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761_HW2

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Basics of finite element methods
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  Zeliha Kilic MATH 761 Homework 2Question 1.  If   w  is continuous on   [0 , 1] ,  and     10  vw  = 0 , ∀ v  ∈  V   =  { v | v  is continuous on   [0 , 1]  v  ∈  PC  [0 , 1] ,v (0) =  v (1) = 0 } ,  then   w  = 0 . Answer 1.  Let’s partition   [0 , 1]  in such a way that   v  will be continuous on each of them. [0 , 1] = [0 ,x 1 ] ∪ [ x 1 ,x 2 ] ∪ ... ∪ [ x i ,x i +1 ] ∪ ... ∪ [ x k , 1] We will prove this by Proof by Contradiction method. Assume that   w   = 0 . ã  Let’s assume that   { i 1  =  i,i 2 ,...,i k }  is the set of the indices of   x,  for which the  following condition holds. If the infimum of   w  1 on   [ x i ,x i +1 ] , inf  [ x i ,x i +1 ] w <  0 ,  then considering the freedom of choice of   v  ∈  V   meaning that we can find   v  ∈  V   s.t. v | [ x i ,x i +1 ]  <  0 ,  hence the infimum of   ( vw ) | [ x i ,x i +1 ] ,  will be greater   0 .    10 vwdx  ≥  L ([ x i ,x i +1 ]) = Σ i k k = i 1 (∆([ x i ,x i +1 ])) Inf  (( wv )( x ) | [ x i ,x i +1] ) dx >  0 , hence we obtained contradiction, therefore   w  has to be   0 ,  on   [0 , 1] . ã  We can easily obtain a similar contradiction when we assume   w | [ x i ,x i +1 ]  >  0 ,  by choosing an element   v | [ x i ,x i +1 ]  >  0 . Question 2.  Show that the stiffness matrix   A  is SPD. Answer 2.  We would like to prove that the stiffness matrix   A  is SPD. Let’s assume that  { Φ i }  be th basis of   V  h  ⊂  V,  then we know that   ∀ v h  ∈  V  h , ∃{ η i } s.t. v h  = Σ M i =1 η i Φ i ( x ) . ã  Symetric property for   A  follows from the the inner product. Because every entries  a ij  of   A  is defined as   a ij  = (Φ  i , Φ   j ) = (Φ   j , Φ  i ) =  a  ji , ∀ i,j. ã  In order to prove the positive definiteness of this matrix we need to prove the follow-ing, v T h Av h  >  0 , ∀ v h  ∈  V  h 1 Since  w  is given as a continuous function, we definitely know that the infimum of   w,  exists on a compact domain. 1  . Considering the above representation of   0   =  v h ,v T h Av h  = Σ M i =1 Σ M  j =1 η i a ij η  j  = Σ M i,j =1 η i ((Φ  i , Φ   j ) η  j ) = (Σ M  j =1 η i Φ  i , Σ M  j =1 Φ   j ) = ( v  h ,v  h )  >  0 , (1) Therefore,  A  is positive definite. Question 3.  Let the nodes be equally spaced with spacing   h,  what is   A ?  DOos it remind you of anything that we talked previously in class?  Answer 3.  Steps size:=  u,  and since we have equally spaced nodes,  { 0 =  x 0 ,x 1 ,x 2 ,...,x m  =1 } ,  it will become   u  =  1 m +1 .  We would like to see what the stiffness matrix   A  = ( a ij )  is.Let’s consider the linear basis   { Φ i } M i =1  for which we have supp Φ  j  ⊂  [ x  j − 1 ,x  j ] , Φ i ( x  j ) =  1 ,  if   i   =  j 0 ,  if   i  =  j as we defined in class. Tberefore we know that every entries of   A  can be written as  follows: a ij  = (Φ  i , Φ   j )Φ   j ( x ) = lim h → 0 Φ  j ( x  + h ) − Φ  j ( x ) h  (2) xx  j − 1 x  j x  j +1 1 Φ  j ( x  j ) = 11Φ  j As we see in the above figure of   Φ  j ,  it is supported in   [ x  j − 1 ,x  j +1 ] , hence we can write explicitly each   Φ  j ,  as follows: Φ  j ( x ) =  ( x − x j − 1 ) u  ,  if   x  ∈  [ x  j − 1 ,x  j )1 −  x − x j u  ,  if   x  ∈  [ x  j ,x  j +1 ) 2  therefore, Φ   j ( x ) =  1 u ,  if   x  ∈  [ x  j − 1 ,x  j ) − 1 u ,  if   x  ∈  [ x  j ,x  j +1 ) Now, we only need to calculate the inner products of   { Φ   j }  s.  