A Finite Element Method for an Ill-Posed Problem. Martin-Luther-Universitat, Fachbereich Mathematik/Informatik,Postfach 8, D Halle, Abstract

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1 A Finite Element Method for an Ill-Posed Problem W. Lucht Martin-Luther-Universitat, Fachbereich Mathematik/Informatik,Postfach 8, D-699 Halle, Germany Abstract For an ill-posed problem which has its origin in several applications (e.g. electrocardiology) a weak formulation is given over a Hilbert space without any constraints. This is achieved by means of Lagrangian multipliers. Beside theoretical questions (e.g. existence and uniqueness of a solutuion) a nite element approximation is considered. Error estimates, an investigation of the condition number of the corresponding nite dimensional system and a numerical example are given.. Introduction This work has its origin in problems appearing in several applications, e. g. in electrocardiology [,2,3]. In this eld an electric potential U is measured on a subset of the surface ( ) of the human body. However what is needed in practice is the potential on the surface of the heart (see Figur which shows a simplied situation). In most cases U can not be measured on. In a classical formulation a mathematical steady state model is Problem div(a grad U) = in R n ; n = 2 or 3 () U(x) = D(x) ; x 2 (x) = ; x 2 (3) is the domain between the boundaries and which later are assumed to be suciently smooth (e.g. Lipschitz means dierentiation with respect to the outer A is the electrical conductivity which is bounded and positive < A A(x) A < for x 2 D on is assumed to be known. The boundary condition (3) means that there is no electric current going from the body on into the surrounding medium. It is well known that Problem is ill-posed [2,3]. In the following we rst consider a weak formulation using a representation with Lagrange multipliers. It is shown that there exists a unique solution in a suitable space. Then for a numerical solution error estimates, the condition number of the nite system and a nite element example are considered.

2 Fig. * * 2. Weak formulation with Lagrange multipliers For convenience we transform Problem posing there is a smooth enough function which is dened on sasying = D(x); x 2. Then u = U must be a solution of Problem 2 div(a grad u) + b u = f in (4) u = on (5) g = on (6) with the abbreviations f = div(a grad ) and g = For generality we have added eq. (4) the term b u where b may be a given smooth function dened on. For b = Problem 2 is equivalent to Problem. In order to get a weak formulation we introduce the space V = H () H =2 () ; = [, with norm k (v; ) k V = k v k 2 H () + k k2 H =2 () =2 where standard notation has been used, see e.g.[4]. Furthermore we dene the bilinear resp. linear forms a(u; v) = A @x i + b u v B(u; ; v; ) = a(u; v) + F (v; ) F (v) =! dx ; dx = dx dx 2 dx n ; (7) v ds + f v dx u ds (8) g v ds (9) (in the following we press the volumelement dx and the surface element ds) to get 2

3 Problem 3 Find (u; ) 2 V such that B(u; ; v; ) = F (v; ) 8 (v; ) 2 V () This is the weak formulation of Problem 2. It reproduces the boundary conditions (5), (6) and in + = on () Thus the Lagrange multiplier gives essentially the normal derivative of u on where no boundary condition is known a priori. Problems of this type were investigated in early times by Babuska [5]. Note that the underlying space V does not contain any constraints. The bilinear form B(; ) is in general not symmetric what is caused by the surface terms in (8). In the next section the existence and uniqueness of a solution of Problem 3 is considered. 3. Existence and uniqueness Since we did not succeed to prove the V-ellipticity of the bilinear form B the Lax-Milgram Lemma can not be applied. To prove existence and uniqueness we therefore use a theorem due to Babuska [6] Theorem Let H and H 2 be two Hilbert spaces with scalar products (; ) H respectively. Further let b(u; v) be a bilinear form on H H 2 ; u 2 H ; v 2 H 2 such that and (; ) H2 j b(u; v) j c k u k H k v k H2 ; c < (2) j b(u; v) j c 2 k v k H2 ; c 2 > (3) kuk H j b(u; v) j c 3 k u k H ; c 3 > (4) kvk H2 Further let F 2 H 2 be a linear form on H 2. Then there exists exactly one element u 2 H such that b(u ; v) = F (v) 8v 2 H 2 and k u k H c 3 k F k H 2 Herewith, H 2 ist the dual space of H 2. The task is to verify the assumptions (2), (3), (4) for b = B and H = H 2 = V We do this by a modication of the method presented in [5]. Some auxiliary results are necessary. 3

