A generalized inf-sup stable variational formulation for the wave equation
aa r X i v : . [ m a t h . NA ] J a n A generalized inf–sup stable variational formulationfor the wave equation
Olaf Steinbach , Marco Zank Institut f¨ur Angewandte Mathematik, TU Graz,Steyrergasse 30, 8010 Graz, Austria [email protected] Fakult¨at f¨ur Mathematik, Universit¨at Wien,Oskar–Morgenstern–Platz 1, 1090 Wien, Austria [email protected]
Abstract
In this paper, we consider a variational formulation for the Dirichlet problem ofthe wave equation with zero boundary and initial conditions, where we use integrationby parts in space and time. To prove unique solvability in a subspace of H ( Q ) with Q being the space–time domain, the classical assumption is to consider the right–handside f in L ( Q ). Here, we analyze a generalized setting of this variational formulation,which allows us to prove unique solvability also for f being in the dual space of thetest space, i.e., the solution operator is an isomorphism between the ansatz spaceand the dual of the test space. This new approach is based on a suitable extensionof the ansatz space to include the information of the differential operator of thewave equation at the initial time t = 0. These results are of utmost importance forthe formulation and numerical analysis of unconditionally stable space–time finiteelement methods, and for the numerical analysis of boundary element methods toovercome the well–known norm gap in the analysis of boundary integral operators. For the analysis of hyperbolic partial differential equations, a variety of approaches suchas Fourier methods, semigroups, or Galerkin methods are available, see, e.g., [18, 19, 20,27, 30]. The theoretical results on existence and uniqueness of solutions also form thebasis for the numerical analysis of related discretization schemes such as finite elementmethods, e.g., [5, 6, 7, 8, 10, 11, 14, 16, 21, 25, 26, 29, 31] or boundary element methods,e.g., [1, 4, 12, 13, 22]. 1s for elliptic second–order partial differential equations, we consider the weak solutionof the inhomogeneous wave equation in the energy space H ( Q ) with respect to the space–time domain Q := Ω × (0 , T ). However, to ensure existence and uniqueness of a weaksolution, we need to assume f ∈ L ( Q ), see, e.g., [26, Theorem 5.1]. While this is astandard assumption to conclude sufficient regularity for the solution u , and therefore, toobtain linear convergence for piecewise linear finite element approximations u h , stabilityof common finite element discretizations require in most cases some CFL condition, whichrelates the spatial and temporal mesh sizes to each other, see, e.g., [25, 26, 29, 31].Although the variational formulation to find u in a suitable subspace of H ( Q ) is well–defined for f being in the dual of the test space, this is not sufficient to establish uniquesolvability. This is mainly due to the missing information about the differential operatorof the wave equation at t = 0 in the standard ansatz space. Hence, we are not able todefine an isomorphism between the ansatz space and the dual of the test space. But suchan isomorphism is an important ingredient in the analysis of equivalent boundary integralformulations for boundary value problems for the wave equation, and the numerical analysisof related boundary and finite element methods.In this paper, we are interested in inf–sup stable variational formulations for the Dirich-let boundary value problem for the wave equation, (cid:3) u ( x, t ) = f ( x, t ) for ( x, t ) ∈ Q := Ω × (0 , T ) ,u ( x, t ) = 0 for ( x, t ) ∈ Σ := Γ × (0 , T ) ,u ( x,
0) = 0 for x ∈ Ω ,∂ t u ( x, t ) | t =0 = 0 for x ∈ Ω , (1.1)where (cid:3) u := ∂ tt u − ∆ x u is the wave operator, Ω ⊂ R d , d = 1 , ,
3, is a bounded domainwith, for d = 2 ,
3, Lipschitz boundary Γ = ∂ Ω, T > f is somegiven source. For simplicity, we assume homogeneous Dirichlet boundary conditions as wellas homogeneous initial conditions, see, e.g., [18, 29, 31] for the treatment of inhomogeneousinitial conditions. A possible variational formulation of (1.1) is to find u ∈ H , , ( Q ) := L (0 , T ; H (Ω)) ∩ H , (0 , T ; L (Ω)) such that a ( u, v ) = Z T Z Ω f ( x, t ) v ( x, t ) d x d t (1.2)is satisfied for all v ∈ H , , ( Q ) := L (0 , T ; H (Ω)) ∩ H , (0 , T ; L (Ω)) with the bilinear form a ( · , · ) : H , , ( Q ) × H , , ( Q ) → R , a ( u, v ) := − Z T Z Ω ∂ t u ( x, t ) ∂ t v ( x, t ) d x d t + Z T Z Ω ∇ x u ( x, t ) · ∇ x v ( x, t ) d x d t (1.3)for u ∈ H , , ( Q ) , v ∈ H , , ( Q ). In addition to the standard Bochner space L (0 , T ; H (Ω)),we use the space H , (0 , T ; L (Ω)) of all v ∈ L ( Q ) with ∂ t v ∈ L ( Q ), and v ( x,
0) = 0 for2 ∈ Ω. Moreover, H , (0 , T ; L (Ω)) is defined analogously with v ( x, T ) = 0 for x ∈ Ω. Thespaces H , , ( Q ), H , , ( Q ) are Hilbert spaces with the inner products h w, z i H , , ( Q ) := h w, z i H , , ( Q ) := h ∂ t w, ∂ t z i L ( Q ) + h∇ x w, ∇ x z i L ( Q ) , and the corresponding induced norms k · k H , , ( Q ) , k · k H , , ( Q ) . The bilinear form a ( · , · ) in(1.3) is continuous, i.e., for all u ∈ H , , ( Q ) and v ∈ H , , ( Q ) we have | a ( u, v ) | ≤ k ∂ t u k L ( Q ) k ∂ t v k L ( Q ) + k∇ x u k L ( Q ) k∇ x v k L ( Q ) ≤ q k ∂ t u k L ( Q ) + k∇ x u k L ( Q ) q k ∂ t v k L ( Q ) + k∇ x v k L ( Q ) = k u k H , , ( Q ) k v k H , , ( Q ) . Note that the first initial condition u ( · ,
0) = 0 in Ω is incorporated in the ansatz space H , , ( Q ), whereas the second initial condition ∂ t u ( · , t ) | t =0 = 0 in Ω is considered as a naturalcondition in the variational formulation. Thus, an inhomogeneous condition ∂ t u ( · , t ) | t =0 = v in Ω with given v could be realized with the right–hand side f v ∈ [ H , , ( Q )] ′ , h f v , v i Q = h v , v ( · , i Ω , v ∈ H , , ( Q ) . (1.4)However, known existence results for the variational formulation (1.2) do not allow right–hand sides in [ H , , ( Q )] ′ , which is the aim of the new approach as given in Section 3. So,when assuming f ∈ L ( Q ), we are able to construct a unique solution u ∈ H , , ( Q ) of thevariational formulation (1.2), satisfying the stability estimate [26, Theorem 5.1], see also[18, 25, 28], k u k H , , ( Q ) = q k ∂ t u k L ( Q ) + k∇ x u k L ( Q ) ≤ √ T k f k L ( Q ) . (1.5)The stability estimate (1.5) does not fit to the situation of the Banach–Neˇcas–Babuˇskatheorem as stated, e.g., in [9, Theorem 2.6], see also [2, 3, 17]. The next theorem statesthat it is not possible to prove the corresponding inf–sup condition, i.e., (1.6), for thebilinear form (1.3). Theorem 1.1 [28, Theorem 4.2.24]
