Evolution semi-linear hyperbolic equations in a ... - Semantic Scholar
Semi-linear Hyperbolic Equation
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Evolution semi-linear hyperbolic equations in a bounded domain Aimin Huang & Du Pham Abstract. In this article, our goal is to prove the existence and uniqueness of solution for 1D and 2D semi-linear hyperbolic equations in a bounded domain with a monotone nonlinear term. We use elliptic regularization and a finite difference scheme in time to build the approximate solutions for the semi-linear hyperbolic equations, and we utilize the regularization method together with the monotonicity and convexity of the nonlinear term to show the existence of the resulting stationary problems. Finally, the existence of the solution for the evolution problem is done by studying the convergence of the approximate solutions and by using the standard Minty method, and the uniqueness is achieved.
1. Introduction We aim to study in this article the initial and boundary value problem in dimension one and two for some first-order semi-linear reaction convection equation ∂t u + Au + P (u) = f,
(1.1)
where Au = ∂x u + ∂y u and P : R → R is a monotone function growing at the rate of a polynomial of odd order; a related problem was studied in space dimension one in [JP07] in the context of singular perturbation and boundary layer theory. The second-order semi-linear hyperbolic equations, in particular the non-linear wave equation, have been studied by many authors, see [Liu03, Sat68, Gla73] and references therein, where the authors considered the Dirichlet boundary conditions. In this article, we study the first-order semi-linear hyperbolic equation (1.1) and consider the dissipative boundary conditions in the sense of [BS07, Chapter 3] (see Remark 3.1), which is a first attempt in order to further study the first-order semilinear hyperbolic systems. We are going to use the finite difference method in time to study the problem (1.1). Finite difference discretization in time for approximating the solutions PDEs has been used since 1970’s to establish existence of their solutions, see e.g. [Tem77] or its latest edition [Tem01] for Navier Stokes equations. This method consists of discretizing the time domain 2010 Mathematics Subject Classification. 35L02, 35L60, 65M06, 65M12. Key words and phrases. semi-linear hyperbolic; nonlinear monotonicity; existence and uniqueness; integration by parts; polynomial growth. The Institute for Scientific Computing and Applied Mathematics, Indiana University, 831 East Third Street, Rawles Hall, Bloomington, Indiana 47405, U.S.A..
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A. Huang and D. Pham
into subintervals and replacing the time derivative in the partial differential equation by a finite difference approximation. The time-dependent parabolic problem is thus transformed into a sequence of stationary problems which can be solved by Galerkin methods. The finite difference discretization in time method applied to some hyperbolic PDEs appeared in the late 1970’s, see e.g. [Mar79, Kac84, Mun87]. The evolution problem is then transformed to a stationary problem but is still of hyperbolic type. In our case, the stationary problems are semi-linear hyperbolic with monotonic nonlinear term. To show the existence of the solutions to these nonlinear problems, we use the perturbation method, see [Lio73, Oma74]. To show the convergence of the approximate solutions, we first carry out a priori estimates to obtain some uniform bounds and hence the weak convergence of the approximate solutions, and then pass to the limit. To deal with the nonlinearity, we use fundamental results in the theory of monotone and accretive operators discovered during the latter half of the century by Minty [Min62, Min63], Browder [Bro65], Lions [Lio69], Brezis [Bre73] and many others. We also use a result in convex analysis, see Theorem B.1 from [ET76], to cope with non zero initial data and still be able to keep the monotonicity of the problem. Note that many stationary or time dependant problems of type (1.1) with a monotone nonlinearity are studied in the last references. However, a major difference with our study is that the operator A + P is not coercive unlike in the previous references. Our work is organized as follows. We first consider a one dimensional problem of the semilinear hyperbolic equation in Section 2. In Section 3, we then use the result in Section 2 by treating the time variable as a spacial variable to construct the solutions for the stationary problem in dimension two, and then use the similar techniques as in Section 2 to obtain the existence and uniqueness of our problem. 2. A one dimensional problem In this section, our aim is to prove the existence and uniqueness of solution of a semilinear hyperbolic equation in space dimension one. We first study the stationary boundary value problem, and then use finite differences in time to study the evolution problem. The semi-linear hyperbolic equation that we consider reads ∂u(x, t) ∂u(x, t) + + P (u(x, t)) = f (x, t), ∂t ∂x m+1
(2.1)
m+1
where x ∈ Ω = (0, 1), t ∈ (0, T ), f ∈ L m (0, T ; L m (Ω)) and P : R → R is monotone and has a polynomial growth at infinity in the following sense: (P (u) − P (v)) (u − v) ≥ 0,
α1 um+1 − β1 ≤ P (u)u ≤ α2 um+1 + β1 , m
|P (u)| ≤ α2 |u| + β2 ,
(2.2a) (2.2b) (2.2c)
for some positive constants α1 , α2 , β1 , β2 and an odd integer m. An example of such a P is an increasing polynomial of odd order m = 2s−1 with a positive leading coefficient a2s−1 > 0 of the form P (u) =
2s−1 X j=0
aj uj .
(2.3)
Semi-linear Hyperbolic Equation
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
We first see that P satisfies (2.2a) since P is increasing, and the estimates in Lemma 2.1 below assert that P also satisfies (2.2b) and (2.2c). Lemma 2.1. Assume that P is of the form (2.3) with the leading coefficient a2s−1 > 0. Then 1 3 a2s−1 u2s − b1 ≤ P (u)u ≤ a2s−1 u2s + b1 (2.4a) 2 2 1 3 a2s−1 u2s − b2 |u| ≤ P (u)u ≤ a2s−1 u2s + b2 |u| (2.4b) 2 2 3 |P (u)| ≤ a2s−1 |u|2s−1 + b2 , (2.4c) 2 where b1 , b2 depend only on the coefficients aj ’s and s. Proof. Using Young’s inequality, we are able to find b1 , ˜b2 > 0 such that 2s−2 X 1 j+1 aj u ≤ a2s−1 u2s + b1 , j=0 2 2s−2 2s−2 X X 1 2s−1 j+1 j a2s−1 |u| + ˜b2 . aj u ≤ |u| |aj ||u| ≤ |u| 2 j=0
j=0
These two inequalities imply (2.4a) and (2.4b), and (2.4c) is a consequence of (2.4b). We endow (2.1) with the following initial and boundary conditions ( u(t, 0) = g(t), u(0, x) = u0 (x).
(2.5)
Our agenda is as follows; we first study in Section 2.1 the stationary problem corresponding to (2.1); see also (2.27). We then use this result to show the existence of solution to a finite difference scheme in time of (2.1), (2.27). The existence of solution of (2.1), (2.27) in L∞ (0, T ; L2 (Ω) ∩ Lm+1 (ΩT ) is achieved by studying the convergence property of the finite difference solutions in Section 2.2. Since the solution is in the weak sense, we develop a trace theorem and an integration by parts result as tools to obtain the existence result. These results are presented in the Appendix A. The uniqueness of solution is done by the monotonicity of P and an integration by parts argument. Throughout this work, we will interchangeably use the notation ΩT = Ω × (0, T ) at convenience. We now study the stationary problem that will help us to build approximate solutions m+1 to the time dependent problem (2.1). We also set V = Lm+1 (Ω) and V 0 = L m (Ω) to be dual spaces of each other. In the following, we denote by h·, ·i the duality pairing between 0 Lr (Ω) and Lr (Ω), where 1 < r < +∞, and r0 is the conjugate exponents of r satisfying that 1/r + 1/r0 = 1. We also use the notations k · k for the norm in L2 (Ω). 2.1. The stationary boundary value problem. In this subsection, we use a regularization and variational inequality method to study the following boundary value problem: ( ∂u in Ω, ∂x + P (u) + λu = f, (2.6) u(0) = g0 , 8
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for some constant λ > 0 and g0 ∈ R. The reason why we introduce an extra term λu in (2.6) is to prepare for using the finite difference scheme for the evolution problem. In order to show the existence, we introduce the following elliptic regularization problem − ∂xx u + ∂x u + P (u ) + λu = f,
in Ω,
(2.7a)
u (0) = g0 , ux (1) = 0. (2.7b) We are going to apply Theorem B.2 to obtain a unique solution u of (2.7a)-(2.7b). Notice that by Sobolev embedding, we have H 1 (Ω) ⊂ C(Ω). We set X = H 1 (Ω) with the usual norm kuk2H 1 = kuk2 + k∂x uk2 , and K = {u ∈ H 1 (Ω) : u(0) = g0 }.
We then define the form a (u, v), for all u, v ∈ H 1 (Ω), Z Z Z Z a (u, v) = ∂x u∂x v dx + ux v dx + P (u)v dx + λ uv dx, Ω
Ω
Ω
(2.8)
Ω
and the operator A : X 7→ X 0 , for all u, v ∈ H 1 (Ω),
hA u, vi = a (u, v).
By definition, it is easy to verify that K is a non-empty convex closed set (u ≡ g0 ∈ K) and that A is weakly continuous over finite dimensional subspaces of X and bounded. For u, v ∈ K, we have Z Z hA u − A v, u − vi = (∂x u − ∂x v)2 dx + (u − v)x (u − v) dx Ω Ω Z Z + (P (u) − P (v))(u − v) dx + λ (u − v)2 dx (2.9) Ω Ω Z 1 ≥ (u − v)x (u − v) dx = (u(1) − v(1))2 2 Ω ≥ 0,
and the equality holds if and only if u = v. We thus deduce that RA is strictly monotone. x To show that A is coercive, we first note that u(x) = g0 + 0 ∂x u(x0 )dx0 , which implies that kuk2 ≤ 2g02 + 2k∂x uk2 ; (2.10) and then for u ∈ K, integration by parts and using (2.2b) and (2.10) yield Z Z Z 1 1 2 2 2 a (u, u) = (∂x u) dx + u(1) − u(0) + P (u)u dx + λ u2 dx 2 2 Ω Ω Ω Z Z Z 1 1 2 2 2 m+1 ≥ (∂x u) dx − g0 + u(1) + α1 u dx − β1 + λ u2 dx (2.11) 2 2 Ω Ω Ω Z Z 1+ 2 1 um+1 dx + λ u2 dx. ≥ kuk2H 1 − g0 − β1 + u(1)2 + α1 4 2 2 Ω Ω Hence, for u ∈ K,
hA u, ui a (u, u) (1 + )g02 + 2β1 = ≥ kukH 1 − → +∞, kukH 1 kukH 1 4 2kukH 1
as kukH 1 → +∞.
(2.12)
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
Applying Theorem B.2 to A , we find that for any f ∈ L(m+1)/m (Ω)(⊂ X 0 = (H 1 (Ω))0 ) there exists a unique u ∈ K such that hA u − f, v − u i ≥ 0,
∀ v ∈ K.
(2.13)
Lemma 2.2. Assume that f ∈ L(m+1)/m (Ω). Then there exists a unique u which satisfies (2.7a)-(2.7b), and the following energy estimates independent of hold: u is bounded in Lm+1 (Ω), √ ∂x u is bounded in L2 (Ω),
u is bounded in W
1,(m+1)/m
(2.14)
(Ω).
Proof. We already showed the existence of u satisfying (2.13), and we now prove that u satisfies (2.7a)-(2.7b). For any w ∈ C ∞ (Ω) with w(0) = 0, choosing v = u ± w ∈ K in (2.13) gives hA u − f, ±wi ≥ 0,
(2.15)
hA u , wi = hf, wi.
(2.16)
which yields
If we further assume that w ∈ Cc∞ (Ω), then integrating by parts in (2.16) gives Z
(−∂xx u + ∂x u + P (u ) + λu )w dx =
Ω
Z f w dx,
(2.17)
Ω
which implies that (2.7a) holds in the sense of distributions on (0, 1). We have u (0) = g0 since u ∈ K, and we are left to recover the boundary condition ux (1) = 0. From (2.7a), we obtain that ∂xx u belongs to L(m+1)/m (Ω), which shows that u belongs to C 1 (Ω) by the Sobolev embedding W 2,(m+1)/m ⊂ C 1 (Ω). Hence the trace ux |x=1 makes sense, and integrating by parts in (2.16) gives Z
(−∂xx u + ∂x u + P (u ) + λu )w dx +
ux (1)w(1)
Ω
Z =
f w dx.
(2.18)
Ω
Combining (2.17) and (2.18) leads to ux (1)w(1) = 0, which implies that ux (1) = 0. Therefore, u satisfies (2.7a)-(2.7b). We now turn to prove the energy estimates (2.14). Choosing v ≡ g0 ∈ K in (2.13) yields hA u , u i ≤ hf, u i + hA u − f, g0 i.
(2.19)
Using H¨ older’s inequality and Young’s inequality, we obtain (m+1)/m
hf, u i ≤ kf kV 0 ku kV ≤ ckf kV 0
+
α1 m+1 ku kV ; 4
(2.20) 8
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and using (2.2c) for P (u ) and Young’s inequality for the term P (u )g0 , we find Z Z Z Z f g0 dx P (u )g0 dx + λ u g0 dx − ux g0 dx + hA u − f, g0 i = Ω Ω Ω Ω Z Z Z f g0 dx P (u )g0 dx + λ u g0 dx − = g0 (u (1) − g0 ) + Ω Ω Ω Z 1 α2 m+1 1 (m+1)/m ≤ (u (1))2 − g02 + (u ) + c(g0m+1 + g0 ) dx 2 2 Ω 4 Z Z 1 2 m 1 +λ ((u ) + g02 ) dx + ( f (m+1)/m + g m+1 ) dx. m+1 0 Ω 2 Ω m+1
(2.21)
for some constant c > 0. Combining (2.19)-(2.21), we find with (2.11) that 1 1 √ α1 m+1 (m+1)/m (m+1)/m ) + λg02 . ≤ c(kf kV 0 + g0m+1 + g0 k ∂x u k2 + ku kV 4 2 2 Hence, we established the first two estimates in (2.14). The last estimate in (2.14) follows by applying Lemma A.1 with λ(x) = 1, g = f − P (u ) − λu , (2.7a) and the first estimate. The first and last estimates in (2.14) show that there exists a subsequence of u , still denoted by u , which weakly converges to some u in Lm+1 (Ω) and in W 1,(m+1)/m (Ω). By the compact Sobolev embedding W 1,(m+1)/m ⊂ C 0,α (Ω) valid for 0 < α < 1/4, we obtain that u strongly converges to u in C(Ω), which, together with the dominated convergence theorem, implies that P (u ) weakly converges to P (u) in L(m+1)/m (Ω). Therefore, passing to the limit in (2.7a), we conclude that the limit u solves (2.6)1 (at least in the sense of distributions). It remains to verify that u satisfies the boundary conditions (2.6)2 . Using the uniform boundedness of (2.14), (2.7a) gives − ∂xx u + ∂x u is uniformly bounded in L(m+1)/m (0, 1). x
(2.22)
Applying Lemma A.1 to u with p = (m+1)/m and X = R, we see that the ∂x u are uniformly (m+1)/m bounded in Lx (0, 1), and the corresponding traces converge, i.e. u(0) = g0 . We thus find a solution u satisfying (2.6). Therefore, we have the following result. Lemma 2.3. Assume that f ∈ L(m+1)/m (Ω) and λ > 0. Then there exists a unique u ∈ Lm+1 (Ω) ∩ W 1,(m+1)/m (Ω) satisfying (2.6). Proof. We only need to prove the uniqueness. Suppose that u1 , u2 both satisfy (2.6) and belong to Lm+1 (Ω) ∩ W 1,(m+1)/m (Ω), and let w = u1 − u2 . Then w satisfies wx + P (u1 ) − P (u2 ) + λw = 0, w(0) = 0.
