Numerical Methods for Partial Differential Equations

Transcription

1 Numerical Methods for Partial Differential Equations Eric de Sturler University of Illinois at Urbana-Champaign Read section 8. to see where equations of type (au x ) x = f show up and their (exact) solution (in nice cases). Read 8..2 for some further background on solutions to the variational problem. The solutions to the variational problem satisfy the differential equation if f(x) and a(x) satisfy the regularity assumptions of the differential equation. f(x) continuous a(x) continuous and differentiable The important aspect of the variational formulation is that is also has solutions if the functions f(x) and a(x) do not satisfy these regularity assumptions. Eric de Sturler 2-2 /2/

2 The variational formulation of (au x ) x = f for x c (, ) and u() = u() = is the following. Find u c V such that au v dx = fvdx for all v c V. Since we want all integrals to be well defined a useful choice for V is V h v : v 2 dx <, ( v ) 2 dx <, v() = v() =. Eric de Sturler 2 If u, v c V, where V h v : v 2 dx <, ( v ) 2 dx <, v() = v() =. and f c L 2 (, ) and a is bounded (and positive) then au v dx [ [ a(u ) 2 dx] /2 [ a(v ) 2 dx] /2 < and fv [ [ f 2 dx] /2 [ v 2 dx] /2 < by Cauchy-Schwartz. Read section Eric de Sturler /2/

3 It is important to realize that solving an equation involving an inner product (linear system) is equivalent to solving a minimization problem. Consider the functional F(w) = 2 a (w ) 2 dx fwdx. The quantity F(w) is the total energy of the function w(x). The space V is the space of functions that satisfy the boundary conditions and have finite energy. Note that for any w c V, we have F(w) = F(u + v) = F(u) + au v dx fvdx + 2 av v dx = F(u) + 2 av v dx m F(u) So, that F(w) m F(u). Eric de Sturler 2 In addition, consider the function g( ) = F(u + v) for arbitrary (but fixed) v. g( ) h F(u + v) = 2 a(u + v ) 2 dx f(u + v)dx = 2 a(u ) 2 dx fudx + au v dx fvdx a(v ) 2 dx = F(u) + [ au v dx fvdx] a(v ) 2 dx Taking the derivative of g with respect to, we see g ( ) = [ au v dx fvdx] + a(v ) 2 dx = g = Since the function v was arbitrary, F attains a minimum at u ( = ). Om the other hand, g ( ) = for = only if au v dx fvdx = (the weak formulation) is satisfied. Eric de Sturler /2/

4 Now we consider the finite element formulation. Given some partition and V h the set of continuous, piecewise linear, functions over that partition that satisfy the boundary conditions: Find the function U c V h such that au v dx = fvdx for all v c V h. Of course, the equivalent relation holds for the exact solution wrt to all v. So we can subtract the exact solution and multiply by v c V h : a(u U) v dx = for all v c V h. So the error satisfies the Galerkin orthogonality condition. I Eric de Sturler 2 Now we write U(x) = j j j (x) Integrating with each basis function j (for v) in turn we obtain a ij a i j dx and b i = f i dx = This leads to a system of equations, A = b, where A is symmetric (or Hermitian) positive definite. Note that for A SPD and b a given vector we have the quadratic system 2 xt Ax x T b + c which takes its minimum for the solution : A = b. The matrix A is symmetric, positive definite, and tridiagonal. I Eric de Sturler /2/

5 So far we have dealt only with homogenous Dirichlet boundary conditions. What if we have a nonhomogeneous Dirichlet boundary cond.? u() = u and u() = Let v be such that v () = u, v () =, v 2 dx <, and (v ) 2 dx <. We define the trial space to be the set of functions V u = v + v : v 2 dx <, (v ) 2 dx <, v() = v() = and the test space to be the set of functions V = v : v 2 dx <, (v ) 2 dx <, v() = v() = Note that u v c V. So, given some v we pick a function v from a space such that the residual is orthogonal to that space. Compare to solving a linear ODE by combining a general and a particular solution. Eric de Sturler 2 Another way to think about this is that considering the problem for u v has reduced the non-homogeneous problem to a homogeneous one. Eric de Sturler 2 9- /2/

6 Now consider a homogeneous Neumann boundary condition a(x)u (x) = (at one of the end points, which corresponds to u = ). More generally we can impose a Robin boundary condition (at x = ) a()u () + (u() u ) = g, with m the boundary heat conductivity, u a given outside temperature, and a given hear flux. g Now consider (au ) = f for x c (, ) u() = and a()u () = g Eric de Sturler 2 (au ) = f for x c (, ) u() = and a()u () = g To derive the variational formulation we multiply by test function and v integrate (by parts): (au x ) x vdx = fvdx g vd(au x ) = fvdx g [(au x )v] + au x v x dx = fvdx g a()u x ()v() a()u x ()v() + au x v x dx = fvdx Now we take v() = to get rid of the unknown boundary values: au x v x dx = fvdx + g v() Eric de Sturler 2-2 /2/