There does not exist a constant c > such that eachright–hand side f ∈ L ( Q ) and the corresponding solution u ∈ H , , ( Q ) of (1.2) satisfy k u k H , , ( Q ) ≤ c k f k [ H , , ( Q )] ′ . In particular, the inf–sup condition c S k u k H , , ( Q ) ≤ sup = v ∈ H , , ( Q ) | a ( u, v ) |k v k H , , ( Q ) for all u ∈ H , , ( Q ) (1.6) with a constant c S > does not hold true. − ∆ x , and theanalysis of the related ordinary differential equation (1.7), which allows us also to presentthe essential ingredients for the new approach. So, for µ >
0, we consider the scalarordinary differential equation (cid:3) µ u ( t ) := ∂ tt u ( t ) + µu ( t ) = f ( t ) for t ∈ (0 , T ) , u (0) = ∂ t u ( t ) | t =0 = 0 . (1.7)The related variational formulation is to find u ∈ H , (0 , T ) for a given right–hand side f ∈ [ H , (0 , T )] ′ such that a µ ( u, v ) = h f, v i (0 ,T ) (1.8)is satisfied for all v ∈ H , (0 , T ). The bilinear form a µ ( · , · ) : H , (0 , T ) × H , (0 , T ) → R isdefined by a µ ( u, v ) := − Z T ∂ t u ( t ) ∂ t v ( t ) d t + µ Z T u ( t ) v ( t ) d t (1.9)for u ∈ H , (0 , T ), v ∈ H , (0 , T ). As before, u ∈ H , (0 , T ) covers the initial condition u (0) = 0, while v ∈ H , (0 , T ) satisfies the terminal condition v ( T ) = 0, and where theinner product h ∂ t ( · ) , ∂ t ( · ) i L (0 ,T ) makes both to Hilbert spaces. Note that the second ini-tial condition ∂ t u ( t ) | t =0 = 0 enters the variational formulation (1.8) as natural condition.The dual space [ H , (0 , T )] ′ is characterized as completion of L (0 , T ) with respect to theHilbertian norm k f k [ H , (0 ,T )] ′ = sup = v ∈ H , (0 ,T ) |h f, v i (0 ,T ) |k ∂ t v k L (0 ,T ) , where h· , ·i (0 ,T ) denotes the duality pairing as extension of the inner product in L (0 , T ) , see, e.g., [27, Satz 17.3]. The continuity of the bilinear form (1.9) follows from | a µ ( u, v ) | ≤ (cid:18) π µT (cid:19) k ∂ t u k L (0 ,T ) k ∂ t v k L (0 ,T ) (1.10)for all u ∈ H , (0 , T ) and v ∈ H , (0 , T ), where the Cauchy–Schwarz and the Poincar´einequalities [28, Lemma 3.4.5] are used. Furthermore, the bilinear form (1.9) satisfies theinf–sup condition [26, Lemma 4.2]22 + √ µT k ∂ t u k L (0 ,T ) ≤ sup = v ∈ H , (0 ,T ) | a µ ( u, v ) |k ∂ t v k L (0 ,T ) for all u ∈ H , (0 , T ) . (1.11)Together with the positivity condition [28, Lemma 4.2.4] a µ ( u v , v ) = h v, v i L (0 ,T ) = k v k L (0 ,T ) > u v ( t ) = 1 √ µ Z t v ( s ) sin( √ µ ( t − s )) d s, = v ∈ L (0 , T ) ,
4e conclude unique solvability of (1.8) as well as the stability estimate k ∂ t u k L (0 ,T ) ≤ (cid:18) √ µT (cid:19) k f k [ H , (0 ,T )] ′ (1.12)with the help of the Banach–Neˇcas–Babuˇska theorem [9, Theorem 2.6]. As discussed in[26, Remark 4.4], the stability estimate (1.12) is sharp in √ µ and T , respectively. It turnsout, however, that the estimate (1.12) is not sufficient to prove a related stability estimatefor the solution of the wave equation (1.1), see Theorem 1.1. This is mainly due to theappearance of √ µ in the stability constant, i.e., (1.12) is not uniform in µ . Instead, whenassuming f ∈ L (0 , T ), we can prove the stability estimate [26, Lemma 4.5] k ∂ t u k L (0 ,T ) + µ k u k L (0 ,T ) ≤ T k f k L (0 ,T ) , (1.13)which allows to prove the stability estimate (1.5) for the solution of the wave equation(1.1), see [26, Theorem 5.1] and [18, 28].Since the variational formulation (1.8) is well–defined also for f ∈ [ H , (0 , T )] ′ , we areinterested to establish, instead of (1.11), an inf–sup stability condition with a constant,which is independent of µ , and which later on can be generalized to the analysis of thevariational problem (1.2).The remainder of this paper is structured as follows: In Section 2, we present themain ideas in order to solve the ordinary differential equation (1.7). For this purpose, weintroduce a suitable function space and prove a related inf–sup stability condition. Then,in Section 3, these results are generalized to analyze a variational formulation for the waveequation (1.1). The main result of this paper is given in Theorem 3.9, where we statebijectivity of the solution operator for the Dirichlet problem of the wave equation withzero boundary and initial conditions. Finally, in Section 4, we give some conclusions andcomment on ongoing work. In this section, we derive a different setting of a variational formulation for (1.7) in order toestablish an inf–sup condition with a constant independent of µ . For this purpose, we firstfollow the approach as for the heat equation, see, e.g., [26]. So, for given u ∈ H , (0 , T ), wedefine h ∂ tt u + µu, v i (0 ,T ) := a µ ( u, v ) for all v ∈ H , (0 , T ) . Since the bilinear form a µ ( · , · ) is continuous, see (1.10), ∂ tt u + µu is a continuous functionalin [ H , (0 , T )] ′ . Therefore, by the Riesz representation theorem there exists a unique element w u ∈ H , (0 , T ), satisfying h ∂ tt u + µu, v i (0 ,T ) = h ∂ t w u , ∂ t v i L (0 ,T ) for all v ∈ H , (0 , T ) , k ∂ tt u + µu k H , (0 ,T )] ′ = a µ ( u, w u ) = k ∂ t w u k L (0 ,T ) . (2.1)At first glance, (2.1) implies the inf–sup condition k ∂ tt u + µu k [ H , (0 ,T )] ′ = sup = v ∈ H , (0 ,T ) | a µ ( u, v ) |k ∂ t v k L (0 ,T ) for u ∈ H , (0 , T ) , but u