Note that outside of the sup-port of these basis functions, clearly we have   0 ,  for the derivatives for all   j  = 1 , 2 ,...,M. (Φ   j , Φ   j ) =   x j +1 x j − 1 ( 1 u ) 2 dx  +   x j +1 x j ( − 1 u  ) 2 dx  =  2 u . (Φ   j − 1 , Φ   j ) =   x j +1 x j − 2 Φ   j − 1 ( x ) · Φ   j ( x ) dx  =  2   x j x j − 1 Φ   j − 1 ( x ) · Φ   j ( x ) dx  =   x j x j − 1 − 1 u  ·  1 u dx  =  − 1 u  . Let’s assume that   | i  −  j |  >  1 ,  meaning that the best case that could happen would be  | i −  j |  = 2 ,  then we have the following figure: x  j − 2 x  j − 1 x  j 1 Φ  j − 1 x  j x  j +1 x  j +2 1 Φ  j +1 10 As we see on the above figure,  supp  Φ i ∩ supp  Φ  j  =  ∅ .  This means that there is no interval on which they are non zero thus,  (Φ  i , Φ   j ) = 0 ,  when   | i −  j |  >  1 . Question 4.  How can you construct   V  h ,  for piecewise quadratic functions? 1. Find the basis functions.2. What is the stiffness matrix for the uniform partition?  Answer 4.  First, we discretize   [0 , 1]  uniformly as follows: [0 , 1] = [0 =  x 0 ,x 1 ] ∪ [ x 1 ,x 2 ] ∪ ... ∪ [ x M  ,x M  +1  = 1] where the step size is   u  =  1 M  +1 .  Considering this discretization we will find a basis   { P  i } M i =1 of quadratic functions for   V  h  satisfying the following condition: P  i ( x  j ) =  1 ,  if   i   =  j 0 ,  if   i  =  j 2 Note the support of these functions 3  P   j ( x ) =  ( x − x j − 1 )( x − x j +1 )( x j − x j − 1 )( x j − x j +1 ) ,  is a quadratic polynomial and satisfies the above condition for all   j  = 1 , 2 ,...,M.  Since we considered uniform partitioning for   [0 , 1] ,P   j ( x ) = ( x  − x  j − 1 )( x  − x  j +1 ) − ( u ) 2  for   j  ∈ { 1 , 2 ,...,M  } .  If we take each   P   j  supported in   [ x  j − 1 ,x  j +1 ]  as the basis element of  V  h ,  we will have the following, for the derivative of these basis functions: P    j ( x ) = lim h → 0 P   j ( x  + h ) − P   j ( x ) h  =  lim h → 0( x + h − x j − 1 )( x + h − x j +1 ) − ( x − x j − 1 )( x − x j +1 ) − ( u ) 2 · h  ,  if   x  ∈  ( x  j − 1 ,x  j +1 )0 ,  if   x  ∈  [ x  j − 1 ,x  j +1 ] c P    j ( x ) =  ( − 2 x + x j +1 + x j − 1 )( u ) 2  ,  if   x  ∈  ( x  j − 1 ,x  j +1 )0 ,  if   x  ∈  [ x  j − 1 ,x  j +1 ] c In order to get the entries of the corresponding stiffness matrix   ˜ A  = ( ˜ a ij )  for considered quadratic polynomial basis   { P   j } M  j =1 ,  that we have the figure below, associated to the   ( V  h ) problem, we need to calculate the following inner products: ˜ a  jj  = ( P    j ,P    j ) =   x j +1 x j − 1 ( − 2 x + x j +1 + x j − 1 ) 2 ( u ) 4  dx  =   a +2 ua ( − 2 x + a +2 u + a ) 2 ( u ) 4  dx  =  3  83 u , ˜ a  jj − 1  = ( P    j ,P    j − 1 ) =  4 ( P    j − 1 ,P    j ) = ˜ a  j − 1  j ;  then  ( P    j − 1 ,P    j ) =   x j +1 x j − 2 ( − 2 x + x j + x j − 2 )( − 2 x + x j +1 + x j − 1 )( u ) 4  dx  =  5   x j x j − 1 ( − 2 x + x j + x j − 2 )( − 2 x + x j +1 + x j − 1 )( u ) 4  dx ( P    j − 1 ,P    j ) =  6   b +2 ub + u ( − 2 x + b +2 u + b )( − 2 x + b +2 u + u + b + u )( u ) 4  dx  =  − 23 u Now, we would like to find the inner product when we have   | i −  j |  >  1˜ a ij  = ( P   i ,P    j ) =   x j +1 x i − 1 ( − 2 x + x i +1 + x i − 1 )( − 2 x + x j +1 + x j − 1 )( u ) 4  dx  =  7 0 . 3 Thanks to Mathematica 4 Because of the symetry property of the inner product 5 Due to the supports of these basis functions 6 Take  x j − 2  =  b 7 Due to the supports of these basis functions 4
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