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3 Coercive Problems 3 Coercive Problems This chapter deals with problems whose weak formulation is endowed with a coercivity property. The key examples investigated henceforth are scalar ellip- tic PDEs, spectral problems associated with the Laplacian, and PDE systems derived from continuum mechanics. The goal is twofold: First, to set up a mathematical framework for well-posedness; then, to investigate conforming and non-conforming ¯nite element approximations based on Galerkin meth- ods. Error estimates are derived from the theoretical results of Chapters 1 and 2 and are illustrated numerically. The last section of this chapter is con- cerned with coercivity loss and is meant to be a transition to Chapters 4 and 5. 3.1 Scalar Elliptic PDEs: Theory Let ­ be a domain in Rd. Consider a di®erential operator in the form L u = (σ u) + ¯ u + ¹u; (3.1) L ¡r¢ ¢r ¢r where σ, ¯, and ¹ are functions de¯ned over ­ and taking their values in Rd;d, Rd, and R, respectively. Given a function f : ­ R, consider the problem of ! ¯nding a function u : ­ R such that ! u = f in ­; L (3.2) u = g on @­; ( B where the operator accounts for boundary conditions. The model problem (3.2) arises in severalB applications: (i) Heat transfer: u is the temperature, σ = · where · is the thermal conductivity, ¯ is the flow ¯eld, ¹ = 0, and fIis the externally supplied heat per unit volume. 112 Chapter 3. Coercive Problems (ii) Advection{di®usion: u is the concentration of a solute transported in a flow ¯eld ¯. The matrix σ models the solute di®usivity resulting from either molecular di®usion or turbulent mixing by the carrier flow. Solute production or destruction by chemical reaction is accounted for by the linear term ¹u, and the right-hand side f models ¯xed sources or sinks. Henceforth, the following assumptions are made on the data: f L2(­), d;d d 2 σ [L1(­)] , ¯ [L1(­)] , ¯ L1(­), and ¹ L1(­). Furthermore, the2 operator is assumed2 to ber¢elliptic2 in the following2 sense: L De¯nition 3.1. The operator de¯ned in (3.1) is said to be elliptic if there L exists σ0 > 0 such that d d 2 » R ; σij »i»j σ0 » d a.e. in ­: (3.3) 8 2 ¸ k k i;j=1 X Equation (3.2) is then called an elliptic PDE. Example 3.2. A fundamental example of an elliptic operator is the Lapla- cian, = ¢, which is obtained for σ = , ¯ = 0, and ¹ = 0. L ¡ I ut 3.1.1 Review of boundary conditions and their weak formulation We ¯rst proceed formally and then specify the mathematical framework for the weak formulation. Homogeneous Dirichlet boundary condition. We want to enforce u = 0 on @­. Multiplying the PDE u = f by a (su±ciently smooth) test function v vanishing at the boundary, inLtegrating over ­, and using the Green formula (σ u) v = v σ u v (n σ u); (3.4) ¡r¢ ¢r r ¢ ¢r ¡ ¢ ¢r Z­ Z­ Z@­ yields v σ u + v(¯ u) + ¹uv = fv: r ¢ ¢r ¢r Z­ Z­ A possible regularity requirement on u and v for the integrals over ­ to be meaningful is u H1(­) and v H1(­): 2 2 Since u H1(­), Theorem B.52 implies that u has a trace at the bound- 2 ary. Because of the boundary condition u @­ = 0, the solution is sought in 1 1j H0 (­). Test functions are also taken in H0 (­), leading to the following weak formulation: 1 Seek u H0 (­) such that 2 1 (3.5) aσ;¯;¹(u; v) = fv; v H (­); ½ ­ 8 2 0 with the bilinear form R aσ;¯;¹(u; v) = v σ u + v(¯ u) + ¹uv: (3.6) r ¢ ¢r ¢r Z­ 3.1. Scalar Elliptic PDEs: Theory 113 Proposition 3.3. If u solves (3.5), then u = f a.e. in ­ and u = 0 a.e. on @­. L Proof. Let ' (­) and let u be a solution to (3.5). Hence, 2 D (σ u); ' 0; = σ u; ' 0; = ' σ u h¡r¢ ¢r iD D h ¢r r iD D r ¢ ¢r Z­ = (f ¯ u ¹u) '; ¡ ¢r ¡ Z­ 2 yielding u; ' 0; = ­ f'. Owing to the density of (­) in L (­), u = f in L2(­).hLTherefore,iD D u = f a.e. in ­. Moreover, Du = 0 a.e. on L@­ by de¯nition of H1(­); seeLR Theorem B.52. 0 tu Non-homogeneous Dirichlet boundary condition. We want to enforce u = g on @­, where g : @­ R is a given function. We assume that g ! 