7→ k ∂ tt u + µu k [ H , (0 ,T )] ′ only defines a semi–norm in H , (0 , T ) since the differentialoperator ∂ tt + µ is treated only in (0 , T ), i.e., its behavior in t = 0 is not covered in (2.1). Asexample, consider, e.g., u ∈ H , (0 , T ) with u ( t ) = sin( √ µt ) and k ∂ tt u + µu k [ H , (0 ,T )] ′ = 0.Hence, we need to modify the ansatz space to determine u in a suitable way. For thispurpose, we first introduce notations for additional function spaces and operators.In this work, C ∞ ( A ) is the set of infinitely differentiable real–valued functions withcompact support in any domain A ⊂ R d , d = 1 , , , . The set C ∞ ( A ) is endowed withthe, usual for distributions, locally convex topology and is called the space of test functionson A. The set of (Schwartz) distributions [ C ∞ ( A )] ′ is given by all linear and sequentiallycontinuous functionals on C ∞ ( A ).For given u ∈ L (0 , T ), we define the extension e u ∈ L ( − T, T ) by e u ( t ) := ( u ( t ) for t ∈ (0 , T ) , t ∈ ( − T, . (2.2)The application of the differential operator (cid:3) µ to e u is defined as distribution on ( − T, T ),i.e., for all test functions ϕ ∈ C ∞ ( − T, T ), we define h (cid:3) µ e u, ϕ i ( − T,T ) := Z T − T e u ( t ) (cid:3) µ ϕ ( t ) d t = Z T u ( t ) (cid:3) µ ϕ ( t ) d t. (2.3)This motivates to consider the dual space [ H ( − T, T )] ′ of H ( − T, T ), which is characterizedas completion of L ( − T, T ) with respect to the Hilbertian norm k g k [ H ( − T,T )] ′ := sup = z ∈ H ( − T,T ) |h g, z i ( − T,T ) |k ∂ t z k L ( − T,T ) , where h· , ·i ( − T,T ) denotes the duality pairing as extension of the inner product in L ( − T, T ),see, e.g., [27, Satz 17.3]. In other words, for [ H ( − T, T )] ′ , there exists an inner product h· , ·i [ H ( − T,T )] ′ , inducing the norm k · k [ H ( − T,T )] ′ = q h· , ·i [ H ( − T,T )] ′ , i.e., with this abstractinner product, [ H ( − T, T )] ′ is a Hilbert space. Additionally, we define the subspace H − ,T ] ( − T, T ) := n g ∈ [ H ( − T, T )] ′ : ∀ z ∈ H ( − T, T ) with supp z ⊂ ( − T,
0) : h g, z i ( − T,T ) = 0 o ⊂ [ H ( − T, T )] ′ , k·k [ H ( − T,T )] ′ . To characterize the subspace H − ,T ] ( − T, T ),we introduce the following notations. Let R : H ( − T, T ) → H , (0 , T ) be the continuousand surjective restriction operator, defined by R z = z | (0 ,T ) for z ∈ H ( − T, T ), with itsadjoint operator R ′ : [ H , (0 , T )] ′ → [ H ( − T, T )] ′ . Furthermore, let E : H , (0 , T ) → H ( − T, T ) be any continuous and injective extension operator with its adjoint operator E ′ : [ H ( − T, T )] ′ → [ H , (0 , T )] ′ , satisfying kE v k H ( − T,T ) ≤ c E k v k H , (0 ,T ) with a constant c E > RE v = v for all v ∈ H , (0 , T ) . An example for such anextension operator is given by reflection in t = 0, i.e., consider the function v , defined for v ∈ H , (0 , T ) by v ( t ) = ( v ( t ) for t ∈ [0 , T ) ,v ( − t ) for t ∈ ( − T, , which leads to the constant c E = 2 in this particular case. With this, we prove the followinglemma. Lemma 2.1
The spaces ( H − ,T ] ( − T, T ) , k · k [ H ( − T,T )] ′ ) and ([ H , (0 , T )] ′ , k · k [ H , (0 ,T )] ′ ) areisometric, i.e., the mapping E ′| H − ,T ] ( − T,T ) : H − ,T ] ( − T, T ) → [ H , (0 , T )] ′ is bijective with k g k [ H ( − T,T )] ′ = kE ′ g k [ H , (0 ,T )] ′ for all g ∈ H − ,T ] ( − T, T ) . In addition, for g ∈ H − ,T ] ( − T, T ) , the relation h g, z i ( − T,T ) = hE ′ g, R z i (0 ,T ) for all z ∈ H ( − T, T ) (2.4) holds true, i.e., R ′ E ′ g = g . In particular, the subspace H − ,T ] ( − T, T ) ⊂ [ H ( − T, T )] ′ isclosed, i.e., complete. Proof.
First, we prove that k g k [ H ( − T,T )] ′ = kE ′ g k [ H , (0 ,T )] ′ for all g ∈ H − ,T ] ( − T, T ).For this purpose, let g ∈ H − ,T ] ( − T, T ) be arbitrary but fixed. The Riesz representationtheorem gives the unique element z g ∈ H ( − T, T ) with h g, z i ( − T,T ) = h ∂ t z g , ∂ t z i L ( − T,T ) for all z ∈ H ( − T, T ) , and k g k [ H ( − T,T )] ′ = k ∂ t z g k L ( − T,T ) . It holds true that z g | ( − T, = 0, since we have0 = h g, z i ( − T,T ) = h ∂ t z g , ∂ t z i L ( − T,T ) = h ∂ t z g | ( − T, , ∂ t z | ( − T, i L ( − T, z ∈ H ( − T, T ) with supp z ⊂ ( − T, h g, z i ( − T,T ) = h ∂ t z g , ∂ t z i L ( − T,T ) = h ∂ t R z g , ∂ t R z i L (0 ,T ) (2.5)for all z ∈ H ( − T, T ). So, using (2.5) with z = E v for v ∈ H , (0 , T ) this gives hE ′ g, v i (0 ,T ) = h g, E v i ( − T,T ) = h ∂ t R z g , ∂ t RE v i L (0 ,T ) = h ∂ t R z g , ∂ t v i L (0 ,T ) , (2.6)i.e., kE ′ g k [ H , (0 ,T )] ′ = k ∂ t R z g k L (0 ,T ) = k ∂ t z g k L ( − T,T ) = k g k [ H ( − T,T )] ′ . Second, we prove that E ′| H − ,T ] ( − T,T ) is surjective. For this purpose, let f ∈ [ H , (0 , T )] ′ begiven. Set g f = R ′ f ∈ [ H ( − T, T )] ′ , i.e., h g f , z i ( − T,T ) = hR ′ f, z i ( − T,T ) = h f, R z i (0 ,T ) for all z ∈ H ( − T, T ). With this it follows immediately that g f ∈ H − ,T ] ( − T, T ). Moreover,we have hE ′ g f , v i (0 ,T ) = h g f , E v i ( − T,T ) = h f, RE v i (0 ,T ) = h f, v i (0 ,T ) for all v ∈ H , (0 , T ), i.e., E ′ g f = f in [ H , (0 , T )] ′ . In other words, E ′| H − ,T ] ( − T,T ) is surjective.Third, the equality (2.4) follows from (2.5) and (2.6) for v = R z for any z ∈ H ( − T, T ).The last assertion of the lemma is straightforward.The last lemma gives immediately the following corollary.
Corollary 2.2
For all g ∈ H − ,T ] ( − T, T ) , the norm representation k g k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) |h g, E v i ( − T,T ) |k ∂ t v k L (0 ,T ) holds true. Proof.
Let g ∈ H − ,T ] ( − T, T ) be arbitrary but fixed. With Lemma 2.1, we have k g k [ H ( − T,T )] ′ = kE ′ g k [ H , (0 ,T )] ′ = sup = v ∈ H , (0 ,T ) |hE ′ g, v i (0 ,T ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) |h g, E v i ( − T,T ) |k ∂ t v k L (0 ,T ) , i.e., the assertion is proven.Next, we introduce H (0 , T ) := n u = e u | (0 ,T ) : e u ∈ L ( − T, T ) , e u | ( − T, = 0 , (cid:3) µ e u ∈ [ H ( − T, T )] ′ o with the norm k u k H (0 ,T ) := q k u k L (0 ,T ) + k (cid:3) µ e u k H ( − T,T )] ′ . u ∈ H (0 , T ), the condition (cid:3) µ e u ∈ [ H ( − T, T )] ′ involves that there existsan element f u ∈ [ H ( − T, T )] ′ with h (cid:3) µ e u, ϕ i ( − T,T ) = h f u , ϕ i ( − T,T ) for all ϕ ∈ C ∞ ( − T, T ) . Note that ϕ ∈ H ( − T, T ) for ϕ ∈ C ∞ ( − T, T ), and that C ∞ ( − T, T ) is dense in H ( − T, T ).Hence, the element f u ∈ [ H ( − T, T )] ′ is unique and therefore, in the following, we identifythe distribution (cid:3) µ e u : C ∞ ( − T, T ) → R with the functional f u : H ( − T, T ) → R .Next, we state properties of the space H (0 , T ). Clearly, ( H (0 , T ) , k · k H (0 ,T ) ) is a normedvector space, and it is even a Banach space. Lemma 2.3
The normed vector space ( H (0 , T ) , k · k H (0 ,T ) ) is a Banach space. Proof.
