FEM for Elliptic problems

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FEM for Elliptic problems presented Introduction; variational formulation; existence of solutions: lax-milgram theorem; FEM problem,... Invite you to consult the documentation

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FEM for Elliptic Problems Sebastian Gonzalez Pintor November, 2016 Proof. Multiply equation (D) by v ∈ V and integrate over the whole domain Z 1 Z 1 00 f v dx, u v dx = − We follow the results from [Joh12] and [And15]. 1 Introduction then we integrate by parts and use the boundary conditions on the left side to obtain Z 1 Z 1 − u00 v dx = − [u0 v]10 + u0 v 0 dx Outline: Variational form. and Minimization prob. • • • • 0 0 Definition of (D), (V) and (M) Equivalence (D) ⇒ (V ) ⇔ (M ) If u ∈ C 2 then (D) ⇐ (V ) Uniqueness of (V ). 0 0 = −u0 (1)v(1) + u0 (0)v(0) +(u0 , v 0 ) {z } | v(1)=v(0)=0 0 0 = (u , v ). We notice that u ∈ C 2 and u(0) = u(1) = 0 implies that u ∈ V , and because the choice of v ∈ V is arbitrary, the function u satisfies the equation Lets consider the following two-point boundary value problem − u00 (x) = f (x) for x ∈ (0, 1) (D) (u0 , v 0 ) = (f, v) ∀v ∈ V, u(0) = u(1) = 0 where u0 = ux and f is a given continuous function. We introduce the notation Z 1 (u, v) = v(x)w(x) dx what is the same as problem (V). Proposition 2 (V ⇒ M ). If u ∈ V is a solution of the variational problem (V), then u is a solution of the minimization problem (M) . 0 for real valued piecewise continuous bounded functions, and the linear space Z 1 V = {v : v ∈ C 0 ([0, 1]), |v 0 |2 dx < ∞, Proof. Let us consider an arbitrary function w ∈ V . We define v = w − u ∈ V , so that using the definition of the functional F we have 1 0 0 (w , w ) − (w, f ) 2 1 = (v 0 + u0 , v 0 + u0 ) − (v + u, f ) 2 1 1 = (v 0 , v 0 ) + (u0 , v 0 ) + (u0 , u0 ) − (v, f ) − (u, f ) 2 2 1 1 = (v 0 , v 0 ) + (u0 , v 0 ) − (v, f ) + (u0 , u0 ) − (u, f ) | {z } |2 2 {z } 0 F (w) = and v(0) = v(1) = 0} and the linear functional F (v) = 1 0 0 (v , v ) − (f, v). 2 We can define the problems (M) and (V) as follows: (I) Find u ∈ V such that F (u) ≤ F (v) ∀v ∈ V, (M) Find u ∈ V such that (u0 , v 0 ) = (f, v) ∀v ∈ V. (II) (V) but because u is solution of (V) we have that (I) = 0, and we can rewrite (II) = F (u), so using that (v 0 , v 0 ) ≥ 0 Now we will show that if u is a solution of the two- we obtain 1 point boundary value problem (D), then u is a solution F (w) = (v 0 , v 0 ) + F (u) ≥ F (u). of the variational problem (V) (D ⇒ V ), and that u is 2 a solution of the variational problem (V) if, and only Because w ∈ V was chosen arbitrarily, it means that u if, u is also a solution of the minimization problem (M) is a solution of (M). (V ⇔ M ). Proposition 1 (D ⇒ V ). If u ∈ C 2 is a solution of Proposition 3 (M ⇒ V ). If u ∈ V is a solution of the the two-point boundary value problem (D), then u is a minimization problem (M), then u is a solution of the solution of the variational problem (V) . variational problem (V) . 