Lecture Notes: African Institute of Mathematics Senegal, January Topic Title: A short introduction to numerical methods for elliptic PDEs

Transcription

1 Lecture Notes: African Institute of Mathematics Senegal, January 26 opic itle: A short introduction to numerical methods for elliptic PDEs Authors and Lecturers: Gerard Awanou (University of Illinois-Chicago) and Johnny Guzmán (Brown University)

2 2. Description of Course In this course we will mostly focus on second-order elliptic Partial Differential Equations (PDEs). We start by developing finite difference methods (a five point stencil method) for Poisson problems in one and two dimensions. hen, we develop finite element methods for general second order problems. Finally, we develop finite element methods for Stokes problems. Students are expected to write computer code to implement the numerical methods. Also, theoretical aspects such as stability and error analysis will be covered and students are expected to be able to follow all the main arguments.

3 CHAPER wo Point Boundary Value Problem: finite difference and finite element methods Before we jump into higher dimensional problems it is instructive to consider a two point boundary value problem. wo Point Boundary Value problem (.a) (.b) (.c) u (x) =f(x) for all < x <, u() =a, u() =b, Here the function f(x) is given. his is commonly known as the source term. he data a, b is also normally given an this is known as the Dirichlet boundary data. he unknown function is u. For now we are assuming that u belongs to C 2 ([, ]) = {u : u, u, and u are continuous on [, ]}. We would like to prove a crucial property of these type of equations weak maximum principle for this problem. Before we do that let us recall basic properties of calculus. Proposition. Suppose that w C 2 ((, )) an suppose that w has a local maximum at < z < then (.2) w (z) = and w (z) Similarly if w has a local minimum at z then (.3) w (z) = and w (z). heorem. Suppose u solves (.) and f(x) for all x (, ) then (.4) max u(x) max{a, b}. x [,] On the other hand, if f(x) for all x (, ) then (.5) min u(x) min{a, b}. x [,] Proof. We only prove (.4) and leave the proof of (.5) to the reader. We start by defining the function φ(x) = (x /2)2 2 and we let w ɛ (x) = u(x)+ɛφ(x) where ɛ >. Noting that φ (x) = for all x. hen we see that w ɛ satisfies (.6a) (.6b) (.6c) w ɛ (x) =f(x) ɛ for all < x <, w ɛ () =a + ɛ 8, w ɛ () =b + ɛ 8, 3

4 4. WO POIN BOUNDARY VALUE PROBLEM: FINIE DIFFERENCE AND FINIE ELEMEN MEHODS Figure. Example: N = 3, h = /4 Since f(x) for all < x < we have that w ɛ (x) = (f(x) ɛ) >. We then see that w ɛ cannot have a maximum at any point < x <. Indeed, if w ɛ had a maximum at x then by (.2) we have w ɛ (x) contradicting that w ɛ (x) >. Hence, the maximum must occur at x = or x =. hat is, u(x) u(x) + ɛφ(x) = w ɛ (x) max{w ɛ (), w ɛ ()} = max{a + ɛ/8, b + ɛ/8} for all x. herefore, Now taking the limit at ɛ we have u(x) max{a, b} + ɛ/8 for all x. u(x) max{a, b} for all x. aking the maximum we get (.4)... Finite Difference method. Now we develop a computational method to approximate (.). o do that, we let N be an integer and define h = N+ and define x n = n h for n =,, 2,... N + (see Figure ). We will approximate u(x n ) for n =,,..., N +. Indeed, the finite difference method will find numbers v, v,..., v N+ such that v n u(x n ) for n =,,..., N +.

5 . WO POIN BOUNDARY VALUE PROBLEM 5 In order to derive the method, we use aylor expansions on the solution u u (x i ) (u (x i+ ) u (x i ))/h ( u(x i+) u(x i )) u(x i) u(x i ) )/h h h = (u(x i+ ) 2u(x i ) + u(x i ))/h 2. In fact, we can show the following lemma (we leave the details to the reader). Lemma. It holds (.7) u (x i ) = (u(x i+ ) 2u(x i ) + u(x i ))/h 2 h2 24 (u(4) (θ ) + u (4) (θ 2 )) where x i θ x i, x i θ 2 x i+. (.8a) (.8b) (.8c) herefore, we define Now, the finite difference method finds v, v,..., v N+ such that: ( v n+ + 2v n v n )/h 2 =f(x n ) for all n =, 2,..., N, v =a, v N+ =b. his will give rise to a system of N equations with N unknowns. h 2 (2v v 2 ) =f(x ) + a h 2 h 2 ( v + 2v 2 v 3 ) =f(x 2 ). h 2 ( v N + 2v N ) =f(x N ) + b h 2 In matrix form we can write 2... v f(x ) a 2. h v 2 f(x 2 ).. v 3 = f(x 3 ) + h v N f(x N ) b You can easily write a computer code to implement this method. Next, we will analyze the finite difference method described above. o do this we use some notation. We let S = {x, x,..., x N+ } and we define all discrete functions as P h = {v : v is a real valued function with domain S}. hat is, a function in P h only has to be defined on the discrete points S. We use the notation v n = v(x n ). We also denote the i-th derivative of a function as D i u(x) = u (i) (x). In particular, we denote the second derivative D 2 u(x) = u (x). We then define the discrete second derivative as D 2 h v(x n) h 2 (v(x n+) 2v(x n ) + v(x n )) for all n =,..., N.

