FEM for Elliptic Problems

FEM for Elliptic Problems Sebastian Gonzalez Pintor November, 2016 We follow the results from [Joh12] and [And15]. 1 Introduction • Definition of (D), (V) and (M) • Equivalence (D)⇒ (V )⇔ (M) • If u ∈ C2 then (D)⇐ (V ) • Uniqueness of (V ). Outline: Variational form. and Minimization prob. Lets consider the following two-point boundary value problem − u′′(x) = f(x) for x ∈ (0, 1) (D) u(0) = u(1) = 0 where u′ = ux and f is a given continuous function. We introduce the notation (u, v) = ∫ 1 0 v(x)w(x) dx for real valued piecewise continuous bounded functions, and the linear space V = {v : v ∈ C0([0, 1]), ∫ 1 0 |v′|2 dx <∞, and v(0) = v(1) = 0} and the linear functional F (v) = 1 2 (v′, v′)− (f, v). We can define the problems (M) and (V) as follows: Find u ∈ V such that F (u) ≤ F (v) ∀v ∈ V, (M) Find u ∈ V such that (u′, v′) = (f, v) ∀v ∈ V. (V) Now we will show that if u is a solution of the two- point boundary value problem (D), then u is a solution of the variational problem (V) (D ⇒ V ), and that u is a solution of the variational problem (V) if, and only if, u is also a solution of the minimization problem (M) (V ⇔M). Proposition 1 (D ⇒ V ). If u ∈ C2 is a solution of the two-point boundary value problem (D), then u is a solution of the variational problem (V) . Proof. Multiply equation (D) by v ∈ V and integrate over the whole domain − ∫ 1 0 u′′v dx = ∫ 1 0 fv dx, then we integrate by parts and use the boundary condi- tions on the left side to obtain − ∫ 1 0 u′′v dx =− [u′v]10 + ∫ 1 0 u′v′ dx = −u′(1)v(1) + u′(0)v(0)︸ ︷︷ ︸ v(1)=v(0)=0 +(u′, v′) = (u′, v′). We notice that u ∈ C2 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 (u′, v′) = (f, v) ∀v ∈ V, 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) . 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 F (w) = 1 2 (w′, w′)− (w, f) = 1 2 (v′ + u′, v′ + u′)− (v + u, f) = 1 2 (v′, v′) + (u′, v′) + 1 2 (u′, u′)− (v, f)− (u, f) = 1 2 (v′, v′) + (u′, v′)− (v, f)︸ ︷︷ ︸ (I) + 1 2 (u′, u′)− (u, f)︸ ︷︷ ︸ (II) but because u is solution of (V) we have that (I) = 0, and we can rewrite (II) = F (u), so using that (v′, v′) ≥ 0 we obtain F (w) = 1 2 (v′, v′) + F (u) ≥ F (u). Because w ∈ V was chosen arbitrarily, it means that u is a solution of (M). Proposition 3 (M ⇒ V ). If u ∈ V is a solution of the minimization problem (M), then u is a solution of the variational problem (V) . 1 Proof. If u is a solution of (M) we have that F (u) ≤ F (u+ εv), for all ε ∈ R, because u+ εv ∈ V for every v ∈ V . Thus, the differentiable function g(ε) ≡ F (u+ εv) = 1 2 (u′, u′) + ε(u′, v′) + ε2 2 (v′, v′)− (u, f)− ε(v, f) = 1 2 (u′, u′)− (u, f) + ε[(u′, v′)− (v, f)] + ε 2 2 (v′, v′), has a minimum at ε = 0 and hence g′(0) = 0. But g′(ε) = (u′, v′)− (v, f) + ε(v′, v′), and evaluating at ε = 0 we get g′(0) = (u′, v′)− (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. Proof. Suppose that u1 and u2 are solutions of (V), i.e., u1, u2 ∈ V and (u′1, v ′) = (f, v) ∀v ∈ V, (u′w, v ′) = (f, v) ∀v ∈ V. Subtracting these two equations and v = u1 − u2 ∈ V gives ∫ 1 0 (u′1 − u′2)2 dx = 0 which shows that u′1(x)− u′2(x) = 0 ∀x ∈ [0, 1], 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. 