7 So, the problem (au ) = f for x c (, ) u() = and a()u () = g is transformed into Find a function such that for all, u c V auxvxdx = fvdx + gv() v c V where V = v : v 2 dx <, (v ) 2 dx <, v() =. So only the (homogeneous) Dirichlet boundary condition is enforced. We refer to Neumann and Robin boundary conditions as natural boundary conditions, because they are taken care of automatically in the weak formulation. Dirichlet boundary conditions are referred to as essential boundary conditions. Eric de Sturler 2 Have we solved our problem? Note we can apply the Galerkin condition with arbitrary value for v() (which is not constrained). The Galerkin condition gives fvdx + g v() = ( au ) vdx + a()u ()v() If we take v() = we get fvdx = ( au ) vdx for x c (, ) which shows that in equation in the interior is satisfied (see section 8..2) If we take v() = and use fvdx = ( au ) vdx for x c (, ) we get g v() = a()u ()v(), which shows the boundary condition is satisfied. Eric de Sturler /2/

8 Study section 8..2, the solution of the homogeneous problem. Let u satisfy the differential equation (au ) = f and u() = u() =. Then, clearly u satisfies the Galerkin equation, since we can integrate by parts and u c V h v : v 2 dx <, (v ) 2 dx <, v() = v() =. For the opposite (the Galerkin solution satisfies the differential equation) we need to prove two things (and make some natural assumptions).. We should show (under regularity assumptions on a(x,t) and f(x, t) ) that u is twice differentiable 2. We should show that (assuming ) the Galerkin solution satisfies the differential equation. Eric de Sturler 2 Assume the Galerkin solution is twice differentiable, then by 'disintegrating by parts we go from the Galerkin condition back to [ (au ) f]vdx = for all v c V. Now the question is whether, if this condition holds for all v c V, it is possible for [ (au ) f] to be non zero on some small part of the domain. Assume this is true for x c (, + ) and on this interval [ (au ) f] > (negative also possible). Note we make no assumption on the value of [ (au ) f] outside this interval. Now we take for v the hat function on [, + ] and this leads to a contradiction. Hence [ (au ) f] must be zero everywhere. Note that the chosen hat function is in V. Eric de Sturler /2/

9 So assuming u is twice differentiable the differential equation is exactly satisfied. Now we must prove condition. The weak solution u is differentiable and satisfies (u ) 2 <. We know that a > is continuous and differentiable; so, w = au is well-defined. Now we can integrate w = f d w(x) = w x f(z)dz. Since f is continuous, w is differentiable. Next, we define u = w a, and since a > and both a and w are differentiable, u is well-defined. x w(z) Now we integrate again to get u(x) = u + a(z)dz We know (b.c.) u =. The value w is chosen such that u() =. The fact that picking the right is hard is not of theoretical significance. w The assumptions on f and a therefore lead to a weak solution u that is twice continuously differentiable, and the differential equation is solved in the classical sense. Eric de Sturler 2 In order to compute the integrals arising in the finite element formulation (inner products) we need quadrature rules, as exact integration is impossible or expensive. Obviously, the accuracy of these quadrature rules affects the accuracy of the solution (see section 6.3, table p. 24). The basic rule is to use a quadrature rule that is sufficiently accurate that the order of convergence of the method is the same as that of the method with exact integration. We will now consider estimates of the error, both a priori error estimates as a posteriori estimates. Eric de Sturler /2/

10 Let u be the exact solution and U its approximation in V h. We define the error e = u U. The error is measured using the energy norm derived from the PDE: ævæ E = a (x)(v (x)) 2 /2 dx (prove this is a norm) /2 Let the weighted L 2 norm with weight a be defined as æwæ a = aw 2 dx. Then ævæ E = æv æ a. For V h v : v (x) 2 dx <, ( v (x)) 2 dx >, v() = v() = the integral v, u = av u dx defines an inner product (why?). Cauchy s inequality gives av u dx [ æv æ a æu æ a = ævæ E æuæ E and vudx [ ævæ a æuæ a Eric de Sturler 2 First, we compute an a priori error estimate. We show that in the energy norm the Galerkin approximation is the best approximation from. V h We have for any v c V h : æ(u U) æ 2 a = a (u U) (u U) dx = a (u U) (u v) dx + a (u U) (v U) dx = a (u U) (u v) dx Note, a (u U) (v U) dx = as (v U) c V h (error orthogonal to V h ). The Cauchy inequality now gives æ(u U) æ a 2 [ æ(u U) æ a æ(u v) æ a g æ(u U) æ a [ æ(u v) æ a for all. Note this is derived from the orthogonality properties. v c V h Eric de Sturler /2/

