On the Three-Phase-Lag Heat Equation with Spatial Dependent Lags

Nonlinear Analysis and Differential Equations, Vol. 5, 07, no., 53-66 HIKARI Ltd, www.m-hikari.com https://doi.org/0.988/nade.07.694 On the Three-Phase-Lag Heat Equation with Spatial Dependent Lags Yang Wang College of Information Science and Technology Donghua University Shanghai 060, P. R. China Copyright c 06 Yang Wang. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper we investigate several qualitative properties of the solutions of three-phase-lag heat equation where two of the phase lags τ T (x) and τν (x) are positive functions of spatial variable and the third one τ q is a positive constant. We prove that where κ τ q < τν (x) < κτ T (x) τ q or κ τ q τν (x) < κτ T (x) τ q or τν (x) < κτ T (x) τ q and τν (x) κ τ q > 0 on a proper subdomain the solutions are exponentially stable. When τν (x) κ τ q > 0 and τν < κτ T (x) τ q on a proper subdomain in one-dimensional problem, we are able to show that the solutions remain being exponentially stable. Finally, when κ τ q < τν (x) κτ T (x) τ q, we obtain the solutions are at most polynomially stable. Keywords: three-phase-lag heat equation, exponential stability, polynomial stability Introduction It is well-known that Fourier s heat equation theory implies that the thermal disturbances at some point will be felt instantly anywhere for every distant. Most known theory is the Maxwell- Cattaneo law that proposes an hyperbolic damped equation for the heat conduction. There are serval models described for the conduction of heat in the thermechanical context. We recall the models proposed by Lord and Shulman [3], the Green and Lindsay [5] and Green and Naghdi [6 0].

54 Yang Wang In 995, Tzou [0] proposed a modification of the Fourier constitutive equation. He suggested a theory of thermal flux with delay. The basic constitutive equation is q(x, t + τ q ) = κ T (x, t + τ T ), κ > 0. (.) Where T is the temperature, q is the heat flux vector and τ T and τ q are two delay parameters. Recently, by introducing a thermal displacement that satisfies ν = T, Choudhuri [3] proposed a theory with three-phase-lag which is a modification of Tzou s constitutive equation, q(x, t + τ q ) = κ T (x, t + τ T ) κ ν(x, t + τ ν ). (.) Here τ ν is a new delay. Taking second-order Taylor approximation to the left-hand side of (.) and first-order approximations to the right-hand side of (.) for the temperature, the heat flux and the thermal displacement, we can get q + τ q q + τ q q = κ T κτ T T κ ν κ τ ν ν. (.3) As in [0], T + divq = 0. Thus, (.3) turns into the following equation where τ ν = κ + κ τ ν. Let T + τ q T + τ q. T = κ T + τν T + κτ T T, (.4) a = κτ T τ q τ ν, b = τ ν κ τ q. In [9] and [], the authors showed the exponential stability in the case that a > 0, b 0, strong but non-exponential stability in the case that a 0, b 0 and instability in the case that a < 0 and/or b < 0. Similarly, by the Taylor approximation, equation (.) turns into a dual-phase-lag equation τ T q + τ q T + T = κ T + κτ T T. (.5) In [8] and [], the authors proved exponential stability of equation (.5) when τ q < τ T, strong but non-exponential stability when τ q = τ T and instability when τ q > τ T. Recently, Liu et al. [4] obtained the polynomial stability for the critical case τ q = τ T. They were the first to introduce the delay τ T depends on the spatial variable and proved the exponential stability of (.5) when τ q < τ T (x) and the partially critical case i.e., τ q < τ T (x) only on a subdomain just in the case of one-dimensional problem in [4]. In this paper we consider equation (.4) in the case that τ T and τ ν are functions depending on the spatial variable. Similar to [4], we get some new phenomena for the critical case and partially critical case. We remark that (.4) is of hyperbolic type in T, hence finite propagation speed [9].