6 6. WO POIN BOUNDARY VALUE PROBLEM: FINIE DIFFERENCE AND FINIE ELEMEN MEHODS We also use the notation D 2 h v n D 2 h v(x n). We can then rewrite (.8) with the new notation. Find v P h satisfying (.9a) (.9b) (.9c) D 2 h v n =f(x n ) for all n =, 2,..., N, v =a, v N+ =b.... Discrete Maximum Principle. We now state and prove the discrete analogue to heorem. heorem 2. Suppose v solves (.9) and f(x n ) for all n =, 2,..., N then (.) max n=,,...,n+ v n max{a, b}. On the other hand, if f(x n ) for all n =, 2,..., N then (.) min n=,,...,n+ v n min{a, b}. Proof. We will prove (.) and leave the proof (.) to the reader. Let M = max n=,,...,n+ v n. First suppose that max n=,...,n v n < M then (.) holds. On the other hand, suppose there that there exists an n N such that v n = M. hen from (.9a) we have M = v n = 2 (v n+ + v n ) + h 2 f(x n ) 2 (v n+ + v n ) M sincef(x n ) and by our hypothesis 2 (v n+ + v n ) M. Hence, we must have M = 2 (v n+ + v n ), or that 2 (M v n ) + 2 (M v n+) =. Since by our hypothesis (M v n ) and (M v n+ ) it must be v n = v n = v n+ = M. We can continue this process to show that v i = M = max n=,,...,n+ v n for all i =,,..., N +. In other words, v M is a constant discrete function. herefore, (.) trivially holds. We can use the discrete maximum principle to prove a stability result. heorem 3. Let v P h solve (.9) then we have (.2) max v n max{ a, b } + n=,,...,n+ 8 max f(x n). n=,...,n Proof. Let Q = max n=,...,n f(x n ) and define φ P h as follows φ n = φ(x n ) Q 2 (x n /2) 2 for all n =,, 2,..., N +. Not difficult to see that D 2 h φ n = Q for n =,..., N. Define w P h as w = v + φ. hen we see that w satisfies (.3a) D 2 h w n =f(x n ) Q for all n =, 2,..., N, (.3b) (.3c) v =a + Q 8, v N+ =b + Q 8.

7 . WO POIN BOUNDARY VALUE PROBLEM 7 Now, note that f(x n ) Q for all n and hence by (.) we have Since φ n we have, Moreover, we have max n=,,...,n+ w n max{a + Q 8, b + Q 8 }. max v n max w n n=,,...,n+ n=,,...,n+ max{a + Q 8, b + Q 8 } max{ a, b } + Q 8. Hence, combining the last two inequalities we get (.4) max n=,,...,n+ v n max{ a, b } + Q 8. Noting that D 2 h ( v n) = f(x n ) for all n =, 2,..., N, v = a, v N+ = b. we can apply the previous argument verbatim to get (.5) max n=,,...,n+ ( v n) max{ a, b } + Q 8. Combining inequality (.4) and (.5) we get (.2)...2. Error Estimate for Finite Difference Method. Now that we have established the stability result (.2), we can prove an error estimate for the finite difference method. Let us define the max norm for functions in w C([, ]) as w C([,]) = max x w(x). heorem 4. Suppose that u solves (.) and u C 4 ([, ]). If v solves (.9) we have max u(x n) v n h2 n=,,...,n+ 4 D4 u C([,]). Proof. Define w P h as w n = u(x n ) v n for n =,,..., N +. hen, we see that D 2 h (w n) =τ n for all n =, 2,..., N, w =, w N+ =. where τ n = D 2 h u(x n) f(x n ) = D 2 u(x n ) D 2 h u(x n). By (.7) we have (.6) τ n h2 2 D4 u C([,]) for all n =, 2,..., N. Now applying (.2) we have Combining this with (.6) proves the result. max w n n=,,...,n+ 8 max τ n. n=,...,n

8 8. WO POIN BOUNDARY VALUE PROBLEM: FINIE DIFFERENCE AND FINIE ELEMEN MEHODS.2. Finite element method. In this section we derive the finite element method for (.). he methodology is quite different from the finite difference method but at the end they give similar methods. We first need to define the weak formulation of (.). Let us recall an integration formula (.7) v (x)w(x)dx = v(x)w (x)dx + v()w() v()w(). Now let v C ([, ]) with v() = and v() = then if we multiply (.a) by v and integrate we get If we apply (.7) we get u (x)v(x)dx = u (x)v(x)dx = f(x)v(x)dx u (x)v (x)dx, where we used that v() = = v(). Hence, we have that if u solves (.) then it solves (.8) u (x)v (x)dx = f(x)v(x)dx he finite element method is based on (.8). he next step is to build a finite dimensional space of functions. We let for all v C ([, ]), v() = = v(). V h = {v C([, ]) : for all i =,,..., N, v (xi,x i+ ) is a linear function}. We can define a basis easily for this space in the following way. For each i =,, 2,... we define ψ i (see Figures and 4) { if j = i ψ i (x j ) = if j i hen we see that if v V h then we have v(x) = N+ i= v(x i )ψ i (x) for x. We see that the dimension of V h is exactly N +. We also need to define (.9) he finite element method is as follows: Find u h V h such that V h = {v V h : v() = = v()}. u h (x)v h (x)dx = f(x)v h (x)dx for all v h Vh, where u h () = u h (x ) = a and u h () = u h (x N+ ) = b. Of course, since the space Vh is finite dimensional then (.2) is equivalent to: (.2) u h (x)ψ j(x)dx = f(x)ψ j (x)dx for all j =, 2,..., N, and u h () = u h (x ) = a and u h () = u h (x N+ ) = b. Writing N+ u h (x) = u h (x i )ψ i(x) for x. i=