Consider a Cauchy sequence ( u n ) n ∈ N ⊂ H (0 , T ). Hence, ( u n ) n ∈ N ⊂ L (0 , T ) is alsoa Cauchy sequence in L (0 , T ), and ( (cid:3) µ e u n ) n ∈ N ⊂ [ H ( − T, T )] ′ is also a Cauchy sequencein [ H ( − T, T )] ′ . So, there exist u ∈ L (0 , T ) and f ∈ [ H ( − T, T )] ′ withlim n →∞ k u n − u k L (0 ,T ) = 0 , lim n →∞ k (cid:3) µ e u n − f k [ H ( − T,T )] ′ = 0 . For ϕ ∈ C ∞ ( − T, T ), we have h (cid:3) µ e u, ϕ i ( − T,T ) = h e u, (cid:3) µ ϕ i L ( − T,T ) = Z T u ( t ) (cid:3) µ ϕ ( t ) d t = lim n →∞ Z T u n ( t ) (cid:3) µ ϕ ( t ) d t = lim n →∞ h e u n , (cid:3) µ ϕ i L ( − T,T ) = lim n →∞ h (cid:3) µ e u n , ϕ i ( − T,T ) = h f, ϕ i ( − T,T ) , i.e., (cid:3) µ e u = f ∈ [ H ( − T, T )] ′ . Hence, u ∈ H (0 , T ) follows.With the abstract inner product h· , ·i [ H ( − T,T )] ′ of [ H ( − T, T )] ′ , the inner product h u, v i H (0 ,T ) := h u, v i L (0 ,T ) + h (cid:3) µ e u, (cid:3) µ e v i [ H ( − T,T )] ′ , u, v ∈ H (0 , T ) , induces the norm k · k H (0 ,T ) . Hence, the space ( H (0 , T ) , h· , ·i H (0 ,T ) ) is even a Hilbert space,but this abstract inner product is not used explicitly in the remainder of this work. Lemma 2.4
For all u ∈ H (0 , T ) there holds (cid:3) µ e u ∈ H − ,T ] ( − T, T ) and k (cid:3) µ e u k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) |h (cid:3) µ e u, E v i ( − T,T ) |k ∂ t v k L (0 ,T ) . (2.7) Proof.
First, we prove that (cid:3) µ e u ∈ H − ,T ] ( − T, T ). For this purpose, let u ∈ H (0 , T ) and z ∈ H ( − T, T ) with supp z ⊂ ( − T,
0) be arbitrary but fixed. Due to z | ( − T, ∈ H ( − T, ψ n ) n ∈ N ⊂ C ∞ ( − T,
0) with k ∂ t z | ( − T, − ∂ t ψ n k L ( − T, → n → ∞ .For n ∈ N , define ϕ n ( t ) = ( ψ n ( t ) for t ∈ ( − T, , t ∈ [0 , T ) , ϕ n ) n ∈ N ⊂ C ∞ ( − T, T ) satisfies k ∂ t z − ∂ t ϕ n k L ( − T,T ) = k ∂ t z | ( − T, − ∂ t ψ n k L ( − T, → n → ∞ . So, it follows that h (cid:3) µ e u, z i ( − T,T ) = lim n →∞ h (cid:3) µ e u, ϕ n i ( − T,T ) = lim n →∞ Z T u ( t ) (cid:3) µ ϕ n ( t ) d t = 0 , and therefore, the assertion. The norm representation follows from (cid:3) µ e u ∈ H − ,T ] ( − T, T )and Corollary 2.2.
Lemma 2.5
It holds true that H , (0 , T ) ⊂ H (0 , T ) . Furthermore, each u ∈ H , (0 , T ) withzero extension e u , as defined in (2.2) , satisfies k (cid:3) µ e u k [ H ( − T,T )] ′ ≤ (cid:18) π µT (cid:19) k ∂ t u k L (0 ,T ) , (2.8) and h (cid:3) µ e u, z i ( − T,T ) = a µ ( u, R z ) = −h ∂ t u, ∂ t R z i L (0 ,T ) + µ h u, R z i L (0 ,T ) (2.9) for all z ∈ H ( − T, T ) , where a µ ( · , · ) is the bilinear form (1.9) . Proof.
First, we prove that H , (0 , T ) ⊂ H (0 , T ). For u ∈ H , (0 , T ), we define the extension e u , see (2.2). By construction, we have e u ∈ L ( − T, T ), and e u | ( − T, = 0. It remains to provethat (cid:3) µ e u ∈ [ H ( − T, T )] ′ . For this purpose, define the functional f u ∈ [ H ( − T, T )] ′ by h f u , z i ( − T,T ) := a µ ( u, R z )for all z ∈ H ( − T, T ), where a µ ( · , · ) is the bilinear form (1.9). The continuity of f u followsfrom |h f u , z i ( − T,T ) | = | a µ ( u, R z ) | ≤ (cid:18) π µT (cid:19) k ∂ t u k L (0 ,T ) k ∂ t R z k L (0 ,T ) ≤ (cid:18) π µT (cid:19) k ∂ t u k L (0 ,T ) k ∂ t z k L ( − T,T ) for all z ∈ H ( − T, T ), where the estimate (1.10) is used. Using the definition (2.3), andintegration by parts, this gives h (cid:3) µ e u, ϕ i ( − T,T ) = Z T u ( t ) (cid:3) µ ϕ ( t ) d t = −h ∂ t u, ∂ t R ϕ i L (0 ,T ) + µ h u, R ϕ i L (0 ,T ) = h f u , ϕ i ( − T,T ) ϕ ∈ C ∞ ( − T, T ), i.e., (cid:3) µ e u = f u ∈ [ H ( − T, T )] ′ . The equality (2.9) follows from thedensity of C ∞ ( − T, T ) in H ( − T, T ). The estimate (2.8) is proven by k (cid:3) µ e u k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) |h (cid:3) µ e u, E v i ( − T,T ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) |h f u , E v i ( − T,T ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) | a µ ( u, RE v ) |k ∂ t v k L (0 ,T ) ≤ (cid:18) π µT (cid:19) k ∂ t u k L (0 ,T ) , when using the norm representation (2.7), the equality (2.9), and the bound (1.10).Next, by completion, we define the Hilbert space H , (0 , T ) := H , (0 , T ) k·k H (0 ,T ) ⊂ H (0 , T ) , endowed with the Hilbertian norm k·k H (0 ,T ) , i.e., H , (0 , T ) = n v ∈ H (0 , T ) : ∃ ( v n ) n ∈ N ⊂ H , (0 , T ) with lim n →∞ k v n − v k H (0 ,T ) = 0 o . Lemma 2.6
For u ∈ H , (0 , T ) there holds k (cid:3) µ e u k [ H ( − T,T )] ′ ≥ √ T k u k L (0 ,T ) . Proof.
For 0 = u ∈ H , (0 , T ), there exists a non–trivial sequence ( u n ) n ∈ N ⊂ H , (0 , T ), u n
0, with lim n →∞ k u − u n k H (0 ,T ) = 0 . For each u n ∈ H , (0 , T ), we define w n ∈ H , (0 , T ) as unique solution of the backwardproblem ∂ tt w n ( t ) + µw n ( t ) = u n ( t ) for t ∈ (0 , T ) , w n ( T ) = ∂ t w n ( t ) | t = T = 0 , (2.10)i.e., w n ∈ H , (0 , T ) solves the variational problem a µ ( v, w n ) = h u n , v i L (0 ,T ) for all v ∈ H , (0 , T )with the bilinear form (1.9). In particular for v = u n , this gives a µ ( u n , w n ) = k u n k L (0 ,T ) . Analogously to the estimate (1.13) for the solution of (1.7), we conclude k ∂ t w n k L (0 ,T ) + µ k w n k L (0 ,T ) ≤ T k u n k L (0 ,T ) w n of (2.10), i.e., k ∂ t w n k L (0 ,T ) ≤ √ T k u n k L (0 ,T ) . For the zero extension e u n ∈ L ( − T, T ) of u n ∈ H , (0 , T ), we obtain, when using the normrepresentation (2.7), and (2.9), that k (cid:3) µ e u n k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) |h (cid:3) µ e u n , E v i ( − T,T ) |k ∂ t v k L (0 ,T ) ≥ |h (cid:3) µ e u n , E w n i ( − T,T ) |k ∂ t w n k L (0 ,T ) = | a µ ( u n , w n ) |k ∂ t w n k L (0 ,T ) = k u n k L (0 ,T ) k ∂ t w n k L (0 ,T ) ≥ √ T k u n k L (0 ,T ) , and the assertion follows by completion for n → ∞ . Corollary 2.7
The inner product space (cid:16) H , (0 , T ) , h (cid:3) µ f ( · ) , (cid:3) µ f ( · ) i [ H ( − T,T )] ′ (cid:17) is complete,i.e., a Hilbert space. Proof.
