DISCRETE AND CONTINUOUS Website: http://AIMsciences.org DYNAMICAL SYSTEMS Volume 7,Number4, October 2001 pp. 763–780
GLOBAL EXISTENCE AND UNIQUENESS FOR A HYPERBOLIC SYSTEM WITH FREE BOUNDARY
Tong Yang
Department of Mathematics City University of Hong Kong Hong Kong Fahuai Yi
Department of Mathematics South China Normal University Guangzhou 510631, China
Abstract. In this paper, we consider a 2 × 2 hyperbolic system originates from the theory of phase dynamics. This one-phase problem can be obtained by using the Catteneo-Fourier law which is a variant of the standard Fourier law in one dimen- sional space. A new classical existence and uniqueness result is established by some a priori estimates using the characteristic method. The convergence of the solutions to the one of classical Stefan problems is also obtained.
1. Introduction. In a number of physical situations the Fourier law
q(x, t)=−kTx(x, t), is replaced by
q(x, t + τ)=−kTx(x, t),τ>0, (1.1) where T,k,q are temperature, conductivity and heat flux respectively. One of the approximation of (1.1) is the Cattaneo-Fourier constitutive equation
τqt + q = −kTx. (1.2) Combining (1.2) with the conservation law of energy
et + qx =0, (1.3) yields a hyperbolic system with finite speed of propagation. Here e denotes the energy. In the simple case, one can take 1+T, in the liquid phase, e = T, in the solid phase. Thus in the liquid phase, we have
Tt + qx =0. (1.4)
1991 Mathematics Subject Classification. 35R35. Key words and phrases. Hyperbolic system, Stefan problem, Classical solution. The project was supported by National Natural Science Foundation of China(No.10071024), Guangdong Provincal Natural Science Foundation of China(No.000671), and the Strategic Re- search Grant of City University of Hong Kong # 7000968.
763 764 T. YANG & F.H. YI
We will henceforth assume that the interface is given by x = s(t) with the liquid in {(x, t):0 If τ = 0, then (1.2) and (1.4) are reduced to the usual heat equation Tt − kTxx =0. (1.5) For the classical heat transfer model (1.5), the one-phase Stefan problem is to find the temperature T (x, t) and a function s(t) > 0 such that (1.5) is satisfied in the domain {(x, t):0 On the fixed boundary x = 0, either the temperature boundary condition T (0,t)=f(t), (1.8) or flux boundary condition q(0,t)=f(t), (1.9) is imposed. If (1.2) is viewed as a physical law which is valid throughout the space in the weak sense, then we obtain the jump relation τqs = kT, on x = s(t). (1.10) And for (1.4), the jump condition on the boundary is (1 + T )s = q, on x = s(t). (1.11) In [2] Friedman and Hu considered (1.2) and (1.4) with free boundary conditions (1.10) and (1.11). They proved that the system has global classical solution under the fixed boundary conditions (1.8) or (1.9). They also obtained a convergence ∗ result for the temperature in (L∞) sense for τ → 0 under the boundary condition (1.8). In [2], they pointed out in the Remark 7.2 (see [2], p.278) that if the free boundary