9 . WO POIN BOUNDARY VALUE PROBLEM x = x =.2 x 2 =.4 x 3 =.6 x 4 =.8 x 5 = Figure 2. Example of a function in V h, h=.2 we get (.2) herefore, we have N u h (x i ) i= u h (x)ψ j(x)dx = ψ i(x)ψ j(x)dx = a N+ i= u h (x i )ψ i(x)ψ j(x)dx = f(x)ψ j (x)dx ψ (x)ψ j(x) b N+ i= u h (x i ) ψ i(x)ψ j(x)dx. ψ N+(x)ψ j(x) for all j =, 2,..., N, o write this method in matrix form we define ψ (x)ψ (x)dx ψ (x)ψ 2 (x)dx... ψ (x)ψ N (x)dx A = ψ 2 (x)ψ (x)dx ψ 2 (x)ψ 2 (x)dx ψ N (x)ψ (x)dx... ψ N (x)ψ N (x)dx

10 . WO POIN BOUNDARY VALUE PROBLEM: FINIE DIFFERENCE AND FINIE ELEMEN MEHODS x x x 2 x 3 x 4 x 5 Figure 3. Graph of ψ hen, U solves u h (x ) u h (x 2 ) U =. u h (x N ) f(x)ψ (x)dx F = f(x)ψ 2(x)dx. f(x)ψ N(x)dx ψ (x)ψ (x)dx ψ N+ (x)ψ (x)dx G = a ψ (x)ψ 2 (x)dx. b ψ N+ (x)ψ 2 (x)dx. ψ (x)ψ N (x)dx ψ N+ (x)ψ N (x)dx AU = F + G.

11 . WO POIN BOUNDARY VALUE PROBLEM x x x 2 x 3 x 4 x 5 Figure 4. Graph of ψ In order to be able to implement the method we need to find need to be able to evaluate the entries of the matrix A and the vectors F and G. Lets start with A. An easy calculation shows that (see Figure 5 x [, x i ) ψ i(x) /h x [x i, x i ) = /h x [x i, x i+ ) x (x i+, ]. Since A ij = ψ i(x)ψ j(x)dx. First note that since the support of ψ i is in (x i, x i+ ) we see that A ij = when j =,..., i 2 and j = i + 2,..., N. We thus only have to consider the case j = i, i, i +. Let us start the j = i, In this case A ii = xi x i (/h) 2 dx + xi+ x i ( /h) 2 dx = 2/h.

12 2. WO POIN BOUNDARY VALUE PROBLEM: FINIE DIFFERENCE AND FINIE ELEMEN MEHODS x x x 2 x 3 x 4 x 5 Figure 5. Graph of ψ, h=.2 For j = i + we have A i,i+ = Finally, for j = i we have A i,i = xi+ x i xi ( /h)(/h)dx = /h. x i (/h)( /h)dx = /h. In other words, we get A = h.... 2

13 . WO POIN BOUNDARY VALUE PROBLEM 3 We can easily calculate G to get a G = h. b Finally, in general we cannot compute exactly F. Instead we approximate F i using the midpoint rule: xi xi+ F i = f(x)ψ i (x)dx = f(x)ψ i (x)dx + f(x)ψ i (x)dx h f(x i) + f(x i+ ). x i x i 2 where x i = x i +x i 2. Hence, computationally we can solve 2... u h (x ) 2. h u h (x 2 ).. u h (x 3 ) = u h (x N ) f(x )+f(x 2 ) 2 f(x 2 )+f(x 3 ) 2 f(x 3 )+f(x 4 ) 2. f(x N )+f(x N+ ) 2 a + h 2. b It is worth noting that the finite element method (where we approximate F with the midpoint rule) is very similar to the finite difference method. he only difference is the right hand side.

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15 CHAPER 2 Finite difference method for poisson problem in two dimensions In this chapter we consider a standard finite difference method for Poisson s problem. For simplicity, we will consider the domain = (, ) 2. We let denote the boundary of. hroughout we let x = (x, x 2 ) R 2. We define the Laplacian applied to a smooth function w: w(x) = 2 x w(x) + 2 x 2 w(x). he Poisson problem is given by (.a) (.b) u(x) =f(x) u(x) =g(x) for x for x. Here the functions f andg are given. he function f is known as the source term and g as the Dirichlet data. We will assume that f and g are smooth functions. he unknown is the function u for this section assume it belongs to C 2 () C(). As we did in the one-dimensional case we can easily prove the weak maximum principle heorem 5. Suppose u solves (.) and f(x) for x then (.2) max u(x) max g(x). x x On the other hand, if f(x) for all x then (.3) min u(x) min g(x). x x Proof. We only prove (.2) and leave (.3) to the reader. Define φ(x) = (x /2) 2 +(x 2 /2) 2 4 and let w ɛ (x) = u(x) + ɛφ(x) where ɛ >. We see φ(x) = for all x and therefore we get w ɛ (x) =f(x) ɛ for x, w ɛ (x) =g(x) + ɛφ(x) for x. Assume that w ɛ attains a local maximum at x. hen by (.2) we must have 2 x w ɛ (x) and 2 x 2 w ɛ (x) which would imply that w ɛ (x). However, w ɛ (x) = (f(x) ɛ) > by our hypothesis which reaches a contradiction. herefore, w ɛ cannot attain a local maximum in let alone its global maximum in. Hence, w ɛ must attain its global maximum on. In other words, u(x) u(x) + ɛφ(x) = w ɛ (x) max y w ɛ(y) for all x. On the other hand, max w ɛ(y) max g(y) + ɛ max φ(y) max g(y) + ɛ y y y x 8. Where we used that max y φ(y) 8. Combining the last two inequalities and taking the limit as ɛ gives u(x) max g(y) for all x. y 5