1 and integrating by parts the first term, using that v(0) = v(1) = 0 and that u00 exists we get, Proof. If u is a solution of (M) we have that F (u) ≤ F (u + εv), (u0 , v 0 ) = (f, v) ∀v ∈ V, for all ε ∈ R, because u + εv ∈ V for every v ∈ V . Thus, the differentiable function − (u00 , v) = (f, v) ∀v ∈ V, (f + u00 , v) = 0 ∀v ∈ V, g(ε) ≡ F (u + εv) but with the assumption that f + u00 is continuous, this can only hold if f (x) + u00 (x) = 0 ∀x ∈ (0, 1) . 2 1 0 0 ε (u , u ) + ε(u0 , v 0 ) + (v 0 , v 0 ) − (u, f ) − ε(v, f ) 2 2 ε2 1 0 0 = (u , u ) − (u, f ) + ε[(u0 , v 0 ) − (v, f )] + (v 0 , v 0 ), 2 2 = 2 Variational Formulation 0 has a minimum at ε = 0 and hence g (0) = 0. But Outline: Hilbert Spaces g 0 (ε) = (u0 , v 0 ) − (v, f ) + ε(v 0 , v 0 ), • Hilbert Spaces (L2 , H1 and H10 ) – Scalar product, Cauchy-Schwarz inequality and Cauchy sequence • Natural and Essential boundary conditions and evaluating at ε = 0 we get g 0 (0) = (u0 , v 0 ) − (v, f ) = 0, hence u is a solution of (V). We can also show that Proposition 4 (Uniqueness). A solution to (V) is uniquely determined. Definition 1. A map a(·, ·) : V × V → R that is symmetric and bilinear, i.e., ∀u, v, w ∈ V , ∀α, β ∈ R a(v, w) = (w, v), Proof. Suppose that u1 and u2 are solutions of (V), i.e., u1 , u2 ∈ V and (u01 , v 0 ) = (f, v) ∀v ∈ V, (u0w , v 0 ) (symmetric) a(αu + βv, w) = αa(u, w) + βa(v, w) (a(·, w) is linear) a(u, αv + βw) = αa(u, v) + βa(u, w) (a(u, ·) is linear) is called a scalar product on V if = (f, v) ∀v ∈ V. a(v, v) > 0, ∀v ∈ V, v 6= 0, Subtracting these two equations and v = u1 − u2 ∈ V gives Z 1 (u01 − u02 )2 dx = 0 and the norm associated with the scalar product is defined by ||v||a = (a(v, v))1/2 , ∀v ∈ V. 0 Moreover, if < ·, · > is a scalar product with corresponding norm || · ||, we have that the following CauchySchwarz’s inequality is satisfied which shows that u01 (x) − u02 (x) = 0 ∀x ∈ [0, 1], | < v, w > | ≤ ||v|| ||w||, ∀v, w ∈ V. so u1 − u2 is constant in [0, 1], which together with the boundary conditions u1 (0) = u2 (0) = 0 gives u1 (x) = u2 (x), ∀x ∈ [0, 1], and the uniqueness follows. Definition 2. A linear space V with a scalar product < ·, · > and the corresponding norm || · || is said to be a Hilbert space if V is complete, i.e., if every Cauchy sequence with respect to || · || is convergent, where Also, because every solution of (D) or (M) is also solution of (V), but the solution of (V) is unique, it follows the uniqueness of the solutions of (D) and (M). We notice here that we have not proved yet that the solution exists, just uniqueness in case of existence. Existence will be shown later for a general variational problem, given that it satisfies some conditions. • a sequence is said to be Cauchy if ∀ε > 0 ∃N ∈ N such that ||vi − vj || < ε if i, j ≥ N, • a sequence is convergent in V if ∃v ∈ V such that ∀ε > 0 ∃N ∈ N such that ||vi − v|| < ε if i ≥ N. Last, we show that Proposition 5. If u00 (x) exists and it is continuous for a solution of (V), then u is also a solution of (D). Example 2.1. We can see that the space V := C 0 ([−1, 1]) is a linear space, and