(2.23)
Since w ∈ Lm+1 (Ω) and P (u1 ), P (u2 ) ∈ L(m+1)/m (Ω), wx ∈ L(m+1)/m (Ω), and it follows that wwx belongs to L1 (Ω) by H¨ older’s inequality, and that Z 1 1 1 (2.24) wwx dx = |w(1)|2 − |w(0)|2 = |w(1)|2 . 2 2 2 Ω
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
We now multiply (2.23)1 by w and use (2.24); we find Z Z 1 2 w(1) + (P (u1 ) − P (u2 ))(u1 − u2 ) dx + λ w2 dx = 0. 2 Ω Ω
(2.25)
The left-hand side of (2.25) is always positive thanks to the monotonicity (2.2a) of P . Hence, we must have w = 0 if (2.25) holds. We thus completed the proof. Remark 2.1. s i) The same result would be true without uniqueness if λ = 0. Uniqueness can be recovered if λ = 0 and P is strictly monotone. ii) With the same method we could replace (2.1)1 by ∂u + P (u) + λu = f, (2.26) ∂x with a ∈ C 1 (Ω) such that a(0)a(1) 6= 0. The boundary conditions would be imposed at x = 0 and/or x = 1, depending on the sign of a(0) and a(1). a(x)
2.2. The evolution problem. In this section, we show the existence and uniqueness of solution to the following semi-linear hyperbolic equation ∂t u + ∂x u + P (u) = f, in Ω × (0, T ), u(0, t) = g(t), (2.27) u(x, 0) = u0 (x), where P : R 7→ R is as above, that is P satisfies the assumptions (2.2a),(2.2b) and (2.2c). We m+1 m+1 suppose that f ∈ L m (0, T ; L m (Ω)), g ∈ L2 (0, T ) and u0 ∈ L2 (Ω). The implicit Euler finite difference scheme for (2.27) is set as follows: we let ∆t = T /N , N ∈ N and we set u0 = u0 , and then construct the un for n = 1, · · · , N by induction solving: n u − un−1 + ∂x un + P (un ) = f n , (2.28a) ∆t un (0) = g n , where 1 f = ∆t n
m+1
Z
n∆t
f (x, t)dt, (n−1)∆t
1 g = ∆t n
Z
n∆t
g(t)dt.
(2.28b)
(n−1)∆t m+1
Given un−1 ∈ L m (Ω), the existence of a unique solution un ∈ W 1, m (Ω) of (2.27)is guaranteed by Lemma 2.3. We now need to prove some estimates independent of N in order to pass to the limit k = 1/N → 0. Remark 2.2. Regarding the boundary value problem (2.26) in Remark 2.1 ii), we could also study the following evolution problem ( ∂t u + a(x, t)∂x u + P (u) + λu = f, (2.29) u(x, 0) = u0 (x), where a ∈ C 1 (ΩT ) with a(0, t)a(1, t) 6= 0, ∀ t ∈ [0, T ]. The boundary conditions would be imposed at x = 0 and/or x = 1, depending on the sign of a(0, t) and a(1, t). 8
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A. Huang and D. Pham
2.2.1. A priori estimates. Taking the inner product of each side of (2.28a) with 2un ∆t, we find that 2hun − un−1 , un i + 2∆th∂x un , un i + 2∆thP (un ), un i = 2∆thf n , un i. Using the identity 2hun − un−1 , un i = kun − un−1 k2 + kun k2 − kun−1 k2 ,
(2.30)
the equation becomes kun − un−1 k2 + kun k2 − kun−1 k2 + 2∆th∂x un , un i + 2∆thP (un ), un i = 2∆thf n , un i. Using the Cauchy-Schwarz inequality and the Young inequality for the right hand side, we obtain, recalling that V = Lm+1 (Ω): kun − un−1 k2 + kun k2 − kun−1 k2 + 2∆th∂x un , un i + 2∆thP (un ), un i m+1
, + α1 ∆tkun km+1 ≤ 2∆tkf n kV 0 kun kV ≤ cm ∆tkf n kV m 0 V
√ √ m where cm := 2 2m/(m + 1) m α1 m. From the assumption (2.2b), we find that hP (un ), un i ≥ − β1 . This implies α1 kun km+1 V kun − un−1 k2 + kun k2 − kun−1 k2 + ∆t((un (1))2 − (g n )2 ) + α1 ∆tkun km+1 V
m+1
≤ cm ∆tkf n kV m + 2β1 ∆t. 0 We sum the above inequalities for n = 1, . . . , n0 and obtain: n0 X n=1
n
n−1 2
ku − u
n0 2
k + ku k +∆t
n0 X
n
2
(u (1)) + α1 ∆t
n=1
≤ cm ∆t
n0 X n=1
N X n=1
m+1
kun km+1 V
+ ∆t kf n kV m 0
N X
(g n )2 + ku0 k2 + 2β1 T.
n=1
We have by (2.28b) ∆t
N X n=1
n
kf k
m+1 m V0
m+1 m m+1 L m (0,T ;V 0 )
≤ kf k
,
∆t
N X
(g n )2 ≤ kgk2L2 (0,T ) ,
n=1
and we conclude that n0 X n=1
n
n−1 2
ku − u
n0 2
k + ku k + ∆t
≤ K0 = K0 (P, T, f, g) := cm kf k for all 1 ≤ n0 ≤ N . We have proven the following bounds:
n0 X
n
2
(u (1)) + α1 ∆t
n=1
m+1 m m+1 L m (0,T ;V 0 )
n0 X n=1
kun km+1 V
+ kgk2L2 (0,T ) + ku0 k2 + 2β1 T,
(2.31)
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation Lemma 2.4.
n 2 ku k ≤ K0 for all 1 ≤ n ≤ N, n0 X K0 , kun km+1 ∆t ≤ V α1 n=1 n0 X ∆t (un (1))2 ≤ K0 , n=1 N X kun − un−1 k2 ≤ K0 ,
(2.32)
n=1
where K0 is defined in (2.31), and depends only on the data. 2.2.2. Passage to the limit . We now introduce two approximate solutions to (2.1), which are denoted by uk and u ˜k with k = ∆t: for each t ∈ In := ((n − 1)∆t, n∆t], n = 1 . . . , N , we set uk (t) = un , u ˜k (t) =
(2.33)
un − un−1 (t − n∆t) + un , ∆t
that is, uk is the step function on the interval (0, T ) with values taken from the right of each interval In , and u ˜k is the piecewise linear function that interpolates un−1 and un on In . We first have an equivalent form of the scheme (2.28a) as follows ˜k + ∂x uk + P (uk ) = fk , in Ω × (0, T ), ∂t u u ˜k (x, 0) = u0 (x), (2.34) uk (0, t) = gk (t), where fk = f n and gk = g n on In for n = 1, . . . , N . We infer from (2.31) the following inequality for the approximate solutions: ≤ K0 (f, g) ∆tk∂t u ˜k k2L2 (0,T ;L2 (Ω)) + kuk (·, t)k2 + kun |x=1 k2L2 (0,T ) + α1 kuk km+1 Lm+1 (0,T,V ) for all 0 ≤ t ≤ T . Using (2.4c) and kuk kLm+1 (ΩT ) being bounded, we find that kP (uk )k
Therefore, as k → 0
L
m+1 m (ΩT )
uk , u ˜k ∈ bounded set of L∞ (0, T ; L2 (Ω)), m+1 (0, T ; V ) = Lm+1 (Ω ), T uk ∈ bounded set of L m+1 P (uk ) ∈ bounded set of L m (0, T ; V 0 ), uk (1) ∈ bounded set of L2 (0, T ), √ ∆t ∂t u ˜k ∈ bounded set of L2 (0, T ; L2 (Ω)).
(2.35)
is also bounded.
(2.36)
As a consequence of the estimates (2.36), we find from equation (2.34) that ∂t u ˜k + ∂x uk ∈ bounded set of L
m+1 m
(0, T ; V 0 ).
(2.37) 8
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We now have an estimate of the distance between the two functions uk and u ˜k : direct computations show that 2 Z n∆t n Z n∆t 1 un − un−1 2 3 u − un−1 k k˜ uk − uk k2 dt = k(t − n∆t) dt = k k ∆t . ∆t 3 ∆t (n−1)∆t (n−1)∆t This implies k˜ uk − uk kL2 (ΩT ) =
√
3 k∂t u ˜k kL2 (ΩT ) ∆t. 3
(2.38)
From (2.38) and (2.36)5 , we find 1
k˜ uk − uk kL2 (0,T ;L2 (Ω)) = O(∆t 2 ).
(2.39)
By (2.36) and (2.39), there exist subsequences of u ˜k and uk which we still denote by k, and m+1 ∞ 2 m+1 u ∈ L (0, T ; L (Ω)) ∩ L (ΩT ), χ ∈ L m (0, T ; V 0 ), κ ∈ L2 (0, T ) such that for k → 0 ˜k −* u weak- ∗ in L∞ (0, T ; L2 (Ω)) and weakly in Lm+1 (ΩT ), uk , u m+1 (2.40) P (uk )−* χ weakly in L m (0, T ; V 0 ), uk (1)−* κ weakly in L2 (0, T ). From (2.37), we are also able to extract a subsequence still denoted by k such that ∂t u ˜k + ∂x uk m+1 0 weakly converges to some ξ in L m (0, T ; V ). From (2.40)1 , we infer that ∂t u ˜k , ∂x uk converge to ∂t u, ∂x u in the sense of distributions respectively. We thus have that ξ = ∂t u + ∂x u, and that m+1 (2.41) ∂t u ˜k + ∂x uk −* ∂t u + ∂x u weakly in L m (0, T ; V 0 ). m+1
m+1
We have u ∈ L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (ΩT ), ∂t u + ∂x u ∈ L m (0, T ; V 0 ) = L m (ΩT ). Thanks to Lemma A.2 with U = ΩT , a = b = 1, p = (m + 1)/m, the traces u(0, t) and u(x, 0) make sense. ¯ We now denote by u ¯k , u ˜k , f¯k the extensions of the functions uk , u ˜k , fk by zero to the domain (−∞, 1) × (−∞, T ); u ¯0 is the extension of u0 to (−∞, 1) and g¯k is the extension of gk to (−∞, T ). We infer from (2.40)2 , (2.41) and (2.34)2,3 that ¯ ∂t u ˜k (x, t) − δ0 (t)¯ u0 (x) + ∂x u ¯k (x, t) − δ0 (x)¯ gk (t) + P (¯ uk (x, t)) = f¯k (x, t) where δ0 is the dirac delta function having mass at 0. Passing to the limit as k → 0, we find in the sense of distributions ∂t u ¯(x, t) − δ0 (t)¯ u0 (x) + ∂x u ¯(x, t) − δ0 (x)¯ g (t) + χ(x, ¯ t) = f¯(x, t)
where u ¯, χ, ¯ f¯ are the extensions of u, χ, f on (−∞, 1) × (−∞, T ) and g¯ is the extension of g to (−∞, T ). This equation and ∂t u + ∂x u + χ = f on ΩT imply ∂t u + ∂x u + χ = f, u(x, 0) = u0 (x), (2.42) u(0, t) = g(t). Remark 2.3. To recover the initial condition and boundary condition for (2.42), we can also take the inner product of (2.34) with ϕ(x)ψ(t) for ϕ(x) ∈ D(Ω), ψ(t) ∈ D([0, T ]), ϕ(1) = ψ(T ) = 0 and carry out the integration by parts to obtain −h˜ uk , ϕψt i − hu0 , ϕψ(0)iL2x (Ω) − huk , ϕx ψi − hgk , ϕ(0)ψiL2t (0,T ) + hP (uk ), ϕψi = hfk , ϕψi.
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation We then let k run to zero and find that
−hu, ϕψt i − hu0 , ϕψ(0)iL2x (Ω) − hu, ϕx ψi − hg, ϕ(0)ψiL2t (0,T ) + hχ, ϕψi = hf, ϕψi. This also implies (2.42).
There remains to show that χ = P (u). For that purpose, let v ∈ Lm+1 (0, T ; V ) so that m+1 P (v) ∈ L m (0, T ; V 0 ); we consider Z
T
Xk = 0
hP (uk ) − P (v), uk − vidt.
We first see that Xk ≥ 0 thanks to the monotonicity (2.2a) of P , and then write Xk = Xk1 + Xk2 + Xk3 , where Xk1 =
T
Z 0
Xk2 =
hP (uk ), uk i dt,
T
Z 0
Xk3 =
hP (uk ), vi dt,
Z 0
T
hP (v), (uk − v)i dt.
The weak convergence in (2.40) immediately shows that as k → 0: Xk2
−→
Z 0
T
hχ, vi dt,
and
Xk3
−→
Z
T
hP (v), (u − v)i dt.