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. Last, we show that Proposition 5. If u′′(x) exists and it is continuous for a solution of (V), then u is also a solution of (D). Proof. For this, assuming that u is a solution of (V), we have (u′, v′) = (f, v) ∀v ∈ V, and integrating by parts the first term, using that v(0) = v(1) = 0 and that u′′ exists we get, (u′, v′) = (f, v) ∀v ∈ V, − (u′′, v) = (f, v) ∀v ∈ V, (f + u′′, v) = 0 ∀v ∈ V, but with the assumption that f + u′′ is continuous, this can only hold if f(x) + u′′(x) = 0 ∀x ∈ (0, 1) . 2 Variational Formulation • Hilbert Spaces (L2, H1 and H10) – Scalar product, Cauchy-Schwarz inequal- ity and Cauchy sequence • Natural and Essential boundary conditions Outline: Hilbert Spaces Definition 1. A map a(·, ·) : V × V → R that is sym- metric and bilinear, i.e., ∀u, v, w ∈ V , ∀α, β ∈ R a(v, w) = (w, v), (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 a(v, v) > 0, ∀v ∈ V, v 6= 0, and the norm associated with the scalar product is defined by ||v||a = (a(v, v))1/2, ∀v ∈ V. Moreover, if is a scalar product with correspond- ing norm || · ||, we have that the following Cauchy- Schwarz’s inequality is satisfied | | ≤ ||v|| ||w||, ∀v, w ∈ V. 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 • 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. Example 2.1. We can see that the space V := C0([−1, 1]) is a linear space, and that (·, ·)V : V ×V → R defined by (v, w)V = ∫ 1 −1 v(x)w(x) dx, 2 -1 -0.5 0 0.5 1 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 v j v i 1/N 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 norm ||v||V = (v, v)1/2V . But V with this norm is not complete, thus it is not a Hilbert space. Proof. To see this we can build a Cauchy sequence that is not convergent inside V . For example, consider the sequence vn =  0 if x ∈ [−1, 0], nx if x ∈ (0, 1/n), 1 if x ∈ [1/n, 1]. Take ε > 0, and 1/ε < N ∈ N, we can see (Figure 1 that the sequence is Cauchy ||vi − vj ||2V = ∫ 1 −1 (vi(x)− vj(x))2 dx = ∫ 1/N 0 0≤·≤1︷ ︸︸ ︷ (vi(x)− vj(x))2 dx ≤ ∫ 1/N 0 dx = 1/N < ε, but the sequence converges to the function H(x) = { 0 if x ∈ [−1, 0], 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: • The linear space of square-integrable functions in a bounded domain Ω ∈ Rd, L2(Ω) = {v : ∫ Ω |v|2 dx <∞}, together with the scalar product and its associated norm as follows (v, w)L2(Ω) = ∫ Ω vw dx, ||v||L2(Ω) = ∫ Ω v2 dx. • We also introduce the linear space of square inte- grable derivatives up to order 1, H1(Ω) = {v ∈ L2(Ω), vxi ∈ L2, i = 1, . . . , d}, together with the scalar product and the associated norm as follows (v, w)H1(Ω) = ∫ Ω vw +∇v · ∇w dx, ||v||H1(Ω) = ∫ Ω v2+|∇v|2 dx = ||v||L2(Ω)+ ||∇v||L2(Ω). • And the space H10(Ω) := {v ∈ H1(Ω) : v = 0 on Γ}, together with the same scalar product and the as- sociated norm as for H1(Ω), where Γ := ∂Ω is the boundary of Ω. 2.1 Boundary Conditions The boundary conditions for the differential equation are usually classified as Dirichlet (when they impose a value for the function on the boundary), Neumann (when they force a value for the derivative of the function on the boundary), or mixed (mixing Dirichlet and Newman), u = g on ∂Ω (Dirichlet) ~n∇u = h on ∂Ω (Neumann) γu+ ~n∇u = g on ∂Ω (Mixed) where g, h, γ are given functions. Building the varia- tional formulation, when integrating by parts the orig- inal 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 • essential boundary conditions will appear in the function space V , while • natural boundary conditions will naturally appear in the variational form. In general, for an equation of order 2k, the variational form needs an space with functions differentiable up to order k, so essential boundary conditions will be inposed 3 in the space for orders 0, . . . , k − 1, and natural bound- ary 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 func- tion where the functions would be zero. For instance, consider the boundary, Γ, split in two parts, Γ1 and Γ2, then the two spaces are • u ∈ V1 = {v : v ∈ H1, and v = g on Γ1}, • v ∈ V2 = {v : v ∈ H1, and v = 0 on Γ1}, 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. Example 2.2. Consider a bounded domain Ω ∈ Rn, with boundary Γ = ∂Ω ∈ C1 piecewise, and the bound- ary is split in two parts Γ1 and Γ2. The problem reads −∆u = f in Ω ⊂ Rn, u = g on Γ1 ~n · ∇u = h on Γ2 Proof. We use the spaces V1 = {v : v ∈ H1, and v = g on Γ1}, V2 = {v : v ∈ H1, and v = 0 on Γ1}, 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 − ∫ Ω ∆uv dx = ∫ Ω fv dx, ∀v ∈ V2, but integrating by parts (Green’s formula) the left part − ∫ Ω ∆uv dx = ∫ Ω ∇u · ∇v dx− ∫ Γ ~n∇uv ds = ∫ Ω ∇u · ∇v dx− ∫ Γ1 ~n∇uv ds︸ ︷︷ ︸ v=0 on Γ1 − ∫ Γ2 ~n∇uv ds︸ ︷︷ ︸ ~n∇u=h on Γ2 = ∫ Ω ∇u · ∇v dx− ∫ Γ2 hv ds, so the variational formulation can be rewritten as Find u ∈ V1 such that∫ Ω ∇u · ∇v dx︸ ︷︷ ︸ a(u,v) = ∫ Ω fv dx+ ∫ Γ2 hv ds︸ ︷︷ ︸ L(v) , ∀v ∈ V2. 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 F (v) = 1 2 a(v, v)− L(v), and it is equivalent to the variational problem above. 3 Existence of solutions: Lax-Milgram theorem • Parallelogram law • Lax-Milgram Theorem • (V )⇔ (M) • Poincare´ Inequalities for H1 and H10 • Trace Inequality Outline: Lax-Milgram The complete proofs of the following results are provided in [And15]. Parallelogram Law: Let || · || be a norm associated to a scalar product . The following equivalence holds ||x+ y||2 + ||x− y||2 = 2||x||2 + 2||y||2. 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: (1) a(·, ·) is symmetric, i.e., a(v, w) = a(w, v), ∀v, w ∈ V ; (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 ; (4) L(·) is continuous, i.e., ∃Λ ∈ R s.t. |L(v)| ≥ Λ||v||V ,∀v ∈ V ; Then there exists a unique function u ∈ V such that a(u, v) = L(v), ∀v ∈ V, and the following estimate holds ||u||V ≤ Λ α . 