Høgskolen i Narvik Sivilingeniørutdanningen

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1 Høgskolen i Narvik Sivilingeniørutdanningen Eksamen i Faget STE6237 ELEMENTMETODEN Klassen: 4.ID 4.IT 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 chechkin@mech.math.msu.su Narvik 25

2 Consider two-point boundary value problem (D): u (x) + 8u(x) = for < x < 3 u () = ; u(3) =. Task. Show that the solution u is also the solution of a variational problem (V) and a minimization problem (M). Derive the integral identity and write the respective functional. Solution. Let us introduce the set of admissible functions in the following way: V = {v C [, 3]; v(3) = }. Multiplying the equation by the test-function v V, integrating over [, 3], we obtain 3 3 u (x)v(x) dx + 8 u(x)v(x) dx =, and finally integrating by parts and using the fact that v(3) = and u () =, we deduce 3 3 u (x)v (x) dx u (3)v(3) + u ()v() + 8 = 3 3 u (x)v (x) dx + 8 which is valid for any function v V. The formulation is: Find u V such that u(x)v(x) dx = u(x)v(x) dx + v() =, (u, v ) + 8(u, v) = v() v V, where (f, g) = 3 f(x)g(x) dx. 2

3 The formulation of a minimization problem is: Find u V such that F(u) F(v) for any v V, where F(v) = 2 (v, v ) + 4(v, v) + v(). Task 2. Prove that the respective bilinear and linear forms satisfy four basic properties (the bilinear form is (i) symmetric, (ii) continuous and (iii) V-elliptic; the linear form is (iv) continuous). Solution. (i) a(u, v) = (u, v ) + 8(u, v) = = 3 3 v (x)u (x) dx u (x)v (x) dx + 8 u(x)v(x) dx = v(x)u(x) dx = (v, u ) + 8(v, u) = a(v, u); The continuity of the bilinear form. By the Cauchy-Schwarz-Bunyakovski s inequality we get 3 3 (ii) a(u, v) = u (x)v (x) dx + 8 u(x)v(x) dx 8 (u, v) H 8 u H v H, i.e. γ = 8; Let us check V -ellipticity: (iii) a(v, v) = (v, v ) + 8(v, v) (v, v) H = v 2 H, i.e. α = ; Now the continuity of the linear form. (iv) L(v) = v() 6 v H and Λ = 6. Here we used the auxiliary inequality v() 6 v H, 3

4 which can be proved in the following way. Consider v(x) v() = x v (x) dx. By the inequalities (a b) 2 2a 2 + 2b 2 and Cauchy-Schwarz-Bunyakovski s inequality we get v 2 () = v(x) x 2 x 2 v (x) dx 2v 2 (x) + 2 v (x) dx x x 3 2v 2 (x) + 2 dx (v (x)) 2 dx 2v 2 (x) (v (x)) 2 dx. Integrating this inequality over the segment [,3], we obtain or 3 3v 2 () 2 v 2 () v 2 (x) dx (v (x)) 2 dx 3 v 2 (x) dx + 6 (v (x)) 2 dx 6 v 2 H. Task 3. Define the space V h of piecewise linear functions, find the basis functions and determine the approximate solution u h in the case of three intervals (uniform partition). Solution. Let us remind that for the problem with Neumann boundary condition in the end point x = there exists three basis functions for the partition onto three subintervals in the space V h (see figures and formulae). Now we calculate the stiffness matrix A = (ϕ, ϕ ) + 8 (ϕ, ϕ ) (ϕ, ϕ 2 ) + 8 (ϕ, ϕ 2 ) (ϕ, ϕ 3 ) + 8 (ϕ, ϕ 3 ) (ϕ, ϕ 2 ) + 8 (ϕ, ϕ 2 ) (ϕ 2, ϕ 2 ) + 8 (ϕ 2, ϕ 2 ) (ϕ 2, ϕ 3 ) + 8 (ϕ 2, ϕ 3 ) (ϕ, ϕ 3) + 8 (ϕ, ϕ 3 ) (ϕ 2, ϕ 3) + 8 (ϕ 2, ϕ 3 ) (ϕ 3, ϕ 3) + 8 (ϕ 3, ϕ 3 ) 4.