16 6 2. FINIE DIFFERENCE MEHOD FOR POISSON PROBLEM IN WO DIMENSIONS x x x 2 x 3 x 4 x 5 Figure. Graph of ψ his proves (.2).. Finite difference method In order to define a finite difference we need a grid defined on. Let N + be given and define h = /(N + ). hen, we define the grid as the collection of points We denote the the interior points as S I S = {(nh, mh) : n, m =,,..., N + }. S I = {(nh, mh) : n, m =,..., N}. and the boundary points as S b = S\S I. As we did in the one-dimensional case we define the space of discrete functions: P h = {v : v is a real valued function with domain S}. We denote the principle directions of R 2 by e = (, ) and e 2 = (, ).

17 . FINIE DIFFERENCE MEHOD 7 Figure 2. Five point stencil We define the discrete Laplacian for every x S I as follows h v(x) h 2 (v(x + he ) 2v(x) + v(x he )) + h 2 (v(x + he 2) 2v(x) + v(x he 2 )) = h 2 ( 4v(x) + v(x he ) + v(x he 2 ) + v(x + he ) + v(x + he 2 )). In fact, we can prove the following result using aylor s theorem Lemma 2. Let u C 4 () (.) max x S I u(x) h u(x) C h 2 max{ 4 x u C(), 4 x 2 u C() } (.2a) (.2b) where the constant C is independent of h and u. Here we define w C() = max w(x). x he standard finite difference method finds v P h such that h v(x) =f(x) for all x S I v(x) =g(x) for all x S b. In order to be able to write a computer code we need to write the equivalent linear system that the finite difference gives rise to. o do that we should enumerate the vertices of S I (note that there are N 2 vertices belonging to S I ). We define z, z 2,..., z N 2 as (see Figure 3) We then define v(z k ) = v k for k =,..., N 2. z n+(m )N = (nh, mh) for all n, m =,..., N.

18 8 2. FINIE DIFFERENCE MEHOD FOR POISSON PROBLEM IN WO DIMENSIONS Figure 3. Grid, N = 4, grid points S b are in red and S I in blue o write the linear system we need to distinguish between vertices that share an edge with a boundary vertex and those that don t. For vertices that don t share an edge with a boundary vertex we have the following linear equation h 2 (4v k v k+ v k+n v k v k N ) = f(z k ) for k = n + (m )N with 2 n N and 2 m N. hen for those interior vertices that share an edge with vertices belonging to line x = we have the following linear system h 2 (4v v 2 v N+ ) =f(z ) + h 2 (g(z he ) + g(z he 2 )) h 2 (4v 2 v 3 v N+2 v ) =f(z 2 ) + h 2 (g(z he 2 )). h 2 (4v N v N v 2N v N 2 ) =f(z N ) + h 2 (g(z N he 2 )) h 2 (4v N v 2N v N ) =f(z N ) + h 2 (g(z N he 2 ) + g(z N + he )) Similarly, we can write the linear equations for the rest of vertices sharing an edge with a boundary vertex.

19 2. ERROR ANALYSIS OF FINIE DIFFERENCE MEHOD 9 2. Error Analysis of Finite difference method In order to obtain the error estimates of the finite difference method we need a stability result. And, before that we need a discrete maximum principle 2.. Discrete Maximum Principle. heorem 6. Suppose v solves (.2) and f(x) for x S I then (2.) max v(x) max g(x). x S x S b On the other hand, if f(x) for all x S I then (2.2) min v(x) min g(x). x S x S b Proof. We will prove (2.) and leave the proof (2.2) to the reader. Let M = max x S v(x). First suppose that then max x SI v(x) < M then (2.) holds. On the other hand, suppose there that there exists x S I such that v(x) = M. hen from (.2a) we have M =v(x) = 4 (v(x + he ) + v(x + he 2 ) + v(x he ) + v(x he 2 )) + h 2 f(x) 4 (v(x + he ) + v(x + he 2 ) + v(x he ) + v(x he 2 )) M sincef(x) and by our hypothesis 4 (v(x + he ) + v(x + he 2 ) + v(x he ) + v(x he 2 )) M. Hence, we must have M = 4 (v(x + he ) + v(x + he 2 ) + v(x he ) + v(x he 2 )), or that 4 (M v(x + he )) + 4 (M v(x + he 2)) + 4 (M v(x he )) + 4 (M v(x he 2)) =. By our hypothesis each term on the left hand side is non-negative as so it must be v(x) = v(x + he ) = v(x + he 2 ) = v(x he ) = v(x he 2 ) = M. We can repeat the argument for the v(x + he ), v(x + he 2 ), v(x he ), v(x he 2 ) and have all its neighbors also equal to M. If we repeat this process we will arrive at v M is a constant discrete function. herefore, (2.) trivially holds Discrete Stability Result. heorem 7. Suppose v solves (.2) then (2.3) max v(x) max g(x) + x S x S b 8 max f(x). x S I Proof. Let φ(x) = Q 4 (x /2) 2 + (x 2 /2) 2 where Q = max x SI f(x). he it is not difficult to show that h φ(x) = Q for all x S I. hen we define w P h as follows w(x) = v(x) + φ(x). We see that w solves (2.4a) (2.4b) h w(x) =f(x) Q for all x S I w(x) =g(x) + φ(x) for all x S b.