4 Proof. A sketch of the proof (complete proof in [And15]) • Build a solution of (M): ∃u ∈ V solving (M) – Norm ||v|| = a(v, v) is equivalent to || · ||V – We see that β = inf v∈V F (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). • (M)⇒(V): Because of proposition 6. • Uniqueness: If u1 and u2 solve (V), subtracting the equations we obtain a(u1 − u2, v) = 0, and using (2) for v = u1 − u2 we get the result. • Estimate: Combining (2) and (4) with (V). Proposition 6 ((V)⇔(M)). Let a and L be such that they satisfy the conditions of Theorem 1. Then, the fol- lowing problems are equivalent: • Find u ∈ V such that a(u, v) = L(v), ∀v ∈ V ; (V) • Find u ∈ V such that F (u) ≤ F (v), ∀v ∈ V ; (M) where F (v) = 1 2 a(v, v)− L(v). Proof. Similar to (V⇔M) before (complete proof in [And15]). 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∫ Ω (∇u · ∇v + uv) dx = ∫ Ω fv dx, ∀v ∈ H1(Ω) Proof. Using a(v, w) := ∫ Ω (∇v · ∇w + vw) dx, L(w) := ∫ Ω fw dx, we check for the different conditions: 1. a is clearly symmetric. 2. a(v, v) = ||v||2H1(Ω) proves V-ellipticity 3. Cauchy-Schwarz in H1(Ω) provides continuity for |a(v, w)| = ||v||H1(Ω)||w||H1(Ω) 4. Cauchy-Schwarz in L2(Ω) provides continuity for |L(v)|= |(f, v)L2(Ω)|= ||f ||L2(Ω)||v||L2(Ω) ≤Λ||v||H1(Ω) with Λ = ||f ||L2(Ω). Thus, using theorem 1 there exists a unique solution. 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 Poincare´ inequalities, always having a mechanism that identify constant functions with the function zero. Theorem 2 (Poincare´ inequality for H10(Ω)). Let Ω be a bounded domain in Rn with boundary ∂Ω ∈ C1 piecewise. Let u ∈ H10(Ω). Then there exists a constant C > 0 such that ||u||L2(Ω) ≤ C||∇u||L2(Ω). Proof. A sketch of the proof (complete proof in [And15]) • We use the function φ(x) = 1/(2n)∑x2i , with ∆φ = 1, with the Green’s formula and bcs. to get ||u||2L2(Ω) = − ∫ Ω 2u(∇u · ∇φ) dx • and we use Cauchy-Schwarz inequality and boundness of φ in Ω∣∣∣∣−∫ Ω 2u(∇u · ∇φ) dx ∣∣∣∣ ≤ C||u||L2(Ω)||∇u||L2(Ω) Theorem 3 (Poincare´ inequality for H1(Ω)). Let Ω be a bounded domain in Rn with boundary ∂Ω ∈ C1 piecewise. Let u ∈ H1(Ω) and let u¯ = 1 |Ω| ∫ Ω u dx, (u¯ is the average over Ω). Then there exists a constant C > 0 such that ||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 in- volved, i.e., when Neumann boundary conditions are used. Theorem 4 (Trace inequality). Let Ω be a bounded do- main in Rn with boundary ∂Ω ∈ C1 piecewise. Then there exists a constant C > 0 such that ||u||L2(∂Ω) ≤ C||u||1/2L2(Ω)||u|| 1/2 H1(Ω) . Proof. We skip the proof. Example 3.2. Prove, using Lax-Milgram theorem, that there exists a unique solution for the variational formu- lation associated to the problem{ −∆u = f in Ω, u = 0 on ∂Ω. 5 Proof. We multiply by a function v ∈ H10(Ω) and, after