simple fact( (ξ2 + η2)θ 1 + k(ξ2 + η2)s )β ≤ k−β(ξ2 + η2)(θ−s)β , 6 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 we can find that∫∫ R2 ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s )β∣∣â(ξ, η)∣∣2 dξ dη ≤ k−β ∫∫ R2 (ξ2 + η2)(θ−s)β ∣∣â(ξ, η)∣∣2 dξ dη. (2.5) To derive the upper bound for the RHS, we make the decomposition∫∫ R2 (ξ2 + η2)(θ−s)β ∣∣â(ξ, η)∣∣2 dξ dη = ∫∫ ξ2+η2>1 (ξ2 + η2)(θ−s)β ∣∣â(ξ, η)∣∣2 dξ dη + ∫∫ ξ2+η2≤1 (ξ2 + η2)(θ−s)β ∣∣â(ξ, η)∣∣2 dξ dη =: J1 + J2. On the one hand, the term J1 can be easily bounded as J1 = ∫∫ ξ2+η2>1 (ξ2 + η2)(θ−s)β ∣∣â(ξ, η)∣∣2 dξ dη = ∫∫ ξ2+η2>1 |â(ξ, η)|2 (ξ2 + η2)(s−θ)β dξ dη ≤ ∫∫ ξ2+η2>1 ∣∣â(ξ, η)∣∣2 dξ dη ≤ ∥a∥2L2(R2). On the other hand, it is easy to see that∣∣â(ξ, η)∣∣ = ∣∣∣ 1 2π ∫∫ R2 e−ixξ−iyηa(x, y) dx dy ∣∣∣ ≤ ∥a∥L1(R2). This estimate implies that J2 = ∫∫ ξ2+η2≤1 (ξ2 + η2)(θ−s)β ∣∣â(ξ, η)∣∣2 dξ dη ≤ ∥a∥2L1(R2) ∫∫ ξ2+η2≤1 1 (ξ2 + η2)(s−θ)β dξ dη. Let us set ξ = r cosφ and η = r sinφ. Then∫∫ ξ2+η2≤1 1 (ξ2 + η2)(s−θ)β dξ dη = ∫∫ (0,2π)×(0,1) rdrdφ r2(s−θ)β = ∫∫ (0,2π)×(0,1) r1−2(s−θ)βdrdφ = π 1− (s− θ)β , where we choose β > 0 such that (s− θ)β < 1. From all the above observations, we obtain∫∫ R2 (ξ2 + η2)(θ−s)β ∣∣â(ξ, η)∣∣2 dξ dη ≤ ( 1 + π 1− (s− θ)β )( ∥a∥2L2(R2) + ∥a∥2L1(R2) ) . (2.6) Combining (2.5) and (2.6), we obtain∫∫ R2 ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s )β∣∣â(ξ, η)∣∣2 dξ dη ≤ ( 1 + π 1− (s− θ)β )( ∥a∥2L2(R2) + ∥a∥2L1(R2) ) . (2.7) Combining (2.2) and (2.7) yields ∥P(t)a∥2L2(R2) ≤ ∫∫ R2 ∣∣â(ξ, η)∣∣2 dξ dη + C2 βT 2β ∫∫ R2 ( (ξ2 + η2)θ 1 + k(ξ2 + η2)s )β∣∣â(ξ, η)∣∣2 dξ dη ≤ Ck,β,s,θ,T ( ∥a∥2L2(R2) + ∥a∥2L1(R2) ) . (2.8) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 7 Next, we derive the estimate for Q(t)b with the assumption that b ∈ L1(R2) ∩ Hµ(R2). The similar techniques as in the first part help us to deduce ∥Q(t)b∥2L2(R2) = ∫∫ R2 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2∣∣̂b(ξ, η)∣∣2 dξ dη ≤ CεT 2ε ∫∫ R2 ( 1 + k(ξ2 + η2)s )1−ε (ξ2 + η2)θ−θε ∣∣̂b(ξ, η)∣∣2 dξ dη, (2.9) for some ε ∈ (0, 1). Using (1 + d)s ≤ 1 + ds, one has( 1 + k(ξ2 + η2)s )1−ε ≤ 1 + k1−ε(ξ2 + η2)s(1−ε). As a consequence, we derive∫∫ R2 ( 1 + k(ξ2 + η2)s )1−ε (ξ2 + η2)θ−θε ∣∣̂b(ξ, η)∣∣2 dξ dη ≤ ∫∫ R2 1 (ξ2 + η2)θ−θε ∣∣̂b(ξ, η)∣∣2 dξ dη + k1−ε ∫∫ R2 1 (ξ2 + η2)(θ−s)(1−ε) ∣∣̂b(ξ, η)∣∣2 dξ dη = J ′ 1 + J ′ 2. (2.10) Using again that ∣∣̂b(ξ, η)∣∣ = ∣∣∣ 1 2π ∫∫ R2 e−ixξ−iyηb(x, y) dx dy ∣∣∣ ≤ ∥b∥L1(R2), one has J ′ 1 = ∫∫ ξ2+η2>1 1 (ξ2 + η2)θ−θε ∣∣̂b(ξ, η)∣∣2 dξ dη + ∫∫ ξ2+η2≤1 1 (ξ2 + η2)θ−θε ∣∣̂b(ξ, η)∣∣2 dξ dη ≤ ∫∫ R2 ∣∣̂b(ξ, η)∣∣2 dξ dη + ∥b∥2L1(R2) ∫∫ ξ2+η2≤1 1 (ξ2 + η2)θ−θε dξ dη ≤ ∥b∥2L2(R2) + ∥b∥2L1(R2) ∫∫ ξ2+η2≤1 1 (ξ2 + η2)θ−θε dξ dη. (2.11) We use the change of variables ξ = r cosφ and η = r sinφ. Then∫∫ ξ2+η2≤1 1 (ξ2 + η2)θ−θε dξ dη = ∫∫ (0,2π)×(0,1) rdrdφ r2θ−2θε = ∫∫ (0,2π)×(0,1) r1−2θ+2θεdrdφ = π 1− θ + θε . It follows from (2.11) that J ′ 1 ≤ ( 1 + π 1− θ + θε )( ∥b∥2L2(R2) + ∥b∥2L1(R2) ) . (2.12) Let us now consider the term J ′ 2. Since µ < (s− θ), we can choose ε > 0 such that ε = 1− µ s− θ . Then, we obtain J ′ 2 = k1−ε ∫∫ R2 1 (ξ2 + η2)(θ−s)(1−ε) ∣∣̂b(ξ, η)∣∣2 dξ dη = k1−ε ∫∫ R2 (ξ2 + η2)(s−θ)(1−ε) ∣∣̂b(ξ, η)∣∣2 dξ dη = k1−ε∥b∥2Hµ(R2). (2.13) Combining (2.9), (2.10), (2.12) and (2.13), we deduce that ∥Q(t)b∥2L2(R2) ≤ CεT 2ε ( 1 + π 1− θ + θε + k1−ε )( ∥b∥2L2(R2) + ∥b∥2L1(R2) + ∥b∥2Hµ(R2) ) . (2.14) 8 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 Combining (2.8) and (2.14) yields ∥u(t)∥L2(R2) ≤ C ( ∥a∥L1(R2) + ∥a∥L2(R2) + ∥b∥L1(R2) + ∥b∥L2(R2) + ∥b∥Hµ(R2) ) . The proof is complete. □ 2.2. Convergence of the mild solution. Through the remainder of this section, we assume that G and b are identically zero. This part includes two main results. Firstly, we show that the mild solution to Problem (1.1) converges to a solution of the homogeneous fractional wave equation with the same initial data. More precisely, suppose that u is the mild solution to Problem (1.1)-(1.2) and w is the mild solution to the wave equation wtt + (−∆)θw = 0, in R2 × (0, T ], (2.15) with initial data w(x, y, 0) = a(x, y), wt(x, y, 0) ≡ 0. (2.16) We show that u converges to w as k → 0+. The second goal is to prove that the mild solution u to Problem (1.1)-(1.2), behaves like the solution v of the homogeneous classical Love equation vtt (I − k∆)−∆v = 0, (2.17) with initial conditions v(x, y, 0) = a(x, y), vt(x, y, 0) ≡ 0, (2.18) as s, θ reach 1−. The first result reads as follows. Theorem 2.2. Let a ∈ Hρ(R2) such that 0 ≤ θ < ρ < θ + 2s. Then ∥u− w∥2L∞(0,T ;L2(R2)) ≤ Tk 2s+θ−ρ 2s ∥a∥Hρ(R2). (2.19) Proof. Note that the mild solution to Problem (1.1)-(1.2), with G, b ≡ 0, satisfies u(x, y, t) = 1 2π ∫∫ R2 cos (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) â(ξ, η)eixξ+iyη dξ dη, (2.20) and the mild solution to Problem (2.15)-(2.16) is given by w(x, y, t) = 1 2π ∫∫ R2 [ cos (√ (ξ2 + η2)θt ) â(ξ, η) ] eixξ+iyη dξ dη. (2.21) In view of the inequality | cos(α1)− cos(α2)| ≤ |α1 − α2| for any α1, α2 ∈ R, we find that∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − cos (√ (ξ2 + η2)θt )∣∣∣ ≤ t ∣∣∣√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s − √ (ξ2 + η2)θ ∣∣∣ ≤ T (ξ2+η2)θ 1+k(ξ2+η2)s − (ξ2 + η2)θ√ (ξ2+η2)θ 1+k(ξ2+η2)s + √ (ξ2 + η2)θ ≤ T k(ξ2 + η2)θ+s( 1 + k(ξ2 + η2)s ) (ξ2 + η2)θ/2 = T k(ξ2 + η2)s+ θ 2 1 + k(ξ2 + η2)s . (2.22) Using the inequality 1 + z ≥ zγ for 0 < γ < 1, we obtain 1 + k(ξ2 + η2)s ≥ kγ(ξ2 + η2)sγ . (2.23) Combining (2.22) and (2.23) gives us∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − cos (√ (ξ2 + η2)θt )∣∣∣ ≤ Tk1−γ(ξ2 + η2)s−sγ+ θ 2 . (2.24) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 9 From (2.21) and (2.20), we obtain ∥u(x, y, t)− w(x, y, t)∥2L2(R2) = ∫∫ R2 ∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) cos (√ (ξ2 + η2)θt )∣∣∣2|â(ξ, η)|2 dξ dη ≤ T 2k2−2γ ∫∫ R2 (ξ2 + η2)2s−2sγ+θ|â(ξ, η)|2 dξ dη. (2.25) Let γ = 2s+θ−ρ 2s , and note that 0 < γ < 1. The estimate (2.25) with the choice γ = 2s+θ−ρ 2s implies the desired result (2.19). □ The next theorem states the second goal of this subsection. Theorem 2.3. Suppose that a ∈ L1(R2) ∩ Ḣ2ε(R2) for all 0 < ε < 1 2 . Then ∥u(t)− v(t)∥2L2(R2) ≤ Cε,T,k,µ,s,θ ( (1− θ)sε + (1− θ)2θε + (1− s)(s+1)ε + (1− s)2µε )( ∥a∥2 Ḣ2ε(R2) + ∥a∥2L1(R2) ) . Here µ is a positive constant