4 Lemma Suppose 2 H =2 () and < b b b on. Let w N; 2 H () be the weak solution of w N; + w N; = in N; = on, and let N; 2 H () be the (weak) solution of a(u; v) = v 8v 2 H () Then there exist positive constants c ; c 2 (c c 2 ) which are independent of such that c k w N; k H () k w N; k H () c 2 k w N; k H () (5) Proof The proof follows immediately from the denition of the weak solutions w N; and w N; Since this proof is elementary we omit it here. Theorem 2 Under the assumptions of Lemma there are constants c ; c 2 ; < c c 2 < such that c w N; k k 2 H c =2 () 2 w N; (6) and min[a; b ] k w N; k 2 H () w N; max[a; b ] k w N; k 2 H (7) () Proof Because the proof is a modication of the proof of Theorem 2.7 in [5] we only sketch it here.the inequalities (7) follow immediately from the variational equation (we press in w N; and w N; ) a(w N ; v) = R v 8v 2 H () by setting v = w N In order to prove (6) we estimate and take into account the norm denition j v jk k H =2 ()k v k H =2 () k k H =2 () = v2h =2 () j R v j k v k H =2 () Now we use [7,p.9] k k H ()=k w =2 N k H ().Combining these relations and applying Lemma we get the upper bound k k 2 c R H =2 () 2 w N ; c 2 > To gain the lower bound we make use of the inquality (7) and Lemma giving c Since furthermore [8,Chapter 8, Section 7.3] w N k w N k H (); c > k w N k H () c k k H =2 () we obtain the nal lower bound. This proves the Theorem. Lemma 2 Let 2 H =2 () and u = v w N; 2 H () Then there is a constant c > such that k u k H () c k (v; ) k V 4

5 Proof The following estimates are valid k u k H ()k v k H() + k w N; k H () k v k H () + q min [A; b ] s ()w N; k v k H () + qc min [A; b ] k k H =2 () The last two inequalities follow from ineqs. (7) and (6). From this and the denition of the norm k k V Lemma 2 follows. Now we can prove existence and uniqueness of Problem 3. Theorem 3 (I) Assume for given (v; ) 2 V; v 6= ; there exists 2 H =2 () such that I = v + w N; (8) v + v + Then the bilinear form B(; ) satises assumption (3) of Theorem. (II) Let (u; ) 2 V; u 6= ; be given and 2 H =2 () such that I 2 = 2 u + (u + w N; ) + Then the bilinear form B(.,.) satises assumption (4) of Theorem. u (9) Before we prove this Theorem we note that the conditions (8) and (9) can be fullled. For example, consider (8) and pose any element (v; ) 2 V; v 6= ; is given. Then one can compute the sum c = v + v + w N; Suppose c Then one can take, e.g., = If c < ; choose an arbitrary 2 H =2 () and compute c = R v If c + c ; (8) is true. If c + c < and c >, replace by c and choose the constant c > such that c + cc If c < ; replace by c ; c < ; and choose c such that c + cc In every case (8) can be fullled. The other cases can be discussed in an analogous manner. Proof of Theorem 3 (I) We take u = v w N; 2 H () and obtain according to (8) B(v w N; ; ; v; ) = a(v; v) a(w N; ; v) + v + (v w N; ) Now a(v; v) min [A; b ] k v k 2 H and a(w () N;; v) = R v Theorem 2 gives c 2 w N; = c 2 w N; + w N; k k 2 H =2 () With these relations we can estimate B(v w N; ; ; v; ) min[a; b ] k v k 2 H () + k k 2 H c +I =2 () 2 5