using Green’s formula we get the variational problem Find u ∈ V := H10(Ω), Ω ⊂ R2, such that∫ Ω ∇u · ∇v dx︸ ︷︷ ︸ a(u,v) = ∫ Ω fv dx︸ ︷︷ ︸ L(v) , ∀v ∈ H10(Ω) We check for the different conditions of the Lax-Milgram theorem: (1) a is clearly symmetric. (2) Using Poincare´ inequality we see that a(v, v) = ||∇v||2L2(Ω) = 1 1 + C ||∇v||2L2(Ω) + C 1 + C ||∇v||2L2(Ω) ≥ 1 1 + C ||∇v||2L2(Ω) + 1 1 + C ||v||2L2(Ω) = α ( ||∇v||2L2(Ω) + ||v||2L2(Ω) ) = α||v||2H1(Ω) which proves V-ellipticity for α = 1/(C + 1). (3) Using Cauchy-Schwarz in L2(Ω) |a(v, w)| = (∇v,∇w)L2(Ω) ≤ ||∇v||L2(Ω)||∇w||L2(Ω) ≤ ||v||H1(Ω)||w||H1(Ω) so we have continuity. (4) Cauchy-Schwarz in L2(Ω) |L(v)| = (f, v)L2(Ω) ≤ ||f ||L2(Ω)||v||L2(Ω) = Λ||v||L2(Ω) ≤ Λ||v||H1(Ω) with || · ||L2(Ω) ≤ ||·||H1(Ω) and ||f ||L2(Ω) <∞ proves continuity with Λ = ||f ||L2(Ω). Thus we can apply theorem 1. Example 3.3. Prove, using Lax-Milgram theorem, that there exists a unique solution for the variational formu- lation associated to the problem −∆u = f in Ω ⊂ Rn, ~n · ∇u = g on ∂Ω u¯ = 0 where u¯ = 1 |Ω| ∫ Ω udx, where Ω is bounded and ∂Ω ∈ C1 piecewise. Proof. First, we notice that the last condition is nec- essary, because otherwise the problem will have infinite solutions u + C for every constant C ∈ R. In order to avoid this, we must impose an extra condition to fix the solution. Now, to get the variational problem, we multi- ply by a function v and, after using Green’s formula and the boundary conditions we get Find u ∈ V := {v ∈ H1(Ω) with v¯ = 0}, such that∫ Ω ∇u · ∇v dx︸ ︷︷ ︸ a(u,v) = ∫ Ω fv dx+ ∫ ∂Ω gv ds︸ ︷︷ ︸ L(v) , ∀v ∈ V We can see that V is a subspace of H1(Ω), and with the same scalar product, (·, ·)H1(Ω), and the corresponding norm, || · ||H1(Ω), it is a Hilbert space. Now we check for the different conditions of the Lax-Milgram theorem: (1) a is clearly symmetric. (2) Using Poincare´ inequality for H1(Ω), and the fact that u¯ = 0 we see that a(v, v) = ||∇v||2L2(Ω) = 1 1 + C ||∇v||2L2(Ω) + C 1 + C ||∇v||2L2(Ω) ≥ 1 1 + C ||∇v||2L2(Ω) + 1 1 + C ||v − v¯||2L2(Ω) = 1 1 + C ( ||∇v||2L2(Ω) + ||v||2L2(Ω) ) = α||v||2H1(Ω) which proves V-ellipticity for α = 1/(C + 1). (3) Using Cauchy-Schwarz in L2(Ω) |a(v, w)| = (∇v,∇w)L2(Ω) ≤ ||∇v||L2(Ω)||∇w||L2(Ω) ≤ ||v||H1(Ω)||w||H1(Ω) so we have continuity. (4) Using Cauchy-Schwarz in L2(Ω) and Trace inequality |L(v)| = (f, v)L2(Ω) + (g, v)L2(∂Ω) ≤ ||f ||L2(Ω)||v||L2(Ω) + ||g||L2(∂Ω)||v||1/2L2(Ω)||v|| 1/2 H1(Ω) ≤ ||f ||L2(Ω)||v||H1(Ω) + ||g||L2(∂Ω)||v||H1(Ω) = (||f ||L2(Ω) + ||g||L2(∂Ω)) ||v||H1(Ω) = Λ||v||H1(Ω) proves continuity with Λ = ||f ||L2(Ω) + ||g||L2(∂Ω) for ||f ||L2(Ω)<∞ and ||g||L2(∂Ω)<∞. Thus we can apply theorem 1. 