0
For Xk1 , we use the scheme (2.34) and write Xk1
Z
T
hfk , uk i dt −
= 0
Z = 0
T
Z 0
h∂t u ˜k , uk i dt −
Z 0
T
h∂x uk , uk i dt
N
T
hfk , uk i dt −
1X n 1 1 ku − un−1 k2 − kuN k2 + ku0 k2 2 2 2 n=1
1 1 − kuk |x=1 k2L2 (0,T ) + kgk k2L2 (0,T ) 2 2 Z T 1 1 1 1 ≤ hfk , uk i dt − kuk |t=T k2 + ku0 k2 − kuk |x=1 k2L2 (0,T ) + kgk k2L2 (0,T ) . 2 2 2 2 0 From Lemma A.3 and since u0 ∈ L2 (Ω) and g ∈ L2 (0, T ), we know that the traces of u at x = 0, 1 and t = 0, T are L2 -functions. We also have that lim inf kuk |t=T k ≥ ku|t=T k, k→0 Z T Z T lim inf kuk |x=1 kL2 (0,T ) ≥ ku|x=1 kL2 (0,T ) , hfk , uk i dt → hf, ui dt, and kgk kL2 (0,T ) → kgkL2 (0,T ) , k→0
0
0
which together yield Z T 1 1 1 1 lim sup Xk1 ≤ hf, ui dt − ku|t=T k2 + ku0 k2 − ku|x=1 k2L2 (0,T ) + kgk2L2 (0,T ) . 2 2 2 2 k→0 0
(2.43) 8
12
A. Huang and D. Pham m+1
Since we have u ∈ Lm+1 (ΩT ) and ∂t u + ∂x u ∈ L m (ΩT ), applying Lemma A.3 to u with y = t, U = ΩT , p = m + 1 and q = m+1 m and using (2.42), we obtain that Z 1 2 2 2 2 ku|t=T k − ku0 k + ku|x=1 kL2 (0,T ) − kgkL2 (0,T ) = hu, ut + ux i dxdt 2 ΩT Z (2.44) = hu, f − χi dxdt. ΩT
Combining (2.43) and (2.44), we find that
We finally conclude that Z T Z lim sup Xk ≤ hχ, ui dt − k→0
0
lim sup Xk1 k→0
≤
Z
T
T
0
hχ, vi dt −
0
Z 0
T
hχ, ui dt.
hP (v), (u − v)i dt =
Z 0
T
hχ − P (v), (u − v)i dt,
which, together with Xk ≥ 0, implies that for all v ∈ Lm+1 (0, T ; V ), Z T hχ − P (v), (u − v)i dt ≥ 0. 0
The standard Minty method (see [Lio69]) then yields χ = P (u). 2.2.3. Main result of Section 2. m+1
m+1
Theorem 2.1. Assume that f ∈ L m (0, T ; L m (Ω)), g ∈ L2 (0, T ) and u0 ∈ L2 (Ω) are given. Then there exists a unique solution u = u(t, x) ∈ L∞ (0, T ; L2 (Ω))∩Lm+1 (0, T ; Lm+1 (Ω)) satisfying (2.1) and (2.5). Proof. The existence of a solution u was proved in Section 2.2.1, and we thus only need to show the uniqueness. Let u1 , u2 belong to L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (0, T ; Lm+1 (Ω)) and both satisfy (2.1)-(2.5). Set w = u1 − u2 ; we then see that w satisfies wt + wx = P (u2 ) − P (u1 ), (2.45) w(t, 0) = 0, w(0, x) = 0. Since u1 , u2 ∈ Lm+1 (ΩT ), we find from (2.45)1 that wt + wx belongs to L(m+1)/m (ΩT ) by assumption (2.2b). Applying Lemma A.3 to w with y = t, U = Ωt := Ω × (0, t), p = m + 1 and q = (m + 1)/m and using (2.45)2,3 , we obtain Z t Z Z w(1, t0 )2 dt0 + w(x, t)2 dx = w(wt + wx ) dxdt0 , ∀ t ∈ [0, T ], 0
Ω
Ωt
which, together with (2.45)1 , implies that Z Z 2 w(x, t) dx + (u1 − u2 )(P (u1 ) − P (u2 )) dxdt0 ≤ 0, Ω
Ωt
∀ t ∈ [0, T ].
(2.46)
Therefore, we can conclude that u1 (·, t) = u2 (·, t) for all t ∈ [0, T ] and hence u1 = u2 , and we thus completed the proof.
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation 3. A two dimensional problem
In this section, we consider the rectangular domain Ω = (0, 1)2 in R2 and our aim is to show the existence and uniqueness of solution to the following semi-linear hyperbolic equation: ∂t u + Au + P (u) = f, in Ω × (0, T ), u(0, y, t) = g1 (y, t), u(x, 0, t) = g2 (x, t) (3.1) u(x, y, 0) = u0 (x, y), where Au = ∂x u + ∂y u and P : R → R is as in Section 2 satisfying (2.2a)-(2.2c). We suppose m+1 m+1 that f ∈ L m (0, T ; L m (Ω)), g1 ∈ L2 (0, T ; L2y (0, 1)), g2 ∈ L2 (0, T ; L2x (0, 1)) and u0 ∈ L2 (Ω). The standard finite difference scheme of (3.1) is set as follows: Let ∆t = T /N ; we set u0 = u0 , and construct the un for n = 1, · · · , N recursively by setting n u − un−1 + Aun + P (un ) = f n , (3.2a) ∆t un (0, y) = g n (y), un (x, 0) = g n (x), 1
2
where 1 f = ∆t n
Z
n∆t
f (x, y, t)dt, (n−1)∆t
g1n (y)
1 = ∆t
Z
n∆t
g1 (y, t)dt,
g2n (x)
(n−1)∆t
1 = ∆t
Z
n∆t
g2 (x, t)dt. (n−1)∆t
(3.2b) We rewrite the equation (3.2a)1 as ( 1 n ∂y un + ∂x un + P (un ) + ∆t u = fn + un (0, y) = g1n (y), un (x, 0) = g2n (x). m+1
1 n−1 , ∆t u
(3.3)
m+1
Given un−1 ∈ L m (Ω), the existence and uniqueness of un in L m (Ω) is guaranteed by 1 1 u and noting that P (u) + ∆t u is applying Theorem 2.1 with t = y and P (u) = P (u) + ∆t 1 strictly monotone thanks to λ = ∆t > 0. 3.1. A priori estimates. We still use the previous notations, i.e. V = Lm+1 (Ω) and m+1 V 0 = L m (Ω) to be dual spaces of each other; h·, ·i is the duality pairing between Lr (Ω) and 0 Lr (Ω), for 1/r + 1/r0 = 1 and k · k for the norm in L2 (Ω). Taking the inner product of each side of (3.2a) with 2un ∆t, and using the identity (2.30), we find that kun − un−1 k2 + kun k2 − kun−1 k2 + 2hAun , un i∆t + 2hP (un ), un i∆t = 2hf n , un i∆t. Using the Cauchy-Schwarz inequality and the Young inequality for the right hand side, we find that kun − un−1 k2 + kun k2 − kun−1 k2 + 2hAun , un i∆t + 2hP (un ), un i∆t m+1
≤ 2kf n kV 0 kun kV ∆t ≤ cm kf n kV m ∆t + α1 kun km+1 ∆t, 0 V
8
14
A. Huang and D. Pham
√ √ m where cm := 2 2m/(m + 1) m α1 . We know from (2.2b) that hP (un ), un i ≥ α1 kun km+1 − β1 and this implies kun − un−1 k2 + kun k2 − kun−1 k2 + 2∆thAun , un i + α1 ∆tkun km+1 V
m+1
≤ cm ∆tkf n kV m + 2β1 ∆t. 0 Since g1n ∈ L2y (0, 1), g2n ∈ L2x (0, 1), and Aun ∈ L and a = b = c = 1 to obtain
m+1 m
(ΩT ), we apply Lemma A.4 with U = Ω
2hAun , un i = 2h∂x un + ∂y un , un i
= kun |x=1 k2L2y (0,1) − kg1n (y)k2L2y (0,1) + kun |y=1 k2L2x (0,1) − kg2n (x)k2L2x (0,1) .
Therefore kun − un−1 k2 + kun k2 − kun−1 k2 + ∆t(kun |x=1 k2L2y (0,1) + kun |y=1 k2L2x (0,1) ) + α1 ∆tkun km+1 V m+1
≤ cm ∆tkf n kV m + 2β1 ∆t + ∆tkg1n (y)k2L2y (0,1) + ∆tkg2n (x)k2L2x (0,1) . 0 We sum the above inequalities for n = 1, . . . , n0 : n0 X n=1
kun − un−1 k2 + kun0 k2 + ∆t ≤ cm ∆t
N X n=1
n0 X
(kun |x=1 k2L2y (0,1) + kun |y=1 k2L2x (0,1) ) + α1 ∆t
n=1
m+1
+ 2β1 T + ∆t kf n kV m 0
N X n=1
kg1n k2L2y (0,1) + ∆t
N X n=1
n0 X n=1
kun km+1 V
kg2n k2L2x (0,1) + ku0 k2 .
Since we have ∆t
∆t
∆t
N X n=1 N X n=1 N X n=1
m+1
kf n kV m ≤ kf k 0
n=1
,
kg1n k2L2y (0,1) ≤ kg1 k2L2 (0,T ;L2y (0,1)) , kg2n k2L2x (0,1) ≤ kg2 k2L2 (0,T ;L2x (0,1)) ,
we conclude that, for all 1 ≤ n0 ≤ N , n0 X
m+1 m m+1 L m (0,T ;V 0 )
kun − un−1 k2 + kun0 k2 + ∆t
n0 X n=1
kun |x=1 k2L2y (0,1) + ∆t
n0 X
kun |y=1 k2L2x (0,1)
n=1 n0 X
+ α1 ∆t
n=1
where K0 := cm kf k
m+1 m m+1 L m (0,T ;V 0 )
(3.4) kun km+1 V
≤ K0 ,
+ kg1 k2L2 (0,T ;L2 (0,1)) + kg2 k2L2 (0,T ;L2 (0,1)) + ku0 k2 + 2β1 T . y
x
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
3.2. Passage to the limit. We introduce two approximate solutions uk and u ˜k defined on t ∈ In := ((n − 1)∆t, n∆t], n = 1 . . . , N as follows: uk is the step function on In with values taken from the right of each interval In , and u ˜k is the piecewise linear function that interpolates un−1 and un on In . We first have an equivalent form of the scheme (3.2a) as follows ˜k + Auk + P (uk ) = fk , in Ω × (0, T ), ∂t u uk (0, y, t) = g1k (y, t), uk (x, 0, t) = g2k (x, t), (3.5) u ˜k (x, y, 0) = u0 (x, y), where fk , g1k and g2k are the step functions on (0, T ), respectively equal to f n , g2n and g2n on the interval In , n = 1 . . . , N . From (3.4), we find ∆tk∂t u ˜k k2L2 (0,T ;L2 (Ω)) + kuk (t)k2 + kun |x=1 k2L2 (0,T ;L2y (0,1))
+ kun |y=1 k2L2 (0,T ;L2x (0,1)) + α1 kuk km+1 ≤ K0 , Lm+1 (0,T,V )
for all 0 ≤ t ≤ T . Therefore, as k → 0:
uk , u ˜k ∈ bounded set of L∞ (0, T ; L2 (Ω)), uk ∈ bounded set of Lm+1 (0, T ; V ) = Lm+1 (ΩT ), m+1 P (uk ) ∈ bounded set of L m (0, T ; V 0 ), uk |∂Ω ∈ bounded set of L2 (0, T ; L2 (∂Ω)), √ ∆t ∂t u ˜k ∈ bounded set of L2 (0, T ; L2 (Ω)).
(3.6)
(3.7)
As a consequence of these bounds, from equation (3.5), we find that ∂t u ˜k + Auk is bounded in m+1 m L (ΩT ) independently of k. From these bounds along with (2.38), we extract a subsequence still denoted by k such that for k → 0 uk , u ˜k −* u weak-* in L∞ (0, T ; L2 (Ω)) and weakly in Lm+1 (ΩT ), P (u )−* χ weakly in L m+1 m (0, T ; V 0 ), k (3.8) uk |∂Ω −* κ weakly in L2 (0, T ; ∂Ω), m+1 ∂t u ˜k + Auk −* ξ = ∂t u + Au weakly in L m (0, T ; V 0 ). m+1
We have u ∈ L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (ΩT ), ∂t u + Au ∈ L m (ΩT ). Thanks to Lemma A.4 with M = ΩT , a = b = c = 1 and p = (m + 1)/m, the traces u|t=0 and u|∂Ω make sense. As in dimensional one case, by extending all functions by zero to the domain (−∞, 1)2 × (−∞, T ), letting k → 0 and combing with (2.40)2 , (2.41), and (3.5)2,3 , we find the initial condition and boundary condition for the limit equation reads ∂t u + Au + χ = f, u(0, y, t) = g1 (y, t), u(x, 0, t) = g2 (x, t), (3.9) u(x, y, 0) = u (x, y). 0 Thanks to Lemma A.5 with a = b = c = 1 and u|x=0 = g1 (y, t) ∈ L2 (0, T ; Ly (0, 1)), u|0=0 = g2 (x, t) ∈ L2 (0, T ; Lx (0, 1)), u|t=0 = u0 (x, y) ∈ L2 (Ω), all the traces u|x=1 , u|y=1 and u|t=T are 8
16
A. Huang and D. Pham
square integrable functions, and the integration by parts is valid for h∂t u + Au, vi; hence we can similarly (as in Section 2.2.2) show that Z T hχ − P (v), (u − v)i dt ≥ 0, 0
Lm+1 (0, T ; V
for all v ∈ ) and thus by the Minty method χ = P (u). This shows that u is a solution of (3.1). We now state the result of existence and uniqueness of a solution u for (3.1): m+1
Theorem 3.1. Assume that f ∈ L m (ΩT ), g1 ∈ L2 (0, T ; L2y (0, 1)), g2 ∈ L2 (0, T ; L2x (0, 1)) and u0 ∈ L2 (Ω) are given. Then there exists a unique function u ∈ L∞ (0, T ; L2 (Ω))∩Lm+1 (ΩT ) satisfying (3.1). Proof. The existence of a solution u was proved in Section 3.2, and we thus only need to show the uniqueness. Let u1 , u2 belong to L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (0, T ; Lm+1 (Ω)) that both satisfy (3.1). Setting w = u1 − u2 , we then see that w satisfies wt + Au = P (u2 ) − P (u1 ), (3.10) w(t, 0, y) = w(t, x, 0) = 0, w(0, x) = 0. Since u1 , u2 ∈ Lm+1 (ΩT ), we find from (3.10)1 that wt + Au belongs to L(m+1)/m (ΩT ) by assumption (2.2b). Applying Lemma A.5 to w with z = t, M = Ωt := Ω × (0, t), p = m + 1 and q = (m + 1)/m and using (3.10)2,3 , we obtain that for all t ∈ [0, T ], Z 1Z t Z 1Z t Z Z w(x, 1, t0 )2 dxdt0 + w(1, y, t0 )2 dydt0 + w(x, y, t)2 dxdy = w(wt + wx ) dxdt0 , 0
0
0
0
Ω
which, together with (3.10)1 , implies that Z Z 2 w(x, y, t) dxdy + (u1 − u2 )(P (u1 ) − P (u2 )) dxdt0 ≤ 0, Ω
Ωt
Ωt
∀ t ∈ [0, T ].