11 Since, we know that æ(u U) æ a [ æ(u v) æ a for all v c V h we can find a bound on the error ( æ(u U) æ a ) by taking an appropriate v c V h. A useful choice is the nodal interpolant v = h u c V h. In chapter 5 we derived (a variation of) the following bound æ(u hu) æ a [ C i æhu æ a, where C i is an interpolation constant depending on a. This leads to the error bound æu Uæ E = æ(u U) æ a [ æ(u h u) æ a [ C i æhu æ a. This proves convergence for mesh size going to zero and bounded u. Eric de Sturler 2 The error bound æu Uæ E = æ(u U) æ a [ Ciæhu æ a. This proves convergence for the mesh size going to zero and bounded u. Note that u() U() =, and so integrating the derivative of the error (u U) given a bound on this derivative gives a bound on the error that is O(æhæ a ). We will show in chapter 5 that the error actually behaves like O(æhæ 2 a ). The a priori error estimate/bound establishes convergence for the mesh width going to zero. Such error bounds are often of limited value in practice, because they are typically pessimistic. Eric de Sturler /2/

12 Now we derive an a posteriori error estimate. Since these take the actual solution into account, they are more useful for (adaptive) error control. æeæ 2 E = æe æ 2 a = ae e dx = au e dx au e dx = fedx au e dx Using the nodal interpolant we get æe æ a 2 = f (e h e)dx au (e h e) dx = M+ Ij f(e h e)dx j= au (e h e) dx, integrating the last term by parts using (e h e) = at the nodes: æe æ 2 a = R (U)(e h e)dx. Eric de Sturler 2 We get æe æ 2 a = R (U)(e h e)dx, where R(U) = f (au ) on (, ). From the Cauchy inequality we get æe æ 2 hr(u) a = a a (e h e) h dx [ hr(u) a a 2 dx /2 a (e h e) h 2 dx /2 = a [hr(u)] 2 dx /2 a (e he) h 2 dx /2 = æhr(u)æ a æh (e h e)æ a As shown in chapter 5 æe æ a =[ æu U æ a [ C i æhr(u)æ a, where C i depends only on a., which then leads to æh (e he)æ a [ C i æe æ a Eric de Sturler /2/

13 We can use the a posteriori error estimate to dynamically adapt the mesh., so that a given error tolerance will be satisfied. In fact, we would like to find the mesh that satisfies the error tolerance with the minimal number of elements (is work). Typically, we start with a (relatively) coarse mesh and refine based on the a posteriori error estimate. We want to equidistribute the error, that is, the contribution from each element to the global error (integral) is about equal. We can use a criterion like C i æhr(u)æ a [ TOL. Eric de Sturler 2 Now consider piecewise quadratic polynomials over each interval/element. The basis functions can be the Lagrange polynomials with nodes at the endpoints and the midpoint. In order to make them continuous we can either add a constraint (additional equation) or we can glue together basis functions from adjacent domains to enforce continuity (as we did to get the hat functions). i /2 (x) i Eric de Sturler /2/

14 Every local quadratic polynomial can be written: p(x) = p(x i ) (x x i /2 )(x x i ) ( h i/2)$( h i ) + p(x i /2 ) (x x i )(x x i ) ( h + p i/2)$(h i/2) (x i ) (x x i /2)(x x i ) (h i/2)$(h i ) Piecing (glueing) the basis functions together over two intervals we get i /2 = 4(x x i )(x x i ) 2, h i for x c I i, otherwise, i =, 2, 3,, M + i = 2(x x i+/2 )(x x i+ ) 2, for x c I i+ h i+ 2(x x i /2 )(x x i ) 2, h i for x c I i, otherwise, i =, 2, 3,, M Eric de Sturler 2 Using these basis functions a function v can be written uniquely as v(x) = i= i= M+ M v (x i /2 ) i /2 (x) + v (x i ) i (x) Hence, we say these basis functions form a nodal basis. We can write our approximate solution as U(x) = /2 /2 (x) + (x) + + M+/2 M+/2 The product with the test functions then leads to the system A = b: a ij = i j dx, b i = i fdx What is the sparsity pattern of the matrix? Eric de Sturler /2/

15 Further, the matrix A is symmetric and positive definite. Using a piecewise quadratic polynomial leads to more accurate approximations and better convergence properties. Why? Eric de Sturler /2/