6 where I is given in (8). The right hand side (RHS) can be estimated further with the help of Lemma 2 RHS c k (v; ) k 2 V +I c c =2 k (v; ) k V k u k H () +I where c is a positive constant. According to Theorem, ineq. (3), we need which can be replaced by Assumption (8) implies j B(u; ; v; ) j k(u;)k V j B(u; ; v; ) j k(u;)k V 6= k (u; ) k V j B(u; ; v; ) j k(u;)k V 6= k (u; ) k V k(u;)k V 6= c c =2 o k (v; ) k k u k! H () k (u; ) k V and this RHS can be estimated by means of a positive constant c; RHS c k (u; ) k V ; which proves part (I) of the Theorem. The second part (II) can be veried in the same manner. The rst assumption of Theorem (continuity of B(; )) is easy to prove. Hence Theorem 3 yields in conjunction with Theorem the unique weak solution of Problem Finite dimensional Approximation (FEM) For numerical approximations we now consider on the basis of Problem 3 a nite dimensional Problem 4 Find (u; ) h 2 V h such that B((u; ) h ; (v; ) h ) = F ((v; ) h ) 8(v; ) 2 V h (2) where V h is a nite dimensional subspace of V. First we are interested in error estimates and the basic result is (as always) a typical Cea-Lemma which is under certain conditions also valid in our case Lemma 3 Let (u; ) 2 V and (u; ) h 2 V h be the solution of Problem 3 and Problem 4 resp. Suppose the constants c and c 3 in Theorem are such that c < c 3 Then k (u; ) (u; ) h k V c inf k (u; ) (v; ) h k V (v;) h 2V h where c = c 3 + c c 3 c (2) 6

7 Proof The proof starts from Theorem, ineq.(4), and uses the triangle inequality c 3 k (u; ) h (v; ) h k V j B((u; ) h (v; ) h ; (w; )) j k(w;)k V = j B((u; ) h (u; ); (w; )) + B((u; ) (v; ) h ; (w; )) j k(w;)k V The RHS can be estimated further by ineq. (2) and the triangle inequality again giving (c 3 c ) k (u; ) (u; ) h k V (c + c 3 ) k (u; ) (v; ) h k V This completes the proof of Lemma 3. Note that c and c 3 in Theorem must be such that c < c 3. Lemma 4 Let (u; ) 2 V and (u; ) h 2 V h be the solution of Problem 3 and Problem 4 resp. Let further s h be the projection of (u; ) onto V h. Then k (u; ) (u; ) h k V ( + c =c 3 ) k (u; ) s h (u; ) k V (22) where c ; c 3 > are the constants in Theorem. Proof The proof may be omitted here since it is similar to the proof of Lemma 3. Note that the parameters c ; c 3 appear explicitly again. Next we consider a nite element approximation of Problem 3. For that we pose that is a polyhedron. This allows us to cover the sets resp. with polygonal n-dimensional resp. (n)-dimensional elements (n 2). In this way we construct a triangulation T h of which satises the usual assumptions [4,p.38]. On this triangulation two nite elements (K; P K ; K ) and (K; P Kb ; Kb ); K 2 [ h T h can be dened. (We use standard theory as developed, e.g., in [4, Chapters 2,3].) As usual the spaces P K and P Kb consist of polynomials. We pose that both of the nite elements satisfy Ciarlets assumptions H ; H 2 ; H 3 [4,p.32], i.e. (H ) T h is a regular family of triangulations; (H 2 ) both of the nite elements are ane families of nite elements; (H 3 ) all nite elements are of class C Since V is dened by V = H () H =2 () beside H () we need H =2 () which is the dual space of the Sobolow space H =2 () and this space is generated by the trace operator p H ()! L 2 () such that p v is the trace of v 2 H () on for every smooth v, see e.g. [7,8]. Let H h() be a nite dimensional subspace of H () In our case the space P Kb = [v h j K ; v h 2 H h()]; K 2 [ h T h ; is chosen to generate a function the p projection of which is in the nite dimensional space H =2 h () H =2 () L 2 () In other words, the numerical approximation for 2 H =2 () H =2 () is constructed by the set of polynomials P Kb ; K 2 T h ; while the numerical approximation u h 2 Hh () H () is dened in the standard way by means of P K ; K 2 T h In general, the order of the polynomials in P K and P Kb may be dierent. 7