(3.11)
Therefore, we can conclude that u1 (·, t) = u2 (·, t) for all t ∈ [0, T ] and hence u1 = u2 , and we thus completed the proof. Remark 3.1 (The generality of operator A and the boundary conditions). We could study the following general form of A = a(x, y, t)∂x u + b(x, y, t)∂y u where a, b ∈ C 1 (ΩT ) are never zero on the boundary ∂Ω. The boundary conditions imposed on the boundary ∂Ω depend on the signs of a, b on the boundary. For example, we could impose the following boundary conditions u(0, y, t) = g1 (y, t) when a(0, y, t) > 0,
u(1, y, t) = g1 (y, t) when a(1, y, t) < 0,
u(x, 0, t) = g2 (x, t) when b(x, 0, t) > 0,
u(x, 1, t) = g2 (x, t) when b(x, 1, t) < 0.
(3.12)
The boundary conditions in (3.12) are strictly dissipative in the sense of [BS07, Chapter 3]. Our main result in Theorem 3.1 is still valid without much changing of its proof. Indeed, Theorem 2.1 or 2.2 can be applied with t = y to the problem (3.3) induced by the finite difference scheme in time in the general case.
Semi-linear Hyperbolic Equation
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Appendix A. Trace theorem In this Appendix we give some trace theorems which we used in the article. See Appendix A in [ST10] for some other trace results. We fix p such that 1 < p < +∞, and we also let q = p0 the conjugate exponent of p (i.e. 1/p + 1/q = 1). Lemma A.1. Let X be a reflexive Banach space, and λ = λ(x) ∈ C 0 ([0, 1]) such that λ(x) ≥ c0 for some positive constant c0 . Assume that two sequences of function u , g ∈ Lpx (0, 1; X) satisfy ( −uxx + λ(x)ux = g , (A.1) u (0) = u0 , ux (1) = 0, with g bounded in Lpx (0, 1; X) independently of , u0 ∈ X and p > 1. Then the ux are bounded in Lpx (0, 1; X) independently of , and, for any subsequence u → u, g → g (strongly or weakly) converging in Lpx (0, 1; X), u (0) converges to u(0) in X (weakly at least), and hence u(0) = u0 . Proof. For simplicity, we drop , and by solving (A.1), we obtain Z 1 1 − R x1 λ(¯x)/ d¯x e x g(x1 )dx1 . ux (x) = x
(A.2)
Taking the X-norm of (A.2) and raising to the p-th power, and using H¨older’s inequality, we find Z 1 p/q Z 1 1 − R x1 λ(¯x)/ d¯x 1 − R x1 λ(¯x)/ d¯x p x kux kX ≤ (A.3) e dx1 e x kg(x1 )kpX dx1 . x x Direct computations show that the first integral in (A.3) is less than Z 1 Z 1 1 − R x1 c0 / d¯x 1 −c0 (x1 −x)/ 1 x e dx1 = e dx1 ≤ . c0 x x Then integrating (A.3) from 0 to 1 with respect to x and switching the order of integration for x, x1 , we finally arrive at Z Z 1 1 p 1 p kux kX dx ≤ ( ) kgkpX dx. (A.4) c 0 0 0 It follows that the ux are uniformly bounded in Lpx (0, 1; X); hence u ∈ Cx ([0, 1]; X). The existence of the trace u(0) and its linear continuous dependence on {u, g} follow easily. Remark A.1. If the coefficient function in the convection term is negative, that is, instead of (A.1) we have ( −uxx − λ(x)ux = g , (A.5) u (1) = u1 , ux (0) = 0, for two sequences u , g ∈ Lpx (0, 1; X), then the same result as in Lemma A.1 follows; that means the ux are bounded in Lpx (0, 1; X) independently of , and also for any subsequence u → u, g → g which converge (strongly or weakly) in Lpx (0, 1; X), u (1) converges to u(1) in X (weakly at least), and hence u(1) = u1 . 8
18
A. Huang and D. Pham Let U = (0, L1 ) × (0, L2 ) be a subset of R2 , and we consider the space: X (U) = {u ∈ Lp (U), T u = aux + buy ∈ Lp (U)},
endowed with its natural norm (kukpLp (U ) + kT ukpLp (U ) )1/p , where L1 and L2 are positive constants, and a and b are non-zero constants. Then we have the following trace result. Lemma A.2. If u ∈ X (U), then the traces of u are defined on all of ∂U, i.e. the traces of u are defined at x = 0, L1 , and y = 0, L2 , and they belong to the respective spaces Wy−1,p (0, L2 ) and Wx−1,p (0, L1 ). Furthermore the trace operators are linear continuous in the corresponding spaces, e.g., u ∈ X (U) → u|x=0 is continuous from X (U) into Wy−1,p (0, L2 ). Proof. Since u ∈ Lp (U), we see that uy belongs to Lpx (0, L1 ; Wy−1,p (0, L2 )), which, together with T u ∈ Lp (U) and a, b are non-zeros, implies that ux ∈ Lpx (0, L1 ; Wy−1,p (0, L2 )). In combination with u ∈ Lpx (0, L1 ; Lpy (0, L2 )), we obtain that u ∈ Cx ([0, L1 ]; Wy−1,p (0, L2 )). Hence, the traces of u are well-defined at x = 0 and L1 , and belong to Wy−1,p (0, L2 ). The continuity of the corresponding mappings is easy. The proof for the traces at y = 0 and L2 is similar. Lemma A.3. Assume that u ∈ Lp (U) and T u ∈ Lq (U) with a, b > 0. Then by Lemma A.2 with p = min{p, q}, we know that the traces of u on ∂U are well-defined. Furthermore, we assume that the traces u(0, y) ∈ L2 (0, L2 ) and u(x, 0) ∈ L2 (0, L1 ). Then the traces u(L1 , y), u(x, L2 ) belong to L2 (0, L2 ), L2 (0, L1 ) respectively, and the identity holds Z u(T u) dxdy = U
a ku(L1 , y)k2L2 (0,L2 ) − ku(0, y)k2L2 (0,L2 ) 2 b + ku(x, L2 )k2L2 (0,L1 ) − ku(x, 0)k2L2 (0,L1 ) . 2
(A.6)
Proof. Without loss of generality, we may assume that b = 1. We first denote by Γ1 , Γ2 , Γ3 , Γ4 the boundaries x = 0, x = L1 , y = 0, y = L2 respectively, and then introduce a new coordinate system (x0 , y 0 ) such that 0 x a 1 x = , y0 −1 a y which is equivalent to 0 1 x a −1 x = . y y0 1 + a2 1 a We denote by Γ0i the image of Γi by this transformation for all i ∈ {1, 2, 3, 4}; we also denote by U 0 the image of U and denote by u0 the transform of u. Let (αy0 , y 0 ) and (βy0 , y 0 ) be the end points of the intersection of U 0 with the line y 0 fixed, (αy0 , y 0 ) ∈ Γ01 or Γ03 , (βy0 , y 0 ) ∈ Γ02 or Γ04 (see Figure A.1).
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
aL2
y′ Γ′1
Γ′4 αy′
y
′
βy ′
Γ′3
x′
Γ′2 b
b
−L1 Figure A.1. The domain U 0 Direct computation shows that ∂u0 1 ∂u ∂u 1 = a + = T u. ∂x0 1 + a2 ∂x ∂y 1 + a2 Therefore, for u ∈ Lp (U) and T u ∈ Lq (U), we see that u0 ∈ Lp (U 0 ) and u0x0 ∈ Lq (U 0 ), which shows that u0 u0x0 belongs to L1 (U 0 ) by H¨older’s inequality. Hence for fixed y 0 , we see that Z βy 0 0 0 2 0 0 2 u0 (x0 , y 0 )u0x0 (x0 , y 0 ) dx0 . u (βy0 , y ) = u (αy0 , y ) + 2 (A.7) αy 0
y0
Integrating (A.7) with respect to from −L1 to aL2 , we obtain Z aL2 Z aL2 Z aL2 Z βy0 0 0 2 0 0 0 2 0 u (βy0 , y ) dy = u0 (x0 , y 0 )u0x0 (x0 , y 0 ) dx0 dy 0 . u (αy0 , y ) dy + 2 −L1
−L1
−L1
(A.8)
αy 0
Since the traces of u at Γ1 = {x = 0} and Γ3 = {y = 0} are L2 functions, we also have that the traces of u0 at Γ01 and Γ03 are L2 functions. Therefore, we infer from (A.8) that the traces of u0 at Γ02 and Γ04 are L2 functions, which shows that the traces of u at Γ2 = {x = L1 } and Γ4 = {y = L2 } are L2 functions. Now, transforming back into the original system (x, y), (A.8) is equivalent to Z L2 Z L1 Z L2 Z L1 2 2 2 a u(L1 , y) dy + u(x, L2 ) dx =a u(0, y) dy + u(x, 0)2 dx 0 0 0 0 (A.9) Z L Z L 1
2
+2
u(T u) dxdy, 0
0
which is (A.6) in the case b = 1. We thus completed the proof.
Lemmas A.2 and A.3 are also valid in the three dimensional case. We state them as follows. We let M = (0, L1 ) × (0, L2 ) × (0, L3 ) be a subset of R3 , and we consider the space: X (M) = {u ∈ Lp (M), T u = aux + buy + cuz ∈ Lp (M)},
endowed with its natural norm (kukpLp (M) + kT ukpLp (M) )1/p , where L1 , L2 , L3 are positive constants, and a, b, c are non-zero constants. Then we have the following trace result. 8
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Lemma A.4. If u ∈ X (M), then the traces of u are defined on all of ∂M, i.e. the traces of u are defined at x = 0, L1 , y = 0, L2 and z = 0, L3 , and they belong to the respective spaces −1,p −1,p −1,p Wy,z ((0, L2 ) × (0, L3 )), Wx,z ((0, L1 ) × (0, L3 )) and Wy,z ((0, L2 ) × (0, L3 )). Furthermore the trace operators are linear continuous in the corresponding spaces. Lemma A.5. Assume that u ∈ Lp (M) and T u ∈ Lq (M) with a, b, c > 0. Then by Lemma A.4 with p = min{p, q}, we know that the traces of u on ∂M are well-defined. Furthermore, we assume that the traces u|x=0 , u|y=0 , u|z=0 are L2 -functions with respect to the corresponding domains. Then the traces u|x=L1 , u|y=L2 , u|z=L3 are also L2 -functions, and the following identity holds: Z a u(T u) dxdydz = ku|x=L1 k2L2 − ku|x=0 k2L2 2 M b (A.10) + ku|y=L2 k2L2 − ku|y=0 k2L2 2 c + ku|z=L3 k2L2 − ku|z=0 k2L2 . 2 The proof of Lemmas A.4 and A.5 are the same as the proof of of Lemmas A.2 and A.3, using the following coordinate transformation adapted for Lemma A.5: 0 a b c x x y 0 = 0 c −b y . b2 +c2 z z0 b c a Appendix B. Convex analysis and Variational Problems The following results are taken from [ET76]. Let X be a reflexive separable Banach space and X 0 be the dual space of X, and let K be a non-empty convex closed subset of X. Theorem B.1. Let J : K → R be a convex and lower semi-continuous functional, and assume that one of the following holds. (1) K is bounded; or (2) J(u) → +∞ as kuk → +∞, and u ∈ K. Then inf v∈K J(v) > −∞, and there exists at least one u ∈ K such that J(u) = inf J(v). v∈K
(B.1)
The set of u satisfying (B.1) is convex and closed. If J is strictly convex, then there exists one and only one u satisfying (B.1). Theorem B.2. Let A be an operator from X to X 0 satisfying the following properties: (1) A is weakly continuous over the finite dimensional subspaces of X. (2) A is bounded: kAukX 0 is bounded if kukX is bounded. (3) A is monotone: hAu − Av, u − vihX 0 ,Xi ≥ 0, ∀ u, v ∈ K. (4) Either one of the following holds: (a) K is bounded; or
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation (b) A is coercive, that is: hAu, uihX 0 ,Xi → +∞, kukX
as kukX → +∞, u ∈ K.
Then for all ` ∈ X 0 , there exists at least one u ∈ K such that hAu − `, v − ui ≥ 0,
∀ v ∈ K.
(B.2)
If furthermore, A is strictly monotone, i.e. hAu − Av, u − vihX 0 ,Xi > 0,
∀ u, v ∈ K, u 6= v,
then there exists one and only one u satisfying (B.2). Acknowledgments This work was partially supported by the National Science Foundation under the grants NSF DMS-0906440 and DMS-1206438, and by the Research Fund of Indiana University. The authors would like to thank Professor Roger Temam for his suggestion of this problem and valuable advice during this research. References [BS07] [Bre73]
[Bro65] [ET76]
[Gla73] [JP07] [Kac84] [Lio69] [Lio73] [Liu03] [Mar79] [Min62] [Min63] [Mun87]
S. Benzoni-Gavage and D. Serre, Multi-dimensional Hyperbolic Partial Differential Equations, Oxford University Press, 2007. H. Br´ezis, Op´erateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert, North-Holland Publishing Co., Amsterdam, 1973, North-Holland Mathematics Studies, No. 5. Notas de Matem´ atica (50). Felix E. Browder, Nonlinear monotone operators and convex sets in Banach spaces, Bull. Amer. Math. Soc. 71 (1965), 780–785. Ivar Ekeland and Roger T´emam, Convex analysis and variational problems, english ed., Classics in Applied Mathematics, vol. 28, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 1999, Translated from the French. RobertT. Glassey, Blow-up theorems for nonlinear wave equations, Mathematische Zeitschrift 132 (1973), 183–203 (English). Chang-Yeol Jung and Du Pham, Singular perturbation of semi-linear reaction-convection equations in a channel and numerical applications, Adv. Differential Equations 12 (2007), no. 3, 265–300. Jozef Kaˇcur, Application of Rothe’s method to perturbed linear hyperbolic equations and variational inequalities, Czechoslovak Math. J. 34(109) (1984), no. 1, 92–106. J.-L. Lions, Quelques m´ethodes de r´esolution des probl`emes aux limites non lin´eaires, Dunod, 1969. , Perturbations singuli`eres dans les probl`emes aux limites et en contrˆ ole optimal, Lecture Notes in Mathematics, Vol. 323, Springer-Verlag, Berlin, 1973. Yacheng Liu, On potential wells and vacuum isolating of solutions for semilinear wave equations, Journal of Differential Equations 192 (2003), no. 1, 155 – 169. Erich Martensen, The Rothe method for the wave equation in several space dimensions, Proc. Roy. Soc. Edinburgh Sect. A 84 (1979), no. 1-2, 1–18. George J. Minty, Monotone (nonlinear) operators in Hilbert space, Duke Math. J. 29 (1962), 341–346. , on a “monotonicity” method for the solution of non-linear equations in Banach spaces, Proc. Nat. Acad. Sci. U.S.A. 50 (1963), 1038–1041. Claus-Dieter Munz, On the convergence of the Rothe method for nonlinear hyperbolic conservation laws, Problems of applied analysis (Oberwolfach, 1985), Methoden Verfahren Math. Phys., vol. 33, Lang, Frankfurt am Main, 1987, pp. 27–40.