6 4 FEM problem • Generate (FEM) formulation – Create triangulation Th – Vh ⊂ V with piecewise linear functions – FEM problem • Galerkin orthogonality and error estimate Outline: FEM We must construct a finite subspace Vh ⊂ V . For this we will assume that the boundary of the domain, ∂Ω, is a polygonal curve. Then we use a partition of the domain Ω into a set of non-overlapping triangles, Th = {{T}i}, such that no vertex of one triangle lies on the edge of another triangle (see Figure 2). Figure 2: Triangulation on a square with a hole (from https://fenicsproject.org) Then, for a particular variational formulation Find u ∈ V such that a(u, v) = L(v), ∀v ∈ V, we can define the finite space for linear functions as Vh := {v : v ∈ C0, v|Ti is linear for Ti ∈ Th}, where Vh is a subspace of V , and the Finite Element problem (FEM) is defined as the restriction to Vh as follows Find u∈Vh such that a(u, v) = (f, v), ∀v∈Vh. (FEM) In general, V will be a Sobolev space like H1(Ω), H10(Ω), or variations of these spaces, depending on the particular variational formulation that fits to our problem. More- over, here we consider linear functions, but we can also consider higher order polynomials of degree k, v|Ti ∈ Pk. We notice that because Vh is finite, we can choose a basis ψj(x), j = 1, . . . ,M such that the (FEM) problem can be reformulated as Find u ∈ Vh such that (u′, ψ′j) = (f, ψj) j = 1, . . . ,M and using the same basis to expand u we have u(x) =∑M i=1 ξiψi(x), where ξi is chosen accordingly, and the problem can be rewritten in matrix form as Find ξ ∈ Rn such that Aξ = b. (FEM) The FEM solution has an interpretation from a geomet- ric point of view. In particular, being uh the solution of the FEM problem, uh ∈ Vh ⊂ V , so if we choose v ∈ Vh ⊂ V we have a(u, v) = L(v), ∀v ∈ Vh, a(uh, v) = L(v), ∀v ∈ Vh, so subtracting the second equation from the first one we get the so called Galerkin Orthogonality a(u− uh, v) = 0, ∀v ∈ Vh, (GO) but because a(·, ·) is a scalar product in V , it means that the error for the solution u − uh is orthogonal to the space Vh. Moreover, in the norm defined by ||v||a := |a(v, v)|1/2, called the energy norm, the solution of the FEM problem, uh, satisfies ||u− uh||2a = a(u− uh, u− uh) = a(u− uh, u−v + v︸ ︷︷ ︸ =0 −uh) = a(u− uh, u− v) + a(u− uh, v − uh)︸ ︷︷ ︸ =0(v−uh∈Vhand (GO)) = a(u− uh, u− v) ≤ ||u− uh||a||u− v||a so for ||u−uh|| 6= 0 we can divide by this term to obtain the estimate ||u− uh||a ≤ ||u− v||a, ∀v ∈ Vh meaning that uh is the one minimizing the distance to the real solution u with respect to this norm inside Vh. If the bilinear form a(·, ·) satisfies the conditions for the Lax-Milgram theorem, then this norm is also equivalent to the norm in the space V , C1||v||V ≤ ||v||a ≤ C2||v||V , ∀v ∈ V, with C1 = √ α and C2 = √ C in the theorem. Here, the following estimate holds for uh with respect to the norm || · ||V ||u− uh||V ≤ C2 C1 ||u− v||V , ∀v ∈ Vh. 7 References [And15] H˚akan Andreasson. Lax-Milgrams sats, Poincare´s olikhet, sp˚arolikhet, 2015. Available at Grundutb/CTH/tma690/1617/HakanFEM.pdf. 1, 4, 5 [Joh12] Claes Johnson. Numerical solution of par- tial differential equations by the finite element method. Courier Corporation, 2012. 1 8