that (·, ·)V : V × V → R Proof. For this, assuming that u is a solution of (V), we defined by have Z 1 (v, w) = v(x)w(x) dx, V (u0 , v 0 ) = (f, v) ∀v ∈ V, −1 2 1.2 together with the scalar product and its associated norm as follows Z (v, w)L2 (Ω) = vw dx, ZΩ ||v||L2 (Ω) = v 2 dx. 1 vi 0.8 vj 0.6 Ω 0.4 • We also introduce the linear space of square integrable derivatives up to order 1, 0.2 H1 (Ω) = {v ∈ L2 (Ω), vxi ∈ L2 , i = 1, . . . , d}, 0 1/N together with the scalar product and the associated norm as follows Z (v, w)H1 (Ω) = vw + ∇v · ∇w dx, Z Ω ||v||H1 (Ω) = v 2 +|∇v|2 dx = ||v||L2 (Ω) + ||∇v||L2 (Ω) . -0.2 -1 -0.5 0 0.5 1 Figure 1: For i, j > N > 1/ε, the square of the area between vi and vj is smaller than ε. is a scalar product, with the corresponding associated 1/2 norm ||v||V = (v, v)V . But V with this norm is not complete, thus it is not a Hilbert space. Ω • And the space H10 (Ω) := {v ∈ H1 (Ω) : v = 0 on Γ}, Proof. To see this we can build a Cauchy sequence that is not convergent inside V . For example, consider the sequence   if x ∈ [−1, 0], 0 vn = nx if x ∈ (0, 1/n),   1 if x ∈ [1/n, 1]. together with the same scalar product and the associated norm as for H1 (Ω), where Γ := ∂Ω is the boundary of Ω. 2.1 Boundary Conditions Take ε > 0, and 1/ε < N ∈ N, we can see (Figure 1 that The boundary conditions for the differential equation are the sequence is Cauchy usually classified as Dirichlet (when they impose a value Z 1 for the function on the boundary), Neumann (when they 2 2 ||vi − vj ||V = (vi (x) − vj (x)) dx force a value for the derivative of the function on the −1 boundary), or mixed (mixing Dirichlet and Newman), 0≤·≤1 Z 1/N z }| { = (vi (x) − vj (x))2 dx u=g on ∂Ω (Dirichlet) 0 Z 1/N ~n∇u = h on ∂Ω (Neumann) ≤ dx = 1/N < ε, γu + ~n∇u = g on ∂Ω (Mixed) 0 where g, h, γ are given functions. Building the variational formulation, when integrating by parts the original equation multiplied by a test function, different boundary conditions will appear in different ways in the variational formulation. In particular, we focus on Dirichlet and Neumann boundary conditions, that for second order differential equations will become essential and natural boundary conditions, meaning that but the sequence converges to the function ( 0 if x ∈ [−1, 0], H(x) = 1 if x ∈ (0, 1], which is not inside V . It means that the space V with this norm is not complete, so it is not a Hilbert space. In particular, for a bounded domain Ω ∈ Rd , we define the following function spaces, with appropriate scalar products and associated norms, which are Hilbert spaces: • essential boundary conditions will appear in the function space V , while • natural boundary conditions will naturally appear • The linear space of square-integrable functions in a in the variational form. bounded domain Ω ∈ Rd , In general, for an equation of order 2k, the variational Z 2 2 form needs an space with functions differentiable up to L (Ω) = {v : |v| dx < ∞}, Ω order k, so essential boundary conditions will be inposed 3 in the space for orders 0, . . . , k − 1, and natural boundary conditions will be built in the variational form for orders k, . . . , 2k − 1. It is also important to notice that essential boundary conditions will generate two spaces, not necessarily the same, one for which the condition is imposed to the solution, and other one for the test function where the functions would be zero. For instance, consider the boundary, Γ, split in two parts, Γ1 and Γ2 , then the two spaces are In addition, we can see that the minimization problem associated to this variational problem can be written by Find u ∈ V1 such that F (u) ≤ F (v)∀v ∈ V1 where 1 a(v, v) − L(v), 2 and it is equivalent to the variational problem above. F (v) = • u ∈ V1 = {v : v ∈ H1 , and v = g on Γ1 }, • v ∈ V2 = {v : v ∈ H1 , and v = 0 on Γ1 }, 3 and they are the same only when g = 0. We notice that the Neumann boundary conditions do not have any effect into the function spaces; instead, they will appear in the variational formulation. Existence of solutions: Lax-Milgram theorem Outline: Lax-Milgram • Parallelogram law • Lax-Milgram Theorem Example 2.2. Consider a bounded domain Ω ∈ Rn , • (V ) ⇔ (M ) with boundary Γ = ∂Ω ∈ C 1 piecewise, and the bound• Poincar´e Inequalities for H1 and H10 ary is split in two parts Γ1 and Γ2 . The problem reads • Trace Inequality  n  −∆u = f in Ω ⊂ R ,  u=g on Γ1   ~n · ∇u = h on Γ2 The complete proofs of the following results are provided in [And15]. Proof. We use the spaces Parallelogram Law: Let ||·|| be a norm associated to a V1 = {v : v ∈ H1 , and v = g on Γ1 }, scalar product < ·, · >. The following equivalence holds V2 = {v : v ∈ H1 , and v = 0 on Γ1 }, ||x + y||2 + ||x − y||2 = 2||x||2 + 2||y||2 . where the solution must be in V1 , and V2 is the space of test functions. Now, multiply the original equation by a test function v and integrate over the whole domain, obtaining Z Z − ∆uv dx = f v dx, ∀v ∈ V2 , Theorem 1 (Lax-Milgram). Let V be a Hilbert space with scalar product < ·, · >V and associated norm || · ||V . Assume that a(·, ·) is a bilinear functional and L a linear functional satisfying: Ω (1) a(·, ·) is symmetric, i.e., Ω but integrating by parts (Green’s formula) the left part Z − ∆uv dx Ω Z Z = ∇u · ∇v dx − ~n∇uv ds Z ZΩ ZΓ ~n∇uv ds = ∇u · ∇v dx − ~n∇uv ds − Γ Ω Γ } | 2 {z } | 1 {z v=0 on Γ1 Z (2) a(·, ·) is V-elliptic, i.e., ∃α > 0 s.t. a(v, v) ≥ α||v||2V , ∀v ∈ V ; (3) a(·, ·) is continuous, i.e., ∃C ∈ R s.t. |a(v, w)| ≤ C||v||V ||w||V , ∀v, w ∈ V ; ~ n∇u=h on Γ2 Z ∇u · ∇v dx − = a(v, w) = a(w, v), ∀v, w ∈ V ; Ω (4) L(·) is continuous, i.e., hv ds, Γ2 ∃Λ ∈ R s.t. |L(v)| ≥ Λ||v||V , ∀v ∈ V ; so the variational formulation can be rewritten as Find u ∈ V1 such that Z Z Z f v dx + hv ds, ∀v ∈ V2 . ∇u · ∇v dx = Γ2 | Ω {z } |Ω {z } a(u,v) Then there exists a unique function u ∈ V such that a(u, v) = L(v), ∀v ∈ V, and the following estimate holds L(v) ||u||V ≤ 4 Λ . α Proof. A sketch of the proof (complete proof in [And15]) It is sometimes useful to be able to bound the norm of the function with the bound of the derivative in order to verify conditions (2) − (4) of the Lax-Milgram theorem. It is easy to show that this can not be done with constant functions, so that we can use the followings Poincar´e inequalities, always having a mechanism that identify constant functions with the function zero. Theorem 2 (Poincar´e inequality for H10 (Ω)). Let Ω be a bounded domain in Rn with boundary ∂Ω ∈ C 1 piecewise. Let u ∈ H10 (Ω). Then there exists a constant C > 0 such that • Build a solution of (M): ∃u ∈ V solving (M) – Norm ||v|| = a(v, v) is equivalent to || · ||V – We see that β = inf F (v) > −∞ v∈V – {vi } ⊂ V , with F (vi ) → β, is Cauchy. – Completeness implies that ∃u ∈ V s.t. vi → u – Show that F (u) = β. Thus u solves (M). ||u||L2 (Ω) ≤ C||∇u||L2 (Ω) . • (M)⇒(V): Because of proposition 6. • Uniqueness: If u1 and u2 solve (V), subtracting the Proof. A sketch of the proof (complete proof in [And15]) P equations we obtain a(u1 − u2 , v) = 0, and using (2) • We use the function φ(x) = 1/(2n) x2i , with ∆φ = for v = u1 − u2 we get the result. 1, with the Green’s formula and bcs. to get Z • Estimate: Combining (2) and (4) with (V). 2u(∇u · ∇φ) dx ||u||2L2 (Ω) = − Ω Proposition 6 ((V)⇔(M)). Let a and L be such that • and we use Cauchy-Schwarz inequality and boundness they satisfy the conditions of Theorem 1. Then, the folof φ in Ω lowing problems are equivalent: Z ≤ C||u||L2 (Ω) ||∇u||L2 (Ω) − 2u(∇u · ∇φ) dx • Find u ∈ V such that a(u, v) = L(v), ∀v ∈ V ; (V) Ω • Find u ∈ V such that F (u) ≤ F (v), ∀v ∈ V ; where F (v) = (M) 1 a(v, v) − L(v). 2 Theorem 3 (Poincar´e inequality for H1 (Ω)). Let Ω be a bounded domain in Rn with boundary ∂Ω ∈ C 1 piecewise. Let u ∈ H1 (Ω) and let Z 1 u ¯= u dx, (¯ u is the average over Ω). |Ω| Ω Proof. Similar to (V⇔M) before (complete proof in [And15]). Then there exists a constant C > 0 such that Example 3.1. Show, using Lax-Milgram theorem, that the following problem has a unique solution: Find u ∈ V := H1 (Ω), Ω ⊂ R2 , such that Z Z (∇u · ∇v + uv) dx = f v dx, ∀v ∈ H1 (Ω) Ω ||u − u ¯||L2 (Ω) ≤ C||∇u||L2 (Ω) . Proof. We skip the proof. The next result (Trace inequality) can be used to prove continuity for a or L when boundary integrals are involved, i.e., when Neumann boundary conditions are used. Theorem 4 (Trace inequality). Let Ω be a bounded domain in Rn with boundary ∂Ω ∈ C 1 piecewise. Then there exists a constant C > 0 such that Ω Proof. Using Z a(v, w) := (∇v · ∇w + vw) dx, Z L(w) := Ω f w dx, Ω we check for the different conditions: 1/2 1/2 ||u||L2 (∂Ω) ≤ C||u||L2 (Ω) ||u||H1 (Ω) . 1. a is clearly symmetric. 2. a(v, v) = ||v||2H1 (Ω) proves V-ellipticity Proof. We skip the proof. 3. Cauchy-Schwarz in H1 (Ω) provides continuity for |a(v, w)| = ||v||H1 (Ω) ||w||H1 (Ω) Example 3.2. Prove, using Lax-Milgram theorem, that 4. Cauchy-Schwarz in L (Ω) provides continuity for there exists a unique solution for the variational formu|L(v)| = |(f, v)L2 (Ω) | = ||f ||L2 (Ω) ||v||L2 (Ω) ≤ Λ||v||H1 (Ω) lation associated to the problem ( with Λ = ||f ||L2 (Ω) . −∆u = f in Ω, Thus, using theorem 1 there exists a unique solution. u=0 on ∂Ω. 2 5