On the three-phase-lag heat equation 55 Main Results For spatial dependent τ T and τν, equation (.4) is modified as the following T + τ T + τ. T = κ T + div(τ 3 (x) T ) + κdiv(τ (x) T ), in Ω (0, ), (.) T (x, 0) = T 0 (x), T (x, 0) = T 0 (x), T (x, 0) = T 0 (x), T (x, 0) = T 0 (x) in Ω, (.) T (, t) Ω = 0, for t [0, ), (.3) where Ω is a bounded domain with smooth boundary Ω, and τ = τ q, τ (x) = τ T (x) and τ 3 (x) = τ ν (x). In reference to the constant coefficient equation (.4), we denote a(x) = κτ (x) τ τ 3 (x) and b(x) = τ 3 (x) κ τ. We only consider a(x) 0 and b(x) 0 on Ω, since our interest is the stability of (.)-(.3) in these cases: (i). a(x) and b(x) are strictly positive, i.e., a(x) a 0 > 0 and b(x) b 0 > 0 on Ω; (ii). critical case: a(x) a 0 > 0 and b(x) 0 on Ω; (iii). critical case: a(x) 0 and b(x) b 0 > 0 on Ω; (iv). partially critical case: a(x) a 0 > 0 and b(x) > 0 only on a subdomain of positive measure Ω 0 Ω; (v). partially critical case: b(x) b 0 > 0 and a(x) > 0 only on a subdomain of positive measure Ω 0 Ω. It is important to identify a proper state space so that the energy of the system (.)-(.3) is dissipative. For this purpose, we take the inner product of T τ + τ T + T with (.) in L (Ω) to get Since d dt T τ + τ T + T = κ T, T τ + τ T + T + div(τ 3 (x) T ), T τ + τ T + + κdiv(τ (x) T ), T + τ T + τ T. κ T, T τ + τ T + T = κ T, ( T τ + τ T + T ) = d dt κ T κ T, τ T κ T, τ T = d dt κ T d dt κ τ T, T + κ τ T + d 4 dt κ τ T, T (.4) d dt κ τ T, T

56 Yang Wang div(τ 3 (x) T ), T τ + τ T + T = τ 3 (x) T, ( T τ + τ T + T ) = τ 3 (x) T d dt τ τ 3 (x) T τ 3 (x) T, τ T, and = τ 3 (x) T + τ τ 3 (x) T d dt τ τ 3 (x) T d dt τ τ 3 (x) T, T κdiv(τ (x) T ), T τ + τ T + T = κ τ (x) T, ( T τ + τ T + T ) = d dt κ τ (x) T κτ τ (x) T 4 d dt κτ τ (x) T, (.4) leads to de(t) = b (x) T τ dt a (x) T. (.5) where the energy of the system (.)-(.3) is E(t) = T τ + τ T + T + κ T + κ τ T, T + κ τ T, T κ τ T +τ τ 3 (x) T + τ τ 3 (x) T, T + κ τ (x) T + κτ τ (x) T = T τ + τ T + T + κ (T + τ T + τ T ) + τ b (x) ( T + τ T ) + τ (a(x) + b(x)) T + τ 3 4 a (x) T. Let H 0 (Ω) = {X H (Ω) : X Ω = 0}, and hence H := H 0 (Ω) H 0 (Ω) H 0 (Ω) L (Ω). Denoting by Z = (Z, Z, Z 3, Z 4 ) T and W = (W, W, W 3, W 4 ) T, we can define the inner product i.e., Z, W H = Z + τ Z 3 + τ Z 4, W + τ W 3 + τ W 4 + κ (Z + τ Z + τ Z 3), (W + τ W + τ W 3) + τ b (x) (Z + τ Z 3), b (x) (W + τ W 3) + τ (a(x) + b(x)) Z, (a(x) + b(x)) W + τ 3 4 a (x) Z3, a (x) W3, Z H = Z + τ Z 3 + τ Z 4 + κ (Z + τ Z + τ Z 3) + τ b (x) (Z + τ Z 3) + τ (a(x) + b(x)) Z + τ 3 4 a (x) Z3.

