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1 Høgskolen i Narvik Sivilingeniørutdanningen Eksamen i Faget STE67 ELEMENTMETODEN Klasse: 4.ID Dato: Tid: Kl Tillatte hjelpemidler under eksamen: Kalkulator. Bok Numerical solution of partial differential equations by the finite element method Bok Eleementmetoder. Forelesningsnotater I, II Forelesningsnotater Engelsk/Norsk, Norsk/Engelsk ordbok Faglig kontaktperson under eksamen: Ekstern Professor Gregory A. Chechkin tel Narvik 23
2 . Consider twopoint boundary value problem (D): u (x) = x 3 + x 2 for < x < 2 u () = ; u(2) =. a) Derive the integral identity and explain why the solution u is also the solution of a variational problem (V). Solution. Let us introduce the set of admissible functions in the following way: V = {v C [, 2]; v(2) = }. Multiplying the equation by the testfunction v V, integrating over [, 2], we obtain u (x)v(x) dx = (x 3 + x 2 )v(x) dx, and finally integrating by parts and using the fact that v(2) = and u () =, we deduce u (x)v (x) dx u (2)v(2)+u ()v() = which is valid for any function v V. The formulation is: Find u V such that u (x)v (x) dx = (u, v ) = ((x 3 + x 2 ), v) v V, (x 3 +x 2 )v(x) dx, where (f, g) = f(x)g(x) dx. b) Find the functional, formulate the minimization problem and explain why the solution u is also the solution of a minimization problem (M). 2
3 Solution. The formulation is: Find u V such that F(u) F(v) for any v V, where F(v) = 2 (v, v ) ((x 3 + x 2 ), v). c) Prove the equivalence of these formulations, i.e. (M) (V ) (D). Solution. By the tasks a) and b) we proved that (D) = (V ), (D) = (M) Now let us check that (V ) = (M). Assume that u is a solution of (V), let v V and set w = v u so that v = u + w and w V. We have F(v) = F(u + w) = 2 (u + w, u + w ) ((x 3 + x 2 ), u + w) = = 2 (u, u ) ((x 3 + x 2 ), u) + (u, w ) ((x 3 + x 2 ), w) + 2 (w, w ) = = F(u) (w, w ) F(u), since (u, w ) ((x 3 + x 2 ), w) = and (w, w ). Let us show that (M) = (V ). Assume that u is a solution of (M), let v V and denote by g(t) the function g(t) = F(u+tv) = 2 (u, u )+t(u, v )+ t2 2 (v, v ) ((x 3 +x 2 ), u) t((x 3 +x 2 ), v). The differentiable function g(t) has a minimum at t = and hence g () =. 3
4 It is easy to see that and hence u is a solution of (V). Summing up, we have shown that g () = (u, v ) ((x 3 + x 2 ), v) (D) = (V ) (M). Finally, if u is a smooth weak solution, then from the integral identity of (V) integrating by parts in the back direction we can obtain the equation of (D). Everything is proved. d) Define the space V h of piecewise linear functions and show that (u u h ) 2 + ((u u h ) ) 2 dx K (u v) 2 + ((u v) ) 2 dx for any v V h. Here u h is the approximate solution of the respective variational problem (V h ). Solution. Recall that u is a solution of (D) ( respectively (V) ) and u h is a solution of (V h ) that is (u h, v ) = ((x 3 + x 2 ), v) v V h. Subtracting the integral identities for (V) and (V h ) we obtain ((u u h ), v ) = v V h. () Let v V h be an arbitrary function and set w = u h v. Then w V h and using () with v replaced by w, we get, using CauchySchwarzBunyakovski s inequality ((u u h ) ) 2 dx = ((u u h ), (u u h ) ) + ((u u h ), w ) = 4
5 Dividing by we obtain = ((u u h ), (u u h + w) ) = ((u u h ), (u v) ) 2 ((u u h ) ) 2 dx ((u v) ) 2 dx ((u u h ) ) 2 dx ((u u h ) ) 2 dx Using the Friedrichs inequality, we get (u u h ) 2 + ((u u h ) ) 2 dx K and, hence, by (2) we deduce (u u h ) 2 + ((u u h ) ) 2 dx K K ((u v) ) 2 dx K 2, 2. ((u v) ) 2 dx. (2) ((u u h ) ) 2 dx ((u u h ) ) 2 dx (u v) 2 + ((u v) ) 2 dx. 2. Explain why v P 3 (, 2) is uniquely determined by the values v(), v (), v(2), v (2). Find the corresponding basis functions. Solution. To determine a cubic polynom, which has the form a 3 x 3 + a 2 x 2 + a x + a, 5
6 in the unique way it is necessary to have four different conditions. Let us write the given conditions and check that they are different. We have a a a + a = v() 3a a 2 + a = v () a a a 2 + a = v(2) 3a a a = v (2) They are different if det A, where A is a matrix of the system. The determinant of the matrix A is equal the the following: = Then there exists only one solution of the system and consequently the cubic polynom is determined uniquely. Let us remind that there exists four basis functions which correspond to the following conditions: and v() =, v () =, v(2) =, v (2) = ; v() =, v () =, v(2) =, v (2) = ; v() =, v () =, v(2) =, v (2) = v() =, v () =, v(2) =, v (2) = ; It is possible to calculate them. They are cubic then they have the form ψ i = a i x 3 + b i x 2 + c i x + d i, i =, 2, 3, 4. Let us find the first function. Consider the system a b c d =.