conditions (1.10) and (1.11) are replaced by (1.6) and (1.7), then the problem is open. If we view (1.2) as a local constitutive law, then it does not yield condition (1.10) on the free boundary. It is then natural to assume that the temperature is continu- ous across the free boundary which gives (1.6). Substituting (1.6) into (1.11) yields (1.7). Li considered the system (1.2) and (1.4) with the free boundary conditions (1.6), (1.7) and fixed boundary condition (1.8) (see [5]). He proved that the problem has a global classical solution, but he did not consider the problem with fixed boundary condition (1.9). In fact, the problem with boundary condition (1.9) is more difficult than the problem with boundary condition (1.8). In this paper, we will consider the system (1.2) and (1.4) with free boundary condition (1.6) and (1.7) in two aspects. On HYPERBOLIC SYSTEM WITH FREE BOUNDARY 765 one hand, we prove global classical existence under the boundary condition (1.9) in next section. On the other hand, we prove the solutions with boundary condition (1.8) obtained in [5] uniformly converge to the solution of classical Stefan problem (1.5)- (1.7) when parameter τ → 0+ in the last section. For the weak solution of the system (1.2) and (1.3), please refer to [7], [8] and the references therein. 2. Global existence. In this section we consider the system Tt + qx =0, 0 T (x, 0) = T0(x), (2.6) q(x, 0) = q0(x), (2.7) s(0) = 1, (2.8) where τ is a positive parameter. T,q and s are unknown functions. We suppose s(0) = 1 for simplicity. First we make a transformation √ A = T + τq, √ B = T − τq. Then the Riemann invariants A and B satisfy √ A − B τAt + Ax + √ =0, 0 1 1 Theorem 2.1. For fixed τ>0, let A0,B0 ∈ C [0, 1],f(t) ∈ C [0, +∞). Suppose the compatible conditions are satisfied at (0,0) and (1,0), and |A0(1)| < 1. (2.17) 766 T. YANG & F.H. YI Then there is a t0 > 0, such that the problem (2.9)-(2.16) has a unique solution A, B, s ∈ C1 Q ×C1 Q ×C2 ,t Q { x, t |s | < √1 . Remark on condition (2.17): This condition implies (0) τ It means that s(t) is between the two characteristic lines starting from (1,0) when t is small. The case when |A0(1)| > 1 will be considered by the authors in the future. From theorem 2.1 we can extend the solution (A,B,s) to any t>0 as long as the following three cases do not occur: There exists t1 > 0 such that lim s(t)=0, (2.18) tt1 1 or lim s (t)=√ , (2.19) tt1 τ or lim (|A(·,t)|C1[0,s(t)] + |B(·,t)|C1[0,s(t)])=+∞. (2.20) tt1 In the following we prove that none of (2.18)-(2.20) occur. And this immediately yields the global existence. Lemma 2.1. Let (A, B) be a C1 solution of (2.9)-(2.16) with (2.17), then |s t | < √1 , |A s t ,t | < , |B s t ,t | < . ( ) τ ( ( ) ) 1 (( ( ) ) 1 Proof: We prove this lemma by contradiction. Suppose that there exists t1 > 0, s t q s t ,t √1 , s t q s t ,t < √1 ≤ t |s t | < √1 |A s t ,t | < , |B s t ,t | < . (2.12) and ( ) τ imply ( ( ) ) 1 then (2.11) implies ( ( ) ) 1 1 Lemma 2.2. (A, B) is a C solution of (2.9)-(2.16) with (2.17). If A0(x) > −1,B0(x) > −1 and f(t) > 0, then A(x, t) > −1,B(x, t) > −1. HYPERBOLIC SYSTEM WITH FREE BOUNDARY 767 Proof: t x = s(t) (x, t) Ω4 (s(η(x, t)),η(x, t)) ∗ t = tτ Ω2 Ω3 Ω 1 x 0 Figure 1 In Ω1 ∪ Ω2, cf. Figure 1, we have t − r t − − t t 1 t − r B x, t e 2τ B x √ , e 2τ A √ x, r dr, ( )= ( + τ 0) + τ ( τ + ) (2.23) 2 0 and in Ω3 ∪ Ω4 t − r t − √ − t−η 1 τ t − r B(x, t)=e 2τ B(s(η),η)+ e A( √ + x, r)dr. (2.24) 2τ η τ Similarly in Ω1 ∪ Ω3,wehave t − r t − − √t t 1 r − t A x, t e 2 τ A x − √ , e 2τ B √ x, r dr, ( )= ( τ 0) + τ ( τ + ) (2.25) 2 0 and in Ω2 ∪ Ω4 t x √ − √ 1 − t−r r − t A x, t e 2 τ A ,t− τx e 2τ B √ x, r dr. ( )= (0 )+ τ ( τ + ) (2.26) 2 √ t− τx We also prove this lemma by contradiction. Suppose that there is a smallest value t1, such that A(x, t) > −1,B(x, t) > −1, for 0 ≤ x ≤ s(t), 0 ≤ t A(x1,t1)=−1, or B(x1,t1)=−1, for some 0 ≤ x1 ≤ s(t1). (2.27) If A(x ,t )=−1 holds, then x > 0 because 1 1 1 √ √ A(0,t1)=B(0,t1)+2 τf(t1) ≥−1+2 τf(t1) > −1. But for any small ε>0, we have t1 − ε ε 1 − t1−r r − t1 A x ,t e 2τ A x − √ ,t − ε e 2τ B √ x ,r dr ( 1 1)= ( 1 τ 1 )+ τ ( τ + 1 ) 2 t1−ε − ε − t1−r t > −{e 2τ e 2τ 1 } − , +[ ]t1−ε = 1 which is a contradiction to A(x1,t1)=−1. Next if B(x1,t1)=−1 holds, then x1 Lemma 2.3 Under the assumptions of Lemma 2.2, if f(t) ≥ δ>0, then √ √ − τ+t A(s(t),t) ≥−1+2 τδe 2τ . Proof: From (2.13) and f(t) ≥ δ>0, we have √ √ A(0,t)=B(0,t)+2 τf(t) ≥−1+2 τδ. √ If δ is small enough, then A(x, 0) > −1+2 τδ. If (s(t),t) ∈ Ω3, then by (2.25) we have t − t t 1 − r−t r − t A(s(t),t)=e 2τ A(s(t) − √ , 0) + e 2τ B( √ + s(t),r)dr τ 2τ τ √ 0 t − t ≥ e 2τ (−1+2 τδ) − (1 − e 2τ ) √ − t = −1+2 τδe 2τ . If (s(t),t) ∈ Ω4, then (2.26) gives t s(t) √ − √ 1 − t−r r − t A s t ,t e 2 τ A ,t− τs t e 2τ B √ s t ,r dr ( ( ) )= (0 ( )) + τ √ ( τ + ( ) ) 2 t− τs(t) s(t) s(t) − √ √ − √ ≥ e 2 τ (−1+2 τδ) − (1 − e 2 τ ) s(t) √ − √ = −1+2 τδe 2 τ √ √ − τ+t ≥−1+2 τδe 2τ , s t ≤ √t . by using ( ) 1+ τ Lemma 2.4. Under the assumptions of Lemma 2.2, we have A(x, t) ≤ C0,B(x, t) ≤ C0, where C0 depends on τ,t,max{B0(x),A0(x)} and max f(t). 0≤σ≤t Proof: Let I(t) = max B(x, t),J(t) = max A(x, t),K(t) = max{I(t),J(t)}, 0≤x≤s(t) 0≤x≤s(t) ∗ ∗ and C˜ = max {B0(x),A0(x)}. Suppose that I(t)=B(x ,t). If (x ,t) ∈ Ω1 ∪ Ω2, 0≤x≤1 then by (2.23) we have, t ∗ − t ∗ t 1 − t−r t − r ∗ I t B x ,t e 2τ B x √ , e 2τ A √ x ,r dr ( )= ( )= ( + τ 0) + τ ( τ + ) 2 0 t − t 1 − t−r ≤ Ce˜ 2τ e 2τ K r dr. + τ ( ) 2 0 ∗ If (x ,t) ∈ Ω3 ∪ Ω4, then (2.24) gives ∗ t − t−η 1 − t−r I(t) ≤ e 2τ + e 2τ K(r)dr. 2τ η∗ Thus t − t−σ 1 − t−r I(t) ≤ max {(1 + C˜)e 2τ + e 2τ K(r)dr}. 