On the three-phase-lag heat equation 57 Denoting Z := (Z, Z, Z 3, Z 4 ) T = (T, T, T, T ) T, we then convert system (.)-(.3) to a first-order evolution equation on Hilbert space H, dz = AZ, (.6) dt Z(0) = Z 0 = (T 0, T 0, T 0, T 0 ) T, (.7) where the operator A is given by Z 3 AZ = Z 4 ( Z3 τ Z 4 + κ Z +div(τ 3 (x) Z ) + κdiv(τ (x) Z 3 ) ) Z (.8) τ and D(A) = {Z = (Z, Z, Z 3, Z 4 ) T H Z, Z, Z 3 H (Ω), Z 4 H 0 (Ω)}. (.9) Theorem.. A is the infinitesimal generator of a C 0 semigroup of contractions on the Hilbert space H. Proof. By (.5), Re AZ, Z H = d dt Z H = b (x) Z τ a (x) Z3 0. (.0) Thus, A is dissipative. Now for F = (f, f, f 3, f 4 ) T H, we look for Z = (Z, Z, Z 3, Z 4 ) T D(A) such that (I A)Z = F. Equivalently, we consider the following system Z Z = f, (.) Z Z 3 = f, (.) Z 3 Z 4 = f 3, (.3) Z 4 ( τ Z3 τ Z 4 + κ Z + div(τ 3 (x) Z ) + κdiv ( τ (x) Z 3 ) ) = f 4, (.4) Z Ω = Z Ω = Z 3 Ω = Z 4 Ω = 0. (.5) From (.), (.), (.3) and (.4), we have ( + τ = f 4 + ( + τ + )Z κ τ τ Z τ + τ )f + ( + τ div(τ 3 (x) Z ) κ τ div(τ (x) Z ) + )f + ( + )f 3 τ τ τ div(τ 3 (x) f ) κ τ div(τ (x) (f + f )). (.6)

58 Yang Wang Let φ H0. Multiplying (.6) by φ, we get the following variational equation ( + τ = f 4, φ + ( + τ + τ + )Z, φ + κ τ τ Z, φ + τ + τ )f, φ + ( + τ τ 3 (x) Z, φ + κ τ τ (x) Z, φ + τ )f, φ + ( + τ )f 3, φ τ 3 (x) f, φ + κ τ τ (x) (f + f ), φ. (.7) It is easy to check that the left-hand of (.7) is a continuous and coercive bilinear form on the space H 0 H 0, and the right-hand side is a continuous linear form on the space H 0 H 0. Then, due to Lax-Milgram Lemma ([], Theorem.9.), (.7) admits a unique solution Z H 0. (.7) also implies that the weak solution Z of (.6) associated with the boundary conditions (.5) belongs to the space H. Therefore, (Z, Z, Z 3, Z 4 ) T D(A) and (I A) is compact in the energy space H. Then, thanks to Lumer-Philips Theorem ([6], Theorem.4.3), we conclude that A generates a C 0 semigroup of contractions on H. Our main results for system (.)-(.3) are stated in the following two theorems. Theorem.. Assume that a(x) C (Ω). Then the semigroup e At is (). exponentially stable for the cases (i), (ii) and (iv), i.e., there exist constants M, ω > 0, such that e AT Z 0 H Me ωt Z 0 H, t > 0, Z 0 H; (). polynomially stable of order for the case (iii), i.e., there exists a constant C > 0, such that e AT Z 0 H C t Z 0 D(A), t > 0, Z 0 D(A). Theorem.3. Let Ω = [0, L], Ω 0 = (x, x ) Ω. If b(x) b 0 > 0 on Ω and a(x) C (Ω), a(x) > 0 on Ω 0 and a(x) = 0 on Ω\Ω 0, i.e., the case (v), then the semigroup e At is exponentially stable. Remark.. Case () in Theorem. extends the corresponding result for the constant coefficient case a > 0, b 0 considered in [9]. Case () in Theorem. improves the slow decay conclusion for the critical case a 0, b 0 in [9] by a specific polynomial decay rate. However, whether this is the best decay rate is still open. Remark.. The result in Theorem.3 reveals a transition process from exponential stability to polynomial stability as a(x) changes from positive to partially positive to zero. By far, we are only able to prove it for one-dimensional problem. The proof of Theorem. and Theorem.3 will be presented in next two sections. Our main tools are the following well-known frequency domain characterization of stability for a semigroup on Hilbert space, combined with contradiction argument in [5].