The assertion follows immediately from Lemma 2.6.In the following, H , (0 , T ) is endowed with the Hilbertian norm k (cid:3) µ f ( · ) k [ H ( − T,T )] ′ . Withthis new Hilbert space, the bilinear form e a µ ( · , · ) : H , (0 , T ) × H , (0 , T ) → R , e a µ ( u, v ) := h (cid:3) µ e u, E v i ( − T,T ) , is continuous, i.e., | e a µ ( u, v ) | = |h (cid:3) µ e u, E v i ( − T,T ) | ≤ k (cid:3) µ e u k [ H ( − T,T )] ′ k ∂ t v k L (0 ,T ) (2.11)for all u ∈ H , (0 , T ) and v ∈ H , (0 , T ), and fulfills the inf–sup condition k (cid:3) µ e u k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) |h (cid:3) µ e u, E v i ( − T,T ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) | e a µ ( u, v ) |k ∂ t v k L (0 ,T ) (2.12)for all u ∈ H , (0 , T ), where the norm representation (2.7) is used. In addition, Lemma 2.5yields the representation e a µ ( u, v ) = a µ ( u, v ) (2.13)for all u ∈ H , (0 , T ) ⊂ H , (0 , T ), v ∈ H , (0 , T ), which is used in the following lemma. Lemma 2.8
For all = v ∈ H , (0 , T ) , there exists a function u v ∈ H , (0 , T ) such that e a µ ( u v , v ) > . roof. For 0 = v ∈ H , (0 , T ), there exists the unique solution u v ∈ H , (0 , T ) satisfying a µ ( u v , w ) = h v, w i L (0 ,T ) for all w ∈ H , (0 , T ) . Using the representation (2.13), this gives e a µ ( u v , w ) = h v, w i L (0 ,T ) for all w ∈ H , (0 , T ) , and in particular for w = v , we obtain e a µ ( u v , v ) = k v k L (0 ,T ) > , i.e., the assertion.Next, we state the new variational setting for the scalar ordinary differential equation (1.7).For given f ∈ [ H , (0 , T )] ′ , we consider the variational formulation to find u ∈ H , (0 , T )such that e a µ ( u, v ) = h f, v i (0 ,T ) for all v ∈ H , (0 , T ) , (2.14)i.e., the operator equation E ′ (cid:3) µ e u = f in [ H , (0 , T )] ′ . With the properties of the bilinear form e a µ ( · , · ), the unique solvability of the variationalformulation (2.14), i.e., the main theorem of this section, is proven. Theorem 2.9
For each given f ∈ [ H , (0 , T )] ′ , there exists a unique solution u ∈ H , (0 , T ) of the variational formulation (2.14) . Furthermore, L µ : [ H , (0 , T )] ′ → H , (0 , T ) , L µ f := u, is an isomorphism satisfying k (cid:3) µ e u k [ H ( − T,T )] ′ = k (cid:3) µ g L µ f k [ H ( − T,T )] ′ = k f k [ H , (0 ,T )] ′ . Proof.
With the help of the Banach–Neˇcas–Babuˇska theorem [9, Theorem 2.6], the resultsin (2.11), (2.12) and Lemma 2.8 yield the existence and uniqueness of a solution u ∈H , (0 , T ). In addition, with the variational formulation (2.14), the equalities k (cid:3) µ e u k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) | e a µ ( u, v ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) |h f, v i (0 ,T ) |k ∂ t v k L (0 ,T ) = k f k [ H , (0 ,T )] ′ hold true, and therefore, the assertion.While the unique solution u of the variational formulation (2.14) is considered in H , (0 , T ),it turns out that indeed u ∈ H , (0 , T ). In fact, the following lemma clarifies the relationbetween H , (0 , T ) and H , (0 , T ). 13 emma 2.10 There holds H , (0 , T ) = H , (0 , T ) with the norm equivalence inequalities (cid:18) π µT (cid:19) − k (cid:3) µ e u k [ H ( − T,T )] ′ ≤ k ∂ t u k L (0 ,T ) ≤ (cid:18) √ µT (cid:19) k (cid:3) µ e u k [ H ( − T,T )] ′ for all u ∈ H , (0 , T ) . Proof.
We first prove that H , (0 , T ) = H , (0 , T ). As H , (0 , T ) ⊃ H , (0 , T ), see Lemma 2.5,it remains to prove that H , (0 , T ) ⊂ H , (0 , T ). For this purpose, let u ∈ H , (0 , T ) be fixed.Consider the unique solution ˆ u ∈ H , (0 , T ) of the variational formulation (1.8) for theright–hand side f = L − µ u ∈ [ H , (0 , T )] ′ , where L µ is the solution operator of Theorem 2.9.So, using Lemma 2.5 and the variational formulations (1.8), (2.14) this yields e a µ (ˆ u, v ) = a µ (ˆ u, v ) = h f, v i (0 ,T ) = e a µ ( u, v )for all v ∈ H , (0 , T ), i.e., u = ˆ u ∈ H , (0 , T ). Thus, we have H , (0 , T ) ⊂ H , (0 , T ). Theupper norm equivalence inequality is proven by k (cid:3) µ e u k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) | e a µ ( u, v ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) | a µ ( u, v ) |k ∂ t v k L (0 ,T ) ≥
22 + √ µT k ∂ t u k L (0 ,T ) for all u ∈ H , (0 , T ) = H , (0 , T ), where the inf–sup conditions (2.12), (1.11) are used. Thelower inequality follows from (2.8). Corollary 2.11
For all u ∈ H , (0 , T ) and all v ∈ H , (0 , T ) , the equality e a µ ( u, v ) = a µ ( u, v ) is valid, i.e., the variational formulations (1.8) and (2.14) are equivalent. Proof.
The assertion follows immediately from Lemma 2.10 and (2.13).