7 The solution is a = 4, b = 3 4, c =, d = or ψ = 4 x3 3 4 x2 +. Let us find the second function. Consider the system The solution is a 2 b 2 c 2 d 2 =. a 2 = 4, b 2 =, c 2 =, d 2 = or ψ 2 = 4 x3 x 2 + x. Let us find the third function. Consider the system The solution is a 3 b 3 c 3 d 3 =. a 3 = 4, b 3 = 3 4, c 3 =, d 3 = or ψ 3 = 4 x x2. Let us find the fourth function. Consider the system a 4 b 4 c 4 d 4 =.
8 The solution is a 4 = 4, b 4 = 2, c 4 =, d 4 = or ψ 4 = 4 x3 2 x2. 3. Consider piecewise linear finite element space V h with basis elements {ϕ j (x)}. Find the element stiffness matrix for the triangle K with vertices at (, ), (2, ), (, ). Solution. Without loss of generality let us denote by ϕ the finction which is equal to in the point (, ), by ϕ 2 the function which is equal to in the point (2, ) and by ϕ 3 the function which is equal to in the point (, ). The element stiffness matrix has the form where a K (ϕ, ϕ ) a K (ϕ, ϕ 2 ) a K (ϕ, ϕ 3 ) a K (ϕ 2, ϕ ) a K (ϕ 2, ϕ 2 ) a K (ϕ 2, ϕ 3 ) a K (ϕ 3, ϕ ) a K (ϕ 3, ϕ 2 ) a K (ϕ 3, ϕ 3 ) a K (ϕ i, ϕ j ) = ϕ i ϕ j dx. Let us calculate the gradient of each basic functions. In fact they are ( ) ( ) ( ) ϕ = 2, ϕ 2 = 2, ϕ 3 =. Finally the element stiffness matrix is equal to Consider the problem (D) K u (x) = sin x for < x < u () = ; u() =. 8,
9 a) Formulate the problem (V h ), which corresponds to problem (D) in terms of stiffness matrix, load vector and coefficients of unknown function. Solution. The variational formulation in V h is (V h ) Find u h V h : (u h, v ) = (sin x, v) v V h. It is easy to prove that instead of considering this identity for any v V h one can consider only M equations with basic functions (u h, ϕ ) = (sin x, ϕ ) (u h, ϕ M ) = (sin x, ϕ M) Let us prove that from the system of equations the integral identity follows for any v V h. Suppose that the representation of v has the form: then v = η ϕ η M ϕ M, (u h, v ) = (u h, η ϕ η M ϕ M) = η (u h, ϕ ) η M (u h, ϕ M) = = η (sin x, ϕ )+...+η M (sin x, ϕ M ) = (sin x, η ϕ +...+η M ϕ M ) = (sin x, v). And we complete the proof. Now let us consider the representation of an unknown function u h = ξ ϕ ξ M ϕ M (3) and substitute it in the system (3). We get (ξ ϕ ξ Mϕ M, ϕ ) = (sin x, ϕ ) (ξ ϕ ξ Mϕ M, ϕ M ) = (sin x, ϕ M) or ξ (ϕ, ϕ ) ξ M(ϕ M, ϕ ) = (sin x, ϕ ) ξ (ϕ, ϕ M) ξ M (ϕ M, ϕ M) = (sin x, ϕ M ). (4) 9
10 Finally the variational problem was reformulated in the form: Find unknown vector ξ, which satisfies the problem Aξ = b, Here the stiffness matrix A is the matrix of system (4) and the load vector b is the vector of the righthandsides of system (4). b) Formulate the problem (M h ), which corresponds to problem (D) in terms of stiffness matrix, load vector and coefficients of unknown function. Solution. From the representation we have v = η ϕ η M ϕ M, a(v, v) = a(η ϕ η M ϕ M, η ϕ η M ϕ M ) = = η a(ϕ, ϕ )η + η a(ϕ, ϕ 2 )η η M a(ϕ M, ϕ M )η M = η Aη, L(v) = (sin x, η ϕ η M ϕ M ) = b η, where the dot denotes the usual scalar (inner) product in IR M : ζ η = ζ η ζ M η M. Minimization problem may be formulated as: Find unknown vector ξ IR M, such that [ ] ξ Aξ b ξ = min 2 η IR M 2 η Aη b η.
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