1. Let a(x) > 0, and assume that u and u h are the solutions of the Dirichlet problem:

Transcription

1 Mathematics Chalmers & GU TMA37/MMG800: Partial Differential Equations, ; kl Telephone: Ida Säfström: Calculators, formula notes and other subject related material are not allowed. Each problem gives max 6p. Valid bonus poits will be added to the scores. Breakings: 3: 15-0p, 4: 1-7p och 5: 8p- For GU studentsg:15-7p, VG: 8p- For solutions and gradings information see the couse diary in: 1. Let ax) > 0, and assume that u and u h are the solutions of the Dirichlet problem: 1) BVP) ax)u x)) = fx), 0 < x < 1 u0) = u1) = 0, and its cg1) finite element FEM) approximation, respectively. Prove that there is a constant C i, depending only on ax), such that ) u u h E C i hu a.. Consider the estimate ). Derive the exact relation that shows how C i depends on ax). 3. Consider the two-dimensional Poisson equation with Neumann boundary condition 3) u = f, in R ; n u = k u, on, where k > 0 and n is the outward unit normal to is the boundary of ). a) Prove the Poincare inequality: u L) C u L ) + u L)). b) Use the inequality in a) and show that u L ) 0 as k. 4. Consider the problem { u = f, in = {x1, x 4) ) : 1 < x 1 <, 0 < x < } u = 0, 0n =, where f = 1 for x 1 < 0 and f = for x 1 > 0. a) Write down the discrete system SU = b S is the stiffness matrix and b is the load vector) in approximating 4) by cg1) FEM in the following triangulation: x x 1 b) Consider the same problem as in a), replacing the Dirichlet data u = 0 only) on x 1 = by the Neumann data: n u = 0 on x 1 =, 0 < x <. 5. a) Formulate a relevant minimization problem for the solution of the Poisson equation 5) u = f, in R, with n u = bg u), on, where f > 0, b > 0 and g are given functions. b) Derive an a priori error estimate for cg1) approximation in the corresponding energy-norm. MA

2 void!

3 TMA37/MMG800: Partial Differential Equations, ; kl Lösningar/Solutions. 1. See Lecture Notes or text book chapter 8.. Note that the interpolation theorem is not in the weighted norm. The ax) dependence of the interpolation constant C i can be shown as follows: Thus 1 ) 1/ u π h u) a = ax)u x) π h u) x)) dx 0 max ax)1/) u π h u) L c i max ax)1/) hu L x [0,1] x [0,1] 1 ) 1/ = c i max ax)1/) hx) u x) dx x [0,1] c i max x [0,1] ax) 1/ ) min x [0,1] ax) 1/ ) 6) C i = c i max x [0,1] ax) 1/ ) min x [0,1] ax) 1/ ), where c i is the interpolation constant independent of ax) /. ax)hx) u x) dx) 0 3. a) There is smooth function φ such that φ = 1 so that, using Greens formula u = u φ = u n φ u u φ C 1 u + C u u C 1 u + 1 u + 1 C u. This yields u C 1 u + C u C u + u ), where C = maxc 1, C ), C 1 = max n φ, and C = max φ ). b) Multiply the equation u = f by u and integrate over. Partial integration together with the boundary data n u = ku and Cauchy s inequality, yields u + k u = u u + u n u) = u u) = fu u f C u + u ) f = u C f + u C f 1 u + 1 u + C f. Subtracting 1 u + 1 u from the both sides, we end up with k 1 ) u 1 u + k 1 ) u C f, which gives that u 0 as k. 1