20 2 2. FINIE DIFFERENCE MEHOD FOR POISSON PROBLEM IN WO DIMENSIONS Since f(x) Q for all x S I we can apply (2.) to get v(x) v(x) + φ(x) max y S b (g(y) + φ(y)) for all x S. However, max(g(y) + φ(y)) max g(y) + max φ(y) max g(y) + Q/8. y S b y S b y S b y S b Hence, we have v(x) max g(y) + Q/8. for all x S. y S b Similarly, we can show that Hence, we have which proves the result. v(x) max y S b g(y) + Q/8. for all x S. v(x) max y S b g(y) + Q/8. for all x S, 2.3. Error Estimate. We can prove the desired error estimate. heorem 8. Suppose v solves (.2) and suppose that u C 4 () solves (.) then we have max x S u(x) v(x) C h2 max{ 4 x u C(), 4 x 2 u C() } where C is independent of h and u Proof. Let w P h and let w(x) = u(x) v(x) for all x S. hen, w solves herefore, we get by (2.3) we get h w(x) = h u(x) + f(x) for all x S I w(x) = for all x S b. max x S w(x) 8 max x S I h u(x) + f(x) = 8 max x S I u(x) h u(x). he result now follows from (.).

21 CHAPER 3 Finite Element Methods for second order elliptic Problems In this chapter we study finite element methods for second order elliptic problems.. Function Spaces We first need to study some function spaces. We let R 2 be a bounded domain with Lipschitz boundary. We recall the space of m-th order continuous functions: C m () = {v : all the partial derivatives of order less than or equal to m of v are continuous in }. C m c () = {v C m () : v vanishes outside of }. Here is a compact domain and means =. We will need to make sense of derivatives of functions that are not differentiable everywhere. For example, the function u(x) = x. o do that we need to define the weakderivative of a function if it exists. his is done via integration by parts. Let us recall integration by parts formula for smooth functions. Let n = (n, n 2 ) be the unit outward pointing normal to. Suppose that u, v C (R 2 ) xi u(x)v(x)dx = u(x) xi v(x)dx u(x)v(x)n i ds for i =, 2. Let u be an integrable function on then we say that xi u exists if there is an integrable function φ such that φ(x)v(x)dx = u(x) xi v(x)dx for all v Cc (). in which case we define xi u(x) = φ(x). In other words, if xi u then (.) xi u(x)v(x)dx = u(x) xi v(x)dx for all v Cc (). Or course xi u(x) are uniquely defined up to a set of measure zero. We will also need to define the following function spaces L 2 () = {v : v is integrable on and v 2 (x)dx < }. We define the Sobolev space H m as follows H m () = {v L 2 () : α x α 2 x 2 u L 2 (), for all α + α 2 m}. he associated norm is given as follows m u 2 H m () = u 2 H j () where u 2 H j () = j= 2 α +α 2 =j α x α 2 x 2 u 2 L 2 ().

22 22 3. FINIE ELEMEN MEHODS FOR SECOND ORDER ELLIPIC PROBLEMS 2. Linear second-order elliptic problems Recall that div F (x) = x F (x) + x2 F 2 (x) where F (x) = (F (x), F 2 (x)). We will study second order elliptic problems of the form (2.a) (2.b) div (A(x) u(x)) =f(x) u(x) =g(x) for x for x. Here we assume that A(x) R 2 2 is a symmetric matrix for every x and is smooth on. Notice that if A(x) is the identity matrix for all x then div (A(x) u(x)) = u(x). We can write out div (A(x) u(x)) div (A(x) u(x)) = x (A (x) x u(x) + A 2 (x) x2 u(x)) + x (A 2 (x) x u(x) + A 22 (x) x2 u(x)). In particular, if A is a constant matrix independent of x we have div (A(x) u(x)) = A 2 x u(x) + 2A 2 x x2 u(x) + A 22 2 x 2 u(x). We assume that A is uniformly elliptic. hat, is there exists a constant θ > such that (2.2) y t A(x)y θy t y for all y R 2 and x. One can easily show that this is equivalent to A(x) having uniformly positive eigenvalues on. Since A is smooth we also have that there exists a constant γ > such that (2.3) y t A(x)z γ y z for all y, z R 2 and x. 2.. Weak Formulation of second-order problems. We define H () = {v H () : v vanishes on }. If we use the integration by parts formula (.) we can derive Gauss integral formula div F (x)v(x)dx = F (x) v(x)dx + F (x) nv(x)ds. If we apply this to (2.a) we have A(x) u(x) v(x)dx = f(x)v(x)dx for all v H (). Hence, we can state the weak formulation of (2.) as follows Find u H () with u = g on such that (2.4) A(x) u(x) v(x)dx = f(x)v(x)dx for all v H (). We use the notation (2.5) a(u, v) = and (f, u) = A(x) u(x) v(x)dx. f(x)v(x)dx. Hence, we can write the weak form as follows Find u H () with u = g on such that (2.6) a(u, v) = (f, v) for all v H (). he nice property of the weak formulation is that solutions could exists even though they do not have second derivatives. hey only require that u H (). herefore, the weak formulation is a generalization of (2.).

23 3. FINIE ELEMEN FORMULAION 23 Figure. Illustration of θ, h 3. Finite Element Formulation In this section we develop finite element method for (2.6). o do that we need to define some concepts. For simplicity we will assume that is a polygonal domain. We will assume that we have a family of triangulations of of { h }. For every h we assume that = h. If, K h and K then K is either empty, a vertex or an edge of the triangulation. We assume that the mesh is shape regular: here exists a constant κ > such that (3.) θ κ for all { h }, where θ denotes the smallest angle of the triangle. We also also define h = diam( ), and We now define finite element spaces: h = sup h h. V h = {v C() : v P ( ) for all h }, where P ( ) is the space of affine function defined on. We also define V h = {v V h, v on }.