satisfying 0 < µ < 1. We momentarily postpone the proof of the theorem to consider the following auxiliary lemma. Lemma 2.4. The following inequalities are satisfied. (1) Let z ≥ 1 and 0 < θ ≤ 1. Then for 0 < β < 1 we obtain |zθ − z| ≤ Cβz 1+β(1− θ)β . (2) Let 0 < z < 1 and 0 < θ < 1. Then for 0 < ϑ ≤ 1 we obtain |zθ − z| ≤ Cϑz θ−ϑ(1− θ)ϑ. Proof. If z ≥ 1 then using the inequality 1− e−y ≤ Cβy β for all 0 < β < 1, we obtain |zθ − z| = z − zθ = z ( 1− z−(1−θ) ) = z ( 1− e−(1−θ) log(z) ) ≤ Cβz(1− θ)β logβ(z) ≤ Cβz 1+β(1− θ)β . If 0 < z < 1, then |zθ − z| = zθ − z = zθ ( 1− z(1−θ) ) = zθ ( 1− e−(1−θ) log( 1 z ) ) ≤ Cϑz θ(1− θ)ϑ logϑ(1/z) ≤ Cϑz θ−ϑ(1− θ)ϑ. □ Proof of Theorem 2.3. In view of the inequality | cos(α1)−cos(α2)| ≤ Cε|α1−α2|ε for any α1, α2 > 0 and 0 < ε ≤ 1, we find that∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − cos (√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣ ≤ Cεt ε ∣∣∣√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣ε ≤ CεT ε ∣∣∣√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2)s ∣∣∣ε + CεT ε ∣∣∣√ (ξ2 + η2) 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣ε. 10 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 Bear in mind that v is the solution to (2.17)-(2.18). Then we have v(x, y, t) = 1 2π ∫∫ R2 cos (√ ξ2 + η2 1 + k(ξ2 + η2) t ) â(ξ, η)eixξ+iyη dξ dη. From this representation and (2.20) we have ∥u(t)− v(t)∥2L2(R2) = ∫∫ R2 ∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − cos (√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣2|â(ξ, η)|2 dξ dη ≤ Cε,T ∫∫ R2 | √ (ξ2 + η2)θ − √ (ξ2 + η2)|2ε( 1 + k(ξ2 + η2)s )ε |â(ξ, η)|2 dξ dη + Cε,T ∫∫ R2 ∣∣∣√ (ξ2 + η2) 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣2ε|â(ξ, η)|2 dξ dη =: M1(ξ, η, t) + M2(ξ, η, t). Step 1. Estimate of M1. By similar techniques as in the previous proof, we have M1(ξ, η, t) = Cε,T ∫∫ R2 ∣∣(ξ2 + η2)θ − (ξ2 + η2) ∣∣2ε∣∣√(ξ2 + η2)θ + √ (ξ2 + η2) ∣∣2ε(1 + k(ξ2 + η2)s )ε ∣∣â(ξ, η)∣∣2 dξ dη = Cε,T ∫∫ ξ2+η2≥1 ∣∣(ξ2 + η2)θ − (ξ2 + η2) ∣∣2ε∣∣√(ξ2 + η2)θ + √ (ξ2 + η2) ∣∣2ε(1 + k(ξ2 + η2)s )ε ∣∣â(ξ, η)∣∣2 dξ dη + Cε,T ∫∫ ξ2+η2<1 ∣∣(ξ2 + η2)θ − (ξ2 + η2) ∣∣2ε∣∣√(ξ2 + η2)θ + √ (ξ2 + η2) ∣∣2ε(1 + k(ξ2 + η2)s )ε ∣∣â(ξ, η)∣∣2 dξ dη =: M1,1(ξ, η, t) + M1,2(ξ, η, t). (2.26) In view of the first inequality of Lemma 2.4, the term M1,1 is controlled in the following manner M1,1(ξ, η, t) ≤ Cε,T ∫∫ ξ2+η2≥1 ∣∣(ξ2 + η2)θ − (ξ2 + η2) ∣∣2ε∣∣√(ξ2 + η2) ∣∣2ε(k(ξ2 + η2)s )ε |â(ξ, η)|2 dξ dη = k−εCε,T ∫∫ ξ2+η2≥1 ∣∣(ξ2 + η2)θ − (ξ2 + η2) ∣∣2ε (ξ2 + η2)s+sε |â(ξ, η)|2 dξ dη ≤ k−εCε,β,T (1− θ)2βε ∫∫ ξ2+η2≥1 (ξ2 + η2)−sε+2βε+2ε|â(ξ, η)|2 dξ dη, for some 0 < β < 1. We also note that∫∫ ξ2+η2≥1 (ξ2 + η2)−sε+2βε+2ε|â(ξ, η)|2 dξ dη ≤ ∫∫ R2 (ξ2 + η2)−sε+2βε+2ε|â(ξ, η)|2 dξ dη ≤ ∥a∥2 Ḣ(2β+2−s)ε(R2) . Hence, choosing β = s 2 ∈ (0, 1) yields M1,1(ξ, η, t) ≤ k−εCε,β,T (1− θ)2βε∥a∥Ḣ(2β+2−s)ε(R2) = k−εCε,s,T (1− θ)sε∥a∥2 Ḣ2ε(R2) . (2.27) For M1,2, we use the second inequality of Lemma 2.4 to obtain∣∣∣√(ξ2 + η2)θ − √ (ξ2 + η2) ∣∣∣ ≤ Cϑ(ξ 2 + η2) θ−ϑ 2 (1− θ)ϑ EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 11 for all 0 < ϑ < 1. This implies that M1,2(ξ, η, t) = Cε,T ∫∫ ξ2+η2<1 ∣∣√(ξ2 + η2)θ − √ (ξ2 + η2) ∣∣2ε( 1 + k(ξ2 + η2)s )ε |â(ξ, η)|2 dξ dη ≤ k−εC(ε, ϑ, T )(1− θ)2ϑε ∫∫ ξ2+η2<1 1 (ξ2 + η2)(ϑ+s−θ)ε |â(ξ, η)|2 dξ dη ≤ k−εC(ε, ϑ, T )(1− θ)2ϑε∥a∥2L1(R2) ∫∫ ξ2+η2<1 1 (ξ2 + η2)(ϑ+s−θ)ε dξ dη. Let us keep discussing the integral in the latter inequality. Set ξ = r cosφ and η = r sinφ. Then, since (ϑ+ s− θ)ε < 1, we obtain∫∫ ξ2+η2≤1 1 (ξ2 + η2)(ϑ+s−θ)ε dξ dη = ∫∫ (0,2π)×(0,1) rdrdφ r2(ϑ+s−θ)ε = ∫∫ (0,2π)×(0,1) r1−2(ϑ+s−θ)εdrdφ = π 1− (ϑ+ s− θ)ε . We choose ϑ = θ. It holds M1,2(ξ, η, t) ≤ k−εCε,θ,s,T (1− θ)2θε∥a∥2L1(R2). (2.28) By collecting (2.26), (2.27) and (2.28), we obtain M1(ξ, η, t) ≤ M1,1(ξ, η, t) + M1,2(ξ, η, t) ≤ k−εCε,s,T (1− θ)sε∥a∥2 Ḣ2ε(R2) + k−εCε,θ,s,T (1− θ)2θε∥a∥2L1(R2). (2.29) Step 2. Estimate of M2. Again, we decompose M2(ξ, η, t) as follows M2(ξ, η, t) = Cε,T ∫∫ R2 (ξ2 + η2)ε (√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε( 1 + k(ξ2 + η2) )ε( 1 + k(ξ2 + η2)s )ε |â(ξ, η)|2 dξ dη = Cε,T ∫∫ ξ2+η2≥1 (ξ2 + η2)ε (√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε( 1 + k(ξ2 + η2) )ε( 1 + k(ξ2 + η2)s )ε |â(ξ, η)|2 dξ dη + Cε,T ∫∫ ξ2+η2<1 (ξ2 + η2)ε (√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε( 1 + k(ξ2 + η2) )ε( 1 + k(ξ2 + η2)s )ε |â(ξ, η)|2 dξ dη = M2,1(ξ, η, t) + M2,2(ξ, η, t). Note that(√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε = k2ε ∣∣(ξ2 + η2)− (ξ2 + η2)s ∣∣2ε(√ 1 + k(ξ2 + η2) + √ 1 + k(ξ2 + η2)s )2ε (2.30) If ξ2 + η2 ≥ 1, then for any 0 < δ < 1, we have∣∣∣(ξ2 + η2)− (ξ2 + η2)s ∣∣∣2ε ≤ Cδ(1− s)2δε(ξ2 + η2)2(1+δ)ε. This allows us to obtain(√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε ≤ kεCδ(1− s)2δε(ξ2 + η2)2(1+δ)ε−ε. Furthermore, ( 1 + k(ξ2 + η2) )ε( 1 + k(ξ2 + η2)s )ε ≥ k2ε(ξ2 + η2)ε+sε. Hence, when ξ2 + η2 ≥ 1 it holds (ξ2 + η2)ε (√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε( 1 + k(ξ2 + η2) )ε( 1 + k(ξ2 + η2)s )ε ≤ k−ε(1− s)2δε(ξ2 + η2)(2δ+1−s)ε. 12 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 Thus, we have M2,1(ξ, η, t) ≤ Cε,T,k,δ(1− s)2δε ∫∫ ξ2+η2≥1 (ξ2 + η2)(2δ+1−s)ε|â(ξ, η)|2 dξ dη ≤ Cε,T,k,δ(1− s)2δε∥a∥2 Ḣ(2δ+1−s)ε(R2) . By setting 2δ = s+ 1, we know that δ ∈ (0, 1), then we obtain M2,1(ξ, η, t) ≤ Cε,T,k,δ(1− s)(s+1)ε∥a∥2 Ḣ2ε(R2) . (2.31) On the other hand, when ξ2 + η2 < 1, for any 0 < µ < 1 we find that∣∣∣(ξ2 + η2)− (ξ2 + η2)s ∣∣∣2ε ≤ Cµ(1− s)2µε(ξ2 + η2)2(s−µ)ε. And by (2.30) we have(√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε ≤ kε(ξ2 + η2)2(s−µ)ε−sε = kε(ξ2 + η2)(s−2µ)ε· Therefore, if ξ2 + η2 < 1, then (ξ2 + η2)ε (√ 1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s )2ε( 1 + k(ξ2 + η2) )ε( 1 + k(ξ2 + η2)s )ε ≤ Cµk −ε(1− s)2µε(ξ2 + η2)−2µε. Consequently, we obtain the estimate M2,2(ξ, η, t) ≤ Cε,T,k,µ(1− s)2µε ∫∫ ξ2+η2<1 (ξ2 + η2)−2µε|â(ξ, η)|2 dξ dη ≤ Cε,T,k,µ∥a∥2L1(R2)(1− s)2µε ∫∫ ξ2+η2<1 (ξ2 + η2)−2µε dξ dη· Let us set ξ = r cosφ and η = r sinφ. Then since 2µε < 1, we obtain∫∫ ξ2+η2<1 (ξ2 + η2)−2µε dξ dη = ∫∫ (0,2π)×(0,1) rdrdφ r2µε = ∫∫ (0,2π)×(0,1) r1−2µεdrdφ = π 1− µε . Thus, we deduce that M2,2(ξ, η, t) ≤ Cε,T,k,µ∥a∥2L1(R2)(1− s)2µε. (2.32) Combining (2.31) and (2.32), we deduce that M2(ξ, η, t) ≤ M2,1(ξ, η, t) + M2,2(ξ, η, t) ≤ Cε,T,k,δ,µ ( (1− s)(s+1)ε + (1− s)2µε )( ∥a∥2 Ḣ2ε(R2) + ∥a∥2L1(R2) ) . (2.33) By combining (2.29) and (2.33), we find that for 0 < µ < 1, ∥u(t)−v(t)∥2L2(R2) ≤ C ( (1−θ)sε+(1−θ)2θε+(1−s)(s+1)ε+(1−s)2µε )( ∥a∥2 Ḣ2ε(R2) +∥a∥2L1(R2) ) . This inequality completes the proof of the Theorem 2.3. □ 3. Nonlinear problem In this section, we focus on the semi-linear case of the Problem (1.1)-(1.2), i.e., G = G(u). It is necessary to introduce the mild formula of solutions for this case. Having found the formula for the homogeneous case, the mild representation for this case is easily derived. In fact, applying the Fourier transform to both sides of the Equation (1.1) yields d2 dt2 û(ξ, η, t) + (ξ2 + η2)θ 1 + k(ξ2 + η2)s û(ξ, η, t) = 1 1 + k(ξ2 + η2)s Ĝ(ξ, η, t) with û(ξ, η, 0) = â(ξ, η), quad d dt û(ξ, η, 0) = b̂(ξ, η). EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 13 This implies that û(ξ, η, t) = cos (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) â(ξ, η) + √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) b̂(ξ, η) + 1 1 + k(ξ2 + η2)s √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∫ t 0 sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s (t− τ) ) Ĝ(ξ, η, τ)dτ. From this equation, one can find that u(t) = P(t)a+Q(t)b+ ∫ t 0 Q(t− τ)G(τ)dτ. Here, we recall from the previous section that P(t)v := F−1 ( cos (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) Q(t)v := F−1 (√1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) and define Q(t)v := F−1 ( 1 1 + k(ξ2 + η2)s √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) . To prove the existence and uniqueness of the global mild solution to Problem (1.1)-(1.2), it is useful to consider smoothing effects of the solution operators P(t), Q(t) and Q(t). Lemma 3.1. Let p ≥ 0. The following results hold. • If f ∈ Hγ(R2) for γ = max{p, (θ− s)β+ p} for s, θ, β > 0 such that (θ− s)β+ p > 0, then we obtain ∥P(t)f∥Hp(R2) ≤ ∥f∥Hp(R2) + CβT β ( 1 min(1, k) )β/2 ∥f∥H(θ−s)β+p(R2). (3.1) • If f ∈ Hp+s−θ(R2) for s ≤ θ ≤ p+ s, then for M ≡ Q or M ≡ Q we obtain ∥M(t)f∥Hp(R2) ≤ √ T 22θ−s + 21−s + 2k∥f∥Hp+s−θ(R2). (3.2) Proof. Estimate (3.1) can be easily obtained by using Part 1 of the proof of Theorem 2.1. Thus, we consider only (3.2). By the Plancherel theorem, we find that ∥M(t)f∥2Hp(R2) ≤ ∫∫ R2 (1 + ξ2 + η2)p 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2∣∣Ĝ(ξ, η) ∣∣2 dξ dη = ∫∫ ξ2+η2<1 (1 + ξ2 + η2)p 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2∣∣Ĝ(ξ, η) ∣∣2 dξ dη + ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)p 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2∣∣Ĝ(ξ, η) ∣∣2 dξ dη =: S1 + S2. For the term S1, we use the fact that∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣2 ≤ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t2 to deduce that S1 ≤ t2 ∫∫ ξ2+η2<1 (1 + ξ2 + η2)p ∣∣Ĝ(ξ, η) ∣∣2 dξ dη 14 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 = t2 ∫∫ ξ2+η2<1 (1 + ξ2 + η2)θ−s(1 + ξ2 + η2)p+s−θ ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ t22θ−s ∫∫ ξ2+η2<1 (1 + ξ2 + η2)p+s−θ ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ T 22θ−s∥f∥2Hp+s−θ(R2). We now find the estimate for S2. Since ξ2 + η2 ≥ 1 and θ ≥ s we can use the inequality (1 + z)θ ≤ 1 + zθ for 0 < θ ≤ 1 to obtain 1 (ξ2 + η2)θ ≤ 2 1 + (ξ2 + η2)θ ≤ 2 (1 + ξ2 + η2)θ , k(ξ2 + η2)s (ξ2 + η2)θ = k (ξ2 + η2)θ−s ≤ 2k 1 + (ξ2 + η2)θ−s ≤ 2k (1 + ξ2 + η2)θ−s . Hence, we have immediately that 1 + k(ξ2 + η2)s (ξ2 + η2)θ ≤ 2 (1 + ξ2 + η2)θ + 2k (1 + ξ2 + η2)θ−s ≤ 21−s + 2k (1 + ξ2 + η2)θ−s . Thus, we obtain S2 ≤ (21−s + 2k) ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)p+s−θ ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ (21−s + 2k)∥f∥2Hp+s−θ(R2). From two above inequalities, we confirm that ∥M(t)f∥Hp(R2) ≤ √ T 22θ−s + 21−s + 2k∥f∥Hp+s−θ(R2). The proof is complete. □ 3.1. Existence and uniqueness of global mild solution. In this subsection, we show the unique existence of the global mild solution to the Problem (1.1)-(1.2). To this end, we first introduce a weighted solution space. For m, l > 0 and d ≥ 0, we denote by Xm,ℓ((0, T ];H d(R2)) the space of functions f : R2 × [0, T ] → R such that ∥G(t)∥Hd(R2) is a.e. bounded and ∥w∥Xm,ℓ((0,T ];Hd(R2)) := sup t∈(0,T ] tme−ℓt∥w(t)∥Hd(R2) < ∞. Theorem 3.2. Let s ≤ θ and η, d ≥ 0 such that 0 ≤ d + s − θ ≤ η ≤ d. Suppose that G(0) = 0 and ∥G(w1)−G(w2)∥Hη(R2) ≤ C∥w1 − w2∥Hd(R2), for all w1, w2 ∈ Hd(R2). (3.3) In addition, we presume that a ∈ H(θ−s)β+d(R2) for some β > 0 and b ∈ Hd+s−θ(R2). Then, Problem (1.1)-(1.2) has a unique global solution in Xm,ℓ0((0, T ];H d(R2)) for some sufficiently large l0 and m ∈ (0, 1). Furthermore, we have ∥u(t)∥Hd(R2) ≤ Ck,T,m,β,st −m ( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) ) . Proof. We define the operator Z as follows Zw(t) = P(t)a+Q(t)b+ ∫ t 0 Q(t− τ)G(w(τ))dτ. (3.4) Let w1, w2 be two arbitrary functions in Xm,ℓ((0, T ];H d(R2)). From (3.4) and (3.3), we find that ∥Zw1(t)− Zw2(t)∥Hd(R2) ≤ ∫ t 0 ∥Q(t− τ) ( G(w1(τ))−G(w2(τ)) ) ∥Hd(R2)dτ ≤ √ T 22θ−s + 21−s + 2k ∫ t 0 ∥G(w1(τ))−G(w2(τ))∥Hd+s−θ(R2) ≤ √ T 22θ−s + 21−s + 2k ∫ t 0 ∥G(w1(τ))−G(w2(τ))∥Hη(R2) (3.5) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 15 where we note that the embedding Hη(R2) ↪→ Hd+s−θ(R2) holds, by the assumption d+s−θ ≤ η. Thanks to the Lipschitz property (3.3) of F , Estimate (3.5) becomes ∥Zw1(t)− Zw2(t)∥Hd(R2) ≤ C √ T 22θ−s + 21−s + 2k ∫ t 0 ∥w1(τ)− w2(τ)∥Hd(R2) = C √ T 22θ−s + 21−s + 2k ∫ t 0 τ−meℓττme−ℓτ∥w1(τ)− w2(τ)∥Hd(R2)dτ ≤ C √ T 22θ−s + 21−s + 2k∥w1 − w2∥Xm,ℓ((0,T ];Hd(R2)) (∫ t 0 τ−meℓτdτ ) . Multiplying both sides of the above equation by tme−ℓt, we obtain tme−ℓt∥Zw1(t)− Zw2(t)∥Hd(R2) ≤ C √ T 22θ−s + 21−s + 2k tm (∫ t 0 τmeℓ(τ−t)dτ ) ∥w1 − w2∥Xm,ℓ((0,T ];Hd(R2)) = C √ T 22θ−s + 21−s + 2k ( t ∫ 1 0 ν−me−ℓt(1−ν)dν ) ∥w1 − w2∥Xm,ℓ((0,T ];Hd(R2)). (3.6) The next step is to control the integral quantity. This can be attained by using the following lemma from [5, Lemma 8]. Lemma 3.3. Let c > −1, d > −1 such that c + d ≥ −1, h > 0 and t ∈ [0, T ]. For h > 0, the following limit holds lim γ→∞ ( sup t∈[0,T ] th ∫ 1 0 rc(1− r)de−γt(1−r)dr ) = 0. Applying the above lemma yields lim ℓ→+∞ sup 0≤t≤T ( t ∫ 1 0 ν−me−ℓt(1−ν)dν ) = 0. Thus, there exists a constant ℓ0 such that