On the three-phase-lag heat equation 59 Theorem.4. [4, 7] Let S(t) = e At be a C 0 semigroup of contractions in a Hilbert space H. Suppose that ir ρ(a). (.8) Then, S(t) is exponentially stable if and only if holds. lim (ii A) H < (.9) Theorem.5. [] Let H be a Hilbert space and A generates a bounded C 0 semigroup in H. Assume that ir ρ(a), (.0) sup > k (i A) < +, for some k > 0. (.) Then, there exists a positive constant C > 0 such that for all Z 0 D(A). 3 Proof of Theorem. e ta Z 0 C( t ) k Z0 D(A), t > 0, (.) Proof. We first verify condition (.8). Assume that it is false, i.e., there is a λ = i σ(a). Then there exist λ n (= i n ) λ and normalized Z n = (Z n, Z n, Z 3n, Z 4n ) T such that (i n A)Z n H 0, (3.) which implies iz Z = o(), in H0 (Ω), (3.) iz Z 3 = o(), in H0 (Ω), (3.3) iz 3 Z 4 = o(), in H 0 (Ω), (3.4) i(z + τ Z 3 + τ Z 4) ( κ Z + div(τ 3 (x) Z ) + κdiv(τ (x) Z 3 ) ) = o(), in L.(3.5) For convenience, we have omitted the subscript n hereafter. Thus Re (i A)Z, Z H = Re AZ, Z H = b (x) Z τ a (x) Z3 = o(). (3.6) Hence b (x) Z = o() and a (x) Z3 = o(). (3.7) If b(x) b 0 > 0, we have Z = o(). Since is finite, we get from (3.) and (3.3) that Z = o() and Z 3 = o(). (3.8)

60 Yang Wang If a(x) a 0 > 0, we have Z 3 = o(). Since is finite, we get from (3.3) and (3.) that Then together with (3.4), we have By the P oincaré inequality, Z = o() and Z = o(). (3.9) Z 4 = o(). Z 4 = o(). (3.0) We conclude that Z H = o(). This is a contradiction with the assumption Z H =. Thus, ir ρ(a). Assume that (.9) and (.) are false. We can combine them into one case. Then by the uniform boundedness theorem, there exists a sequence and a unit sequence Z = (Z, Z, Z 3, Z 4 ) T D(A) such that k (ii A)Z H 0, (3.) which implies that k (iz Z ) = o(), in H0 (Ω), (3.) k (iz Z 3 ) = o(), in H0 (Ω), (3.3) k (iz 3 Z 4 ) = o(), in H0 (Ω), (3.4) ( k i(z + τ Z 3 + τ ) Z 4) κ Z div(τ 3 (x) Z ) κdiv(τ (x) Z 3 ) = o(), in L. (3.5) From dissipation, we have k b (x) Z = o(), and k a (x) Z3 = o(). (3.6) If a(x) 0 and b(x) b 0 > 0 on Ω, by (3.) and (3.6), we can also obtain k + Z = o(). (3.7) Taking k =, we get By (3.3), (3.4) and (3.8), we have Z = o(). (3.8) Z 3 = o() and Z 4 = o(). (3.9) Take the inner product of Z 4 with (3.5) in L (Ω), that is i(z +τ Z 3 + τ Z 4), Z 4 κ Z, Z 4 div(τ 3(x) Z ), Z 4 κdiv(τ (x) Z 3 ), Z 4 = o(). Integrating by parts, we rewrite (3.0) as i(z + τ Z 3 + τ Z 4), Z 4 + κ Z, Z 4 + τ 3(x) Z, Z 4 + κτ (x) Z 3, Z 4 (3.0) = o(). (3.)