Remark 2.12
Functions u ∈ C ([0 , T ]) with u (0) = 0 are contained in H , (0 , T ) , sincesuch functions are in H , (0 , T ) . Note that the second initial condition ∂ t u ( t ) | t =0 = 0 is notincorporated in the ansatz space H , (0 , T ) . Remark 2.13
The function u , defined by u ( t ) = sin( √ µt ) for t ∈ (0 , T ) , is obviously in H , (0 , T ) and so, in H , (0 , T ) . For this function, we have k ∂ tt u + µu k [ H , (0 ,T )] ′ = 0 . or e u ( t ) = ( for t ∈ ( − T, , sin( √ µt ) for t ∈ (0 , T ) , the first–order distributional derivative is identified with the function ∂ t e u ( t ) = ( for t ∈ ( − T, , √ µ cos( √ µt ) for t ∈ (0 , T ) , i.e., it is a regular distribution. To compute the second–order distributional derivative of e u , we consider h ∂ tt e u, ϕ i ( − T,T ) = − Z T − T ∂ t e u ( t ) ∂ t ϕ ( t ) d t = − Z T √ µ cos( √ µt ) ∂ t ϕ ( t ) d t = −√ µ cos( √ µt ) ϕ ( t ) (cid:12)(cid:12)(cid:12) T + Z T (cid:16) − µ sin( √ µt ) (cid:17) ϕ ( t ) d t = √ µ ϕ (0) − µ h e u, ϕ i ( − T,T ) for all ϕ ∈ C ∞ ( − T, T ) . Hence, (cid:3) µ e u = ∂ tt e u + µ e u = √ µ δ is a singular distribution with the Dirac distribution δ ∈ H − ,T ] ( − T, T ) ⊂ [ H ( − T, T )] ′ .Furthermore, it follows that k (cid:3) µ e u k [ H ( − T,T )] ′ = sup = v ∈ H , (0 ,T ) |h (cid:3) µ e u, E v i ( − T,T ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) √ µ |h δ , E v i ( − T,T ) |k ∂ t v k L (0 ,T ) = sup = v ∈ H , (0 ,T ) √ µ | v (0) |k ∂ t v k L (0 ,T ) > for, e.g., v ( t ) = T ( T − t ) , where the norm representation (2.7) is used. To summarize, thefunction u ( t ) = sin( √ µt ) for t ∈ (0 , T ) with k (cid:3) µ e u k [ H ( − T,T )] ′ > and k ∂ tt u + µu k [ H , (0 ,T )] ′ = 0 solves the variational formulations (1.8) and (2.14) for the right–hand side f v ∈ [ H , (0 , T )] ′ , h f v , v i (0 ,T ) = √ µv (0) , v ∈ H , (0 , T ) , which realizes the initial condition ∂ t u ( t ) | t =0 = v := √ µ . emark 2.14 The variational formulation (2.14) is the weak formulation of the differen-tial equation ∂ tt e u ( t ) + µ e u ( t ) = e f ( t ) = ( for t ∈ ( − T, ,f ( t ) for t ∈ [0 , T ) , which can be written as coupled system, using u ( t ) = e u ( t ) for t ∈ (0 , T ) , and u − ( t ) = e u ( t ) for t ∈ ( − T, , ∂ tt u ( t ) + µu ( t ) = f ( t ) for t ∈ (0 , T ) ,∂ tt u − ( t ) + µu − ( t ) = 0 for t ∈ ( − T, , u − ( − T ) = ∂ t u − ( t ) | t = − T = 0 together with the transmission interface conditions u (0) = u − (0) , ∂ t u ( t ) | t =0 − ∂ t u − ( t ) | t =0 = v with given v ∈ R , satisfying h e f , z i ( − T,T ) = v z (0) for all z ∈ H ( − T, T ) . The conditions u − ( − T ) = ∂ t u − ( t ) | t = − T = 0 lead to u − ( t ) = e u ( t ) = 0 for t ∈ ( − T, , which finally impliesthe initial conditions u (0) = 0 , ∂ t u ( t ) | t =0 = v . In this section, we generalize the approach, as introduced for the solution of the ordinarydifferential equation (1.7), to end up with a generalized inf–sup stable variational formula-tion for the Dirichlet boundary value problem for the wave equation (1.1). For this purpose,we first introduce notations analogously to them of Section 2.In addition to the space–time domain Q := Ω × (0 , T ) we consider the extended domain Q − := Ω × ( − T, T ). The dual space [ H , , ( Q )] ′ is characterized as completion of L ( Q )with respect to the Hilbertian norm k f k [ H , , ( Q )] ′ = sup = v ∈ H , , ( Q ) |h f, v i Q |k v k H , , ( Q ) , where h· , ·i Q denotes the duality pairing as extension of the inner product in L ( Q ). Notethat [ H , , ( Q )] ′ is a Hilbert space, see Section 2. For given u ∈ L ( Q ), we define theextension e u ∈ L ( Q − ) by e u ( x, t ) := ( u ( x, t ) for ( x, t ) ∈ Q, x, t ) ∈ Q − \ Q. (3.1)The application of the wave operator (cid:3) := ∂ tt − ∆ x to e u is defined as a distribution on Q − , i.e., for all test functions ϕ ∈ C ∞ ( Q − ), we define h (cid:3) e u, ϕ i Q − := Z T − T Z Ω e u ( x, t ) (cid:3) ϕ ( x, t ) d x d t = Z T Z Ω u ( x, t ) (cid:3) ϕ ( x, t ) d x d t. (3.2)16his motivates to consider the dual space [ H ( Q − )] ′ of H ( Q − ), which is characterized ascompletion of L ( Q − ) with respect to the Hilbertian norm k g k [ H ( Q − )] ′ := sup = z ∈ H ( Q − ) |h g, z i Q − |k z k H ( Q − ) , where the inner product h z , z i H ( Q − ) = h ∂ t z , ∂ t z i L ( Q − ) + h∇ x z , ∇ x z i L ( Q − ) , z , z ∈ H ( Q − ) , induces the norm k · k H ( Q − ) , and h· , ·i Q − denotes the duality pairing as extension of theinner product in L ( Q − ), see [27, Satz 17.3]. Note that [ H ( Q − )] ′ is a Hilbert space, seeSection 2. In addition we define the subspace H − | Q ( Q − ) := n g ∈ [ H ( Q − )] ′ : ∀ z ∈ H ( Q − ) with supp z ⊂ Ω × ( − T,
0) : h g, z i Q − = 0 o of [ H ( Q − )] ′ , endowed with the Hilbertian norm k·k [ H ( Q − )] ′ . To characterize the subspace H − | Q ( Q − ), we introduce the following notations. Let R : H ( Q − ) → H , , ( Q ) be the con-tinuous and surjective restriction operator, defined by R z = z | Q for z ∈ H ( Q − ), with itsadjoint operator R ′ : [ H , , ( Q )] ′ → [ H ( Q − )] ′ . Furthermore, let E : H , , ( Q ) → H ( Q − ) beany continuous and injective extension operator with its adjoint operator E ′ : [ H ( Q − )] ′ → [ H , , ( Q )] ′ , satisfying kE v k H ( Q − ) ≤ c E k v k H , , ( Q ) with a constant c E >
0, and RE v = v for all v ∈ H , , ( Q ). An example for such anextension operator is given by reflection in Ω × { } , i.e., consider the function v , definedby v ( x, t ) = ( v ( x, t ) for ( x, t ) ∈ Ω × [0 , T ) ,v ( x, − t ) for ( x, t ) ∈ Ω × ( − T, x, t ) ∈ Q − , and a given function v ∈ H , , ( Q ), which leads to a constant c E = 2 in thisparticular case. With this, we prove the following lemma as the counter part of Lemma2.1. Lemma 3.1