0≤σ≤t 2τ σ HYPERBOLIC SYSTEM WITH FREE BOUNDARY 769 We now estimate J(t) as follows. Suppose that J(t)=A(x0,t). If (x0,t) ∈ Ω2 ∪ Ω3, then by (2.25) we have t − t 1 − t−r J t ≤ Ce˜ 2τ e 2τ K r dr. ( ) + τ ( ) 2 0 If (x0,t) ∈ Ω2 ∪ Ω4, then (2.26) yields t x √ − 0 1 − t−r J t ≤ e 2τ A ,t− τx e 2τ K r dr. ( ) (0 0)+ τ √ ( ) 2 t− τx0 Since √ √ √ √ A(0,t− τx0)=B(0,t− τx0)+2 τf(t − τx0) √ √ t− τx0 √ − t− τx0−σ 1 − t− τx0−r ≤ max√ {(1 + C˜)e 2τ + e 2τ K(r)dr} 0≤σ≤t− τx0 2τ √ √ σ +2τf(t − τx0), we have √ t− τx0 − t−σ 1 − t−r J(t) ≤ max√ {(1 + C˜)e 2τ + e 2τ K(r)dr} 0≤σ≤t− τx0 2τ σ t √ x √ − √0 1 − t−r τe 2 τ f t − τx e 2τ K r dr +2 ( 0)+ τ √ ( ) 2 t− τx0 t √ x √ − t−σ 1 − t−r − √0 { C˜ e 2τ e 2τ K r dr} τe 2 τ f t − τx . = max√ (1 + ) + τ ( ) +2 ( 0) 0≤σ≤t− τx0 2 0 Therefore t t−σ 1 t−r √ √ − 2τ − 2τ K(t) ≤ max {(1 + C˜)e + e K(r)dr} +2 τf(t − τx0) 0≤σ≤t 2τ 0 t √ 1 − t−r ≤ 1+C˜ + e 2τ K(r)dr +2 τ max f(t). τ ≤σ≤t 2 0 0 Using Gronwall inequality, we obtain 1 t √ K(t) ≤ [1 + (e 2τ − 1)][1 + C˜ +2 τ max f(t)]. 4τ 2 0≤σ≤t Lemma 2.5. Under the assumptions of Lemma 2.2, we have s(t) ≥ σ>0, where σ depends on C0,T0(x) and f(t). Proof: By integrating the equation Tt + qx =0wehave s(t) Tt(x, t)dx + q(s(t),t)=q(0,t). 0 It follows that d s(t) T x, t dx s t f t . dt[ ( ) + ( )] = ( ) 0 Integrating the above equality with respect to t yields s(t) 1 t T (x, t)dx + s(t)= T0(x)dx +1+ f(λ)dλ. (2.28) 0 0 0 770 T. YANG & F.H. YI 1 By Lemma 2.4, we have T = (A + B) ≤ C0. Substituting it into (2.28) we get 2 1 t (C0 +1)s(t) ≥ T0(x)dx +1+ f(λ)dλ. 0 0 Therefore 1 t s t ≥ 1 T x dx f λ dλ σ. ( ) C [ 0( ) +1+ ( ) ]:= 0 +1 0 0 Lemma2.6. Under the assumptions of Lemma 2.2, we have |Ax(x, t)|≤C1, |Bx(x, t)|≤C1, where C1 depends on τ and t, Proof: Differentiating (2.9) and (2.10) with respect to x formally, we have √ Ax − Bx τ∂tAx + ∂xAx + √ =0, (2.29) 2 τ √ Bx − Ax τ∂tBx − ∂xBx + √ =0, (2.30) 2 τ and Ax = −Bx − 2f(t) − 2τf (t), on x =0, (2.31) 1 − A Bx = Ax, on x = s(t). (2.32) 1+A Let I(t) = max |Bx(x, t)|,J(t) = max |Ax(x, t)| and K(t) = max{I(t),J(t)}. 0≤x≤s(t) 0≤x≤s(t) By Lemma 2.5 we only need to prove the estimates in Lemma 2.6 in the interval ∗ [0,t ] (see Figure 1). If (x, t) ∈ Ω1 ∪ Ω2, then t − t t 1 − t−r t − r B x, t e 2τ B x √ , e 2τ A √ x, r dr. x( )= x( + τ 0) + τ ( τ + ) 2 0 Thus t t 1 − t−r I t ≤ C˜ e 2τ e 2τ J r dr, ( ) 1 + τ ( ) 2 0 C˜ {|A | ∞ , |B | ∞ } x, t ∈ where 1 = max 0 L 0 L .If( ) Ω3 t t−η(x,t) 1 − t−r |Bx(x, t)|≤e 2τ |Bx(s(η),η)| + e 2τ J(r)dr 2τ η t − t−η 1 − A 1 − t−r = e 2τ |Ax(s(η),η)| + e 2τ J(r)dr. 1+A 2τ η Since η − η 1 − η−r |A s η ,η |≤e 2τ |A ., | e 2τ I r dr x( ( ) ) x( 0) + τ ( ) 2 0 η − η 1 − η−r ≤ C˜ e 2τ e 2τ I r dr, 1 + τ ( ) 2 0 and from Lemma 2.3 √ √ − τ+η 1+A ≥ 2 τδe 2τ , we have by Lemma 2.1 √ √ 1 − A − τ+η ≤ ( τδ) 1e 2τ := C. 1+A HYPERBOLIC SYSTEM WITH FREE BOUNDARY 771 Thus η t − t−η − η 1 − η−r 1 − t−r |B x, t |≤Ce 2τ C˜ e 2τ e 2τ I r dr e 2τ J r dr x( ) [ 1 + τ ( ) ]+ τ ( ) 2 0 2 η t − t 1 − t−r ≤ C C˜ e 2τ e 2τ K r dr . [ 1 + τ ( ) ] 2 0 Therefore t − t 1 − t−r I t ≤ C C˜ e 2τ e 2τ K r dr . ( ) ( 1 + τ ( ) ) 2 0 We can estimate J(t) in the same way and then get the estimate for K(t). Lemma 2.7. Under the assumptions of Lemma 2.2, we have − ε |s t |≤1√ , ( ) τ √ −1 where ε =(1+C1 τ) . Proof: We prove this by contradiction. Suppose there exits a t1 > 0, such that − δ − δ − δ |s t | 1√ |s t | < 1√ ≤ t Tt + qx =0 at(s(t1),t1) √ − δ τq √1 T 1 s t ,t t + τ x + τ =0 at( ( 1) 1) these two equality yield √ − δ T √1 T τq q 1 s t ,t ( t + τ x)+( τ + x)+ τ =0 at( ( 1) 1) (2.35) From (2.33)-(2.35) we have √ − δ δ T τq 1√ ≤ . ( x + x)+ τ 0 √ − δ T τq A ≥−C −C δ 1√ ≤ , i.e., By Lemma 2.6, x + x = x 1 which implies that 1 + τ 0 √ −1 δ ≥ (1 + C1 τ) = ε. This contradicts to the assumption that δ<ε. − δ s t −1√ The case when ( 0)= τ can be discussed similarly. Theorem 2.2. (Global existence) Under the assumptions of Theorem 2.1, if A0(x) > −1,B0(x) > −1 and f(t) ≥ δ>0, then for any t0 > 0, the system (2.9)- A, B, s ∈ C1 Q × C1 Q × C2 ,t . (2.16) has a unique solution ( ) ( s,t0 ) ( s,t0 ) [0 0] 772 T. YANG & F.H. YI t > , A, B, s ∈ C1 Q \{t Proof: Assume that there exists 1 0 such that ( ) ( s,t1 = t } × C1 Q \{t t } × C2 ,t . s t ≥ σ> ≤ t lim (|Ax(·,t)|C1[0,s(t)] + |Bx(·,t)|C1[0,s(t)]) ≤ 2C1. tt1 The theorem follows from Theorem 2.1 and these estimates. Remark: We tried to prove the convergence result of (2.1)-(2.8) with respect to τ → 0, but we did not succeed. In the next section, we will prove a convergence result of (2.1)-(2.8) in which the flux boundary condition (2.5) is replaced by tem- perature boundary condition T = f(t). 3. Uniform Convergence. If we impose a temperature boundary condition (1.8) on x =0, the problem (2.9)-(2.16) becomes √ A − B τAt + Ax + √ =0, 0 A(x, 0) = A0(x), (3.6) B(x, 0) = B0(x), (3.7) s(0) = 1. (3.8) For the problem (3.1)-(3.8), there is a global existence theorem obtained in [5] as stated in the following. ∞ ∞ Theorem 3.1. (D. Li, [5]) Assume A0,B0 ∈ C [0, 1],f(t) ∈ C [0, ∞),f (t) ≥ ,A < ,B < , ( sτ ,t0 ) [0 0] for all 0 0 and the solution satisfies A s t ,t > , ∂xAτ (x, t) < 0,∂xBτ (x, t) < 0. 1 1 Remark on Theorem 3.1: If we only assume that A0,B0,f ∈ C and C com- patible conditions hold at (0, 0) and (1, 0), then there exists a unique solution HYPERBOLIC SYSTEM WITH FREE BOUNDARY 773 1 1 2 (Aτ ,Bτ ,sτ ) ∈ C × C × C . Corresponding to (Tτ ,qτ ,sτ ), the problem (3.1)-(3.8) is ∂tTτ + ∂xqτ =0, 0 Tτ (x, 0) = T0(x), (3.14) qτ (x, 0) = q0(x), (3.15) sτ (0) = 1. (3.16) In this section we prove the solutions (Tτ ,sτ ) of the problem (3.9) -(3.16) converge to the solution (T,s) of the Stefan problem 2 ∂tT − ∂xT =0, 0 T (x, 0) = T0(x), (3.21) s(0) = 1. (3.22) The classical existence and uniqueness for Stefan problem (3.17)-(3.22) was ∞ ∞ solved by Jiang (see [3] and [4]). If T0 ∈ C [0, 1],f(t) ∈ C [0, +∞),T0,f ≥ 0, and the compatible conditions are satisfied at (0,0), then the Stefan problem (3.17)- T,s ∈ C∞ Q × C∞ ,t t > (3.22) has a unique solution ( ) ( s,t0 ) [0 0] for all 0 0. Moreover s(t) is monotone increasing. We now present a weak formulation for the Stefan problem (3.17)-(3.22) which is in the framework of parabolic variational inequality (see [1]). (T (x, t),s(t)) is called a weak solution to the Stefan problem (3.17)-(3.22) if the following holds: t > T x, t ,s t ∈ H1 Q × L∞ ,t ∩ BV ,t For any 0 0, ( ( ) ( )) ( s,t0 ) (0 0) (0 0), such that for ψ ∈ C∞ Q ,ψ x, t ψ ,t any ( t0 ) ( 0)= (0 )=0, t0 (∂tTψ+ ∂xT∂xψ)+ ψ(s(t),t)s (t)dt =0, (3.23) 0 Q s,t0 Q ,M × ,t ,M |s t | ∞ ,t where t0 =(0 ) (0 0) = ( ) L (0 0)and T (0,t)=f(t),T(x, 0) = T0(x). (3.24) In order to prove the main convergence result, we first prove the following lem- mas. A ,B ∈ C1 , ,f t ∈ C1 , ∞ ,f t ≥ ,A < Lemma 3.1. Assume that 0 0 [0 1] ( ) [0 ) ( ) 0 0 ,B < , 0, we have ∂ T x, t < , Q , x τ ( ) 0 in sτ ,t0 (3.25) Proof: ∂xAτ < 0and∂xBτ < 0 imply (3.25). (3.11), (3.13) and (3.25) yield (3.26). It is clear Aτ (x, t) > 0byAτ (sτ (t),t) > 0and∂xAτ (x, t) < 0. Also Bτ (x, t) > −1 by Bτ (sτ (t),t)=−Aτ (sτ (t),t) > −1and∂xBτ (x, t) < 0. Moreover, we have A x, t T − B ≤ C , Q , τ ( )=2 τ τ 2 f +1 in sτ ,t0 B x, t T − A ≤ C , Q . τ ( )=2 τ τ 2 f in sτ ,t0 Hence √ 1 τ |qτ (x, t)| = |Aτ − Bτ |≤Cf +1 2 This completes the proof of the lemma. Lemma 3.2. Under the assumption of Lemma 3.1, for any t>0, we have sτ (t) ≤ M(t), where M(t) is independent of τ>0. Proof: Integrating the equation (3.10) with respect to x gives sτ (t) sτ (t) τ ∂tqτ (x, t)dx + qτ (x, t)dx = f(t). 0 0 It follows that d sτ (t) sτ (t) τ q x, t q x, t dx f t τq2 s t ,t dt τ ( )+ τ ( ) = ( )+ τ ( τ ( ) ) 0 0 2 ≤ Cf +(Cf +1) . Therefore, we have sτ (t) 1 t − t 1 − t−r q x, t dx e τ q x dx C C 2 e τ dr τ ( ) = 0( ) +[ f +( f +1) ] τ 0 0 0 1 2 ≤ q0(x)dx + Cf +(Cf +1) := C. (3.30) 0 Next multiplying the equation (3.9) by x and integrating it with respect to x give d sτ (t) sτ (t) xT x, t dx 1s2 t q x, t dx ≤ C, dt[ τ ( ) + τ ( )] = τ ( ) (3.31) 0 2 0 by using (3.30). Integrating (3.31) with respect to t yields sτ (t) 1 1 2 1 xTτ (x, t)dx + sτ (t) ≤ xT0(x)dx + + Ct. 0 2 0 2 HYPERBOLIC SYSTEM WITH FREE BOUNDARY 775 Since Tτ (x, t) ≥ 0, we have 1 2 2 sτ (t) ≤ 2 xT0(x)dx +1+2Ct := M (t). 0 This completes the proof of the lemma. For any t0 > 0, we denote Q { x, t , We extend the solution by letting T x, t x, t ∈ Q \Q . τ ( )=0 if( ) t0 sτ ,t0 Lemma 3.3. Under the assumption of Lemma 3.1, we have M(t0) 2 2 2 [τ(∂tTτ ) +(∂xTτ ) ](x, t)dx + (∂tTτ ) dxdt ≤ C, 0 Q t0 where C is independent of τ>0. Proof: It is easy to check τ∂2T ∂ T − ∂2T , Q . t τ + t τ x τ =0 in sτ ,t0 (3.32) φ x ∈ C∞ , ∞ ≤ φ x ≤ ,φ x x< 1 , Let ( ) [0 ) be a cutoff function, 0 ( ) 1 ( )=1if 3 and φ x x> 2 . ∂ T − f t φ x , ( )=0if 3 By multiplying (3.30) by t τ ( ) ( ) integrating it over Q τ sτ ,t and dropping the subindex without any ambiguity, we have 2 2 (τ∂t T + ∂tT − ∂xT )(∂tT − f (t)φ(x))dxdt =0. (3.33) Qs,t 776 T. YANG & F.H. YI Now we calculate each term in (3.33) as follows: 2 τ∂t T (∂tT − f (t)φ(x))dxdt (3.34) Q s,t 1 2 2 = τ∂t(∂tT ) dxdt − τ∂ Tf φdxdt 2 t Qs,t Qs,t s(t) 1 1 2 1 2 = τ (∂tT ) (x, t)dx − τ (∂tT ) (x, 0)dx 2 0 2 0 t 1 2 − τ (∂tT ) (s(t),t)s (t)dt + τ∂tTf (t)φ(x)dxdt 2 0 Qs,t s(t) 1 −τ ∂tTf (t)φ(x)dx + τ ∂tT (x, 0)f (0)φ(x)dx 0 0 s(t) 1 t 1 2 1 2 1 2 3 ≥ τ (∂tT ) (x, t)dx − τ (∂xq ) dx − τ (∂xT ) (s(t),t)[s (t)] dt 2 2 0 2 0 0 0 s(t) 1 2 2 2 1 2 − τ (∂tT ) dxdt − τ [f”(t)φ(x)] dxdt − τ (∂tT ) (x, t)dx 