24 24 3. FINIE ELEMEN MEHODS FOR SECOND ORDER ELLIPIC PROBLEMS Figure 2. Example of a triangulation of a polygonal domain the vertices S b are in green, and S I in blue We can also define a basis for the spaces V h. Let us denote by S ={ set of all vertices of h } S I ={x S : x / } S b =S\S I. Let us enumerate the vertices in S, S I, S b : S ={z, z 2,..., z M } S I ={z j, z j2,..., z jq } S b ={z l, z l2,..., z lr } here j j Q M, l l R M, and Q + R = M. For each i =,..., M we define ψ i V h { if j = i ψ i (z j ) = if j i hen we see that if v V h we have v(x) = i v(z i )ψ i (x) for all x.

25 3. FINIE ELEMEN FORMULAION 25 Figure 3. An example of an ψ i basis function he finite element method solves: Find u h V h with u(z) = g(z) for all z S b such that (3.2) A(x) u h (x) v h (x)dx = f(x)v h (x)dx for all v h Vh. or using the a(, ) notation Find u h V h with u(z) = g(z) for all z S b such that (3.3) a(u h, v h ) = (f, v) for all v h Vh. We can easily show, using that a(u h, ), (f, ) are linear forms that (3.4) a(u h, ψ jk ) = (f, ψ jk ) k Q. If we write u h (x) = u h (z i )ψ i (x) for all x. i M we have (3.5) u h (z i )a(ψ i, ψ jk ) = (f, ψ jk ) k Q. i M or using that u h (z) = g(z) for z S b we get: (3.6) u h (z js )a(ψ js, ψ jk ) = (f, ψ jk ) g(z ls )a(ψ ls, ψ jk ) k Q. s Q s R We then have the matrix system

26 26 3. FINIE ELEMEN MEHODS FOR SECOND ORDER ELLIPIC PROBLEMS where a(ψ j, ψ j ) a(ψ j, ψ j2 )... a(ψ j, ψ jq ) u h (z j ) (f, ψ j ) a(ψ j2, ψ j ) a(ψ j2, ψ j2 )... a(ψ j2, ψ jq ) u h (z j2 ) = (f, ψ j2 ). + G, a(ψ jq, ψ j )... a(ψ jq, ψ jq ) u h (z jq ) (f, ψ jq ) a(ψ j, ψ l ) a(ψ j, ψ l2 )... a(ψ j, ψ lr ) a(ψ j2, ψ l ) a(ψ j2, ψ l2 )... a(ψ j2, ψ lr ) g(z.... l ). G = g(z l2 ) g(z lr ) a(ψ jq, ψ l )... a(ψ jq, ψ lr ) 3.. Implemenation of FEM Error Analysis of FEM. In this section we perform an error analysis of the finite element method. We will see that error estimates follows easily from, coercivity and boundedness of bilinear form and from Galerkin Orthogonality of the FEM. Lemma 3. Coercivity: It holds (3.7) θ v 2 L 2 () a(v, v) for all v H (). Proof. Recall that a(v, v) = then the result follows from (2.2). Lemma 4. Boundedness: It holds, A(x) v(x) v(x)dx = ( v(x)) t A(x) v(x)dx (3.8) a(v, w) γ v L 2 () w L 2 () for all v, w H (). Proof. Similar to the previous proof the result follows from (2.3). Lemma 5. Galerkin Orthogonality Let u solve (2.5) and let u h solve (3.3) then we have (3.9) a(u u h, v h ) = for all v h V h. Proof. Since Vh H () we have by (2.5) However, (3.3) states that herefore, we have Using that a(, v h ) is linear we get he result now follows. a(u, v h ) = (f, v h ) for all v h V h. a(u h, v h ) = (f, v h ) for all v h V h. a(u, v h ) a(u h, v h ) = for all v h V h. a(u, v h ) a(u h, v h ) = a(u u h, v h ) for all v h V h. Now we can easily prove the following result.

27 3. FINIE ELEMEN FORMULAION 27 heorem 9. Let u solve (2.5) and let u h solve (3.3) then we have (3.) (u u h ) L 2 () γ θ min (u v h ) L v V g 2 (). h where V g h = {v h V h : v h (z) = g(z) for all z S b }. Proof. By Coercivity (3.7) we obtain Now by linearity of a(u u h, ) we have θ (u u h ) 2 L 2 () a(u u h, u u h ). a(u u h, u u h ) = a(u u h, u v h ) a(u u h, u h v h ) for all v h V g h. Since u h, v h V g h we have u h v h Vh and therefore, we get By Galerkin orthogonality (3.9) Combining the above results we get Using (3.8) we get a(u u h, u h v h ) = for all v h V g h. θ (u u h ) 2 L 2 () a(u u h, u v h ) for all v h V g h. θ (u u h ) 2 L 2 () γ (u u h) L 2 () (u v h ) L 2 () for all v h V g h. Dividing by (u u h ) L 2 () we obtain θ (u u h ) L 2 () γ θ (u v h) L 2 () for all v h V g h. he result follows by taking the minimum over v h V g h. he result (3.2) says that the FEM approximation u h is almost the best approximation to u in the space V h when measured in the L 2 () norm. Next, we give an approximation result in terms of the mesh size h. he proof of this result which is quite technical will proved in the next section. Proposition 2. If u H 2 () then (3.) min (u v h ) L 2 () Ch u H 2 (). v h V g h hen the following corollary follows from (3.) and (3.2). Corollary. Let u solve (2.5) and let u h solve (3.3) and assume that u H 2 (), then (3.2) (u u h ) L 2 () C h u H 2 (). where the constant C is independent of u and h. We now wan to study the error estimates u u h L 2 () instead. For simplicity we assume that g. o do this we need a result from PDE theory. he following is a standard H 2 regularity resullt. Proposition 3. Let u solve (2.5) with g and suppose that is a a convex polygon then (3.3) u H 2 () C f L 2 (). he result seems very reasonable since formally u = f. So, we that some combination of second derivatives are controlled by f. he result says that all the second derivatives of u are controlled by f. We can now state the desired estimate.