sup 0≤t≤T ( t ∫ 1 0 ν−me−ℓ0t(1−ν)dν ) ≤ 1 2C √ T 22θ−s + 21−s + 2k . (3.7) Combining (3.6) and (3.7), we deduce that ∥Zw1 − Zw2∥Xm,ℓ0 ((0,T ];Hd(R2)) ≤ 1 2 ∥w1 − w2∥Xm,ℓ0 ((0,T ];Hd(R2)). (3.8) We also need to deal with the term Zin(t) := P(t)a+Q(t)b. Since a ∈ H(θ−s)β+d(R2) and b ∈ Hd+s−θ(R2), we apply Lemma 3.1 to obtain ∥Zin(t)∥Hd(R2) ≤ ∥P(t)a∥Hd(R2) + ∥Q(t)b∥Hd(R2) ≤ ∥a∥Hd(R2) + CβT β ( 1 min(1, k) )β/2∥a∥H(θ−s)β+d(R2) + √ T 22θ−s + 21−s + 2k∥b∥Hd+s−θ(R2). (3.9) Combining all the above estimates allows us to conclude that Z is a contraction mapping from the space Xm,ℓ0((0, T ];H d(R2)) to the space Xm,ℓ0((0, T ];H d(R2)). Thus, applying the Banach fixed point theory, we deduce that Z has a fixed point u ∈ Xm,ℓ0((0, T ];H d(R2)) which satisfies the integral equation u(t) = P(t)a+Q(t)b+ ∫ t 0 Q(t− τ)G(u(τ))dτ. Next, using (3.8), (3.9), and the triangle inequality, we derive that ∥u∥Xm,ℓ0 ((0,T ];Hd(R2)) ≤ 1 2 ∥u∥Xm,ℓ0 ((0,T ];Hd(R2)) + sup 0≤t≤T tme−ℓ0t∥Zin(t)∥Hd(R2) 16 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 ≤ 1 2 ∥u∥Xm,ℓ0 ((0,T ];Hd(R2)) + Tm∥a∥Hd(R2) + CβT β+m ( 1 min(1, k) )β/2∥a∥H(θ−s)β+d(R2) + √ T 22θ−s + 21−s + 2kTm∥b∥Hd+s−θ(R2). Hence, we arrive at the bound ∥u∥Xm,ℓ0 ((0,T ];Hd(R2)) ≤ Ck,T,m,β,s ( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) ) . From the definition of the space Xm,ℓ0((0, T ];H d(R2)), it follows from the above estimate that ∥u(t)∥Hd(R2) ≤ Ck,T,m,β,st −m ( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) ) . (3.10) □ 3.2. Convergence of mild solutions. Let us consider the nonlinear wave equation utt + (−∆)θu = G(u) (3.11) and the classical Love equation utt −∆u− k∆utt = G(u(x, y, t)), (3.12) associated with the initial data (1.2). In this subsection, we examine the convergence of the mild solution to Problem (1.1)-(1.2) to the mild solution to Problem (3.11)-(1.2) as k → 0 and to Problem (1.1)-(1.2) to the mild solution to Problem (3.12)-(1.2) as θ, s → 1. The first convergence result is stated in the following theorem. Theorem 3.4. Let s < θ, ρ ∈ (θ, 2s + θ) and η, d ≥ 0 satisfy the assumption of Theorem 3.2. For β > 0, α ≥ max(d + ρ, (θ − s)β + d). Furthermore, suppose that a, d is large enough that Problem (3.11)-(1.2) possesses a unique global mild solution. Then, if uk and u∗ are the mild solution, respectively, to Problem (1.1)-(1.2) and Problem (3.11)-(1.2) under the assumption that (a, b) ∈ Hα(R2)×Hd+s(R2) and G satisfies (3.3), G(0) = 0, for k ∈ (0, 1) we obtain ∥uk(t)− u∗(t)∥Hd(R2) ≤ C exp ( C √ T 22θ−s + 21−s + 2kt )( k ρ−θ 2s + k + k s θ+s ) × ( ∥a∥Hd+ρ(R2) + ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s(R2) ) . Remark 3.5. By Theorem 3.2, the conditions a ∈ Hα(R2) and b ∈ Hd+s(R2) as introduced in Theorem 3.4 ensure the global existence and uniqueness of the mild solution of Problem (1.1)- (1.2). Also, we note that the global existence and uniqueness of the mild solution to (3.11)-(1.2) can be obtained by similar arguments as in Theorem 3.2. Therefore, we omit the proof here and restrict our attention to the convergence problem. Proof. In this proof, we define the mild solution to Problem (3.11)-(1.2) as a function u∗ satisfying u∗(t) = P (t)a+Q(t)b+ ∫ t 0 Q(t− τ)G(u∗(τ))dτ, (3.13) where P (t)v = F−1 ( cos (√ (ξ2 + η2)θt ) v̂(ξ, η) ) , Q(t)v = F−1 (√ 1 (ξ2 + η2)θ sin (√ (ξ2 + η2)θt ) v̂(ξ, η) ) . EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 17 Using (2.24), we obtain that for any d ≥ 0 and γ > 0, ∥P(t)a− P (t)a∥2Hd(R2) = ∫∫ R2 (1 + ξ2 + η2)d ∣∣∣ cos(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − cos (√ (ξ2 + η2)θt )∣∣∣2|â(ξ, η)|2 dξ dη ≤ T 2k2−2γ ∫∫ R2 (1 + ξ2 + η2)d(ξ2 + η2)2s−2sγ+θ|â(ξ, η)|2 dξ dη. (3.14) Let γ = 2s+θ−ρ 2s where θ < ρ < 2s+ θ. We follows from (3.14) that ∥P(t)a− P (t)a∥Hd(R2) ≤ TCk,γ,εk ρ−θ 2s √∫∫ R2 (1 + ξ2 + η2)d+ρ|â(ξ, η)|2 dξ dη = Tk ρ−θ 2s ∥a∥Hd+ρ(R2). (3.15) Next, we deal with the difference between Q(t)b and Q(t)b. To this end, we introduce the following functions J1(ξ, η, t) := √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ ( sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − sin (√ (ξ2 + η2)θt )) , J2(ξ, η, t) := (√ 1 (ξ2 + η2)θ √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ ) sin (√ (ξ2 + η2)θt ) We first consider J1(ξ, η, t). Using the inequality | sin(α1)−sin(α2)| ≤ |α1−α2| for any α1, α2 > 0, one can derive∣∣∣J1(ξ, η, t)∣∣∣ ≤ √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t− √ (ξ2 + η2)θt ∣∣∣ = t √ 1 + k(ξ2 + η2)s ∣∣∣1−√ 1 + k(ξ2 + η2)s√ 1 + k(ξ2 + η2)s ∣∣∣ = kt √ 1 + k(ξ2 + η2)s (ξ2 + η2)s 1 + k(ξ2 + η2)s + √ 1 + k(ξ2 + η2)s . This immediately yields ∣∣J1(ξ, η, t)∣∣ ≤ Ctk1/2(ξ2 + η2)s/2. At this point, we can deduce that∫∫ R2 (1 + ξ2 + η2)d ∣∣∣J1(ξ, η, t)∣∣∣2∣∣̂b(ξ, η)∣∣2 dξ dη ≤ Ct2k ∫∫ R2 (1 + ξ2 + η2)d+s ∣∣̂b(ξ, η)∣∣2 dξ dη. We proceed to estimate the term J2. It can be handled by some basic calculations. In fact, we first see that∣∣∣√ 1 (ξ2 + η2)θ − √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ ∣∣∣ = (ξ2 + η2)− θ 2 ∣∣∣1−√ 1 + k(ξ2 + η2)s ∣∣∣ = k(ξ2 + η2)s− θ 2 1 + √ 1 + k(ξ2 + η2)s ≤ k1/2(ξ2 + η2) s−θ 2 . Then, if ξ2 + η2 ≥ 1, the RHS of the above inequality is obviously bounded. If ξ2 + η2 < 1, we use the basic inequality sin (√ (ξ2 + η2)θt ) ≤ √ (ξ2 + η2)θt to find that ∣∣J2(ξ, η, t)∣∣ ≤ k1/2t(ξ2 + η2)s/2. 18 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 Based on this result, we obtain∫∫ R2 (1 + ξ2 + η2)d ∣∣∣J2(ξ, η, t)∣∣∣2∣∣̂b(ξ, η)∣∣2 dξ dη ≤ t2k ∫∫ R2 (1 + ξ2 + η2)d+s ∣∣̂b(ξ, η)∣∣2 dξ dη. Combining estimates for J1 and J2 yields ∥Q(t)a−Q(t)a∥Hd(R2) ≤ CTk1/2∥b∥Hd+s(R2). (3.16) Let us estimate the term ∥Q(t)f − Q(t)f∥Hd(R2). This can be achieved by using again the Plancherel theorem. Indeed, we have 1 1 + k(ξ2 + η2)s √ 1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − √ 1 (ξ2 + η2)θ sin (√ (ξ2 + η2)θt ) = √ 1 (ξ2 + η2)θ [ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − sin (√ (ξ2 + η2)θt )] + (√1 + k(ξ2 + η2)s (ξ2 + η2)θ − √ 1 (ξ2 + η2)θ ) sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) + ( 1 1 + k(ξ2 + η2)s − 1 )√1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) =: J3(ξ, η) + J4(ξ, η) + J5(ξ, η). (3.17) Let us first to treat the term J3(ξ, η). In view of the inequality | sin(α1)− sin(α2)| ≤ Cε|α1 −α2|ε for any α1, α2 > 0 and 0 < ε ≤ 