8 With the assumptions concerning a triangulation of gives automatically a triangulation T hb of the boundary. According to assumption (H 2 ) reference elements K and K b exist for every K 2 T h and T hb; respectively. We need some denitions. We denote by s h the projection s h H ()! Hh () and by s h2 the projection s h2 H =2 ()! L 2 () Further, for an integer k P k is the set of polynomials of order k j v j m; is dened by j v j m; X jj=m j 2 dx where is a multi index. With these denitions we cite for later use Lemma 5 which is a special case of Theorem 3.2. in [4]. Lemma 5 Suppose (H ); (H 2 ); (H 3 ) are valid. (I) (Triangulation of ) Assume there is an integer k such that the following inclusions are satised P k (K) P H (K) and H k+ (K) C (K) ( means continuous injection). Then there exists a constant c independent of h such that, for any function v 2 H k+ () (II) (Triangulation of ) Assume there exists an integer k A =2 k v s h v k H () ch k j v j k+; (23) such that P k (K b ) P b H (K b ) and H k + (K b ) C (K b ) Then there exists a constant c independent of h such that for 2 H k + () For later use we note that H (K b ) H =2 (K b ) k s h2 k L 2 () c h k+ j j k +; (24) Theorem 4 Let u 2 H k+ () and 2 H k + () \ L 2 () Suppose the assumptions of Lemmata 4 and 5 are valid. Then there are constants c > and c > such that k (u; ) (u; ) h k V ( + c =c 3 ) ch k k u k H k+ () +c ch k +3=2 k k H k + () where c ; c 3 are dened in Theorem. c and c are given in eqs. (23) and (24), respectively. Proof The starting point is ineq. (22). Using s h (u; ) = (s h u; s h2) and Lemma 5, (I), ineq. (22) can be written k (u; ) (u; ) h k V ( + c =c 3 ) ch k k u k H k+ () + k s h2 k H =2 () An estimate of the second term on the right hand side is based on the norm denition and the L 2 ()orthogonality of s h2 on s h2 w k s h2 k H =2 () = j ( s h2; w) j L 2 () w2h =2 () k w k H =2 () 8

9 j ( s = h2; w s h2w) L 2 () w2h =2 () k w k H =2 () k s h2 k L 2 () w2h =2 () k w s h2w k L 2 () k w k H =2 () Now we use the interpolation inequality in the spaces H s () k w s h2w k L 2 () ch s k w k H s () ; s 2 [; k + ] Then Lemma 5, (II), can be applied for s=/2 to give k s h2 k H =2 () c h k + k k H k + () ch =2 k w k H =2 () w2h =2 () k w k H =2 () which proves Theorem Numerical solution = c ch k +3=2 k k H k + () Now we consider the nite element solution of Problem 4, eq. (2), for a situation in two dimensions (n = 2) illustrated in Figur. is assumed to be a polygon. Since V h = Hh() H =2 h (); eq. (2) splits into and a(u h ; v h ) + h v h = F (v h ) 8v h 2 Hh () (25) u h h = 8 h 2 H =2 h () (26) where F () has been given in eq. (9). First we dene the spaces Hh() and H =2 h () with dimension M and M b resp. The triangulation (e.g. by triangles) of is posed to have M vertices. The corresponding basis functions are [w j ] M j=. On each of the boundary parts and p vertices (2p < M) may be situated. The vertices on are numbered by P ; P 2 ; ; P p and on by P p+ ; P p+2 ; ; P 2p We assume that the points P ; P p and P p+ ; P 2p are on the border of and ; resp. Here we already use the fact that the number of vertices on and must be the same, see eq. (3). The basis functions corresponding to the boundary points on and are [t j ] 2p j=. The space Hh =2 () can be constructed by standard methods. The setting of H h () H =2 () L 2 () shall be considered shortly. Although dierent nite dimensional subspaces of H =2 () can be dened we will base the numerical solution on the subspace h i H =2 h () = v h 2 C (); v h j K 2 P k (K); K 2 T hb (k = ; 2) where is written as the union = [ K2Thb K To get the nal nite dimensional equations the unknown functions u h in and h on (only this is needed) are represented in the form u h = MX i= (27) u i w i in (28) 9