On the three-phase-lag heat equation 6 Then Z 4 = o(), (3.) since the other terms on the left-hand of (3.) converge to zero by (3.6), (3.8) and (3.9). Combining (3.6), (3.8) and (3.), we have Z H = o(). This is a contradiction with the assumption Z H =. If a(x) a 0 > 0 on Ω, b(x) > 0 on Ω 0 Ω and b(x) = 0 on Ω \ Ω 0. We take k = 0 and obtain Z = o() and Z = o(). (3.3) Taking the inner product of Z 4 with (3.5) in L (Ω) to get i(z + τ Z 3 + τ Z 4), Z 4 κ Z, Z 4 div(τ 3 (x) Z ), Z 4 κdiv(τ (x) Z 3 ), Z 4 = o(). Integrating by parts, we rewrite (3.4) as (3.4) i(z + τ Z 3 + τ Z 4), Z 4 + κ Z, Z 4 + τ 3(x) Z, Z 4 + κτ (x) Z 3, Z 4 = o(). (3.5) By (3.4), we have Z 4 = O(). Then by (3.3), (3.5) is i.e., i(z + τ Z 3 + τ Z 4), Z 4 + o() = i(z + τ Z 3 ), Z 4 + i τ Z 4, Z 4 + o() = i τ Z 4 + o() (by (3.6)) = o(), (3.6) Z 4 = o(). (3.7) Combining (3.6), (3.3) and (3.7), we have Z H = o(). This is a contradiction with the assumption Z H =. 4 Proof of Theorem.3 Proof. Since the proof of (.8) is similar to the proof in Section 3, we will only check the condition (.9). Here we will use special multipliers introduced in []. Assume that (.9) is false. Then by the uniform boundedness theorem, there exist a sequence and a unit sequence Z = (Z, Z, Z 3, Z 4 ) T D(A) such that (ii A)Z H 0. (4.) We rewrite (4.) as iz Z = o(), in H0 (Ω), (4.) iz Z 3 = o(), in H0 (Ω), (4.3) iz 3 Z 4 = o(), in H 0 (Ω), (4.4) i(z + τ Z 3 + τ Z 4) κ Z (τ 3 (x)z ) κ(τ (x)z 3) = o(), in L (Ω). (4.5)

6 Yang Wang Therefore, Re AZ, Z H = d dt Z H = b (x)z τ a (x)z 3 = o(). (4.6) Taking the inner product of a(x)z 4 with (4.5) in L (Ω), we get i(z + τ Z 3 ), a(x)z 4 + τ iz 4, a(x)z 4 κ Z, a(x)z 4 (τ 3 (x)z ), a(x)z 4 κ (τ (x)z 3), a(x)z 4 = o(). (4.7) By (4.3) and (4.4), i(z + τ Z 3 ) = O() and a(x)z 4 = o(). Integrating by parts, we rewrite (4.7) as τ i a (x)z4 +κ Z, (a(x)z 4) + τ 3 (x)z, (a(x)z 4) +κ τ (x)z 3, (a(x)z 4) = o(). (4.8) As Z = o(), Z = o(), Z 4 = O() and Z 4 = Z 3 + o() = O(), we have κ Z, (a(x)z 4) + τ 3 (x)z, (a(x)z 4) = o(). (4.9) As for the last term on the left-hand side of (4.7), by (4.4) and (4.6), we can obtain κ τ (x)z 3, (a(x)z 4) = κ τ (x)z 3, a (x)z 4 + a(x)z 4 = κ τ (x)z 3, a(x)z 4 + o() = κ τ (x)a(x)z 3, Z 3 + o() = o(). (4.0) Combination of (4.8), (4.9) and (4.0) yields a (x)z4 = o(). (4.) Which further leads to, due to (4.4), that a (x)z3 = o(). (4.) Take q(x) C, ([0, L]; R) and q(0) = 0. It follows from the inner product of (4.5) with q(x)(κ Z + τ 3(x)Z + κτ (x)z 3 ) in L (Ω) that i(z + τ Z 3 + τ Z 4), q(x)(κ Z + τ 3 (x)z + κτ (x)z 3) (κ Z + τ 3 (x)z + κτ (x)z 3), q(x)(κ Z + τ 3 (x)z + κτ (x)z 3) = o(). (4.3)