4 4 0 Qs,t Qs,t s(t) 1 2 −τ [f (t)φ(x)] dx − |∂xq0|f (0)dx, 0 0 and ∂tT [∂tT − f (t)φ(x)]dxdt (3.35) Q s,t 1 2 2 ≥ (∂tT ) dxdt − [f (t)φ(x)] dxdt, 2 Qs,t Qs,t and 2 −∂xT [∂tT − f (t)φ(x)]dxdt (3.36) Qs,t t = − ∂xT (s(t),t)∂tT (s(t),t)dt + ∂xT [∂x∂tT − f (t)φ (x)]dxdt 0 Qs,t t s(t) 1 2 1 2 1 2 ≥ (∂xT ) (s(t),t)s (t)dt + (∂xT ) (x, t)dx − (∂xT0) dx 0 2 0 2 0 t 1 2 2 2 − (∂xT ) (s(t),t)s (t)dt − (∂xT ) dxdt − (f φ ) dxdt. 2 0 Qs,t Qs,t 2 2 By substituting (3.34)-(3.36) into (3.33) and noticing that τ[sτ (t)] = Aτ (sτ (t),t) ≤ 1, we have t t 1 2 3 1 2 − τ (∂xT ) (s(t),t)[s (t)] dt ≥− (∂xT ) (s(t),t)s (t)dt. 2 0 2 0 HYPERBOLIC SYSTEM WITH FREE BOUNDARY 777 Therefore, we have when τ ≤ 1, s(t) s(t) 1 2 1 2 1 2 τ (∂tT ) dx + (∂tT ) dxdt + (∂xT ) dx (3.37) 4 0 4 2 0 Q s,t 2 ≤ C + (∂xT ) dxdt. Qs,t Notice that Tτ (x, t)=0ifx ≥ sτ (t), we can rewrite (3.35) as M M 1 2 1 2 1 2 2 τ (∂tT ) dx + (∂tT ) dxdt + (∂xT ) dx ≤ C + (∂xT ) dxdt. 4 0 4 2 0 Qt Qt This implies Lemma 3.3 by Gronwall inequality. Up to now, we have proved ∞ 1 1 2 Tτ ∈ L (0,t0; H (0,M(t0))) ∩ H (0,t0; L (0,M(t0))), 1 2 where the bounds are independent of τ.LetX = H (0,M(t0)),Y = L (0,M(t0)) and Z = C[0,M(t0)]. Then we have that X ⊂ Z ⊂ Y, and that X → Z is compact. Thus, by Simon’s Compactness Theorem (see [9]), we know that {Tτ }τ≥0 is rela- C ,t C ,M t T ∗ x, t ∈ C Q tively compact in ([0 0]; [0 ( 0)]). That is, there exits ( ) ( t0 )and {T } {T } a subsequence τn ( still denoted by τ for simplicity), such that T → T ∗, Q τ → , τ uniformly in t0 ( 0) (3.38) moreover T → T ∗, H1 Q τ → . τ weakly in ( t0 )( 0) (3.39) In the following, we will consider the convergence property of {sτ (t)}. By Helly’s ∗ Theorem there exists a s (t) which is increasing in [0,t0], such that ∗ sτ (t) → s (t), for each t ∈ [0,t0](τ → 0). (3.40) Hence, By Lebesgue Dominated Convergence Theorem ∗ p sτ (t) → s (t), in L (0,t0), ∀1 ≤ p<+∞. (3.41) ∗ ∞ 1,1 It is obviously that s (t) ∈ L (0,t0) ∩ W (0,t0). Theorem 3.2. Under the assumption of Lemma 3.1, for all t0 > 0, the subse- quence of the solutions (Tτ ,qτ ,sτ ) to the problem (3.9)-(3.16) converges to the solution (T,s) of Stefan problem in the following sense, T → T, Q , H1 Q , τ uniformly in t0 and weakly in ( t0 ) and p sτ → s, pointwise in [0,t0], and in L (0,t0), ∀1 ≤ p ≤ +∞, T Q \Q . where we extend =0 in t0 s,t0 778 T. YANG & F.H. YI Proof: From (3.38)-(3.41) we only need to prove that (T ∗,s∗) is the solution of the ψ x, t ∈ C∞ Q Stefan problem (3.17)-(3.22). For this, by multiplying (3.9) by ( ) ( t0 ) ψ x, t ψ ,t ≡ , Q , with ( 0)= (0 ) 0 and integrating over sτ ,t0 we have (∂tTτ + ∂xqτ )ψdxdt (3.42) Q sτ ,t0 = ∂tTτ ψdxdt − qτ ∂xψdxdt Qs ,t Qs ,t τ 0 τ 0 t0 + qτ (sτ (t),t)ψ(sτ (t),t)dt =0. 0 ∂ ψ Q Multiplying the equation (3.10) by x and integrating it over sτ ,t0 give (τ∂tqτ + ∂xTτ + qτ )∂xψdxdt (3.43) Qsτ ,t 0 1 = −τ qτ ∂t∂xψdxdt − τ qτ (x, 0)∂xψ(x, 0)dx 0 Qs ,t τ 0 t0 −τ qτ (sτ (t),t)∂xψ(sτ (t),t)sτ (t)dt + (∂xTτ + qτ )∂xψdxdt =0. 