The spaces ( H − | Q ( Q − ) , k · k [ H ( Q − )] ′ ) and ([ H , , ( Q )] ′ , k · k [ H , , ( Q )] ′ ) are isomet-ric, i.e., the mapping E ′| H − | Q ( Q − ) : H − | Q ( Q − ) → [ H , , ( Q )] ′ is bijective with k g k [ H ( Q − )] ′ = kE ′ g k [ H , , ( Q )] ′ for all g ∈ H − | Q ( Q − ) . In addition, for g ∈ H − | Q ( Q − ) , the relation h g, z i Q − = hE ′ g, R z i Q for all z ∈ H ( Q − ) (3.3) i.e., R ′ E ′ g = g , holds true. In particular, the subspace H − | Q ( Q − ) ⊂ [ H ( Q − )] ′ is closed,i.e., complete. roof. First, we prove that k g k [ H ( Q − )] ′ = kE ′ g k [ H , , ( Q )] ′ for all functionals g ∈ H − | Q ( Q − ).For this purpose, let g ∈ H − | Q ( Q − ) be arbitrary but fixed. The Riesz representation theoremgives the unique element z g ∈ H ( Q − ) with h g, z i Q − = h z g , z i H ( Q − ) for all z ∈ H ( Q − ) , and k g k [ H ( Q − )] ′ = k z g k H ( Q − ) . It holds true that z g | Ω × ( − T, = 0, since we have0 = h g, z i Q − = h z g , z i H ( Q − ) = Z − T Z Ω h ∂ t z g ( x, t ) ∂ t z ( x, t ) + ∇ x z g ( x, t ) · ∇ x z ( x, t ) i d x d t for all z ∈ H ( Q − ) with supp z ⊂ Ω × ( − T, h g, z i Q − = h z g , z i H ( Q − ) = hR z g , R z i H , , ( Q ) (3.4)for all z ∈ H ( Q − ). So, using (3.4) with z = E v for v ∈ H , , ( Q ) this gives hE ′ g, v i Q = h g, E v i Q − = hR z g , RE v i H , , ( Q ) = hR z g , v i H , , ( Q ) , (3.5)i.e., kE ′ g k [ H , , ( Q )] ′ = kR z g k H , , ( Q ) = k z g k H ( Q − ) = k g k [ H ( Q − )] ′ . Next, we prove that E ′| H − | Q ( Q − ) is surjective. For this purpose, let f ∈ [ H , , ( Q )] ′ be given.Set g f = R ′ f , i.e., h g f , z i Q − = hR ′ f, z i Q − = h f, R z i Q for all z ∈ H ( Q − ). With this it follows immediately that g f ∈ H − | Q ( Q − ). Moreover, wehave hE ′ g f , v i Q = h g f , E v i Q − = h f, RE v i Q = h f, v i Q for all v ∈ H , , ( Q ), i.e., E ′ g f = f in [ H , , ( Q )] ′ . In other words, E ′| H − | Q ( Q − ) is surjective.Finally, (3.3) follows from (3.4) and (3.5) for v = R z for any z ∈ H ( Q − ). The lastassertion of the lemma is straightforward.The last lemma gives immediately the following corollary. Corollary 3.2
For all g ∈ H − | Q ( Q − ) , the norm representation k g k [ H ( Q − )] ′ = sup = v ∈ H , , ( Q ) |h g, E v i Q − |k v k H , , ( Q ) holds true. roof. Let g ∈ H − | Q ( Q − ) be arbitrary but fixed. With Lemma 3.1, we have k g k [ H ( Q − )] ′ = kE ′ g k [ H , , ( Q )] ′ = sup = v ∈ H , , ( Q ) |hE ′ g, v i Q |k v k H , , ( Q ) = sup = v ∈ H , , ( Q ) |h g, E v i Q − |k v k H , , ( Q ) , i.e., the assertion is proven.Next, we introduce H ( Q ) := n u = e u | Q : e u ∈ L ( Q − ) , e u | Ω × ( − T, = 0 , (cid:3) e u ∈ [ H ( Q − )] ′ o , with the norm k u k H ( Q ) := q k u k L ( Q ) + k (cid:3) e u k H ( Q − )] ′ . For a function u ∈ H ( Q ), the condition (cid:3) e u ∈ [ H ( Q − )] ′ involves that there exists anelement f u ∈ [ H ( Q − )] ′ with h (cid:3) e u, ϕ i Q − = h f u , ϕ i Q − for all ϕ ∈ C ∞ ( Q − ) . Note that ϕ ∈ H ( Q − ) for ϕ ∈ C ∞ ( Q − ), and that C ∞ ( Q − ) is dense in H ( Q − ). Hence,the element f u ∈ [ H ( Q − )] ′ is unique and therefore, in the following, we identify thedistribution (cid:3) e u : C ∞ ( Q − ) → R with the functional f u : H ( Q − ) → R .Next, we state properties of the space H ( Q ). Clearly, ( H ( Q ) , k · k H ( Q ) ) is a normedvector space and it is even a Banach space. Lemma 3.3
The normed vector space ( H ( Q ) , k · k H ( Q ) ) is a Banach space. Proof.
Consider a Cauchy sequence ( u n ) n ∈ N ⊂ H ( Q ). Hence, ( u n ) n ∈ N ⊂ L ( Q ) is alsoa Cauchy sequence in L ( Q ), and ( (cid:3) e u n ) n ∈ N ⊂ [ H ( Q − )] ′ is also a Cauchy sequence in[ H ( Q − )] ′ . So, there exist u ∈ L ( Q ) and f ∈ [ H ( Q − )] ′ withlim n →∞ k u n − u k L ( Q ) = 0 , lim n →∞ k (cid:3) e u n − f k [ H ( Q − )] ′ = 0 . For ϕ ∈ C ∞ ( Q − ), we have h (cid:3) e u, ϕ i Q − = h e u, (cid:3) ϕ i L ( Q − ) = Z T Z Ω u ( x, t ) (cid:3) ϕ ( x, t ) d x d t = lim n →∞ Z T Z Ω u n ( x, t ) (cid:3) ϕ ( x, t ) d x d t = lim n →∞ h e u n , (cid:3) ϕ i L ( Q − ) = lim n →∞ h (cid:3) e u n , ϕ i Q − = h f, ϕ i Q − , i.e., (cid:3) e u = f ∈ [ H ( Q − )] ′ . Hence, u ∈ H ( Q ) follows.With the abstract inner product h· , ·i [ H ( Q − )] ′ of [ H ( Q − )] ′ , the inner product h· , ·i H ( Q ) := h· , ·i L ( Q ) + h (cid:3) f ( · ) , (cid:3) f ( · ) i [ H ( Q − )] ′ induces the norm k · k H ( Q ) . Hence, the space ( H ( Q ) , h· , ·i H ( Q ) ) is even a Hilbert space, butthis abstract inner product is not used explicitly in the remainder of this work.19 emma 3.4 For all u ∈ H ( Q ) there holds (cid:3) e u ∈ H − | Q ( Q − ) with k (cid:3) e u k [ H ( Q − )] ′ = sup = v ∈ H , , ( Q ) |h (cid:3) e u, E v i Q − |k v k H , , ( Q ) . (3.6) Proof.
First, we prove that (cid:3) e u ∈ H − | Q ( Q − ). For this purpose, let u ∈ H ( Q ) and z ∈ H ( Q − ) with supp z ⊂ Ω × ( − T,
0) be arbitrary but fixed. Due to z | Ω × ( − T, ∈ H (Ω × ( − T, ψ n ) n ∈ N ⊂ C ∞ (Ω × ( − T, k z | Ω × ( − T, − ψ n k H (Ω × ( − T, → n → ∞ , where k w k H (Ω × ( − T, = (cid:18)Z − T Z Ω h | ∂ t w ( x, t ) | + |∇ x w ( x, t ) | i d x d t (cid:19) / for w ∈ H (Ω × ( − T, n ∈ N , define ϕ n ( x, t ) = ( ψ n ( x, t ) for ( x, t ) ∈ Ω × ( − T, , x, t ) ∈ Ω × [0 , T ) , i.e., ( ϕ n ) n ∈ N ⊂ C ∞ ( Q − ) satisfies k z − ϕ n k H ( Q − ) = k z | Ω × ( − T, − ψ n k H (Ω × ( − T, → n → ∞ . So, it follows that h (cid:3) e u, z i Q − = lim n →∞ h (cid:3) e u, ϕ n i Q − = lim n →∞ Z T Z Ω u ( x, t ) (cid:3) ϕ n ( x, t ) d x d t = 0and therefore, the assertion. The norm representation follows from (cid:3) e u ∈ H − | Q ( Q − ) andCorollary 3.2. Lemma 3.5
It holds true that H , , ( Q ) ⊂ H ( Q ) . Furthermore, each u ∈ H , , ( Q ) withzero extension e u , as defined in (3.1) , satisfies k (cid:3) e u k [ H ( Q − )] ′ ≤ k u k H , , ( Q ) , (3.7) and h (cid:3) e u, z i Q − = a ( u, R z ) = −h ∂ t u, ∂ t R z i L ( Q ) + h∇ x u, ∇ x R z i L ( Q ) (3.8) for all z ∈ H ( Q − ) , where a ( · , · ) is the bilinear form (1.3) . Proof.
First, we prove that H , , ( Q ) ⊂ H ( Q ). For u ∈ H , , ( Q ), we define the extension e u , see (3.1). By construction, we have e u ∈ L ( Q − ), and e u | Ω × ( − T, = 0. It remains to provethat (cid:3) e u ∈ [ H ( Q − )] ′ . For this purpose, define the functional f u ∈ [ H ( Q − )] ′ by h f u , z i Q − = a ( u, R z )20or all z ∈ H ( Q − ), where a ( · , · ) is the bilinear form (1.3). The continuity of f u followsfrom |h f u , z i Q − | = | a ( u, R z ) | ≤ k u k H , , ( Q ) kR z k H , , ( Q ) ≤ k u k H , , ( Q ) k z k H ( Q − ) for all z ∈ H ( Q − ), where the estimate (1.4) is used. Using the definition (3.2) andintegration by parts, this gives h (cid:3) e u, ϕ i Q − = Z T Z Ω u ( x, t ) (cid:3) ϕ ( x, t ) d x d t = −h ∂ t u, ∂ t R ϕ i L ( Q ) + h∇ x u, ∇ x R ϕ i L ( Q ) = h f u , ϕ i Q − for all ϕ ∈ C ∞ ( Q − ), i.e., (cid:3) e u = f u ∈ [ H ( Q − )] ′ . The equality (3.8) follows from the densityof C ∞ ( Q − ) in H ( Q − ). The estimate (3.7) is proven by k (cid:3) e u k [ H ( Q − )] ′ = sup = v ∈ H , , ( Q ) |h (cid:3) e u, E v i Q − |k v k H , , ( Q ) = sup = v ∈ H , , ( Q ) |h f u , E v i Q − |k v k H , , ( Q ) = sup = v ∈ H , , ( Q ) | a ( u, RE v ) |k v k H , , ( Q ) ≤ k u k H , , ( Q ) when using the norm representation (3.6), the equality (3.8), and (1.4).Next, by completion, we define the Hilbert space H , ( Q ) := H , , ( Q ) k·k H ( Q ) ⊂ H ( Q ) , endowed with the Hilbertian norm k·k H ( Q ) , i.e., H , ( Q ) = n v ∈ H ( Q ) : ∃ ( v n ) n ∈ N ⊂ H , , ( Q ) with k v n − v k H ( Q ) → o . Lemma 3.6
For u ∈ H , ( Q ) there holds k (cid:3) e u k [ H ( Q − )] ′ ≥ √ T k u k L ( Q ) . Proof.