1 is su±ciently smooth so that there exists a lifting ug of g in H (­), i.e., a 1 function ug H (­) such that ug = g on @­; see 2.1.4. We obtain the weak formulation:2 x Seek u H1(­) such that 2 1 8 u = ug + Á; Á H0 (­); (3.7) 2 1 <> aσ;¯;¹(Á; v) = fv aσ;¯;¹(ug; v); v H (­): ­ ¡ 8 2 0 > R1 Proposition: 3.4. Let g H 2 (@­). If u solves (3.7), then u = f a.e. in ­ and u = g a.e. on @­. 2 L Proof. Similar to that of Proposition 3.3. tu When the operator is the Laplacian, (3.7) is called a Poisson problem. L Neumann boundary condition. Given a function g : @­ R, we want to enforce n σ u = g on @­. Note that in the case σ = ,!the Neumann ¢ ¢r I condition speci¯es the normal derivative of u since n u = @nu. Proceeding as before and using the Neumann condition in the surface¢r integral in (3.4) yields the weak formulation: Seek u H1(­) such that 2 1 (3.8) aσ;¯;¹(u; v) = fv + gv; v H (­): ½ ­ @­ 8 2 Proposition 3.5. Let g L2(@R­). If uR solves (3.8), then u = f a.e. in ­ and n σ u = g a.e. on @2­. L ¢ ¢r Proof. Taking test functions in (­) readily implies u = f a.e. in ­. There- 2 D L 1 fore, (σ u) L (­). Corollary B.59 implies n σ u H 2 (@­)0 = 1 ¡r¢ ¢r 2 ¢ ¢r 2 H¡ 2 (@­) since 114 Chapter 3. Coercive Problems 1 2 Á H (@­); n σ u; Á ¡ 1 1 = (σ u)uÁ + uÁ σ u; 8 2 h ¢ ¢r iH 2 ;H 2 ¡r¢ ¢r r ¢ ¢r Z­ Z­ 1 1 where uÁ H (­) is a lifting of Á in H (­). Then, (3.8) yields 2 1 2 Á H (@­); n σ u; Á ¡ 1 1 = gÁ; 8 2 h ¢ ¢r iH 2 ;H 2 Z@­ 1 2 showing that n σ u = g in H ¡ 2 (@­) and, therefore, in L (@­) since g belongs to this space.¢ ¢r tu Mixed Dirichlet{Neumann boundary conditions. Consider a partition of the boundary in the form @­ = @­ @­ . Impose a Dirichlet con- D [ N dition on @­D and a Neumann condition on @­N. If the Dirichlet condi- tion is non-homogeneous, assume that @­D is smooth enough so that, for all 1 1 g H 2 (@­D), there exists an extension g H 2 (@­) such that g @­ = g 2 2 j D and g 1 c g 1 uniformly in g. Then, using the lifting of g k kH 2 (@­) · k kH 2 (@­D) in H1(­), one can assume that the Dirichlete condition is homogeneous.e The boundarye conditions are thus e u = 0 on @­D; n σ u = g on @­ ; ( ¢ ¢r N with a given function g : @­N R. Proceeding as before, we split! the boundary integral in (3.4) into its con- tributions over @­D and @­N. Taking the solution and the test function in the functional space H1 (­) = u H1(­); u = 0 on @­ ; @­D f 2 Dg the surface integral over @­D vanishes. Furthermore, using the Neumann con- dition in the surface integral over @­N yields the weak formulation: Seek u H1 (­) such that @­D 2 1 (3.9) aσ;¯;¹(u; v) = fv + gv; v H (­): ( ­ @­N 8 2 @­D R R Proposition 3.6. Let @­D @­, assume meas(@­D) > 0, and set @­N = 2 ½ @­ @­D. Let g L (@­N). If u solves (3.9), then u = f a.e. in ­, u = 0 a.e.non @­ , and2 (n σ u) = g a.e. on @­ . L D ¢ ¢r N Proof. Proceed as in the previous proofs. tu Robin boundary condition. Given two functions g; γ : @­ R, we want to enforce γu+n σ u = g on @­. Using this condition in the surface! integral in (3.4) yields the¢ w¢reak formulation: Seek u H1(­) such that 2 1 (3.10) aσ;¯;¹(u; v) + γuv = fv + gv; v H (­): ½ @­ ­ @­ 8 2 R R R 3.1. Scalar Elliptic PDEs: Theory 115 Problem V a(u; v) f(v) 1 Homogeneous Dirichlet H0 (­) aσ;¯;¹(u; v) ­ fv 1 Neumann H (­) aσ;¯;¹(u; v) ­ fvR + @­ gv 1 Dirichlet{Neumann H@­ (­) aσ;¯;¹(u; v) fv + gv D R­ R@­N 1 Robin H (­) aσ;¯;¹(u; v) + @­ γuv R ­ fv +R @­ gv Table 3.1. Weak formulation corresponding to the variousR boundaryR conditionsR for the second-order PDE (3.2). The bilinear form aσ;¯;¹(u; v) is de¯ned in (3.6). 2 Proposition 3.7. Let g L (@­) and let γ L1(@­). If u solves (3.10), then u = f a.e. in ­ and2 γu + n σ u = g a.e.2 on @­. L ¢ ¢r Proof. Proceed as in the previous proofs. tu Summary. Except for the non-homogeneous Dirichlet problem, all the prob- lems considered herein take the generic form: Seek u V such that 2 (3.11) a(u; v) = f(v); v V; ½ 8 2 where V is a Hilbert space satisfying H1(­) V H1(­): 0 ½ ½ Moreover, a is a bilinear form de¯ned on V V , and f is a linear form de¯ned on V ; see Table 3.1.
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