On the three-phase-lag heat equation 63 For the terms on the left-hand side of (4.3), we have i(z + τ Z 3 + τ Z 4), q(x)(κ Z + τ 3 (x)z + κτ (x)z 3) = i(z + τ Z 3 ), q(x)(κ Z + τ 3 (x)z ) + κ iz, q(x)τ (x)z 3 + κ iτ Z 3, q(x)τ (x)z 3 τ Z 3, q(x)(κ Z + τ 3 (x)z + κτ (x)z 3) + o() (by (4.4)) = κ Z 3, q(x)τ (x)z 3 + κ iτ Z 3, q(x)τ (x)z 3 τ Z 3, q(x)τ 3 (x)z τ κ Z 3, q(x)τ (x)z 3 + o() (by (4.3), (4.4) and dissipation) = iτ Z 3, q(x) κτ (x) Z τ 3 + τ Z 3, iq(x)τ 3 (x)z 3 τ κ Z 3, q(x)τ (x)z 3 +o() (by (4.3) and Z 3 = o()) = iτ Z 3, q(x) κτ (x) τ Z 3 iτ Z 3, q(x)τ 3 (x)z 3 τ κ Z 3, q(x)τ (x)z 3 + o() = iτ Z 3, q(x)a(x)z 3 τ κ Z 3, q(x)τ (x)z 3 + o(). (4.4) By dissipation iτ Z 3, q(x)a(x)z 3 converges to zero. Thus, (4.4) is i(z + τ Z 3 + τ Z 4), q(x)(κ Z + τ 3 (x)z + κτ (x)z 3) = τ κ Z 3, q(x)τ (x)z 3 + o(). Then, take the real part for the right-hand side of (4.5), i.e., (4.5) On the other hand, Re i(z + τ Z 3 + τ Z 4), q(x)(κ Z + τ 3 (x)z + κτ (x)z 3) = τ κre Z 3, q(x)τ (x)z 3 + o() = τ 4 κ Z 3, (q(x)τ (x)) Z 3 + o() ( Z 3 = o()). (4.6) Re (κ Z + τ 3 (x)z + τ (x)z 3), q(x)(κ Z + τ 3 (x)z + κτ (x)z 3) = ( q (x) ) (κ Z + τ 3 (x)z + τ (x)z 3) q(x) κ Z + τ 3 (x)z + κτ (x)z 3 L 0 = κ ( q (x) ) τ (x)z 3 q(l) κ Z (L) + τ 3 (L)Z (L) + κτ (L)Z 3(L) + o(). (4.7) Thus, (4.3) can be written as τ 4 κ Z 3, (q(x)τ (x)) Z 3 + κ ( q (x) ) τ (x)z 3 q(l) κ Z (L) + τ 3 (L)Z (L) +κτ (L)Z 3(L) = o(). (4.8)