28 28 3. FINIE ELEMEN MEHODS FOR SECOND ORDER ELLIPIC PROBLEMS heorem. Let u solve (2.5) with g and let u h solve (3.3) and assume that is a a convex polygon, then u u h L 2 () Ch 2 f L 2 (). Please note that this result says that u u h L 2 () converges to zero with one order higher than (u u h ) L 2 (). Proof. Let φ H () solve the following problem hen, setting v = u u h we get a(φ, v) = (u u h, v) If we use Galerkin Orthogonality (3.9) we get By (3.8) we get for all v H (). u u h 2 L 2 () = a(φ, u u h) = a(u u h, φ). u u h 2 L 2 () = a(u u h, φ v h ) for all v h V h. u u h 2 L 2 () γ (u u h) L 2 () (φ v h ) L 2 () for all v h V h. Using (3.) and (3.3) we obtain u u h 2 L 2 () C h (u u h) L 2 () u u h L 2 (). After dividing by u u h L 2 (), using (3.2) and (3.3) we obtain the result Approximation of Piecewise linear functions. We are left to prove (3.). o be precise we will use standard results that are used in the analysis of elliptic PDEs. hese results are useful in many other context and not only important for the approximation theory we will give here. We will fix a reference triangle: ˆ which is the triangle with vertices (,), (, ), (,). We then state a few results. he first is Poincare s inequality. Proposition 4. Poincare s inequality: Suppose that w H ( ˆ ) and let w = ˆ ˆ w(x)dx then, (3.4) w w L 2 ( ˆ ) C w H ( ˆ ), where C is independent of w. he next result is one type Sobolev inequality. here are many Sobolev embedding and inequalities. Proposition 5. A Sobolev inequality: If v H 2 ( ˆ ) then v C( ˆ ) and the following inequality holds (3.5) v L ( ˆ ) C v H 2 ( ˆ ). We will are going to transfer estimates from h to ˆ. he important point here is that ˆ is fixed. o do this, we need to define an affine transformation from F : ˆ define as F (ˆx) = B ˆx + b.

29 3. FINIE ELEMEN FORMULAION 29 Figure 4. An example of an ψ i basis function Not that if has vertices z, z, z 2 then the first column of B is z z while the second column is z 2 z and finally b = z. hat is, B = [ z z z 2 z ], b = z. We will state a proof of the size of entries of B and B. Here we use strongly the shape regularity, (3.) Lemma 6. here exists a constant C independent of h and h such that (3.6) (B ) ij C h (B ) ij Ch for all h, i, j =, 2. We also use the following notation. Let w be defined as a function on then define the function ŵ as function on ˆ given by ŵ(ˆx) = w(f (ˆx)). Using a change of variable formula we have (3.7) w(x)dx = w(f (ˆx)) det(b )dˆx. or (3.8) ˆ w(x)dx = Now notice that using the chain rule we have ˆ ŵ(ˆx) det(b )dˆx. (3.9) x w(x) = x ŵ(f (x)) = B t ˆxŵ(F (x)) = B t ˆxŵ(ˆx),

30 3 3. FINIE ELEMEN MEHODS FOR SECOND ORDER ELLIPIC PROBLEMS where x = F (ˆx). Note that we write x to denote the gradient with respect to the variable x. Whereas, ˆx is gradient with respect to ˆx. Hence using (3.7) and (3.9) we obtain (3.2) x w(x) x v(x)dx = (B t ˆxŵ(ˆx)) (B t ˆxˆv(ˆx)) det(b )dˆx. ˆ We can then prove the following results (3.2) Lemma 7. It holds, c h w L 2 ( ) ŵ L 2 ( ˆ ) c 2 h w L 2 ( ), (3.22) c w H ( ) ŵ H ( ˆ ) c 2 w H ( ), (3.23) c h w H 2 ( ) ŵ H 2 ( ˆ ) c 2h w L 2 ( ). Proof. We only prove (3.22). he proof of the other two proofs follow a similar line of argument. By (3.2) we have By (3.6) we have and he result now follows. w 2 H ( ) = w 2 L 2 ( ) = det B B t ŵ 2 L 2 ( ˆ ). det B C h 2, B t ŵ 2 L 2 ( ˆ ) Ch 2 ŵ 2 L 2 ( ˆ ). We can now state a local approximation result known as the Bramble-Hilbert lemma. he result will follow Lemma 8. Bramble Hilbert: Let w H 2 ( ) then there exists a v P ( ) such that w v L 2 ( ) + h (w v) L 2 ( ) Ch 2 w H 2 ( ). Proof. Let v(x) = c + c x + c 2 x 2 where x = (x, x 2 ). We define first c, c 2 c i = xi w(x)dx for all i =, 2. Note that since c i = xi v(x) we have that (3.24) xi v(x)dx = Now that we have define c, c 2 we define c so that (3.25) v(x)dx = xi w(x)dx for all i =, 2. w(x)dx. hat is, c = (w(x) (c x + c 2 x 2 ))dx. If we define q(x) = w(x) v(x) and consider ˆq(ˆx). Note that by (3.24) xi q(x) = for i =, 2