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[Oma74] Robert E. O’Malley, Jr., Introduction to singular perturbations, Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], New York-London, 1974, Applied Mathematics and Mechanics, Vol. 14. [Sat68] D.H. Sattinger, On global solution of nonlinear hyperbolic equations, Archive for Rational Mechanics and Analysis 30 (1968), 148–172 (English). [ST10] Jean-Claude Saut and Roger Temam, An initial boundary-value problem for the Zakharov-Kuznetsov equation, Adv. Differential Equations 15 (2010), no. 11-12, 1001–1031. [Tem77] Roger Temam, Navier-Stokes equations. Theory and numerical analysis, North-Holland Publishing Co., Amsterdam, 1977, Studies in Mathematics and its Applications, Vol. 2. [Tem01] , Navier-Stokes equations, AMS Chelsea Publishing, Providence, RI, 2001, Theory and numerical analysis, Reprint of the 1984 edition.
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Evolution semi-linear hyperbolic equations in a bounded domain Aimin Huang & Du Pham Abstract. In this article, our goal is to prove the existence and uniqueness of solution for 1D and 2D semi-linear hyperbolic equations in a bounded domain with a monotone nonlinear term. We use elliptic regularization and a finite difference scheme in time to build the approximate solutions for the semi-linear hyperbolic equations, and we utilize the regularization method together with the monotonicity and convexity of the nonlinear term to show the existence of the resulting stationary problems. Finally, the existence of the solution for the evolution problem is done by studying the convergence of the approximate solutions and by using the standard Minty method, and the uniqueness is achieved.
1. Introduction We aim to study in this article the initial and boundary value problem in dimension one and two for some first-order semi-linear reaction convection equation ∂t u + Au + P (u) = f,
(1.1)
where Au = ∂x u + ∂y u and P : R → R is a monotone function growing at the rate of a polynomial of odd order; a related problem was studied in space dimension one in [JP07] in the context of singular perturbation and boundary layer theory. The second-order semi-linear hyperbolic equations, in particular the non-linear wave equation, have been studied by many authors, see [Liu03, Sat68, Gla73] and references therein, where the authors considered the Dirichlet boundary conditions. In this article, we study the first-order semi-linear hyperbolic equation (1.1) and consider the dissipative boundary conditions in the sense of [BS07, Chapter 3] (see Remark 3.1), which is a first attempt in order to further study the first-order semilinear hyperbolic systems. We are going to use the finite difference method in time to study the problem (1.1). Finite difference discretization in time for approximating the solutions PDEs has been used since 1970’s to establish existence of their solutions, see e.g. [Tem77] or its latest edition [Tem01] for Navier Stokes equations. This method consists of discretizing the time domain 2010 Mathematics Subject Classification. 35L02, 35L60, 65M06, 65M12. Key words and phrases. semi-linear hyperbolic; nonlinear monotonicity; existence and uniqueness; integration by parts; polynomial growth. The Institute for Scientific Computing and Applied Mathematics, Indiana University, 831 East Third Street, Rawles Hall, Bloomington, Indiana 47405, U.S.A..
8
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into subintervals and replacing the time derivative in the partial differential equation by a finite difference approximation. The time-dependent parabolic problem is thus transformed into a sequence of stationary problems which can be solved by Galerkin methods. The finite difference discretization in time method applied to some hyperbolic PDEs appeared in the late 1970’s, see e.g. [Mar79, Kac84, Mun87]. The evolution problem is then transformed to a stationary problem but is still of hyperbolic type. In our case, the stationary problems are semi-linear hyperbolic with monotonic nonlinear term. To show the existence of the solutions to these nonlinear problems, we use the perturbation method, see [Lio73, Oma74]. To show the convergence of the approximate solutions, we first carry out a priori estimates to obtain some uniform bounds and hence the weak convergence of the approximate solutions, and then pass to the limit. To deal with the nonlinearity, we use fundamental results in the theory of monotone and accretive operators discovered during the latter half of the century by Minty [Min62, Min63], Browder [Bro65], Lions [Lio69], Brezis [Bre73] and many others. We also use a result in convex analysis, see Theorem B.1 from [ET76], to cope with non zero initial data and still be able to keep the monotonicity of the problem. Note that many stationary or time dependant problems of type (1.1) with a monotone nonlinearity are studied in the last references. However, a major difference with our study is that the operator A + P is not coercive unlike in the previous references. Our work is organized as follows. We first consider a one dimensional problem of the semilinear hyperbolic equation in Section 2. In Section 3, we then use the result in Section 2 by treating the time variable as a spacial variable to construct the solutions for the stationary problem in dimension two, and then use the similar techniques as in Section 2 to obtain the existence and uniqueness of our problem. 2. A one dimensional problem In this section, our aim is to prove the existence and uniqueness of solution of a semilinear hyperbolic equation in space dimension one. We first study the stationary boundary value problem, and then use finite differences in time to study the evolution problem. The semi-linear hyperbolic equation that we consider reads ∂u(x, t) ∂u(x, t) + + P (u(x, t)) = f (x, t), ∂t ∂x m+1
(2.1)
m+1
where x ∈ Ω = (0, 1), t ∈ (0, T ), f ∈ L m (0, T ; L m (Ω)) and P : R → R is monotone and has a polynomial growth at infinity in the following sense: (P (u) − P (v)) (u − v) ≥ 0,
α1 um+1 − β1 ≤ P (u)u ≤ α2 um+1 + β1 , m
|P (u)| ≤ α2 |u| + β2 ,
(2.2a) (2.2b) (2.2c)
for some positive constants α1 , α2 , β1 , β2 and an odd integer m. An example of such a P is an increasing polynomial of odd order m = 2s−1 with a positive leading coefficient a2s−1 > 0 of the form P (u) =
2s−1 X j=0
aj uj .
(2.3)
Semi-linear Hyperbolic Equation
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
We first see that P satisfies (2.2a) since P is increasing, and the estimates in Lemma 2.1 below assert that P also satisfies (2.2b) and (2.2c). Lemma 2.1. Assume that P is of the form (2.3) with the leading coefficient a2s−1 > 0. Then 1 3 a2s−1 u2s − b1 ≤ P (u)u ≤ a2s−1 u2s + b1 (2.4a) 2 2 1 3 a2s−1 u2s − b2 |u| ≤ P (u)u ≤ a2s−1 u2s + b2 |u| (2.4b) 2 2 3 |P (u)| ≤ a2s−1 |u|2s−1 + b2 , (2.4c) 2 where b1 , b2 depend only on the coefficients aj ’s and s. Proof. Using Young’s inequality, we are able to find b1 , ˜b2 > 0 such that 2s−2 X 1 j+1 aj u ≤ a2s−1 u2s + b1 , j=0 2 2s−2 2s−2 X X 1 2s−1 j+1 j a2s−1 |u| + ˜b2 . aj u ≤ |u| |aj ||u| ≤ |u| 2 j=0
j=0
These two inequalities imply (2.4a) and (2.4b), and (2.4c) is a consequence of (2.4b). We endow (2.1) with the following initial and boundary conditions ( u(t, 0) = g(t), u(0, x) = u0 (x).
(2.5)
Our agenda is as follows; we first study in Section 2.1 the stationary problem corresponding to (2.1); see also (2.27). We then use this result to show the existence of solution to a finite difference scheme in time of (2.1), (2.27). The existence of solution of (2.1), (2.27) in L∞ (0, T ; L2 (Ω) ∩ Lm+1 (ΩT ) is achieved by studying the convergence property of the finite difference solutions in Section 2.2. Since the solution is in the weak sense, we develop a trace theorem and an integration by parts result as tools to obtain the existence result. These results are presented in the Appendix A. The uniqueness of solution is done by the monotonicity of P and an integration by parts argument. Throughout this work, we will interchangeably use the notation ΩT = Ω × (0, T ) at convenience. We now study the stationary problem that will help us to build approximate solutions m+1 to the time dependent problem (2.1). We also set V = Lm+1 (Ω) and V 0 = L m (Ω) to be dual spaces of each other. In the following, we denote by h·, ·i the duality pairing between 0 Lr (Ω) and Lr (Ω), where 1 < r < +∞, and r0 is the conjugate exponents of r satisfying that 1/r + 1/r0 = 1. We also use the notations k · k for the norm in L2 (Ω). 2.1. The stationary boundary value problem. In this subsection, we use a regularization and variational inequality method to study the following boundary value problem: ( ∂u in Ω, ∂x + P (u) + λu = f, (2.6) u(0) = g0 , 8
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for some constant λ > 0 and g0 ∈ R. The reason why we introduce an extra term λu in (2.6) is to prepare for using the finite difference scheme for the evolution problem. In order to show the existence, we introduce the following elliptic regularization problem − ∂xx u + ∂x u + P (u ) + λu = f,
in Ω,
(2.7a)
u (0) = g0 , ux (1) = 0. (2.7b) We are going to apply Theorem B.2 to obtain a unique solution u of (2.7a)-(2.7b). Notice that by Sobolev embedding, we have H 1 (Ω) ⊂ C(Ω). We set X = H 1 (Ω) with the usual norm kuk2H 1 = kuk2 + k∂x uk2 , and K = {u ∈ H 1 (Ω) : u(0) = g0 }.
We then define the form a (u, v), for all u, v ∈ H 1 (Ω), Z Z Z Z a (u, v) = ∂x u∂x v dx + ux v dx + P (u)v dx + λ uv dx, Ω
Ω
Ω
(2.8)
Ω
and the operator A : X 7→ X 0 , for all u, v ∈ H 1 (Ω),
hA u, vi = a (u, v).
By definition, it is easy to verify that K is a non-empty convex closed set (u ≡ g0 ∈ K) and that A is weakly continuous over finite dimensional subspaces of X and bounded. For u, v ∈ K, we have Z Z hA u − A v, u − vi = (∂x u − ∂x v)2 dx + (u − v)x (u − v) dx Ω Ω Z Z + (P (u) − P (v))(u − v) dx + λ (u − v)2 dx (2.9) Ω Ω Z 1 ≥ (u − v)x (u − v) dx = (u(1) − v(1))2 2 Ω ≥ 0,
and the equality holds if and only if u = v. We thus deduce that RA is strictly monotone. x To show that A is coercive, we first note that u(x) = g0 + 0 ∂x u(x0 )dx0 , which implies that kuk2 ≤ 2g02 + 2k∂x uk2 ; (2.10) and then for u ∈ K, integration by parts and using (2.2b) and (2.10) yield Z Z Z 1 1 2 2 2 a (u, u) = (∂x u) dx + u(1) − u(0) + P (u)u dx + λ u2 dx 2 2 Ω Ω Ω Z Z Z 1 1 2 2 2 m+1 ≥ (∂x u) dx − g0 + u(1) + α1 u dx − β1 + λ u2 dx (2.11) 2 2 Ω Ω Ω Z Z 1+ 2 1 um+1 dx + λ u2 dx. ≥ kuk2H 1 − g0 − β1 + u(1)2 + α1 4 2 2 Ω Ω Hence, for u ∈ K,
hA u, ui a (u, u) (1 + )g02 + 2β1 = ≥ kukH 1 − → +∞, kukH 1 kukH 1 4 2kukH 1
as kukH 1 → +∞.
(2.12)
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
Applying Theorem B.2 to A , we find that for any f ∈ L(m+1)/m (Ω)(⊂ X 0 = (H 1 (Ω))0 ) there exists a unique u ∈ K such that hA u − f, v − u i ≥ 0,
∀ v ∈ K.
(2.13)
Lemma 2.2. Assume that f ∈ L(m+1)/m (Ω). Then there exists a unique u which satisfies (2.7a)-(2.7b), and the following energy estimates independent of hold: u is bounded in Lm+1 (Ω), √ ∂x u is bounded in L2 (Ω),
u is bounded in W
1,(m+1)/m
(2.14)
(Ω).
Proof. We already showed the existence of u satisfying (2.13), and we now prove that u satisfies (2.7a)-(2.7b). For any w ∈ C ∞ (Ω) with w(0) = 0, choosing v = u ± w ∈ K in (2.13) gives hA u − f, ±wi ≥ 0,
(2.15)
hA u , wi = hf, wi.
(2.16)
which yields
If we further assume that w ∈ Cc∞ (Ω), then integrating by parts in (2.16) gives Z
(−∂xx u + ∂x u + P (u ) + λu )w dx =
Ω
Z f w dx,
(2.17)
Ω
which implies that (2.7a) holds in the sense of distributions on (0, 1). We have u (0) = g0 since u ∈ K, and we are left to recover the boundary condition ux (1) = 0. From (2.7a), we obtain that ∂xx u belongs to L(m+1)/m (Ω), which shows that u belongs to C 1 (Ω) by the Sobolev embedding W 2,(m+1)/m ⊂ C 1 (Ω). Hence the trace ux |x=1 makes sense, and integrating by parts in (2.16) gives Z
(−∂xx u + ∂x u + P (u ) + λu )w dx +
ux (1)w(1)
Ω
Z =
f w dx.
(2.18)
Ω
Combining (2.17) and (2.18) leads to ux (1)w(1) = 0, which implies that ux (1) = 0. Therefore, u satisfies (2.7a)-(2.7b). We now turn to prove the energy estimates (2.14). Choosing v ≡ g0 ∈ K in (2.13) yields hA u , u i ≤ hf, u i + hA u − f, g0 i.