0 Q sτ ,t0 Combining (3.42) and (3.43) yields t0 (∂tTτ ψ + ∂xTτ ∂xψ)dxdt + qτ (sτ (t),t)ψ(sτ (t),t)dt (3.44) 0 Qsτ ,t 0 1 = τ qτ ∂t∂xψdxdt + τ q0(x)∂xψ(x, 0)dx 0 Qs ,t τ 0 t0 +τ qτ (sτ (t),t)∂xψ(sτ (t),t)sτ (t)dt. 0 Since t0 sτ (t) ∂tTτ ψdxdt = ∂tTτ ψdxdt + ∂tTτ ψdxdt, 0 s∗(t) Q Q ∗ sτ ,t0 s ,t0 and t0 sτ (t) ∂tTτ ψdxdt 0 s∗(t) t0 sτ (t) t0 2 1/2 ∗ 1/2 ≤ max |ψ|( (∂tTτ ) dxdt) ( |sτ (t) − s (t)|dt) 0 s∗(t) 0 → 0forτ → 0, we have ∗ ∂tTτ ψdxdt → ∂tT ψdxdt. Q Q ∗ sτ ,t0 s ,t0 HYPERBOLIC SYSTEM WITH FREE BOUNDARY 779 Similarly, we have ∗ ∂xTτ ∂xψdxdt → ∂xT ∂xψdxdt. Q Q ∗ sτ ,t0 s ,t0 Furthermore, we get t0 t0 qτ (sτ (t),t)ψ(sτ (t),t)dt = sτ (t)ψ(sτ (t),t)dt 0 0 t0 d sτ (t) sτ (t) ψ x, t dx − ∂ ψ x, t dx dt = [dt ( ) t ( ) ] 0 0 0 sτ (t0) 1 t0 sτ (t) = ψ(x, t0)dx − ψ(x, 0)dx − ∂tψdxdt, 0 0 0 0 which converges to ∗ ∗ s (t0) 1 t0 s (t) ψ(x, t0)dx − ψ(x, 0)dx − ∂tψdxdt 0 0 0 0 ∗ ∗ t0 d s (t) s (t) ψ x, t dx − ∂ ψ x, t dx dt = [dt ( ) t ( ) ] 0 0 0 t0 = (s∗) (t)ψ(s∗(t),t)dt. 0 It is clear that, for the three terms on the right hand side of (3.44), we have √ √ τ| qt∂t∂xψ|dxdt ≤ τ max | τqτ | |∂t∂xψ|dxdt → 0, Q Q sτ ,t0 sτ ,t0 1 τ q0(x)∂xψ(x, 0)dx → 0, 0 t0 τ| qτ (sτ (t),t)∂xψ(sτ (t),t)sτ (t)dt| 0 t0 √ √ ≤ τ max | τqτ |·max |∂xψ| sτ (t)dt √ √ 0 = τ max | τqτ |·max |∂xψ|(sτ (t0) − 1) → 0. By letting τ → 0 in (3.44), we obtain t0 ∗ ∗ ∗ ∗ (∂tT ψ + ∂xT ∂xψ)dxdt + ψ(s (t),t)(s ) (t)dt =0. 0 Q ∗ s ,t0 We also have ∗ ∗ T (0,t)=f(t),T(x, 0) = T0(x), T → T ∗ Q¯ . T ∗,s∗ by using the result that τ uniformly in t0 Hence, ( ) satisfies (3.23) and (3.24). The proof of the Theorem 3.2 is completed. 780 T. YANG & F.H. YI REFERENCES [1] Friedman, A. (1982): Variational Principles and Free Boundary Problems, Wiley, New York. [2] Friedman, A.; Hu, B. (1989): The Stefan problem for a hyperbolic heat equation, J. Math. Anal. Appl. 138, 249-279. [3] Jiang, L. (1963): Two-phase Stefan problem (I), Acta Math. Sinica 13, 631-646. [4] Jiang, L. (1964): Two-phase Stefan problem (II), Acta Math. Sinica 14, 33-49. [5] Li, D. (1989): The well-posedness of a hyperbolic Stefan problem,, Quart. Appl. Math. XLVI I, 221-231. [6] Li, T.; Yu, W. (1985): Boundary Value Problem for Quasilinear Hyperbolic Systems, Duke Univ. Math. Series V. [7] Shemetov, N.V. (1991): The Stefan problem for a hyperbolic heat equation, International Series of Numerical Mathematics, 99, 365-376. [8] Shemetov, N.V. (1995): Existence and stability results for the hyperbolic Stefan Problem with relaxation, Ann. Mat. Pura Appl., CLXVIII, 301-316. [9] Simon, J. (1987): Compact sets in the space Lp(0,T; B), Ann. Mat. Appl., 146, 997-1037. [10] Varlamov, V. (1998): On the initial boundary value peoblem for the damped Boussinesq equation, Discrete and Continuous Dynamical Systems, 4, 431-444. [11] Yin, H.-M. (1996): A free boundary problem arising from a stree-driven diffusion in polymers, Discrete and Continuous Dynamical Systems, 2, 191-202. Received December 2000; revised June 2001. E-mail address: [email protected] E-mail address: [email protected]