5 First, we construct the basis functions. They are linear on each subinterval and have the following values in the nodes: i.e. ϕ () =, ϕ () =, ϕ (2) =, ϕ 2 () =, ϕ 2 () =, ϕ 2 (2) =, ϕ 3 () =, ϕ 3 () =, ϕ 3 (2) =, Figure : First basis function. ϕ (x) = x +, x [, ];, x [, 2];, x [2, 3]; Figure 2: Second basis function. x, x [, ]; ϕ 2 (x) = x + 2, x [, 2];, x [2, 3]. 5

6 Figure 3: Third basis function. Hence,, x [, ]; ϕ 3 (x) = x, x [, 2]; x + 3, x [2, 3]. (ϕ, ϕ ) + 8 (ϕ, ϕ ) = (ϕ, ϕ 2 ) + 8 (ϕ, ϕ 2 ) = (ϕ 2, ϕ 2 ) + 8 (ϕ 2, ϕ 2 ) = dx + 8 ( x) 2 dx = 3, dx + 8 (ϕ, ϕ 3 ) + 8 (ϕ, ϕ 3 ) =, 2 2 (ϕ 2, ϕ 3 ) + 8 (ϕ 2, ϕ 3 ) = (ϕ 3, ϕ 3 ) + 8 (ϕ 3, ϕ 3 ) = and the load vector 3 b = dx dx x 2 dx + 8 x( x)dx = 3, (2 x) 2 dx = 22 3, (x )(2 x)dx = 3, 2 3 dx + 8 (x ) 2 dx + 8 (3 x) 2 dx = 22 3, ϕ () ϕ 2 () ϕ 3 () = 6. 2

7 The corresponding linear system of equations has the form and hence A ξ = b ξ ξ 2 ξ 3 = ξ = , 27, ξ 2 = 6 48,, ξ 3 = Substituting these constants (obtained coefficients) into the linear combination u h = ξ ϕ + ξ 2 ϕ 2 + ξ 3 ϕ 3, we get, 28x +, 27 on (, ), u h, x +, 2 on (, 2), on (2, 3), The exact solution is hence, λ =, λ,2 = ±2 2, u(x) = C e 2 2x + C 2 e 2 2x. Keeping in mind that u () =, u(3) =, we deduce u(x) = 2 ( 2( + e 2 2 e 2 2x e ) 2x. ) Task 4. Consider the Neumann problem u = in ; u = n on Γ; u dx =. 7

8 Derive the variational formulation (V), which corresponds to this problem, using the space V = {v W,2 (); u dx = } and prove that the conditions (i) (iv) are satisfied, using the Poincaré inequality the inequality v 2 dx K v 2 ds C v dx 2 + ( v 2 + v 2) dx. v 2 dx and Solution. Multiplying the equation by a test function v V = W,2 (), integrating over and using the Green s formula as usual, we get then v u dx =, u v dx = v ds. Respectively, the variational formulation has the form where The conditions (i) (iv): Find u V such that a(u, v) = L(v) v V, a(u, v) = u v dx, L(v) = v ds. (i) a(u, v) = u v dx = v u dx = a(v, u); 8

9 The continuity of the bilinear form. By the Cauchy-Schwarz-Bunyakovski s inequality we get (ii) a(u, v) = u v dx = ( u, v) u v u H v H, i.e. γ = ; Let us check V -ellipticity. Keeping in mind that v dx = by means of the Poincaré inequality we have. (iii) a(v, v) = v 2 dx = v dx 2 + v 2 dx K v 2 H, i.e. α = K ; Now the continuity of the linear form. Using the Cauchy-Schwarz-Bunyakovski s inequality and the second inequality from the statement of the problem, we get and Λ = (iv) L(v) = C. v ds ds v 2 ds C v H 9