(2.19)
Using H¨ older’s inequality and Young’s inequality, we obtain (m+1)/m
hf, u i ≤ kf kV 0 ku kV ≤ ckf kV 0
+
α1 m+1 ku kV ; 4
(2.20) 8
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and using (2.2c) for P (u ) and Young’s inequality for the term P (u )g0 , we find Z Z Z Z f g0 dx P (u )g0 dx + λ u g0 dx − ux g0 dx + hA u − f, g0 i = Ω Ω Ω Ω Z Z Z f g0 dx P (u )g0 dx + λ u g0 dx − = g0 (u (1) − g0 ) + Ω Ω Ω Z 1 α2 m+1 1 (m+1)/m ≤ (u (1))2 − g02 + (u ) + c(g0m+1 + g0 ) dx 2 2 Ω 4 Z Z 1 2 m 1 +λ ((u ) + g02 ) dx + ( f (m+1)/m + g m+1 ) dx. m+1 0 Ω 2 Ω m+1
(2.21)
for some constant c > 0. Combining (2.19)-(2.21), we find with (2.11) that 1 1 √ α1 m+1 (m+1)/m (m+1)/m ) + λg02 . ≤ c(kf kV 0 + g0m+1 + g0 k ∂x u k2 + ku kV 4 2 2 Hence, we established the first two estimates in (2.14). The last estimate in (2.14) follows by applying Lemma A.1 with λ(x) = 1, g = f − P (u ) − λu , (2.7a) and the first estimate. The first and last estimates in (2.14) show that there exists a subsequence of u , still denoted by u , which weakly converges to some u in Lm+1 (Ω) and in W 1,(m+1)/m (Ω). By the compact Sobolev embedding W 1,(m+1)/m ⊂ C 0,α (Ω) valid for 0 < α < 1/4, we obtain that u strongly converges to u in C(Ω), which, together with the dominated convergence theorem, implies that P (u ) weakly converges to P (u) in L(m+1)/m (Ω). Therefore, passing to the limit in (2.7a), we conclude that the limit u solves (2.6)1 (at least in the sense of distributions). It remains to verify that u satisfies the boundary conditions (2.6)2 . Using the uniform boundedness of (2.14), (2.7a) gives − ∂xx u + ∂x u is uniformly bounded in L(m+1)/m (0, 1). x
(2.22)
Applying Lemma A.1 to u with p = (m+1)/m and X = R, we see that the ∂x u are uniformly (m+1)/m bounded in Lx (0, 1), and the corresponding traces converge, i.e. u(0) = g0 . We thus find a solution u satisfying (2.6). Therefore, we have the following result. Lemma 2.3. Assume that f ∈ L(m+1)/m (Ω) and λ > 0. Then there exists a unique u ∈ Lm+1 (Ω) ∩ W 1,(m+1)/m (Ω) satisfying (2.6). Proof. We only need to prove the uniqueness. Suppose that u1 , u2 both satisfy (2.6) and belong to Lm+1 (Ω) ∩ W 1,(m+1)/m (Ω), and let w = u1 − u2 . Then w satisfies wx + P (u1 ) − P (u2 ) + λw = 0, w(0) = 0.
(2.23)
Since w ∈ Lm+1 (Ω) and P (u1 ), P (u2 ) ∈ L(m+1)/m (Ω), wx ∈ L(m+1)/m (Ω), and it follows that wwx belongs to L1 (Ω) by H¨ older’s inequality, and that Z 1 1 1 (2.24) wwx dx = |w(1)|2 − |w(0)|2 = |w(1)|2 . 2 2 2 Ω
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
We now multiply (2.23)1 by w and use (2.24); we find Z Z 1 2 w(1) + (P (u1 ) − P (u2 ))(u1 − u2 ) dx + λ w2 dx = 0. 2 Ω Ω
(2.25)
The left-hand side of (2.25) is always positive thanks to the monotonicity (2.2a) of P . Hence, we must have w = 0 if (2.25) holds. We thus completed the proof. Remark 2.1. s i) The same result would be true without uniqueness if λ = 0. Uniqueness can be recovered if λ = 0 and P is strictly monotone. ii) With the same method we could replace (2.1)1 by ∂u + P (u) + λu = f, (2.26) ∂x with a ∈ C 1 (Ω) such that a(0)a(1) 6= 0. The boundary conditions would be imposed at x = 0 and/or x = 1, depending on the sign of a(0) and a(1). a(x)
2.2. The evolution problem. In this section, we show the existence and uniqueness of solution to the following semi-linear hyperbolic equation ∂t u + ∂x u + P (u) = f, in Ω × (0, T ), u(0, t) = g(t), (2.27) u(x, 0) = u0 (x), where P : R 7→ R is as above, that is P satisfies the assumptions (2.2a),(2.2b) and (2.2c). We m+1 m+1 suppose that f ∈ L m (0, T ; L m (Ω)), g ∈ L2 (0, T ) and u0 ∈ L2 (Ω). The implicit Euler finite difference scheme for (2.27) is set as follows: we let ∆t = T /N , N ∈ N and we set u0 = u0 , and then construct the un for n = 1, · · · , N by induction solving: n u − un−1 + ∂x un + P (un ) = f n , (2.28a) ∆t un (0) = g n , where 1 f = ∆t n
m+1
Z
n∆t
f (x, t)dt, (n−1)∆t
1 g = ∆t n
Z
n∆t
g(t)dt.
(2.28b)
(n−1)∆t m+1
Given un−1 ∈ L m (Ω), the existence of a unique solution un ∈ W 1, m (Ω) of (2.27)is guaranteed by Lemma 2.3. We now need to prove some estimates independent of N in order to pass to the limit k = 1/N → 0. Remark 2.2. Regarding the boundary value problem (2.26) in Remark 2.1 ii), we could also study the following evolution problem ( ∂t u + a(x, t)∂x u + P (u) + λu = f, (2.29) u(x, 0) = u0 (x), where a ∈ C 1 (ΩT ) with a(0, t)a(1, t) 6= 0, ∀ t ∈ [0, T ]. The boundary conditions would be imposed at x = 0 and/or x = 1, depending on the sign of a(0, t) and a(1, t). 8
8
A. Huang and D. Pham
2.2.1. A priori estimates. Taking the inner product of each side of (2.28a) with 2un ∆t, we find that 2hun − un−1 , un i + 2∆th∂x un , un i + 2∆thP (un ), un i = 2∆thf n , un i. Using the identity 2hun − un−1 , un i = kun − un−1 k2 + kun k2 − kun−1 k2 ,
(2.30)
the equation becomes kun − un−1 k2 + kun k2 − kun−1 k2 + 2∆th∂x un , un i + 2∆thP (un ), un i = 2∆thf n , un i. Using the Cauchy-Schwarz inequality and the Young inequality for the right hand side, we obtain, recalling that V = Lm+1 (Ω): kun − un−1 k2 + kun k2 − kun−1 k2 + 2∆th∂x un , un i + 2∆thP (un ), un i m+1
, + α1 ∆tkun km+1 ≤ 2∆tkf n kV 0 kun kV ≤ cm ∆tkf n kV m 0 V
√ √ m where cm := 2 2m/(m + 1) m α1 m. From the assumption (2.2b), we find that hP (un ), un i ≥ − β1 . This implies α1 kun km+1 V kun − un−1 k2 + kun k2 − kun−1 k2 + ∆t((un (1))2 − (g n )2 ) + α1 ∆tkun km+1 V
m+1
≤ cm ∆tkf n kV m + 2β1 ∆t. 0 We sum the above inequalities for n = 1, . . . , n0 and obtain: n0 X n=1
n
n−1 2
ku − u
n0 2
k + ku k +∆t
n0 X
n
2
(u (1)) + α1 ∆t
n=1
≤ cm ∆t
n0 X n=1
N X n=1
m+1
kun km+1 V
+ ∆t kf n kV m 0
N X
(g n )2 + ku0 k2 + 2β1 T.
n=1
We have by (2.28b) ∆t
N X n=1
n
kf k
m+1 m V0
m+1 m m+1 L m (0,T ;V 0 )
≤ kf k
,
∆t
N X
(g n )2 ≤ kgk2L2 (0,T ) ,
n=1
and we conclude that n0 X n=1
n
n−1 2
ku − u
n0 2
k + ku k + ∆t
≤ K0 = K0 (P, T, f, g) := cm kf k for all 1 ≤ n0 ≤ N . We have proven the following bounds:
n0 X
n
2
(u (1)) + α1 ∆t
n=1
m+1 m m+1 L m (0,T ;V 0 )
n0 X n=1
kun km+1 V
+ kgk2L2 (0,T ) + ku0 k2 + 2β1 T,
(2.31)
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation Lemma 2.4.
n 2 ku k ≤ K0 for all 1 ≤ n ≤ N, n0 X K0 , kun km+1 ∆t ≤ V α1 n=1 n0 X ∆t (un (1))2 ≤ K0 , n=1 N X kun − un−1 k2 ≤ K0 ,
(2.32)
n=1
where K0 is defined in (2.31), and depends only on the data. 2.2.2. Passage to the limit . We now introduce two approximate solutions to (2.1), which are denoted by uk and u ˜k with k = ∆t: for each t ∈ In := ((n − 1)∆t, n∆t], n = 1 . . . , N , we set uk (t) = un , u ˜k (t) =
(2.33)
un − un−1 (t − n∆t) + un , ∆t
that is, uk is the step function on the interval (0, T ) with values taken from the right of each interval In , and u ˜k is the piecewise linear function that interpolates un−1 and un on In . We first have an equivalent form of the scheme (2.28a) as follows ˜k + ∂x uk + P (uk ) = fk , in Ω × (0, T ), ∂t u u ˜k (x, 0) = u0 (x), (2.34) uk (0, t) = gk (t), where fk = f n and gk = g n on In for n = 1, . . . , N . We infer from (2.31) the following inequality for the approximate solutions: ≤ K0 (f, g) ∆tk∂t u ˜k k2L2 (0,T ;L2 (Ω)) + kuk (·, t)k2 + kun |x=1 k2L2 (0,T ) + α1 kuk km+1 Lm+1 (0,T,V ) for all 0 ≤ t ≤ T . Using (2.4c) and kuk kLm+1 (ΩT ) being bounded, we find that kP (uk )k
Therefore, as k → 0
L
m+1 m (ΩT )
uk , u ˜k ∈ bounded set of L∞ (0, T ; L2 (Ω)), m+1 (0, T ; V ) = Lm+1 (Ω ), T uk ∈ bounded set of L m+1 P (uk ) ∈ bounded set of L m (0, T ; V 0 ), uk (1) ∈ bounded set of L2 (0, T ), √ ∆t ∂t u ˜k ∈ bounded set of L2 (0, T ; L2 (Ω)).
(2.35)
is also bounded.
(2.36)
As a consequence of the estimates (2.36), we find from equation (2.34) that ∂t u ˜k + ∂x uk ∈ bounded set of L
m+1 m
(0, T ; V 0 ).
(2.37) 8
10
A. Huang and D. Pham
We now have an estimate of the distance between the two functions uk and u ˜k : direct computations show that 2 Z n∆t n Z n∆t 1 un − un−1 2 3 u − un−1 k k˜ uk − uk k2 dt = k(t − n∆t) dt = k k ∆t . ∆t 3 ∆t (n−1)∆t (n−1)∆t This implies k˜ uk − uk kL2 (ΩT ) =
√
3 k∂t u ˜k kL2 (ΩT ) ∆t. 3
(2.38)
From (2.38) and (2.36)5 , we find 1
k˜ uk − uk kL2 (0,T ;L2 (Ω)) = O(∆t 2 ).
(2.39)
By (2.36) and (2.39), there exist subsequences of u ˜k and uk which we still denote by k, and m+1 ∞ 2 m+1 u ∈ L (0, T ; L (Ω)) ∩ L (ΩT ), χ ∈ L m (0, T ; V 0 ), κ ∈ L2 (0, T ) such that for k → 0 ˜k −* u weak- ∗ in L∞ (0, T ; L2 (Ω)) and weakly in Lm+1 (ΩT ), uk , u m+1 (2.40) P (uk )−* χ weakly in L m (0, T ; V 0 ), uk (1)−* κ weakly in L2 (0, T ). From (2.37), we are also able to extract a subsequence still denoted by k such that ∂t u ˜k + ∂x uk m+1 0 weakly converges to some ξ in L m (0, T ; V ). From (2.40)1 , we infer that ∂t u ˜k , ∂x uk converge to ∂t u, ∂x u in the sense of distributions respectively. We thus have that ξ = ∂t u + ∂x u, and that m+1 (2.41) ∂t u ˜k + ∂x uk −* ∂t u + ∂x u weakly in L m (0, T ; V 0 ). m+1
m+1
We have u ∈ L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (ΩT ), ∂t u + ∂x u ∈ L m (0, T ; V 0 ) = L m (ΩT ). Thanks to Lemma A.2 with U = ΩT , a = b = 1, p = (m + 1)/m, the traces u(0, t) and u(x, 0) make sense. ¯ We now denote by u ¯k , u ˜k , f¯k the extensions of the functions uk , u ˜k , fk by zero to the domain (−∞, 1) × (−∞, T ); u ¯0 is the extension of u0 to (−∞, 1) and g¯k is the extension of gk to (−∞, T ). We infer from (2.40)2 , (2.41) and (2.34)2,3 that ¯ ∂t u ˜k (x, t) − δ0 (t)¯ u0 (x) + ∂x u ¯k (x, t) − δ0 (x)¯ gk (t) + P (¯ uk (x, t)) = f¯k (x, t) where δ0 is the dirac delta function having mass at 0. Passing to the limit as k → 0, we find in the sense of distributions ∂t u ¯(x, t) − δ0 (t)¯ u0 (x) + ∂x u ¯(x, t) − δ0 (x)¯ g (t) + χ(x, ¯ t) = f¯(x, t)
where u ¯, χ, ¯ f¯ are the extensions of u, χ, f on (−∞, 1) × (−∞, T ) and g¯ is the extension of g to (−∞, T ). This equation and ∂t u + ∂x u + χ = f on ΩT imply ∂t u + ∂x u + χ = f, u(x, 0) = u0 (x), (2.42) u(0, t) = g(t). Remark 2.3. To recover the initial condition and boundary condition for (2.42), we can also take the inner product of (2.34) with ϕ(x)ψ(t) for ϕ(x) ∈ D(Ω), ψ(t) ∈ D([0, T ]), ϕ(1) = ψ(T ) = 0 and carry out the integration by parts to obtain −h˜ uk , ϕψt i − hu0 , ϕψ(0)iL2x (Ω) − huk , ϕx ψi − hgk , ϕ(0)ψiL2t (0,T ) + hP (uk ), ϕψi = hfk , ϕψi.
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation We then let k run to zero and find that
−hu, ϕψt i − hu0 , ϕψ(0)iL2x (Ω) − hu, ϕx ψi − hg, ϕ(0)ψiL2t (0,T ) + hχ, ϕψi = hf, ϕψi. This also implies (2.42).
There remains to show that χ = P (u). For that purpose, let v ∈ Lm+1 (0, T ; V ) so that m+1 P (v) ∈ L m (0, T ; V 0 ); we consider Z
T
Xk = 0
hP (uk ) − P (v), uk − vidt.
We first see that Xk ≥ 0 thanks to the monotonicity (2.2a) of P , and then write Xk = Xk1 + Xk2 + Xk3 , where Xk1 =
T
Z 0
Xk2 =
hP (uk ), uk i dt,
T
Z 0
Xk3 =
hP (uk ), vi dt,
Z 0
T
hP (v), (uk − v)i dt.
The weak convergence in (2.40) immediately shows that as k → 0: Xk2
−→
Z 0
T
hχ, vi dt,
and
Xk3
−→
Z
T
hP (v), (u − v)i dt.