10 and h = 2pX i=p+ i t i on (29) where u i ; i must be determined. We introduce the following Denition Let D be a (n; m)matrix and I; J; K; L; (I < J; K < L) be four positive integers. Then D(I; J; K; L); n = J I + ; m = L K + means such a (n; m)matrix where the row index runs from I to J and the column index runs from K to L. Then we consider six matrices A = A (; p; p + ; M); A 2 = A 2 (p + ; 2p; p + ; M); A 3 = A 3 (2p + ; M; p + ; M); O = O (; p; ; p); S = S(p + ; 2p; p + ; 2p); O 3 = O 3 (2p + ; M; ; p) A ; A 2 ; A 3 and S have elements a i;j = a(w j ; w i ) and s i;j = R t j w i resp. The elements of O and O 3 are all zero. Furthermore, according to eqs. (9) and (25) we dene the (M; )right hand side vector b with elements b j = fw j and the (M; )vector of the unknowns X with elements gw j ; j = ; ; M (3) X i = p+i for i = ; ; p and X i = u i for i = p + ; ; M With these denitions the rst nite dimensional system, eq. (25), takes the form O A C S A 2 A X = b or for short BX = b (3) O 3 A 3 The second nite dimensional equation, eq. (26), is a linear homogeneous system with a quadratic regular matrix which gives u = on, i.e. u i = for i = ; 2; ; p Therefore, this equation need not be considered further. In the following, eq. (3) is investigated. If the number of unknowns shall be equal to the number of equations determining the unknowns (we claim this) the number of vertices on must be equal to the number of vertices on. This is a remarkable result. Essentially this means that the accuracy of u (and too) on is directly proportional to the number of measurements on As mentioned in section the problem considered in this work is ill-posed. Therefore it is interesting to investigate eq. (3) under this aspect. In fact, the condition number =k B kk B k of the matrix B, eq. (3), is in most cases large. In typical numerical examples we found is of order 3 until 7. In this manner the ill-posedness of the problem appears. We will estimate in terms of the usual parameter h (' maximum diameter of the geometrical nite elements). In order to do this we note that a i;j is determined by a(w j ; w i ) (i.e. it is a volume integral and it only depends on the basis functions [w k ]) whereas s i;j is dened by the surface integral

11 s i;j = R t j w i (it depends on the basisfunctions [t k ] too). From this we can pose that the matrices A i ; i = ; 2; 3; may be of a dierent order of h than the matrix S. Let therefore A i be of the asymptotic form A i ' h m A i ; i = ; 2; 3; where A i does not depend on h for h!. Analogously S may be assumed to be of the form S ' h q S for h! (in general m 6= q). Thus B may be written B ' With these assumptions we have O h m A C h q S h m A 2 A for h! (32) O 3 h m A 3 Theorem 5 The condition number of the matrix B is for h! of order = O (33) h jm qj Proof It is easy to see from eq. (32) that O A det B ' h mm +(qm)p S A 2 O 3 A 3 Let the elements of B be b i;j and the corresponding algebraic complements be B i;j Then one can see that B i;j (which is a determinant of order M ) is of horder B i;j ' h q(p)+m(m p) for i = ; ; M and j = ; ; p; B i;j ' h qp+m(m p) for i = ; ; M and j = p + ; ; M The inverse matrix B can be written as B = det B B.. B M. B M B MM C A ' h q(p)+m(m p) B detb h qp+m(m p) B 2 for h! where the matrices B and B do not depend on h and are of the form 2 B = B (; p; ; M); B = 2 B(p + ; M; ; M) The norm of 2 B can now be estimated k B k= j detb M X i;j= =2 j B ij j 2 A ' O (h 2q + h 2m ) =2 Since according to eq. (32) k B k' O (h 2q + h 2m ) =2 we get for h! = O (2 + h 2(mq) + h 2(qm) ) =2 which for h! proves the order relation of Theorem 5. Theorem 5 shows that calculations of the elements a i;j and s i;j should be made very carefully. Furthermore a preconditioning and/or a regularization of eq. (3) may be necessary. The matrix B of the system (3) is neither symmetric nor denite. Therefore, in order to

