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Part IV, Chapter 18 Weak Formulation of Model Problems 18.1 a Second-Order

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Part IV, Chapter 18 Weak Formulation of Model Problems 18.1 a Second-Order Part IV, Chapter 18 Weak formulation of model problems We consider in this chapter a few simple PDEs posed over a domain D in Rd. (JLG) Dec 29 ♥♥ Our goal is to reformulate these problems in weak forms using the important 2016 notion of test functions. We show that, in general, there are many ways to write weak formulations. The choice can be guided, e.g., by the regularity of the data and the quantities of interest (e.g., the solution or its gradient). Weak formulation are the starting point for building finite element approxi- mations. The scope of this chapter is restricted to time-independent problems for simplicity. 18.1 A second-order PDE 18.1.1 Model problem Let D be a domain in Rd and consider f : D R. The problem we want to solve consists of finding a function u : D → R with some appropriate regularity such that → ∆u = f in D u =0 on ∂D, (18.1) − where we recall that, in Cartesian coordinates, the Laplace operator acts on functions as follows: d ∂2u ∆u := u = . (18.2) ∇·∇ ∂x2 i=1 i X The PDE ∆u = f in D is called the Poisson equation (the PDE is called the Laplace− equation when f = 0). The Laplace operator is ubiqui- tous in physics. For instance it is the prototypical operator encountered in the modelling of physical processes involving diffusion. Simplified settings for application include heat transfer (where u is the temperature and f is the heat source), mass transfer (where u is the concentration of a species and f 236 Chapter18. Weakformulationofmodelproblems is the mass source), porous media flow (where u is the hydraulic head and f is the mass source), electrostatics (where u is the electrostatic potential and f is the charge density), and static equilibria of membranes (where u is the transverse membrane displacement and f is the transverse load). The condition enforced on ∂D in (18.1) is called a boundary condition. In the context of the above models, this condition means that the temperature (the concentration, the electrostatic potential, or the transverse membrane displacement) is fixed at a constant value on ∂D, and without loss of general- ity we can assume that this constant is zero. A condition prescribing the value of the solution at the boundary is called a Dirichlet condition, and when the prescribed value is zero, the condition is called an homogeneous Dirichlet con- dition. Other boundary conditions can be prescribed for the Poisson equation; those are reviewed in Chapter 24 in the more general context of second-order elliptic PDEs. To sum up, (18.1) is the Poisson problem with an homogeneous Dirichlet condition. We now present three weak formulations of (18.1). 18.1.2 First weak formulation We derive a weak formulation for (18.1) by proceeding formally. Consider an arbitrary test function ϕ C0∞(D), where C0∞(D) is the space of infinitely differentiable functions compactly∈ supported in D. As a first step, we multiply the PDE in (18.1) by ϕ and integrate over D to obtain (∆u)ϕ dx = fϕ dx. (18.3) − ZD ZD Equation (18.3) is equivalent to the PDE in (18.1) if ∆u is smooth enough (e.g., integrable over D); indeed, if an integrable function g satisfies D gϕ dx = 0 for all ϕ C0∞(D), then g = 0 a.e. in D. As a second∈ step, we use the divergence formula stating that, forR a smooth vector-valued function Φ, Φ dx = Φ n ds, (18.4) ∇· · ZD Z∂D where n is the outward unit normal to D. We apply this formula to the function Φ = w v, where v and w are two scalar-valued, smooth functions. Since Φ = w∇ v + w∆v, we infer that ∇· ∇ ·∇ (∆v)w dx = v w dx (n v)w ds. (18.5) − ∇ ·∇ − ·∇ ZD ZD Z∂D This is Green’s formula, which is a very useful tool to derive weak formulations of PDEs involving the Laplace operator. This formula is valid, for instance, if v C2(D) C1(D) and w C1(D) C0(D). The formula can be extended to∈ functions∩ in the usual Sobolev∈ spaces.