0
For Xk1 , we use the scheme (2.34) and write Xk1
Z
T
hfk , uk i dt −
= 0
Z = 0
T
Z 0
h∂t u ˜k , uk i dt −
Z 0
T
h∂x uk , uk i dt
N
T
hfk , uk i dt −
1X n 1 1 ku − un−1 k2 − kuN k2 + ku0 k2 2 2 2 n=1
1 1 − kuk |x=1 k2L2 (0,T ) + kgk k2L2 (0,T ) 2 2 Z T 1 1 1 1 ≤ hfk , uk i dt − kuk |t=T k2 + ku0 k2 − kuk |x=1 k2L2 (0,T ) + kgk k2L2 (0,T ) . 2 2 2 2 0 From Lemma A.3 and since u0 ∈ L2 (Ω) and g ∈ L2 (0, T ), we know that the traces of u at x = 0, 1 and t = 0, T are L2 -functions. We also have that lim inf kuk |t=T k ≥ ku|t=T k, k→0 Z T Z T lim inf kuk |x=1 kL2 (0,T ) ≥ ku|x=1 kL2 (0,T ) , hfk , uk i dt → hf, ui dt, and kgk kL2 (0,T ) → kgkL2 (0,T ) , k→0
0
0
which together yield Z T 1 1 1 1 lim sup Xk1 ≤ hf, ui dt − ku|t=T k2 + ku0 k2 − ku|x=1 k2L2 (0,T ) + kgk2L2 (0,T ) . 2 2 2 2 k→0 0
(2.43) 8
12
A. Huang and D. Pham m+1
Since we have u ∈ Lm+1 (ΩT ) and ∂t u + ∂x u ∈ L m (ΩT ), applying Lemma A.3 to u with y = t, U = ΩT , p = m + 1 and q = m+1 m and using (2.42), we obtain that Z 1 2 2 2 2 ku|t=T k − ku0 k + ku|x=1 kL2 (0,T ) − kgkL2 (0,T ) = hu, ut + ux i dxdt 2 ΩT Z (2.44) = hu, f − χi dxdt. ΩT
Combining (2.43) and (2.44), we find that
We finally conclude that Z T Z lim sup Xk ≤ hχ, ui dt − k→0
0
lim sup Xk1 k→0
≤
Z
T
T
0
hχ, vi dt −
0
Z 0
T
hχ, ui dt.
hP (v), (u − v)i dt =
Z 0
T
hχ − P (v), (u − v)i dt,
which, together with Xk ≥ 0, implies that for all v ∈ Lm+1 (0, T ; V ), Z T hχ − P (v), (u − v)i dt ≥ 0. 0
The standard Minty method (see [Lio69]) then yields χ = P (u). 2.2.3. Main result of Section 2. m+1
m+1
Theorem 2.1. Assume that f ∈ L m (0, T ; L m (Ω)), g ∈ L2 (0, T ) and u0 ∈ L2 (Ω) are given. Then there exists a unique solution u = u(t, x) ∈ L∞ (0, T ; L2 (Ω))∩Lm+1 (0, T ; Lm+1 (Ω)) satisfying (2.1) and (2.5). Proof. The existence of a solution u was proved in Section 2.2.1, and we thus only need to show the uniqueness. Let u1 , u2 belong to L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (0, T ; Lm+1 (Ω)) and both satisfy (2.1)-(2.5). Set w = u1 − u2 ; we then see that w satisfies wt + wx = P (u2 ) − P (u1 ), (2.45) w(t, 0) = 0, w(0, x) = 0. Since u1 , u2 ∈ Lm+1 (ΩT ), we find from (2.45)1 that wt + wx belongs to L(m+1)/m (ΩT ) by assumption (2.2b). Applying Lemma A.3 to w with y = t, U = Ωt := Ω × (0, t), p = m + 1 and q = (m + 1)/m and using (2.45)2,3 , we obtain Z t Z Z w(1, t0 )2 dt0 + w(x, t)2 dx = w(wt + wx ) dxdt0 , ∀ t ∈ [0, T ], 0
Ω
Ωt
which, together with (2.45)1 , implies that Z Z 2 w(x, t) dx + (u1 − u2 )(P (u1 ) − P (u2 )) dxdt0 ≤ 0, Ω
Ωt
∀ t ∈ [0, T ].
(2.46)
Therefore, we can conclude that u1 (·, t) = u2 (·, t) for all t ∈ [0, T ] and hence u1 = u2 , and we thus completed the proof.
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation 3. A two dimensional problem
In this section, we consider the rectangular domain Ω = (0, 1)2 in R2 and our aim is to show the existence and uniqueness of solution to the following semi-linear hyperbolic equation: ∂t u + Au + P (u) = f, in Ω × (0, T ), u(0, y, t) = g1 (y, t), u(x, 0, t) = g2 (x, t) (3.1) u(x, y, 0) = u0 (x, y), where Au = ∂x u + ∂y u and P : R → R is as in Section 2 satisfying (2.2a)-(2.2c). We suppose m+1 m+1 that f ∈ L m (0, T ; L m (Ω)), g1 ∈ L2 (0, T ; L2y (0, 1)), g2 ∈ L2 (0, T ; L2x (0, 1)) and u0 ∈ L2 (Ω). The standard finite difference scheme of (3.1) is set as follows: Let ∆t = T /N ; we set u0 = u0 , and construct the un for n = 1, · · · , N recursively by setting n u − un−1 + Aun + P (un ) = f n , (3.2a) ∆t un (0, y) = g n (y), un (x, 0) = g n (x), 1
2
where 1 f = ∆t n
Z
n∆t
f (x, y, t)dt, (n−1)∆t
g1n (y)
1 = ∆t
Z
n∆t
g1 (y, t)dt,
g2n (x)
(n−1)∆t
1 = ∆t
Z
n∆t
g2 (x, t)dt. (n−1)∆t
(3.2b) We rewrite the equation (3.2a)1 as ( 1 n ∂y un + ∂x un + P (un ) + ∆t u = fn + un (0, y) = g1n (y), un (x, 0) = g2n (x). m+1
1 n−1 , ∆t u
(3.3)
m+1
Given un−1 ∈ L m (Ω), the existence and uniqueness of un in L m (Ω) is guaranteed by 1 1 u and noting that P (u) + ∆t u is applying Theorem 2.1 with t = y and P (u) = P (u) + ∆t 1 strictly monotone thanks to λ = ∆t > 0. 3.1. A priori estimates. We still use the previous notations, i.e. V = Lm+1 (Ω) and m+1 V 0 = L m (Ω) to be dual spaces of each other; h·, ·i is the duality pairing between Lr (Ω) and 0 Lr (Ω), for 1/r + 1/r0 = 1 and k · k for the norm in L2 (Ω). Taking the inner product of each side of (3.2a) with 2un ∆t, and using the identity (2.30), we find that kun − un−1 k2 + kun k2 − kun−1 k2 + 2hAun , un i∆t + 2hP (un ), un i∆t = 2hf n , un i∆t. Using the Cauchy-Schwarz inequality and the Young inequality for the right hand side, we find that kun − un−1 k2 + kun k2 − kun−1 k2 + 2hAun , un i∆t + 2hP (un ), un i∆t m+1
≤ 2kf n kV 0 kun kV ∆t ≤ cm kf n kV m ∆t + α1 kun km+1 ∆t, 0 V
8
14
A. Huang and D. Pham
√ √ m where cm := 2 2m/(m + 1) m α1 . We know from (2.2b) that hP (un ), un i ≥ α1 kun km+1 − β1 and this implies kun − un−1 k2 + kun k2 − kun−1 k2 + 2∆thAun , un i + α1 ∆tkun km+1 V
m+1
≤ cm ∆tkf n kV m + 2β1 ∆t. 0 Since g1n ∈ L2y (0, 1), g2n ∈ L2x (0, 1), and Aun ∈ L and a = b = c = 1 to obtain
m+1 m
(ΩT ), we apply Lemma A.4 with U = Ω
2hAun , un i = 2h∂x un + ∂y un , un i
= kun |x=1 k2L2y (0,1) − kg1n (y)k2L2y (0,1) + kun |y=1 k2L2x (0,1) − kg2n (x)k2L2x (0,1) .
Therefore kun − un−1 k2 + kun k2 − kun−1 k2 + ∆t(kun |x=1 k2L2y (0,1) + kun |y=1 k2L2x (0,1) ) + α1 ∆tkun km+1 V m+1
≤ cm ∆tkf n kV m + 2β1 ∆t + ∆tkg1n (y)k2L2y (0,1) + ∆tkg2n (x)k2L2x (0,1) . 0 We sum the above inequalities for n = 1, . . . , n0 : n0 X n=1
kun − un−1 k2 + kun0 k2 + ∆t ≤ cm ∆t
N X n=1
n0 X
(kun |x=1 k2L2y (0,1) + kun |y=1 k2L2x (0,1) ) + α1 ∆t
n=1
m+1
+ 2β1 T + ∆t kf n kV m 0
N X n=1
kg1n k2L2y (0,1) + ∆t
N X n=1
n0 X n=1
kun km+1 V
kg2n k2L2x (0,1) + ku0 k2 .
Since we have ∆t
∆t
∆t
N X n=1 N X n=1 N X n=1
m+1
kf n kV m ≤ kf k 0
n=1
,
kg1n k2L2y (0,1) ≤ kg1 k2L2 (0,T ;L2y (0,1)) , kg2n k2L2x (0,1) ≤ kg2 k2L2 (0,T ;L2x (0,1)) ,
we conclude that, for all 1 ≤ n0 ≤ N , n0 X
m+1 m m+1 L m (0,T ;V 0 )
kun − un−1 k2 + kun0 k2 + ∆t
n0 X n=1
kun |x=1 k2L2y (0,1) + ∆t
n0 X
kun |y=1 k2L2x (0,1)
n=1 n0 X
+ α1 ∆t
n=1
where K0 := cm kf k
m+1 m m+1 L m (0,T ;V 0 )
(3.4) kun km+1 V
≤ K0 ,
+ kg1 k2L2 (0,T ;L2 (0,1)) + kg2 k2L2 (0,T ;L2 (0,1)) + ku0 k2 + 2β1 T . y
x
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
3.2. Passage to the limit. We introduce two approximate solutions uk and u ˜k defined on t ∈ In := ((n − 1)∆t, n∆t], n = 1 . . . , N as follows: uk is the step function on In with values taken from the right of each interval In , and u ˜k is the piecewise linear function that interpolates un−1 and un on In . We first have an equivalent form of the scheme (3.2a) as follows ˜k + Auk + P (uk ) = fk , in Ω × (0, T ), ∂t u uk (0, y, t) = g1k (y, t), uk (x, 0, t) = g2k (x, t), (3.5) u ˜k (x, y, 0) = u0 (x, y), where fk , g1k and g2k are the step functions on (0, T ), respectively equal to f n , g2n and g2n on the interval In , n = 1 . . . , N . From (3.4), we find ∆tk∂t u ˜k k2L2 (0,T ;L2 (Ω)) + kuk (t)k2 + kun |x=1 k2L2 (0,T ;L2y (0,1))
+ kun |y=1 k2L2 (0,T ;L2x (0,1)) + α1 kuk km+1 ≤ K0 , Lm+1 (0,T,V )
for all 0 ≤ t ≤ T . Therefore, as k → 0:
uk , u ˜k ∈ bounded set of L∞ (0, T ; L2 (Ω)), uk ∈ bounded set of Lm+1 (0, T ; V ) = Lm+1 (ΩT ), m+1 P (uk ) ∈ bounded set of L m (0, T ; V 0 ), uk |∂Ω ∈ bounded set of L2 (0, T ; L2 (∂Ω)), √ ∆t ∂t u ˜k ∈ bounded set of L2 (0, T ; L2 (Ω)).
(3.6)
(3.7)
As a consequence of these bounds, from equation (3.5), we find that ∂t u ˜k + Auk is bounded in m+1 m L (ΩT ) independently of k. From these bounds along with (2.38), we extract a subsequence still denoted by k such that for k → 0 uk , u ˜k −* u weak-* in L∞ (0, T ; L2 (Ω)) and weakly in Lm+1 (ΩT ), P (u )−* χ weakly in L m+1 m (0, T ; V 0 ), k (3.8) uk |∂Ω −* κ weakly in L2 (0, T ; ∂Ω), m+1 ∂t u ˜k + Auk −* ξ = ∂t u + Au weakly in L m (0, T ; V 0 ). m+1
We have u ∈ L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (ΩT ), ∂t u + Au ∈ L m (ΩT ). Thanks to Lemma A.4 with M = ΩT , a = b = c = 1 and p = (m + 1)/m, the traces u|t=0 and u|∂Ω make sense. As in dimensional one case, by extending all functions by zero to the domain (−∞, 1)2 × (−∞, T ), letting k → 0 and combing with (2.40)2 , (2.41), and (3.5)2,3 , we find the initial condition and boundary condition for the limit equation reads ∂t u + Au + χ = f, u(0, y, t) = g1 (y, t), u(x, 0, t) = g2 (x, t), (3.9) u(x, y, 0) = u (x, y). 0 Thanks to Lemma A.5 with a = b = c = 1 and u|x=0 = g1 (y, t) ∈ L2 (0, T ; Ly (0, 1)), u|0=0 = g2 (x, t) ∈ L2 (0, T ; Lx (0, 1)), u|t=0 = u0 (x, y) ∈ L2 (Ω), all the traces u|x=1 , u|y=1 and u|t=T are 8
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A. Huang and D. Pham
square integrable functions, and the integration by parts is valid for h∂t u + Au, vi; hence we can similarly (as in Section 2.2.2) show that Z T hχ − P (v), (u − v)i dt ≥ 0, 0
Lm+1 (0, T ; V
for all v ∈ ) and thus by the Minty method χ = P (u). This shows that u is a solution of (3.1). We now state the result of existence and uniqueness of a solution u for (3.1): m+1
Theorem 3.1. Assume that f ∈ L m (ΩT ), g1 ∈ L2 (0, T ; L2y (0, 1)), g2 ∈ L2 (0, T ; L2x (0, 1)) and u0 ∈ L2 (Ω) are given. Then there exists a unique function u ∈ L∞ (0, T ; L2 (Ω))∩Lm+1 (ΩT ) satisfying (3.1). Proof. The existence of a solution u was proved in Section 3.2, and we thus only need to show the uniqueness. Let u1 , u2 belong to L∞ (0, T ; L2 (Ω)) ∩ Lm+1 (0, T ; Lm+1 (Ω)) that both satisfy (3.1). Setting w = u1 − u2 , we then see that w satisfies wt + Au = P (u2 ) − P (u1 ), (3.10) w(t, 0, y) = w(t, x, 0) = 0, w(0, x) = 0. Since u1 , u2 ∈ Lm+1 (ΩT ), we find from (3.10)1 that wt + Au belongs to L(m+1)/m (ΩT ) by assumption (2.2b). Applying Lemma A.5 to w with z = t, M = Ωt := Ω × (0, t), p = m + 1 and q = (m + 1)/m and using (3.10)2,3 , we obtain that for all t ∈ [0, T ], Z 1Z t Z 1Z t Z Z w(x, 1, t0 )2 dxdt0 + w(1, y, t0 )2 dydt0 + w(x, y, t)2 dxdy = w(wt + wx ) dxdt0 , 0
0
0
0
Ω
which, together with (3.10)1 , implies that Z Z 2 w(x, y, t) dxdy + (u1 − u2 )(P (u1 ) − P (u2 )) dxdt0 ≤ 0, Ω
Ωt
Ωt
∀ t ∈ [0, T ].