For 0 = u ∈ H , ( Q ), there exists a non–trivial sequence ( u n ) n ∈ N ⊂ H , , ( Q ), u n
0, with lim n →∞ k u − u n k H ( Q ) = 0 . For each u n ∈ H , , ( Q ), we define w n ∈ H , , ( Q ) as unique solution of the variationalformulation a ( v, w n ) = h u n , v i L ( Q ) for all v ∈ H , , ( Q )with the bilinear form (1.3). In particular for v = u n , this gives a ( u n , w n ) = k u n k L ( Q ) . k w n k H , , ( Q ) ≤ √ T k u n k L ( Q ) . For the zero extension e u n ∈ L ( Q − ) of u n ∈ H , , ( Q ), we obtain, when using the normrepresentation (3.6) and (3.8), that k (cid:3) e u n k [ H ( Q − )] ′ = sup = v ∈ H , , ( Q ) |h (cid:3) e u n , E v i Q − |k v k H , , ( Q ) ≥ |h (cid:3) e u n , E w n i Q − |k w n k H , , ( Q ) = | a ( u n , w n ) |k w n k H , , ( Q ) = k u n k L ( Q ) k w n k H , , ( Q ) ≥ √ T k u n k L ( Q ) , and the assertion follows by completion for n → ∞ . Corollary 3.7
The inner product space (cid:16) H , ( Q ) , h (cid:3) f ( · ) , (cid:3) f ( · ) i [ H ( Q − )] ′ (cid:17) is complete, i.e., aHilbert space. Proof.
The assertion follows immediately from Lemma 3.6.In the following, H , ( Q ) is endowed with the Hilbertian norm k (cid:3) f ( · ) k [ H ( Q − )] ′ . With thisnew Hilbert space, the bilinear form e a ( · , · ) : H , ( Q ) × H , , ( Q ) → R , e a ( u, v ) := h (cid:3) e u, E v i Q − , is continuous, i.e., | e a ( u, v ) | = |h (cid:3) e u, E v i Q − | ≤ k (cid:3) e u k [ H ( Q − )] ′ k v k H , , ( Q ) (3.9)for all u ∈ H , ( Q ) and v ∈ H , , ( Q ), and fulfills the inf–sup condition k (cid:3) e u k [ H ( Q − )] ′ = sup = v ∈ H , , ( Q ) |h (cid:3) e u, E v i Q − |k v k H , , ( Q ) = sup = v ∈ H , , ( Q ) | e a ( u, v ) |k v k H , , ( Q ) (3.10)for all u ∈ H , ( Q ), where the norm representation (3.6) is used. In addition, Lemma 3.5yields the representation e a ( u, v ) = a ( u, v ) (3.11)for all u ∈ H , , ( Q ) ⊂ H , ( Q ), v ∈ H , , ( Q ), which is used in the following lemma. Lemma 3.8
For all = v ∈ H , , ( Q ) , there exists a function u v ∈ H , ( Q ) such that e a ( u v , v ) > . roof. For 0 = v ∈ H , , ( Q ), there exists a unique solution u v ∈ H , , ( Q ) ⊂ H , ( Q ),satisfying a ( u v , w ) = h v, w i L ( Q ) for all w ∈ H , , ( Q ) . Using the representation (3.11), this gives e a ( u v , w ) = h v, w i L ( Q ) for all w ∈ H , , ( Q ) , and in particular for w = v , we obtain e a ( u v , v ) = k v k L ( Q ) > , i.e., the assertion.Next, we state the new variational setting for the wave equation (1.1). For given f ∈ [ H , , ( Q )] ′ , we consider the variational formulation to find u ∈ H , ( Q ) such that e a ( u, v ) = h f, v i Q for all v ∈ H , , ( Q ) , (3.12)i.e., the operator equation E ′ (cid:3) e u = f in [ H , , ( Q )] ′ . With the properties of the bilinear form e a ( · , · ), the unique solvability of the variationalformulation (3.12), i.e., the main theorem of this paper, is proven. Theorem 3.9
For each given f ∈ [ H , , ( Q )] ′ , there exists a unique solution u ∈ H , ( Q ) of the variational formulation (3.12) . Furthermore, L : [ H , , ( Q )] ′ → H , ( Q ) , L f = u, is an isomorphism satisfying k (cid:3) e u k [ H ( Q − )] ′ = k (cid:3) f L f k [ H ( Q − )] ′ = k f k [ H , , ( Q )] ′ . Proof.
With the help of the Banach–Neˇcas–Babuˇska theorem [9, Theorem 2.6], the resultsin (3.9), (3.10) and Lemma 3.8 yield the existence and uniqueness of the solution u ∈H , ( Q ). In addition, with the variational formulation (3.12), the equalities k (cid:3) e u k [ H ( Q − )] ′ = sup = v ∈ H , , ( Q ) | e a ( u, v ) |k v k H , , ( Q ) = sup = v ∈ H , , ( Q ) |h f, v i Q |k v k H , , ( Q ) = k f k [ H , , ( Q )] ′ . hold true and, therefore, the assertion.While for the ordinary differential equation (1.7), this new approach leads to the samevariational setting as already considered in [26, Section 4] and [25], see Lemma 2.10 andCorollary 2.11, the situation is different for the wave equation (1.1). In greater detail, forthe variational formulation (3.12), the Banach–Neˇcas–Babuˇska theorem [9, Theorem 2.6]is applicable, whereas the variational formulation (1.2) does not fit in this framework, seeTheorem 1.1. Additionally, Lemma 3.5 and (3.11) show that the new variational formu-lation (3.12) is a generalization of the variational formulation (1.2). Next, the followingfunctions are given to get a first impression of the solution space H , ( Q ).23 emark 3.10 For u ∈ C ( Q ) with u | Ω ×{ } = u | Σ = 0 there holds u ∈ H , , ( Q ) ⊂ H , ( Q ) .Note that the second initial condition ∂ t u ( · , t ) | t =0 = 0 in Ω is not incorporated in the ansatz space H , ( Q ) , see (1.4) . Remark 3.11
Consider the smooth function u ( x, t ) = sin( πx ) sin( πt ) for ( x, t ) ∈ (0 , × (0 ,
1) = Q, satisfying u | Ω ×{ } = u | Σ = 0 and (cid:3) u = 0 in Q . But there is (cid:3) e u = 0 with e u ( x, t ) = ( u ( x, t ) for ( x, t ) ∈ Q, for ( x, t ) ∈ Q − \ Q, since the distributional derivative fulfills h (cid:3) e u, ϕ i Q − = Z T Z Ω u ( x, t ) (cid:3) ϕ ( x, t ) d x d t = π Z Ω sin( πx ) ϕ ( x,
0) d x for all ϕ ∈ C ∞ ( Q − ) . Thus, the function u ∈ H , , ( Q ) ⊂ H , ( Q ) solves the variationalformulation (3.12) with the right–hand side f v ∈ [ H , , ( Q )] ′ , h f v , v i Q = π Z Ω sin( πx ) v ( x,