∩ In particular, it remains valid for Part IV. Galerkin Approximation 237 all w H1(D) and all v H2(D), and even for all v H1(D) such that ∆v L∈2(D) in which case∈ the boundary integral in (18.5)∈ is understood in a weak∈ sense; see Corollary B.114. We apply Green’s formula to the functions v := u and w := ϕ, assuming enough regularity for u. Since ϕ vanishes at the boundary, we transform (18.3) into u ϕ dx = fϕ dx, ϕ C∞(D). (18.6) ∇ ·∇ ∀ ∈ 0 ZD ZD We now recast (18.6) into a functional framework. Let us take f L2(D). We observe that a natural solution space is ∈ H1(D)= v L2(D) v L2(D) . (18.7) { ∈ | ∇ ∈ } Recall from Proposition B.48 that H1(D) is a Hilbert space when equipped with the inner product (u,v)H1(D) = D uv dx + D u v dx with associ- 2 2 1 ∇ ·∇ ated norm v 1 = ( v dx + v 2 dx) 2 , where 2 denotes the k kH (D) D D k∇R kℓ R k·kℓ Euclidean norm in Rd. Furthermore, in order to account for the boundary R R condition in (18.1), we consider the subspace spanned by those functions in 1 1 H (D) that vanish at the boundary. It turns out that this space is H0 (D); see Theorem B.106. Finally, we can extend the space of test functions in (18.6) 1 1 to the closure of C0∞(D) in H (D), which is by definition H0 (D) (see Def- 1 inition B.62). Indeed, for any test function w H0 (D), there is a sequence 1 ∈ (ϕn)n N in C0∞(D) converging to w in H0 (D). Then, we can pass to the limit ∈ in (18.6) where ϕn is the test function. To sum up, a weak formulation of the Poisson problem with homogeneous Dirichlet condition is as follows: Find u H1(D) such that ∈ 0 (18.8) u w dx = fw dx, w H1(D). D ∇ ·∇ D ∀ ∈ 0 A function u solvingR (18.8) is calledR a weak solution. We now investigate whether a weak solution of (18.8) satisfies the PDE and the boundary condition in (18.1). In what follows, we identify L2(D) 2 1 1 with its dual space L (D)′ so that, using the notation H− (D) = H0 (D)′, 1 2 2 1 we are in the situation where H0 (D) ֒ L (D) L (D)′ ֒ H− (D) with dense (and continuous) embeddings. Notice→ that if≡v H1(D→) (that is to say 2 1 ∈ v L (D)), then ∆v = ( v) H− (D); see Example B.70. ∇ ∈ ∇· ∇ ∈ Proposition 18.1 (Weak solution). Assume that u solves (18.8) with f L2(D). Then, the PDE in (18.1) holds a.e. in D, and the boundary condition∈ holds a.e. in ∂D. 1 Proof. Let u be a weak solution. Then, ∆u H− (D). Consider an ar- ∈ 2 bitrary function ϕ C∞(D) in (18.8). Since f L (D), we infer that ∈ 0 ∈ ∆u, ϕ 1 −1 = u ϕ dx = fϕ dx C ϕ 2 with C = H0 (D),H (D) D D L (D) h− i ∇ ·∇ 2 ≤ k k f 2 . Invoking the density of C∞(D) in L (D) (see Theorem B.36), we k kL (D) R 0 R 238 Chapter18. Weakformulationofmodelproblems infer that ∆u defines a bounded linear form on L2(D) with Riesz–Fr´echet − representative equal to f. Owing to the identification L2(D) L2(D)′, we can ≡ 1 write ∆u = f. Hence, the PDE holds a.e. in D. Finally, since u H0 (D), u vanishes− a.e. in ∂D owing to the Trace Theorem B.106. ∈ ⊓⊔ The crucial advantage of the weak formulation (18.8) with respect to the original formulation (18.1) is that, as we will see in the next chapter, there exists powerful tools that will allow us to assert the existence and uniqueness of weak solutions. Incidentally, note that uniqueness is not a trivial property in spaces larger than H1(D). For instance, one can construct domains in which this property does not hold in L2(D); see Exercise 18.1. 18.1.3 Second weak formulation To derive our second formulation, we introduce the vector-valued function σ = u. To avoid notational collisions, we use the letter p instead of u to denote−∇ the scalar-valued dependent function, and we use the symbol u to denote the dependent pair (σ,p). In many applications, p plays the role of a potential and σ plays the role of a diffusive flux. Since σ = p and ∆p = f, we obtain σ = f. Therefore, the model problem is now−∇ written as− follows: ∇· σ + p =0 in D, σ = f in D, p =0 on ∂D. (18.9) ∇ ∇· This is the mixed formulation of the original problem (18.1). The PDEs in (18.9) are often called Darcy’s equations (in the context of porous media flows, p is the hydraulic head and σ the filtration velocity). We multiply the first PDE in (18.9) by a vector-valued test function τ and integrate over D to obtain σ τ dx + p τ dx = 0. (18.10) · ∇ · ZD ZD We multiply the second PDE in (18.9) by a scalar-valued test function q and integrate over D to obtain ( σ)q dx = fq dx.
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