1, we arrive at |J3(ξ, η)| ≤ Cε √ 1 (ξ2 + η2)θ ∣∣∣√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t− √ (ξ2 + η2)θt ∣∣∣ε = Cεt ε(ξ2 + η2) εθ−θ 2 (1−√ 1 + k(ξ2 + η2)s√ 1 + k(ξ2 + η2)s )ε = Cεt ε(ξ2 + η2) εθ−θ 2 kε (ξ2 + η2)εs( 1 + k(ξ2 + η2)s + √ 1 + k(ξ2 + η2)s )ε . (3.18) Using the inequality( 1 + k(ξ2 + η2)s + √ 1 + k(ξ2 + η2)s )ε > k ε 2 (ξ2 + η2) εs 2 , it follows from (3.18) that |J3(ξ, η)| ≤ Cεt εk ε 2 (ξ2 + η2) εθ+εs−θ 2 , 0 < ε ≤ 1. (3.19) Let us choose ε = 1 then since (3.19), we obtain |J3(ξ, η)| ≤ Tk1/2(ξ2 + η2)s/2, (3.20) where we note that Cε = 1 if ε = 1. Let us choose ε = s θ+s ∈ (0, 1) and if ξ2 + η2 ≥ 1, then since (3.19), we have immediately that∣∣J3(ξ, η)∣∣ ≤ Cs,θT s θ+s k s 2θ+2s (ξ2 + η2) s−θ 2 ≤ 2Cs,θT s θ+s k s 2θ+2s (1 + ξ2 + η2) θ−s 2 , (3.21) where we have used that 1 (ξ2 + η2) θ−s 2 ≤ 2 1 + (ξ2 + η2) θ−s 2 ≤ 2 (1 + ξ2 + η2) θ−s 2 . EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 19 By combining (3.20) and (3.21), one obtains the bound∫∫ R2 (1 + ξ2 + η2)d|J3(ξ, η)|2 ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ ∫∫ ξ2+η2≤1 (1 + ξ2 + η2)θ−s(1 + ξ2 + η2)d+s−θkT 2(ξ2 + η2)s ∣∣Ĝ(ξ, η) ∣∣2 dξ dη + 4Cs,θT 2s θ+s k s θ+s ∫∫ ξ2+η2>1 (1 + ξ2 + η2)d+s−θ ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ ( 2θ−sT 2k + 4Cs,θT 2s θ+s k s θ+s )∫∫ R2 (1 + ξ2 + η2)d+s−θ ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ ( 2θ−sT 2k + 4Cs,θT 2s θ+s k s θ+s ) ∥G∥2Hd+s−θ(R2). (3.22) Let us consider the term J4(ξ, η). Using the inequality sin(z) ≤ z, we know that∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣ ≤ √ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ≤ √ (ξ2 + η2)θt. From the above it follows that |J4(ξ, η)| ≤ t (√ 1 + k(ξ2 + η2)s − 1 ) = t k(ξ2 + η2)s√ 1 + k(ξ2 + η2)s + 1 ≤ T √ k(ξ2 + η2)s/2. (3.23) Using the inequality | sin(z)| ≤ 1, we obtain |J4(ξ, η)| ≤ √ 1 + k(ξ2 + η2)s − 1√ (ξ2 + η2)θ = k(ξ2 + η2)s(√ 1 + k(ξ2 + η2)s + 1 )√ (ξ2 + η2)θ ≤ √ k(ξ2 + η2) s−θ 2 . (3.24) If ξ2 + η2 ≥ 1, then we obtain |J4(ξ, η)| ≤ √ k(ξ2 + η2) s−θ 2 = √ k 1 (ξ2 + η2) θ−s 2 ≤ 2 √ k (1 + ξ2 + η2) θ−s 2 . (3.25) Hence, using (3.23) and (3.25), we obtain∫∫ R2 (1 + ξ2 + η2)d|J4(ξ, η)|2 ∣∣Ĝ(ξ, η) ∣∣2 dξ dη = ∫∫ ξ2+η2≤1 (1 + ξ2 + η2)θ−s(1 + ξ2 + η2)d+s−θkT 2(ξ2 + η2)s ∣∣Ĝ(ξ, η) ∣∣2 dξ dη + ∫∫ ξ2+η2>1 4k(1 + ξ2 + η2)d+s−θ ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ (2θ−sT 2 + 4)k ∫∫ R2 (1 + ξ2 + η2)d+s−θ ∣∣Ĝ(ξ, η) ∣∣2 dξ dη = (2θ−sT 2 + 4)k∥G∥2Hp+s−θ(R2). (3.26) We now consider the term J5(η, ξ). Indeed, using the inequality | sin(z)| ≤ 1, we obtain ∣∣J5(η, ξ)∣∣ = k(ξ2 + η2)s√ 1 + k(ξ2 + η2)s √ 1 (ξ2 + η2)θ ∣∣∣ sin(√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t )∣∣∣ ≤ √ k(ξ2 + η2) s−θ 2 . 20 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 This term can be treated by the same arguments as in (3.24). Thus, we also deduce that∫∫ R2 (1 + ξ2 + η2)d|J5(ξ, η)|2 ∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ (2θ−sT 2 + 4)k∥G∥2Hd+s−θ(R2). (3.27) Combining (3.17), (3.22), (3.26), and (3.27), we obtain ∥Q(t)G−Q(t)G∥2Hd(R2) = ∫∫ R2 (1 + ξ2 + η2)d ∣∣∣√1 + k(ξ2 + η2)s (ξ2 + η2)θ sin (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) − √ 1 (ξ2 + η2)θ sin (√ (ξ2 + η2)θt )∣∣∣2∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ 3 ∫∫ R2 (1 + ξ2 + η2)d ( |J3(ξ, η)|2 + |J4(ξ, η)|2 + |J5(ξ, η)|2 )∣∣Ĝ(ξ, η) ∣∣2 dξ dη ≤ ( 9.2θ−sT 2k + 24k + 12Cs,θT 2s θ+s k s θ+s ) ∥G∥2Hd+s−θ(R2). Thus, we find that ∥Q(t)f −Q(t)f∥Hp(R2) ≤ Cs,θ,T ( k + k s θ+s ) ∥G∥Hd+s−θ(R2). (3.28) We recall that uk(t) = P(t)a+Q(t)b+ ∫ t 0 Q(t− τ)G(uk(τ))dτ. (3.29) By (3.13) and (3.29), we infer that uk(t)− u∗(t) = P(t)a− P (t)a+Q(t)b−Q(t)b + ∫ t 0 Q(t− τ) ( G(uk(τ))−G(u∗(τ)) ) dτ + ∫ t 0 ( Q(t− τ)−Q(t− τ) ) G(u∗(τ))dτ. The inequality implies that ∥uk(t)− u∗(t)∥Hd(R2) ≤ ∥P(t)a− P (t)a∥Hd(R2) + ∥Q(t)b−Q(t)b∥Hd(R2) + ∥∥∫ t 0 Q(t− τ) ( G(uk(τ))−G(u∗(τ)) ) dτ ∥∥ Hd(R2) + ∥∥∫ t 0 ( Q(t− τ)−Q(t− τ) ) G(u∗(τ))dτ ∥∥ Hd(R2) = K1 + K2 + K3 + K4. (3.30) First, in order to bound the term K1, we use the inequality (3.15) to obtain K1 = ∥P(t)a− P (t)a∥Hd(R2) ≤ Tk ρ−θ 2s ∥a∥Hd+ρ(R2), (3.31) for all θ < ρ < 2s+ θ. Second, using Estimate (3.16), we control the term K2 as follows K2 = ∥Q(t)b−Q(t)b∥Hd(R2) ≤ CTk1/2∥b∥Hd+s(R2). (3.32) Let us treat the term K3. By applying Estimate (3.2) of Lemma 3.1, we find that K3 ≤ √ T 22θ−s + 21−s + 2k ∫ t 0 ∥G(uk(τ))−G(u∗(τ))∥Hd+s−θ(R2)dτ ≤ √ T 22θ−s + 21−s + 2k ∫ t 0 ∥G(uk(τ))−G(u∗(τ))∥Hb(R2)dτ ≤ C √ T 22θ−s + 21−s + 2k ∫ t 0 ∥uk(τ)− u∗(τ)∥Hd(R2)dτ, (3.33) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 21 where we used the global Lipschitz of F and the assumption d + s − θ ≤ b. Now we study the term K4. In view of the inequality (3.28), we obtain K4 ≤ Cs,θ,T ( k + k s θ+s )∫ t 0 ∥G(u∗(τ))∥Hd+s−θ(R2)dτ ≤ Cs,θ,T ( k + k s θ+s )∫ t 0 ∥G(u∗(τ))∥Hb(R2)dτ ≤ Cs,θ,T ( k + k s θ+s ) ∫ t 0 ∥u∗(τ)∥Hd(R2)dτ. (3.34) Here in the last estimate, we have used the globally Lipschitz property (3.3). By a similar tech- niques as the proof of (3.10), we obtain that ∥u∗(t)∥Hd(R2) ≤ CT,m,β,st −m ( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) ) . Hence, we have∫ t 0 ∥u∗(τ)∥Hd(R2)dτ ≤ CT,m,β,s ( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) )(∫ t 0 τ−mdτ ) ≤ CT,m,β,s ( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) ) . (3.35) It follows from (3.34) that K4 ≤ CT,s,θ,m,β ( k + k s θ+s )( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) ) . In view of (3.30), (3.31), (3.32), and (3.33), we derive that ∥uk(t)− u∗(t)∥Hd(R2) ≤ Tk ρ−θ 2s ∥a∥Hd+ρ(R2) + CTk1/2∥b∥Hd+s(R2) + CT,s,θ,m,β ( k + k s θ+s )( ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s−θ(R2) ) + C √ T 22θ−s + 21−s + 2k ∫ t 0 ∥uk(τ)− u∗(τ)∥Hd(R2). By simplifying the above expression we have ∥uk(t)− u∗(t)∥Hd(R2) ≤ C ( k ρ−θ 2s + k + k1/2 + k s θ+s )( ∥a∥Hd+ρ(R2) + ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hp+s(R2) ) + C √ T 22θ−s + 21−s + 2k ∫ t 0 ∥uk(τ)− u∗(τ)∥Hd(R2)dτ. By applying Grönwall’s inequality, we deduce that ∥uk(t)− u∗(t)∥Hd(R2) ≤ C exp ( C √ T 22θ−s + 21−s + 2kt )( k ρ−θ 2s + k + k1/2 + k s θ+s ) × ( ∥a∥Hd+ρ(R2) + ∥a∥Hd(R2) + ∥a∥H(θ−s)β+d(R2) + ∥b∥Hd+s(R2) ) . The proof of Theorem 3.4 is complete. □ We now focus on the second main result of the subsection. Suppose that G(0) = 0 and ∥G(w1)−G(w2)∥Hd(R2) ≤ C∥w1 − w2∥Hd(R2). (3.36) Since for η ≤ d, Hd(R2) ↪→ Hη(R2), it is obvious that (3.36) implies (3.3). Therefore, with sufficiently smooth initial data, one can still apply the argument of Theorem 3.2, and to derive the unique global mild solution to Problem (1.1)-(1.2) with the above Lipschitz property for G. By a very similar argument, one can also prove the existence and uniqueness of the global mild solution 22 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 to Problem (3.12)-(1.2). And again, since the main goal of this subsection is the convergence behavior of the mild solution to Problem (1.1)-(1.2), we omit the proof here. Theorem 3.6. Let a ∈ Hd+2s(R2) for some θ = s ∈ ( 12 , 1) and for some d > 1. Let b = 0. Let us assume that some conditions of Theorem 3.4 holds. Suppose that us and u∗∗ are, respectively, the global mild solutions to Problem (1.1)-(1.2) and Problem (3.12)-(1.2). Then we obtain the estimate ∥us(t)− u∗∗(t)∥Hd(R2) ≤ Cγ,s,T,ε,k ( ∥a∥Hd+2s(R2) + ∥u∗∗∥L1(0,T ;Hd(R2)) ) E(s, ε, γ). (3.37) where E(s, ε, γ) = √ (1− s)s + (1− s)2sε + (1− s) 2s−1 4 + (1− s)γε + (1− s) ε+sε−s 2 , for some ε ∈ ( s s+1 , 1) and γ ∈ (0, 1). By similar arguments of (3.35), we can find the upper bound of the term ∥u∗∗∥L1(0,T ;Hd(R2)) on RHS of (3.37). Proof. To make our representation clear, we emphasize the dependence of us on the parameter s by denoting us(t) = Ps(t)a+ ∫ t 0 Qs(t− τ)G(us(τ))dτ. where Ps(t)v := F−1 ( cos (√ (ξ2 + η2)θ 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) Qs(t)v := F−1 (√1 + k(ξ2 + η2)s (ξ2 + η2)s sin (√ (ξ2 + η2)s 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) Qs(t)v := F−1 ( 1 1 + k(ξ2 + η2)s √ 1 + k(ξ2 + η2)s (ξ2 + η2)s sin (√ (ξ2 + η2)s 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) . Also, it is useful to rewrite Qs(t)v = F−1 ( 1√ 1 + k(ξ2 + η2)s √ (ξ2 + η2)s sin (√ (ξ2 + η2)s 1 + k(ξ2 + η2)s t ) v̂(ξ, η) ) . (3.38) By an obvious modification of Theorem 3.2, we also obtain the existence and uniqueness of the mild solution u∗∗ to Problem (3.12)-(1.2). This solution satisfies the integral equation u∗∗(t) = P1(t)a+ ∫ t 0 Q1(t− τ)G(u∗∗(τ))dτ, (3.39) where P1(t)v := F−1 ( cos (√ (ξ2 + η2) 1 + k(ξ2 + η2) t ) v̂(ξ, η) ) Q1(t)v := F−1 (√1 + k(ξ2 + η2) (ξ2 + η2) sin (√ (ξ2 + η2) 1 + k(ξ2 + η2) t ) v̂(ξ, η) ) , and Q1(t)v := F−1 ( 1√ 1 + k(ξ2 + η2) √ (ξ2 + η2) sin (√ (ξ2 + η2) 1 + k(ξ2 + η2) t ) v̂(ξ, η) ) . (3.40) Subtracting (3.29) into (3.39), we obtain us(t)− u∗∗(t) = [ Ps(t)a− P1(t)a ] + ∫ t 0 Qs(t− τ) ( G(us(τ))−G(u∗∗(τ)) ) dτ + ∫ t 0 ( Qs(t− τ)−Q1(t− τ) ) G(u∗∗(τ))dτ. (3.41) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 23 Considering the second term on RHS, we obtain the equality 1√ 1 + k(ξ2 + η2)s √ (ξ2 + η2)s sin (√ (ξ2 + η2)s 1 + k(ξ2 + η2)s t ) − 1√ 1 + k(ξ2 + η2) √ (ξ2 + η2) sin (√ (ξ2 + η2) 1 + k(ξ2 + η2) t ) = 1√ 1 + k(ξ2 + η2)s √ (ξ2 + η2)s ( sin (√ (ξ2 + η2)s 1 + k(ξ2 + η2)s t ) − sin (√ (ξ2 + η2) 1 + k(ξ2 + η2) t )) + ( 1√ 1 + k(ξ2 + η2)s √ (ξ2 + η2)s − 1√ 1 + k(ξ2 + η2) √ (ξ2 + η2) ) × sin (√ (ξ2 + η2) 1 + k(ξ2 + η2) t ) =: Q1(ξ, η, t) + Q2(ξ, η, t). (3.42) In view of the inequality | sin(α1)− sin(α2)| ≤ Cε|α1 − α2|ε for any 0 < ε ≤ 1, we find that ∣∣∣ sin(√ (ξ2 + η2)s 1 + k(ξ2 + η2)s t ) − sin (√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣ ≤ Cεt ε ∣∣∣√ (ξ2 + η2)s 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣ε ≤ CεT ε ∣∣∣√ (ξ2 + η2)s 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2)s ∣∣∣ε + CεT ε ∣∣∣√ (ξ2 + η2) 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣ε. (3.43) This implies that ∫∫ R2 (1 + ξ2 + η2)d ∣∣∣Q1(ξ, η, t) ∣∣∣2|v̂(ξ, η)|2 dξ dη ≤ Cε,T ∫∫ R2 (1 + ξ2 + η2)d( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s ∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣2ε( 1 + k(ξ2 + η2)s )ε |v̂(ξ, η)|2 dξ dη + Cε,T ∫∫ R2 (1 + ξ2 + η2)d( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s ∣∣∣ √ (ξ2 + η2) 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣2ε|v̂(ξ, η)|2 dξ dη = Q3(ξ, η, t) + Q4(ξ, η, t). (3.44) 24 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 The term Q3(ξ, η, t) is rewritten as follows Q3(ξ, η, t) = Cε,T ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)d( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s ∣∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣∣2ε( 1 + k(ξ2 + η2)s )ε |v̂(ξ, η)|2 dξ dη + Cε,T ∫∫ ξ2+η2<1 (1 + ξ2 + η2)d( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s ∣∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣∣2ε( 1 + k(ξ2 + η2)s )ε |v̂(ξ, η)|2 dξ dη = Q3,1(ξ, η, t) + Q3,2(ξ, η, t). (3.45) In Q3,1, we note that √ ξ2 + η2 ≥ 1. In view of this observation and Lemma 2.4, we obtain∣∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣∣ ≤ Cs( √ ξ2 + η2)1+s(1− s)s. Let us choose 0 < ε < 1, we obtain immediately that (1 + s)ε < sε + 2s since 2s > 1. Thus, we obtain Q3,1(ξ, η, t) ≤ Cε,s,T (1− β)2βε ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)d (ξ2 + η2)(1+s)ε−sε−2s kε+1 |v̂(ξ, η)|2 dξ dη ≤ C1(ε, s, T, k)(1− s)2sε∥v∥2Hd(R2), 0 < ε < 1. (3.46) In Q3,2(ξ, η, t), we note that √ ξ2 + η2 < 1. Another application of Lemma 2.4 yields∣∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣∣ ≤ Cρ( √ ξ2 + η2)s−ρ(1− s)ρ, for some 0 < ρ ≤ 1. This together with the observation ( 1 + k(ξ2 + η2)s > 1 and ( 1 + k(ξ2 + η2)s )ε > 1 yields Q3,2(ξ, η, t) ≤ Cε,ρ,s,T (1− s)2ρ ∫∫ ξ2+η2<1 (1 + ξ2 + η2)d(ξ2 + η2)(s−ρ)ε−s|v̂(ξ, η)|2 dξ dη ≤ Cε,ρ,s,T 2 d(1− s)2ρ ∫∫ ξ2+η2<1 (ξ2 + η2)(s−ρ)ε−s|v̂(ξ, η)|2 dξ dη (3.47) Here we choose 0 < ρ < s− 1 2 < 1. In view of the inequality |v̂(ξ, η)| ≤ ∥v∥L∞(R2), (3.48) for v ∈ L∞(R2), we find that∫∫ ξ2+η2<1 (ξ2 + η2)(s−ρ)ε−s|v̂(ξ, η)|2 dξ dη ≤ ∥v∥2L∞(R2) ∫∫ ξ2+η2<1 (ξ2 + η2)(s−ρ)ε−s dξ dη (3.49) Since s < 1, ε < 1 and 0 < ρ < s− 1 2 < 1, we know that s− (s−ρ)ε < 1. Thus, the proper integral equation ∫∫ ξ2+η2<1 (ξ2 + η2)(s−ρ)ε−s dξ dη is convergent. It follows from (3.47) and (3.49) that Q3,2(ξ, η, t) ≤ Cε,ρ,s,T 2 d(1− s)2ρ∥v∥2L∞(R2). It is noteworthy that if d > 1, we have the embedding Hd(R2) ↪→ L∞(R2). (3.50) Thus, we obtain immediately that Q3,2(ξ, η, t) ≤ Cε,s,ρ,d,T (1− s)2ρ∥v∥2Hd(R2). (3.51) Combining (3.45), (3.46), (3.51) and choosing ρ = 2s−1 4 , we deduce that Q3(ξ, η, t) ≤ Q3,1(ξ, η, t) + Q3,2(ξ, η, t) ≤ Cε,s,T,d,k ( (1− s)2sε + (1− s) 2s−1 2 ) ∥v∥2Hd(R2). (3.52) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 25 where 0 < ε < 1, For the term Q4(ξ, η, t) on RHS of (3.44), we have Q5(ξ, η) := ∣∣∣√ (ξ2 + η2) 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣2ε = (ξ2 + η2)ε ∣∣√1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s ∣∣ε( 1 + k(ξ2 + η2)s )ε( 1 + k(ξ2 + η2) )ε = (ξ2 + η2)εkε ∣∣(ξ2 + η2)− (ξ2 + η2)s ∣∣ε( 1 + k(ξ2 + η2)s )ε( 1 + k(ξ2 + η2) )ε(√ 1 + k(ξ2 + η2)s + √ 1 + k(ξ2 + η2) )ε If ξ2 + η2 ≥ 1 then using Lemma 2.4, we find that∣∣(ξ2 + η2)− (ξ2 + η2)s ∣∣ε ≤ Cγ, ε(ξ2 + η2)(1+γ)ε(1− s)γε, 0 < γ < 1. It is obvious to see that( 1 + k(ξ2 + η2)s )1+ε( ξ2 + η2 )s( 1 + k(ξ2 + η2) )ε (√ 1 + k(ξ2 + η2)s + √ 1 + k(ξ2 + η2) )ε > k1+ 5 2 ε(ξ2 + η2)2s+sε+ 3ε 2 . Therefore,∫∫ ξ2+η2≥1 (1 + ξ2 + η2)dQ5(ξ, η)( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s |v̂(ξ, η)|2 dξ dη ≤ Cγ,ε,k(1− s)γε ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)d ( ξ2 + η2 )γε+ ε 2−2s−sε|v̂(ξ, η)|2 dξ dη ≤ Cγ,ε,k(1− s)γε ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)d|v̂(ξ, η)|2 dξ dη ≤ Cγ,ε,k(1− s)γε∥v∥2Hd(R2), (3.53) where we note that γε+ ε 2 − 2s− sε < 0 since the condition 0 < γ < 1 and s > 1 2 . If ξ2 + η2 < 1 then using Lemma 2.4, we find that∣∣(ξ2 + η2)− (ξ2 + η2)s ∣∣ε ≤ Cδ,ε(ξ 2 + η2)(s−δ)ε(1− s)δε, 0 < δ < 1, which allows us to obtain that Q5(ξ, η) ≤ Cρ,εk ε(ξ2 + η2)ε+(s−δ)ε(1− s)δε. Then, we find that∫∫ ξ2+η2<1 (1 + ξ2 + η2)dQ5(ξ, η)( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s |v̂(ξ, η)|2 dξ dη ≤ Cδ,εk ε(1− s)δε ∫∫ ξ2+η2<1 (1 + ξ2 + η2)d ( ξ2 + η2 )ε+(s−δ)ε−s|v̂(ξ, η)|2 dξ dη. (3.54) Since s s+1 < ε < 1 < 2s, we can choose δ such that δ = ε+sε−s 2ε ∈ (0, 1). Then ( ξ2+η2 )ε+(s−δ)ε−s < 1 if ξ2 + η2 < 1 and we follows from (3.54) that∫∫ ξ2+η2<1 (1 + ξ2 + η2)dQ5(ξ, η)( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s |v̂(ξ, η)|2 dξ dη ≤ C(s, ε, k)(1− s) ε+sε−s 2 . (3.55) Combining (3.53) and (3.55), we find that Q4(ξ, η, t) ≤ Cγ,s,ε,k ( (1− s)γε + (1− s) ε+sε−s 2 ) ∥v∥2Hd(R2) (3.56) 26 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 where we remind that 0 < γ < 1. By connecting (3.44), (3.52) and (3.56), we deduce that∫∫ R2 (1 + ξ2 + η2)d ∣∣Q1(ξ, η, t) ∣∣2|v̂(ξ, η)|2 dξ dη ≤ Q3(ξ, η, t) + Q4(ξ, η, t) ≤ Cγ,s,T,ε,k ( (1− s)2sε + (1− s) 2s−1 4 + (1− s)γε + (1− s) ε+sε−s 2 ) ∥v∥2Hd(R2), (3.57) for s s+1 < ε < 1 and 0 < γ < 1. Step 2. Estimation of Q2(ξ, η, t). It is easy to find that Q2(ξ, η, t) = ∣∣∣√1 + k(ξ2 + η2) √ (ξ2 + η2)− √ 1 + k(ξ2 + η2)s √ (ξ2 + η2)s√ 1 + k(ξ2 + η2)s √ 1 + k(ξ2 + η2)(ξ2 + η2) s+1 2 ∣∣∣ × ∣∣∣ sin(√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣ ≤ ∣∣√(ξ2 + η2)− √ (ξ2 + η2)s ∣∣√ 1 + k(ξ2 + η2)s ( ξ2 + η2 ) s+1 2 ∣∣∣ sin(√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣ + ∣∣√1 + k(ξ2 + η2)− √ 1 + k(ξ2 + η2)s ∣∣√ 1 + k(ξ2 + η2) ( ξ2 + η2 )1/2 ∣∣∣ sin(√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣ = Q7(ξ, η, t) + Q8(ξ, η, t). (3.58) Let us consider the quantity ∣∣Q7(ξ, η, t) ∣∣. If ξ2 + η2 ≥ 1, then using Lemma 2.4, we obtain∣∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣∣ ≤ Cs( √ ξ2 + η2)1+ s 2 (1− s)s/2. (3.59) This implies that∫∫ ξ2+η2≥1 (1 + ξ2 + η2)d ∣∣Q7(ξ, η, t) ∣∣2|v̂(ξ, η)|2 dξ dη ≤ Cs(1− s)s ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)d(ξ2 + η2) −s 4 |v̂(ξ, η)|2 dξ dη ≤ Cs(1− s)s∥v∥2Hd(R2). (3.60) If ξ2 + η2 < 1 then using Lemma 2.4, we obtain∣∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣∣ ≤ Cs( √ ξ2 + η2)s/2(1− s)s/2. (3.61) Also, in view of the inequality | sin(z)| ≤ z, we find that∣∣∣ sin(√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣ ≤ T ( ξ2 + η2 )1/2 . (3.62) By the above two estimates, we arrive at∫∫ ξ2+η2<1 (1 + ξ2 + η2)d ∣∣Q7(ξ, η, t) ∣∣2|v̂(ξ, η)|2 dξ dη ≤ Cs2 d(1− s)s ∫∫ ξ2+η2<1 (ξ2 + η2) −s 2 |v̂(ξ, η)|2 dξ dη ≤ Cs2 d(1− s)s∥v∥2L∞(R2) ∫∫ ξ2+η2<1 (ξ2 + η2) −s 2 dξ dη ≤ Cs,d(1− s)s∥v∥2Hd(R2). (3.63) where we have used (3.48) and (3.50). In the above estimate, we also note that the integral∫∫ ξ2+η2<1 (ξ2 + η2) −s 2 dξ dη is convergence since s < 2. By combining (3.60) and (3.63), we find EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 27 that ∫∫ R2 (1 + ξ2 + η2)d ∣∣Q7(ξ, η, t) ∣∣2|v̂(ξ, η)|2 dξ dη ≤ Cs,d(1− s)s∥v∥2Hd(R2). (3.64) Let us deal with the term Q8(ξ, η, t). It is obvious that Q8(ξ, η, t) = k ∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣ ∣∣∣ sin(√ (ξ2+η2) 1+k(ξ2+η2) t )∣∣∣(∣∣∣√1 + k(ξ2 + η2) + √ 1 + k(ξ2 + η2)s ∣∣∣)√1 + k(ξ2 + η2) ( ξ2 + η2 )1/2 If ξ2 + η2 ≥ 1, then (ξ2 + η2) s−2 4 < 1, by using (3.59), we find that Q8(ξ, η, t) ≤ Cs √ k(1− s)s/2(ξ2 + η2) s−2 4 ≤ Cs √ k(1− s)s/2. (3.65) If ξ2 + η2 < 1, then using (3.61) one obtains Q8(ξ, η, t) ≤ Csk(1− s)s/2 (ξ2 + η2) s 4 ∣∣∣ sin(√ (ξ2+η2) 1+k(ξ2+η2) t ) ∣∣∣( ξ2 + η2 )1/2 . (3.66) Here we note that the denominator of Q8(ξ, η, t) is greater than ( ξ2+η2 )1/2 . Using the inequality | sin(z)| ≤ Cεz ε with ε = 1− s 4 ∈ (0, 1), we obtain that∣∣∣ sin(√ (ξ2 + η2) 1 + k(ξ2 + η2) t )∣∣∣ ≤ CsT 1− s 4 ( (ξ2 + η2) 1 + k(ξ2 + η2) ) 1 2− s 8 ≤ CsT 1− s 4 (ξ2 + η2) 1 2− s 8 . (3.67) By collecting (3.66) and (3.67), we infer that Q8(ξ, η, t) ≤ CskT 1− s 4 (1− s)s/2(ξ2 + η2) s 8 ≤ CskT 1− s 4 (1− s)s/2 (3.68) if ξ2 + η2 < 1. Combining (3.65), (3.68), we deduce that∫∫ R2 (1 + ξ2 + η2)d ∣∣Q8(ξ, η, t) ∣∣2|v̂(ξ, η)|2 dξ dη = ∫∫ ξ2+η2≥1 (1 + ξ2 + η2)d ∣∣Q8(ξ, η, t) ∣∣2|v̂(ξ, η)|2 dξ dη + ∫∫ ξ2+η2<1 (1 + ξ2 + η2)d ∣∣Q8(ξ, η, t) ∣∣2|v̂(ξ, η)|2 dξ dη ≤ Cs,k,T (1− s)s∥v∥2Hd(R2). (3.69) Combining (3.58), (3.64) and (3.69), we derive that∫∫ R2 (1 + ξ2 + η2)d ∣∣Q2(ξ, η, t) ∣∣2 dξ dη ≤ 2 ∫∫ R2 (1 + ξ2 + η2)d ∣∣Q7(ξ, η, t) ∣∣2 dξ dη + 2 ∫∫ R2 (1 + ξ2 + η2)d ∣∣Q8(ξ, η, t) ∣∣2 dξ dη ≤ Cs,k,T,d(1− s)s∥v∥2Hd(R2). This estimate together with (3.38), (3.40), (3.42), and (3.57) yields ∥Qs(t)v −Q1(t)v∥2Hd(R2) = ∫∫ R2 (1 + ξ2 + η2)d ( Q1(ξ, η, t) + Q2(ξ, η, t) )2 dξ dη ≤ Cγ,s,d,T,ε,k ( (1− s)s + (1− s)2sε + (1− s) 2s−1 4 + (1− s)γε + (1− s) ε+sε−s 2 ) ∥v∥2Hd(R2), for s s+1 < ε < 1 and 0 < γ < 1. Based on this result, we have ∥Qs(t)v −Q1(t)v∥Hd(R2) ≤ Cγ,s,d,T,ε,kE(s, ε, γ)∥v∥Hd(R2), (3.70) for any v ∈ Hd(R2), d > 1. 