12 solve eq. (3), a stable procedure should be chosen. In [9,] several such algorithms (conjugate gradient methods and preconditioning techniques) are presented. Especially we found the Orthodir-algorithm discussed in [9] is a suitable one. It is guaranted to converge even if the symmetric part of B is not positive denite. 6. Example We choose a two dimensional example. Let be a square from which a disc is removed = [(x; y) 2 R 2 ; < x < ; 5 < y < 5] n [(x; y) 2 R 2 ; (x 5) 2 + y 2 9] and are given by = [(x; y) 2 R 2 ; (x 5) 2 + y 2 = 9] and = [x 2 [; ] and 5 y 5; x and y 2 [5; 5]] respectively. is dened by = [x = and y 5; x = and y 5] is triangulated into N e elements (triangles; for the table below N e = 38). Given such a triangulation we take the corresponding convex hull of the vertices as the new domain. This then is a polygon as assumed in section 5. All numerical calculations are based on this polygon. The space P k is chosen to be of rst order (k = ). For P k b we take rst and second order elements (k = or 2) The data in the Problem 2 are A = ; f = ; g = 5; b = Since B is expected to have a large condition number we used Tychonovs method to regularize eq. (3), i.e. instead of eq. (3) we solved C X = B b with C = B B + I (34) where B is the transposed of B. I is the unit matrix and > a small parameter. Of course, the case! is of special interest and is considered (among others) in the following table. The solution of the linear system (34) was done by means of the Orthodir-algorithm [9]. In the table, the condition number of C ; (C ); and two of the a priori unknown function values of the electric potential are given in terms of for k =. u ist the potential in the point (5;-3) which is situated on. u 2 is u h in the point (5;-5) which is on. (C ) j k = u u For k = 2 the condition number C is larger by about one order than for k =. The function values u h are practically not signicantly dierent from the values for k =. The convergence for! is of the same manner as in the table. 2

13 References [] R. Plonsey and D. Fleming, Bioelectric Phenomena, MacGraw-Hill, New York, 969. [2] P. Colli Franzone, Some inverse problems in electrocardiology, in P. Deuhard and E. Hairer, eds., Numerical treatment of inverse problems in dierential and integral equations, Birkhauser, Basel, 983. [3] P. Colli Franzone, Regularization methods applied to an inverse problem in electrocardiology, in R. Glowinski and J. L. Lions, eds., Computing methods in applied sciences and engineering, North- Holland, 98. [4] P. G. Ciarlet, Finite element method for elliptic problems, North-Holland, 978. [5] I. Babuska, The nite element method with Lagrangian multipliers, Num. Math. 2 (973), [6] I. Babuska, Error-bounds for nite element method, Num. Math. 6 (97), [7] F. Brezzi and M. Fortin, Mixed and hybrid nite element methods, Springer-Verlag, New York, 99. [8] J. L. Lions and E. Magenes, Non-homogeneous boundary value problems and applications, Vol I, Springer-Verlag, Berlin, 972. [9] H. C. Elman, Iterative methods for large, sparse, nonsymmetric systems of linear equations, Research report 229 (Thesis), Yale University, 982. [] G. Birkho, A. Schoenstadt, eds., Elliptic Problem Solvers II, Academic Press, New York, 984 3