(3.11)
Therefore, we can conclude that u1 (·, t) = u2 (·, t) for all t ∈ [0, T ] and hence u1 = u2 , and we thus completed the proof. Remark 3.1 (The generality of operator A and the boundary conditions). We could study the following general form of A = a(x, y, t)∂x u + b(x, y, t)∂y u where a, b ∈ C 1 (ΩT ) are never zero on the boundary ∂Ω. The boundary conditions imposed on the boundary ∂Ω depend on the signs of a, b on the boundary. For example, we could impose the following boundary conditions u(0, y, t) = g1 (y, t) when a(0, y, t) > 0,
u(1, y, t) = g1 (y, t) when a(1, y, t) < 0,
u(x, 0, t) = g2 (x, t) when b(x, 0, t) > 0,
u(x, 1, t) = g2 (x, t) when b(x, 1, t) < 0.
(3.12)
The boundary conditions in (3.12) are strictly dissipative in the sense of [BS07, Chapter 3]. Our main result in Theorem 3.1 is still valid without much changing of its proof. Indeed, Theorem 2.1 or 2.2 can be applied with t = y to the problem (3.3) induced by the finite difference scheme in time in the general case.
Semi-linear Hyperbolic Equation
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Appendix A. Trace theorem In this Appendix we give some trace theorems which we used in the article. See Appendix A in [ST10] for some other trace results. We fix p such that 1 < p < +∞, and we also let q = p0 the conjugate exponent of p (i.e. 1/p + 1/q = 1). Lemma A.1. Let X be a reflexive Banach space, and λ = λ(x) ∈ C 0 ([0, 1]) such that λ(x) ≥ c0 for some positive constant c0 . Assume that two sequences of function u , g ∈ Lpx (0, 1; X) satisfy ( −uxx + λ(x)ux = g , (A.1) u (0) = u0 , ux (1) = 0, with g bounded in Lpx (0, 1; X) independently of , u0 ∈ X and p > 1. Then the ux are bounded in Lpx (0, 1; X) independently of , and, for any subsequence u → u, g → g (strongly or weakly) converging in Lpx (0, 1; X), u (0) converges to u(0) in X (weakly at least), and hence u(0) = u0 . Proof. For simplicity, we drop , and by solving (A.1), we obtain Z 1 1 − R x1 λ(¯x)/ d¯x e x g(x1 )dx1 . ux (x) = x
(A.2)
Taking the X-norm of (A.2) and raising to the p-th power, and using H¨older’s inequality, we find Z 1 p/q Z 1 1 − R x1 λ(¯x)/ d¯x 1 − R x1 λ(¯x)/ d¯x p x kux kX ≤ (A.3) e dx1 e x kg(x1 )kpX dx1 . x x Direct computations show that the first integral in (A.3) is less than Z 1 Z 1 1 − R x1 c0 / d¯x 1 −c0 (x1 −x)/ 1 x e dx1 = e dx1 ≤ . c0 x x Then integrating (A.3) from 0 to 1 with respect to x and switching the order of integration for x, x1 , we finally arrive at Z Z 1 1 p 1 p kux kX dx ≤ ( ) kgkpX dx. (A.4) c 0 0 0 It follows that the ux are uniformly bounded in Lpx (0, 1; X); hence u ∈ Cx ([0, 1]; X). The existence of the trace u(0) and its linear continuous dependence on {u, g} follow easily. Remark A.1. If the coefficient function in the convection term is negative, that is, instead of (A.1) we have ( −uxx − λ(x)ux = g , (A.5) u (1) = u1 , ux (0) = 0, for two sequences u , g ∈ Lpx (0, 1; X), then the same result as in Lemma A.1 follows; that means the ux are bounded in Lpx (0, 1; X) independently of , and also for any subsequence u → u, g → g which converge (strongly or weakly) in Lpx (0, 1; X), u (1) converges to u(1) in X (weakly at least), and hence u(1) = u1 . 8
18
A. Huang and D. Pham Let U = (0, L1 ) × (0, L2 ) be a subset of R2 , and we consider the space: X (U) = {u ∈ Lp (U), T u = aux + buy ∈ Lp (U)},
endowed with its natural norm (kukpLp (U ) + kT ukpLp (U ) )1/p , where L1 and L2 are positive constants, and a and b are non-zero constants. Then we have the following trace result. Lemma A.2. If u ∈ X (U), then the traces of u are defined on all of ∂U, i.e. the traces of u are defined at x = 0, L1 , and y = 0, L2 , and they belong to the respective spaces Wy−1,p (0, L2 ) and Wx−1,p (0, L1 ). Furthermore the trace operators are linear continuous in the corresponding spaces, e.g., u ∈ X (U) → u|x=0 is continuous from X (U) into Wy−1,p (0, L2 ). Proof. Since u ∈ Lp (U), we see that uy belongs to Lpx (0, L1 ; Wy−1,p (0, L2 )), which, together with T u ∈ Lp (U) and a, b are non-zeros, implies that ux ∈ Lpx (0, L1 ; Wy−1,p (0, L2 )). In combination with u ∈ Lpx (0, L1 ; Lpy (0, L2 )), we obtain that u ∈ Cx ([0, L1 ]; Wy−1,p (0, L2 )). Hence, the traces of u are well-defined at x = 0 and L1 , and belong to Wy−1,p (0, L2 ). The continuity of the corresponding mappings is easy. The proof for the traces at y = 0 and L2 is similar. Lemma A.3. Assume that u ∈ Lp (U) and T u ∈ Lq (U) with a, b > 0. Then by Lemma A.2 with p = min{p, q}, we know that the traces of u on ∂U are well-defined. Furthermore, we assume that the traces u(0, y) ∈ L2 (0, L2 ) and u(x, 0) ∈ L2 (0, L1 ). Then the traces u(L1 , y), u(x, L2 ) belong to L2 (0, L2 ), L2 (0, L1 ) respectively, and the identity holds Z u(T u) dxdy = U
a ku(L1 , y)k2L2 (0,L2 ) − ku(0, y)k2L2 (0,L2 ) 2 b + ku(x, L2 )k2L2 (0,L1 ) − ku(x, 0)k2L2 (0,L1 ) . 2
(A.6)
Proof. Without loss of generality, we may assume that b = 1. We first denote by Γ1 , Γ2 , Γ3 , Γ4 the boundaries x = 0, x = L1 , y = 0, y = L2 respectively, and then introduce a new coordinate system (x0 , y 0 ) such that 0 x a 1 x = , y0 −1 a y which is equivalent to 0 1 x a −1 x = . y y0 1 + a2 1 a We denote by Γ0i the image of Γi by this transformation for all i ∈ {1, 2, 3, 4}; we also denote by U 0 the image of U and denote by u0 the transform of u. Let (αy0 , y 0 ) and (βy0 , y 0 ) be the end points of the intersection of U 0 with the line y 0 fixed, (αy0 , y 0 ) ∈ Γ01 or Γ03 , (βy0 , y 0 ) ∈ Γ02 or Γ04 (see Figure A.1).
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation
aL2
y′ Γ′1
Γ′4 αy′
y
′
βy ′
Γ′3
x′
Γ′2 b
b
−L1 Figure A.1. The domain U 0 Direct computation shows that ∂u0 1 ∂u ∂u 1 = a + = T u. ∂x0 1 + a2 ∂x ∂y 1 + a2 Therefore, for u ∈ Lp (U) and T u ∈ Lq (U), we see that u0 ∈ Lp (U 0 ) and u0x0 ∈ Lq (U 0 ), which shows that u0 u0x0 belongs to L1 (U 0 ) by H¨older’s inequality. Hence for fixed y 0 , we see that Z βy 0 0 0 2 0 0 2 u0 (x0 , y 0 )u0x0 (x0 , y 0 ) dx0 . u (βy0 , y ) = u (αy0 , y ) + 2 (A.7) αy 0
y0
Integrating (A.7) with respect to from −L1 to aL2 , we obtain Z aL2 Z aL2 Z aL2 Z βy0 0 0 2 0 0 0 2 0 u (βy0 , y ) dy = u0 (x0 , y 0 )u0x0 (x0 , y 0 ) dx0 dy 0 . u (αy0 , y ) dy + 2 −L1
−L1
−L1
(A.8)
αy 0
Since the traces of u at Γ1 = {x = 0} and Γ3 = {y = 0} are L2 functions, we also have that the traces of u0 at Γ01 and Γ03 are L2 functions. Therefore, we infer from (A.8) that the traces of u0 at Γ02 and Γ04 are L2 functions, which shows that the traces of u at Γ2 = {x = L1 } and Γ4 = {y = L2 } are L2 functions. Now, transforming back into the original system (x, y), (A.8) is equivalent to Z L2 Z L1 Z L2 Z L1 2 2 2 a u(L1 , y) dy + u(x, L2 ) dx =a u(0, y) dy + u(x, 0)2 dx 0 0 0 0 (A.9) Z L Z L 1
2
+2
u(T u) dxdy, 0
0
which is (A.6) in the case b = 1. We thus completed the proof.
Lemmas A.2 and A.3 are also valid in the three dimensional case. We state them as follows. We let M = (0, L1 ) × (0, L2 ) × (0, L3 ) be a subset of R3 , and we consider the space: X (M) = {u ∈ Lp (M), T u = aux + buy + cuz ∈ Lp (M)},
endowed with its natural norm (kukpLp (M) + kT ukpLp (M) )1/p , where L1 , L2 , L3 are positive constants, and a, b, c are non-zero constants. Then we have the following trace result. 8
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Lemma A.4. If u ∈ X (M), then the traces of u are defined on all of ∂M, i.e. the traces of u are defined at x = 0, L1 , y = 0, L2 and z = 0, L3 , and they belong to the respective spaces −1,p −1,p −1,p Wy,z ((0, L2 ) × (0, L3 )), Wx,z ((0, L1 ) × (0, L3 )) and Wy,z ((0, L2 ) × (0, L3 )). Furthermore the trace operators are linear continuous in the corresponding spaces. Lemma A.5. Assume that u ∈ Lp (M) and T u ∈ Lq (M) with a, b, c > 0. Then by Lemma A.4 with p = min{p, q}, we know that the traces of u on ∂M are well-defined. Furthermore, we assume that the traces u|x=0 , u|y=0 , u|z=0 are L2 -functions with respect to the corresponding domains. Then the traces u|x=L1 , u|y=L2 , u|z=L3 are also L2 -functions, and the following identity holds: Z a u(T u) dxdydz = ku|x=L1 k2L2 − ku|x=0 k2L2 2 M b (A.10) + ku|y=L2 k2L2 − ku|y=0 k2L2 2 c + ku|z=L3 k2L2 − ku|z=0 k2L2 . 2 The proof of Lemmas A.4 and A.5 are the same as the proof of of Lemmas A.2 and A.3, using the following coordinate transformation adapted for Lemma A.5: 0 a b c x x y 0 = 0 c −b y . b2 +c2 z z0 b c a Appendix B. Convex analysis and Variational Problems The following results are taken from [ET76]. Let X be a reflexive separable Banach space and X 0 be the dual space of X, and let K be a non-empty convex closed subset of X. Theorem B.1. Let J : K → R be a convex and lower semi-continuous functional, and assume that one of the following holds. (1) K is bounded; or (2) J(u) → +∞ as kuk → +∞, and u ∈ K. Then inf v∈K J(v) > −∞, and there exists at least one u ∈ K such that J(u) = inf J(v). v∈K
(B.1)
The set of u satisfying (B.1) is convex and closed. If J is strictly convex, then there exists one and only one u satisfying (B.1). Theorem B.2. Let A be an operator from X to X 0 satisfying the following properties: (1) A is weakly continuous over the finite dimensional subspaces of X. (2) A is bounded: kAukX 0 is bounded if kukX is bounded. (3) A is monotone: hAu − Av, u − vihX 0 ,Xi ≥ 0, ∀ u, v ∈ K. (4) Either one of the following holds: (a) K is bounded; or
Asymptotic Analysis 84 (2013), no. 3-4, 123-146
Semi-linear Hyperbolic Equation (b) A is coercive, that is: hAu, uihX 0 ,Xi → +∞, kukX
as kukX → +∞, u ∈ K.
Then for all ` ∈ X 0 , there exists at least one u ∈ K such that hAu − `, v − ui ≥ 0,
∀ v ∈ K.
(B.2)
If furthermore, A is strictly monotone, i.e. hAu − Av, u − vihX 0 ,Xi > 0,
∀ u, v ∈ K, u 6= v,
then there exists one and only one u satisfying (B.2). Acknowledgments This work was partially supported by the National Science Foundation under the grants NSF DMS-0906440 and DMS-1206438, and by the Research Fund of Indiana University. The authors would like to thank Professor Roger Temam for his suggestion of this problem and valuable advice during this research. References [BS07] [Bre73]
[Bro65] [ET76]
[Gla73] [JP07] [Kac84] [Lio69] [Lio73] [Liu03] [Mar79] [Min62] [Min63] [Mun87]
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[Oma74] Robert E. O’Malley, Jr., Introduction to singular perturbations, Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], New York-London, 1974, Applied Mathematics and Mechanics, Vol. 14. [Sat68] D.H. Sattinger, On global solution of nonlinear hyperbolic equations, Archive for Rational Mechanics and Analysis 30 (1968), 148–172 (English). [ST10] Jean-Claude Saut and Roger Temam, An initial boundary-value problem for the Zakharov-Kuznetsov equation, Adv. Differential Equations 15 (2010), no. 11-12, 1001–1031. [Tem77] Roger Temam, Navier-Stokes equations. Theory and numerical analysis, North-Holland Publishing Co., Amsterdam, 1977, Studies in Mathematics and its Applications, Vol. 2. [Tem01] , Navier-Stokes equations, AMS Chelsea Publishing, Providence, RI, 2001, Theory and numerical analysis, Reprint of the 1984 edition.