28 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 Now, we turn to estimate the error ∥Ps(t)w − P1(t)w∥Hd(R2). We modify the proof of (3.43) and (3.44). We have ∥Ps(t)w − P1(t)w∥2Hd(R2) ≤ Cε,T ∫∫ R2 (1 + ξ2 + η2)d ∣∣√(ξ2 + η2)s − √ (ξ2 + η2) ∣∣2ε( 1 + k(ξ2 + η2)s )ε |ŵ(ξ, η)|2 dξ dη + Cε,T ∫∫ R2 (1 + ξ2 + η2)d ∣∣∣√ (ξ2 + η2) 1 + k(ξ2 + η2)s − √ (ξ2 + η2) 1 + k(ξ2 + η2) ∣∣∣2ε|ŵ(ξ, η)|2 dξ dη. If we consider a function v such that |v̂(ξ, η)|2 = ( 1 + k(ξ2 + η2)s )( ξ2 + η2 )s|ŵ(ξ, η)|2. (3.71) By the two latter observations and (3.57), we derive ∥Ps(t)w − P1(t)w∥2Hd(R2) ≤ Cγ,s,T,ε,k ( (1− s)2sε + (1− s) 2s−1 4 + (1− s)γε + (1− s) ε+sε−s 2 ) ∥v∥2Hd(R2) (3.72) From (3.71), we find that ∥v∥2Hd(R2) = ∫∫ R2 (1 + ξ2 + η2)d|v̂(ξ, η)|2 dξ dη ≤ (1 + k) ∫∫ R2 (1 + ξ2 + η2)d+2s|ŵ(ξ, η)|2 dξ dη ≤ (1 + k)∥w∥2Hd+2s(R2), (3.73) where we have used that (1 + k(ξ2 + η2)s )( ξ2 + η2)s ≤ (1 + k)(1 + ξ2 + η2)2s. Combining (3.72), (3.73), we obtain ∥Ps(t)w − P1(t)w∥Hd(R2) ≤ Cγ,s,T,ε,kE(s, ε, γ)∥w∥Hd+2s(R2). (3.74) Let us return to the third term on the right- hand side of (3.41). By applying estimate (3.2) of Lemma 3.1, we find that∥∥∫ t 0 Qs(t− τ) ( G(us(τ))−G(u∗∗(τ)) ) dτ ∥∥ Hd(R2) ≤ √ 2T 2 + 2 + 2k ∫ t 0 ∥G(us(τ))−G(u∗∗(τ))∥Hd(R2)dτ ≤ C √ 2T 2 + 2 + 2k ∫ t 0 ∥us(τ)− u∗∗(τ)∥Hd(R2)dτ, (3.75) where we have used the global Lipschitz (3.36) of F . Let us now consider the fourth term on RHS of (3.41). In view of the inequality (3.70) and the global Lipschitz property (3.36) of F , we obtain ∥ ∫ t 0 ( Qs(t− τ)−Q1(t− τ) ) G(u∗∗(τ))dτ∥Hd(R2) ≤ Cγ,s,T,ε,kE(s, ε, γ) ∫ t 0 ∥G(u∗∗(τ))∥Hd(R2)dτ ≤ CCγ,s,T,ε,kE(s, ε, γ) ∫ t 0 ∥u∗∗(τ)∥Hd(R2)dτ ≤ CCγ,s,T,ε,kE(s, ε, γ)∥u∗∗∥L1(0,T ;Hd(R2)). (3.76) Combining (3.41), (3.70), (3.74), (3.75), and (3.76), we verify that ∥us(t)− u∗∗(t)∥Hd(R2) ≤ Cγ,s,T,ε,k,dE(s, ε, γ)∥a∥Hd+2s(R2) + Cγ,s,T,ε,kE(s, ε, γ)∥b∥Hd(R2) + CCγ,s,T,ε,k,dE(s, ε, γ)∥u∗∗∥L1(0,T ;Hd(R2)) EJDE-2025/76 MILD SOLUTIONS TO LOVE-TYPE EQUATIONS 29 + C √ 2T 2 + 2 + 2k ∫ t 0 ∥us(τ)− u∗∗(τ)∥Hd(R2)dτ. By applying Grönwall’s inequality, we deduce that ∥us(t)− u∗∗(t)∥Hd(R2) ≤ Cγ,s,T,ε,k,C,d ( ∥a∥Hd+2s(R2) + ∥u∗∗∥L1(0,T ;Hd(R2)) ) E(s, ε, γ) exp ( C √ 2T 2 + 2 + 2kt ) ≤ Cγ,s,T,ε,k,C,d ( ∥a∥Hd+2s(R2) + ∥u∗∗∥L1(0,T ;Hd(R2)) ) E(s, ε, γ). The proof of Theorem 3.6 is complete. □ Acknowledgements. Nguyen Anh Tuan wants to expresses his gratitude to Van Lang University. References [1] C. Babaoglu, H. Erbay, A. Erkip; Global existence and blow-up of solutions for a general class of doubly dispersive nonlocal nonlinear wave equations, Nonlinear Anal., 77 (2013), 82–93. [2] M. Biomy, K. Zennir, A. Himadan; Local and Global Existence of Solution for Love Type Waves with Past History, Mathematics, 8 (2020), no. 11, 1998. [3] T. Caraballo, B. Guo, N .H. Tuan, R. Wang; Asymptotically autonomous robustness of random attractors for a class of weakly dissipative stochastic wave equations on unbounded domains, Proceedings of the Royal Society of Edinburgh Section A: Mathematics, 151 (2021), no. 6, 1700–1730. [4] T. Caraballo, A. N. Carvalho, J. A. Langa, F. Rivero; A non-autonomous strongly damped wave equation: Existence and continuity of the pullback attractor, Nonlinear Analysis: Theory, Methods & Applications, 74 (2011), no. 6, 2272–2283. [5] Y. Chen, H. Gao, M. J. Garrido-Atienza, B. Schmalfuss; Pathwise solutions of SPDEs driven by Hölder- continuous integrators with exponent larger than 1/2 and random dynamical systems, Discrete and Continuous Dynamical Systems, 34 (2014), 1, 79–98. [6] N. Duruk, H. A. Erbay, A. Erkip; Global existence and blow-up for a class of nonlocal nonlinear Cauchy problems arising in elasticity, Nonlinearity, 23 (2010), no. 1, 107–118. [7] M. Kwaśnicki; Ten equivalent definitions of the fractional Laplace operator, Fractional Calculus and Applied Analysis, 20 (2017), no. 1, 7–51. [8] A. E. H. Love; A Treatise on the Mathematical Theory of Elasticity, Cambridge, 1952. [9] Y. Liu, R. Xu; Potential well method for Cauchy problem of generalized double dispersion equations, J. Math. Anal. Appl., 338 (2008), no. 2, 1169–1187. [10] L. T. P. Ngoc, N. T. Duy, N. T. Long; Existence and properties of solutions of a boundary problem for a Love’s equation, Bull. Malays. Math. Sci. Soc., 37 (2014), no. 4, 997–1016. [11] L. T. P. Ngoc, N. T. Long; Existence, blow-up and exponential decay for a nonlinear Love equation associated with Dirichlet conditions, Appl. Math., 61 (2016), no. 2, 165–196. [12] L. T. P. Ngoc, N. T. Duy, N. A. Triet, N. T. Long; An N-order iterative scheme for a nonlinear Love equation, Vietnam J. Math., 44 (2016), no. 4, 801–816. [13] J. Oscar, G. Loachamı́n; From non-local to local Navier-Stokes equations, Appl. Math. Optim., 89 (2024), 20 pp. [14] B. D. Nam, B. D. Nghia, N. D. Phuong; Regularity and convergent result of mild solution of Love equation, Discrete Contin. Dyn. Syst. Ser. S, 17 (2024), no. 3, 1178–1194. [15] V. Radochova; Remark to the comparison of solution properties of Love’s equation with those of wave equation, Apl. Mat., 23 (1978), 199—207. [16] K. Zennir, M. Biomy; General Decay Rate of Solution for Love-Equation with Past History and Absorption, Mathematics, 8 (2020), no. 9, 1632. [17] K. Zennir, T. Miyasita, P. Papadopoulos; Local existence and global nonexistence of a solution for a Love equation with infinite memory, J. Integral Equations Appl., 33 (2021), no. 1, 117–136. Bui Duc Nam Ho Chi Minh City University of Industry and Trade, Vietnam Email address: nambd@huit.edu.vn Bui Dai Nghia Faculty of Mathematics and Computer Science, University of Science, Ho Chi Minh City, Vietnam. Vietnam National University, Ho Chi Minh City, Vietnam Department of Mathematics, Faculty of Science, Nong Lam University, Ho Chi Minh City, Vietnam Email address: dainghia2008@hcmuaf.edu.vn 30 B. D. NAM, B. D. NGHIA, N. A. TUAN EJDE-2025/76 Nguyen Anh Tuan (corresponding author) Division of Applied Mathematics, Science and Technology Advanced Institute, Van Lang University, Ho Chi Minh City, Vietnam. Faculty of Applied Technology, School of Technology, Van Lang University, Ho Chi Minh City, Viet- nam Email address: nguyenanhtuan@vlu.edu.vn 1. Introduction 1.1. State of the art and main contributions 1.2. Notation and outline 2. Homogeneous case 2.1. Regularity of mild solution 2.2. Convergence of the mild solution 3. Nonlinear problem 3.1. Existence and uniqueness of global mild solution 3.2. Convergence of mild solutions Acknowledgements References