31 Again, using (3.7) and (3.9) we get det B ((B t ) i ˆx ˆq(ˆx)dˆx + (B t ) i2 for i =, 2. Hence, we get Hence, we get Similarly, we have ˆ 3. FINIE ELEMEN FORMULAION 3 [ ] (det B )B ˆx ˆq(ˆx)dˆx ˆx 2 ˆq(ˆx)dˆx ˆ ˆ ˆx2 ˆq(ˆx)dˆx) = [ = ] ˆxi ˆq(ˆx)dˆx = for i =, 2 ˆq(ˆx)dˆx =. xi q(x)dx, hus, ˆxi ˆq and ˆq and by Poincare s inequality we get ˆq 2 H ( ˆ ) = ˆx ˆq 2 L 2 ( ˆ ) + ˆx 2 ˆq 2 L 2 ( ˆ ) C α ˆx α 2 ˆx 2 ˆq 2 L 2 ( ˆ ) = C ˆq 2 H 2 ( ˆ ). Using this result with (3.22) gives α +α 2 =2 q H ( ) C ˆq H ( ˆ ) C ˆq 2 H 2 ( ˆ ) h q H 2 ( ). Also, using (3.2), Poincare s inequality and the previous inequality we get q L 2 ( ) Ch ˆq L 2 ( ˆ ) Ch ˆq H ( ˆ ) Ch q L 2 ( ) Ch 2 q H 2 ( ). We will also need an inverse estimate. We can prove this result with more elementary results. We give a proof that is generalizable to other finite element spaces. Lemma 9. Inverse inequality : Let h then for every v P ( ) v H ( ) C h v L 2 ( ) where the constant C is independent of and v. Proof. By (3.22) (3.26) v H ( ) C ˆv H ( ˆ ) Note that ˆv P ( ˆ ). equivalent. Hence, Since P ( ˆ ) is a finite dimensional space we know that norms are ˆv H ( ˆ ) Ĉ ˆv L 2 ( ˆ ). where the constant Ĉ is independent ˆv. Note that we are using that ˆ is fixed. Using (3.2) we get ˆv L 2 ( ˆ ) C h v L 2 ( ). Combining the above inequalities gives the result. We can state a simple corollary. Corollary 2. Let h with vertices z i (i =,, 2) then for every v P ( ) (3.27) v H ( ) C max{v(z ), v(z ), v(z 2 )}.

32 32 3. FINIE ELEMEN MEHODS FOR SECOND ORDER ELLIPIC PROBLEMS Hence, Proof. We easily see that v 2 L 2 ( ) = v 2 (x)dx v 2 L ( ). v L 2 ( ) v L ( ). Since the maximum of v occurs at the vertices and so we have v L 2 ( ) max{v(z ), v(z ), v(z 2 )}. Using that the mesh is shape regular (3.) we easily have that Ch. Hence, combining the above inequality with (3.26). In order to prove our main result (3.) we need to define the interpolant operator. Definition 3.. For w C() we define I h w = V h by I h w(z) = w(z) for all z S Proof of (3.). First we see that I h u V g h Let us write and so inf (u v) L v V g 2 () (u I h u) L 2 (). h (3.28) (u I h u) 2 L 2 () = h (u I h u) 2 L 2 ( ). From the Bramble-Hilbert Lemma we have there exists q P ( ) so that (3.29) u q L 2 ( ) + h (u q) L 2 ( ) C h 2 u H 2 ( ). Using the triangle inequality we get (3.3) (u I h u) L 2 ( ) C( (u q) L 2 ( ) + (q I h u) L 2 ( )). We let w = (q I h u) and use (3.27) to get (q I h u) L 2 ( ) = w L 2 ( ) C max{w(z ), w(z ), w(z 2 )}, where z i (i =,, 2) are the vertices of. However, w(z i ) = q(z i ) I h u(z i ) = q(z i ) u(z i ) by the definition of I h. Hence, herefore, we have max{w(z ), w(z ), w(z 2 )} u q L ( ). (3.3) (q I h u) L 2 ( ) C u q L ( ). Clearly we have By (3.5) and definition of H 2 -norm u q L ( ) û ˆq L ( ˆ ). û ˆq L ( ˆ ) C û ˆq H 2 ( ˆ ) C( û ˆq H 2 ( ˆ ) + û ˆq H ( ˆ ) + û ˆq L 2 ( ˆ ) ). hen, combining the above inequalities and using (3.2), (3.22), (3.23) we have u q L ( ) C(h u q H 2 ( ) + u q H ( ) + h u q L 2 ( )).

33 3. FINIE ELEMEN FORMULAION 33 If we combine this inequality with this inequality with (3.3) and (3.3) we get (u I h u) L 2 ( ) C(h u q H 2 ( ) + u q H ( ) + h u q L 2 ( )). Using (3.29) we get (u I h u) L 2 ( ) C h u H 2 ( ), where we also used that the second derivatives of q P ( ) are zero since. Hence, using (3.28) we get (u I h u) 2 L 2 () C h2 h u 2 H 2 ( ) = Ch2 u 2 H 2 (). he result now follows after taking the square root of both sides.

34

35 Bibliography [] C. Johnson, Numerical solution of partial differential equations by the finite element method, Cambridge University Press, Cambridge, 987. [2] D. Gilbarg and N. rudinger, Elliptic Partial Differential Equations of Second Order, 2, Springer. 35