4 4. Let V be the linear function space defined by V h := {v : v is continuous in, v = 0, on }. Multiplying the differential equation by v V and integrating over we get that Now using Green s formula we have that u, v) = u, v) Thus the variational formulation is: u, v) = f, v), v V. n u)v ds = u, v), v V. u, v) = f, v), v V. Let V h be the usual finite element space consisting of continuous piecewise linear functions satisfying the boundary condition v = 0 on : The cg1) method is: Find U V h such that U, v) = f, v) v V h With this boundary conditions we have the internal nodes N 1 and N. Making the Ansatz x N N 1 x 1 3 T 1 Ux) = j=1 ξ jϕ j x), where ϕ i are the standard basis functions, we obtain the system of equations ξ j ϕ i ϕ j dx = fϕ j dx, i = 1,, i=1 or, in matrix form, Sξ = F, where S ij = ϕ i, ϕ j ) is the stiffness matrix and F j = f, ϕ j ) is the load vector. We first compute the mass and stiffness matrix for the reference triangle T. The local basis functions are φ 1 x 1, x ) = 1 x 1 h x h, φ 1x 1, x ) = 1 [ ] 1, h 1 φ x 1, x ) = x 1 h, φ x 1, x ) = 1 [ ] 1, h 0 φ 3 x 1, x ) = x h, φ 3x 1, x ) = 1 [ ] 0. h 1 Hence, with T = T dz = h /, s 11 = φ 1, φ 1 ) = T φ 1 dx = T = 1. h Similarly we can compute the other elements and obtain s = We can now assemble the global matrix S from the local one s: S 11 = 8s = 4, S 1 = s 1 = 1, S 1 = s 1 = 1, S = s s = + = 4

5 As for the load vector we have fϕ 1 = ϕ 1 + ϕ 1 = 4 1 x 1<0 x 1> = /3 + 4/3 =. fϕ = ϕ = = Thus the equation system is given by [ x 1>0 ][ ξ1 ] [ = ξ b) With the Neumann boundary data we obtain an addition node at N 3 =, 1) with the obvious corresponding basis function ϕ 3 which gives rise to an additional row and an additional column viz, ϕ 3 ϕ 3 =, ϕ ϕ 3 = ϕ 3 ϕ = 1 fϕ 3 = 1. Consequently the corresponding system reads as ξ 1 ξ ξ 3 = 5. a) Multiply the equation by v, integrate over, partial integrate, and use the boundary data to obtain fv = u)v = n u)v + u v = buv bgv + u v, where :=. This can be rewritten as u v + buv = } {{ } :=au,v) Let now Fw) = 1 = aw, w) lw) = 1 and choose w = u + v, then Fw) = Fu + v) = Fu)+ + u v + buv fv + bgv } {{ } =0 This gives Fu) Fw) for arbitrary w. fv + ]. /3. bgv. } {{ } :=lv) w w + bww fv + bgv, + 1 v v + 1 bvv Fu). }{{} 0 b) Make the discrete ansatz U = M j=1 U jϕ j, and set v = ϕ i, i = 1,,...,M in the variational formulation. Then we get the equation system AU = B, where U is the column vector with entries U j, B is the load vector with elements B j = fϕ i + bgϕ i, and A is the matrix with elements A ij = ϕ i ϕ j + bϕ i ϕ j. Here ϕ j = ϕ j x) is the basis function hat-functions) for the set of all piecewise linear polynomials functions on a triangulation of the domain. 3

6 Finally for the energy-norm v = av, v) 1/, using the definition for U = Ux), and the Galerkin orthogonality, we estimate the error e = u U as e = ae, e) = ae, u U) = ae, u) ae, U) = ae, u) = ae, u) ae, v) = ae, u v) e u v. This gives u U = e u v, for arbitrary piecewise linear function v, due to the fact that for such U and v Galerkin orthogonality gives ae, U) = 0 and ae, v) = 0: Just notice that both U and v are the linear combination of the basis functions ϕ j for which according to the definition of U we have that ae, ϕ j ) = au, ϕ j ) au, ϕ j ) = lϕ j ) lϕ j ) = 0. In particular, we may chose the piecewise linear function v to be the interpolant u and hence get u U u v C hd u, where h is the mesh size and C is an interpolation constant independent of h and u. MA 4