64 Yang Wang Let us also take the inner product of (4.5) with (q(x)τ (x)) Z 3 in L (Ω) to get i(z + τ Z 3 ), (q(x)τ (x)) Z 3 + τ iz 4, (q(x)τ (x)) Z 3 (κ Z + τ 3 (x)z + τ (x)z 3), (q(x)τ (x)) Z 3 = o(). (4.9) For the terms on the left-hand side of (4.9), we have i(z + τ Z 3 ), (q(x)τ (x)) Z 3 = Z 3 + τ Z 4, (q(x)τ (x)) Z 3 + o() (by (4.3) and (4.4)) = o() ( Z 3 = o()), (4.0) and τ iz 4, (q(x)τ (x)) Z 3 = τ Z 3, (q(x)τ (x)) Z 3 + o() (by (4.4)), (4.) (κ Z + τ 3 (x)z + κτ (x)z 3), (q(x)τ (x)) Z 3 = κ Z + τ 3 (x)z + κτ (x)z 3, (q(x)τ (x)) Z 3 + κ Z + τ 3 (x)z + κτ (x)z 3, (q(x)τ (x)) Z 3 = κ τ (x)z 3, (q(x)τ (x)) Z 3 + κ τ (x)z 3, (q(x)τ (x)) Z 3 + o() (by dissipation) = κ τ (x)z 3, (q(x)τ (x)) Z 3 + o() ( Z 3 = o() and τ (x)z 3 = O()), (4.) we rewrite (4.9) as τ Z 3, (q(x)τ (x)) Z + κ τ (x)z 3, (q(x)τ (x)) Z 3 = o(). (4.3) Combination of (4.8) and (4.3) yields κ L 0 τ (x) ( (q(x)τ (x)) +q (x)τ (x) ) Z 3 dx q(l) κ Z (L)+τ 3 (L)Z (L)+κτ (L)Z 3(L) = o(). Take q(x) = τ (x) x 0 (4.4) a(t)dt, which is a solution to the first-order differential equation τ 3 (t) τ (x) ( (q(x)τ (x)) + q (x)τ (x) ) = a(x). (4.4) then is κ L 0 a(x) Z 3 dx q(l) κ Z (L) + τ 3 (L)Z (L) + κτ (L)Z 3(L) = o(). (4.5) The first term in (4.5) converges to zero by dissipation. Therefore, q(l) κ Z (L) + τ 3 (L)Z (L) + κτ (L)Z 3(L) = o(). (4.6) x Take q(x) = dt in (4.4), which is a solution to the first-order differential τ (x) 0 τ (t) equation (q(x)τ (x)) + q (x)τ (x) =. Together with (4.6), we thus have κ L 0 τ (x) Z 3 dx = o(), i.e., κ τ (x)z 3 dx = o(). (4.7)

On the three-phase-lag heat equation 65 Take the inner product of Z 4 with (4.5) in L (Ω), we have i(z + τ Z 3 + τ Z 4), Z 4 κ Z, Z 4 (τ 3(x)Z ), Z 4 κ(τ (x)z 3), Z 4 = o(). (4.8) Integrating by parts, we rewrite (4.8) as i(z + τ Z 3 ), Z 4 + i τ Z 4, Z 4 + κ Z, Z 4 + τ 3(x)Z, Z 4 + κτ (x)z 3, Z 4 = o(), (4.9) The first term on the left-side of (4.9) converges to zero by (4.6) and (4.7). Since Z 4 = O(), which together with dissipation and (4.7), (4.9) indicates Z 4 = o(). (4.30) Then by (4.6), (4.7) and (4.30), we have Z H assumption Z H =. = o(). This is a contradiction with the Acknowledgements. Y. W. is supported by China Scholarship Council. References [] K. Borgmeyer, R. Quintanilla and R. Racke, Phase-lag heat condition: decay rates for limit problems and well-posedness, J. Evol. Equ., 4 (04), 863-884. https://doi.org/0.007/s0008-04-04-6 [] A. Borichev and Y. Tomilov, Optimal polynomial decay of functions and operator semigroups, Math. Annal., 347 (00), 455-478. https://doi.org/0.007/s0008-009-0439-0 [3] S. K. R. Choudhuri, On a thermoelastic three-phase-lag model, J. Thermal Stresses, 30 (007), 3-38. https://doi.org/0.080/0495730603099 [4] F. L. Huang, Characteristic condition for exponential stability of linear dynamical systems in Hilbert spaces, Ann. Differential Equations, (985), 43-56. [5] A. E. Green and K. A. Lindsay, Thermoelasticity, J. Elasticity, (97), -7. https://doi.org/0.007/bf00045689 [6] A. E. Green and P. M. Naghdi, On undamped heat waves in an elastic solid, J. Thermal Stresses, 5 (99), 53-64. https://doi.org/0.080/04957390894636 [7] A. E. Green and P. M. Naghdi, Thermoelasticity without energy dissipation, J. Elasticity, 3 (993), 89-08. https://doi.org/0.007/bf00044969 [8] A. E. Green and P. M. Naghdi, A unified procedure for contruction of theories of deformable media. I. Classical continuum physics, Proc. Royal Society London A, 448 (995), 335 356. https://doi.org/0.098/rspa.995.000

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