First name Grade - Seminar for Applied Mathematicsgrsam/HS16/NumCSE/old_exams/final_exam... · 1 Vector solveLSE ( const Vector & c, const Vector & b, 2 unsignedint i 0 , unsignedint

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Final examNumerical Methods for CSE

R. Hiptmair, G. Alberti, F. Leonardi

August 12, 2016

A “CMake” file is provided with the templates. To generate a “Makefile” for allproblems, type “cmake .” in the folder “~/results”. Compile your programs with“make”.

In order to compile and run the C++ code related to a single problem, e.g. Problem 3, type“make problem3”. Execute the program using “./problem3”.

If you want to manually compile your code, use:

1 g++ − I / u s r / i n c l u d e / e i g e n 3 − s t d =c ++11−Wno−d e p r e c a t e d − d e c l a r a t i o n s −Wno− i g n o r e d − a t t r i b u t e sf i l e n a m e . cpp −o programname

or

1 c l a n g ++ − I / u s r / i n c l u d e / e i g e n 3 − s t d =c ++11−Wno−d e p r e c a t e d − d e c l a r a t i o n s −Wno− i g n o r e d − a t t r i b u t e sf i l e n a m e . cpp −o programname

The flags -Wno-deprecated-declarations -Wno-ignored-attributes sup-press some unwanted EIGEN warning.

For each problem requiring C++ implementation, a template file named problemX.cppis provided. For your own convenience, there is a marker TODO in the places where you aresupposed to write your own code. All templates should compile even if left unchanged.

1

Problem 1 Symmetric Gauss-Seidel iteration (20 pts.)For a square matrix A ∈ Rn,n, define DA,LA,UA ∈ Rn,n by:

(DA)i,j ∶=

⎧⎪⎪⎨⎪⎪⎩

(A)i,j, i = j,

0, i ≠ j, (LA)i,j ∶=

⎧⎪⎪⎨⎪⎪⎩

(A)i,j, i > j,

0, i ≤ j, (UA)i,j ∶=

⎧⎪⎪⎨⎪⎪⎩

(A)i,j, i < j,

0, i ≥ j.

(1)

The symmetric Gauss-Seidel iteration associated with the linear system of equations Ax =

b is defined as

x(k+1) = (UA +DA)−1b − (UA +DA)−1LA(LA +DA)−1(b −UAx(k)), (2)

where x(k),x,b ∈ Rn (k ≥ 0) and x(0) is a given initial guess.

(1a) Give a necessary and sufficient condition on A such that the iteration (2) is well-defined.

Solution: The matrices UA+DA and LA+DA must be invertible, which is equivalent toDA being invertible. This is equivalent to the matrix A having no zeros on the diagonal.

(1b) Assume that (2) is well-defined. Show that a fixed point x ∈ Rn of the iteration (2)is a solution of the linear system of equations Ax = b.

Solution: Let x be such fixed point, then

x = (UA +DA)−1b − (UA +DA)−1LA(LA +DA)−1(b −UAx) (3)

Left-multiply by UA +DA =A −LA (invertible!):

(UA +DA)x = b −LA(LA +DA)−1(b −UAx) (4)= b − (LA +DA −DA)(LA +DA)−1(b −UAx) (5)=UAx +DA(LA +DA)−1(b −UAx) (6)

Hence,

DAx =DA(LA +DA)−1(b −UAx) (7)x = (LA +DA)−1(b −UAx) (8)

(LA +DA)x = b −UAx, (9)

which is equivalent to Ax = b, since LA +DA +UA =A by construction.

2

(1c) Implement a C++ function (in problem1.cpp)

1 us ing Ma t r ix = Eigen : : MatrixXd ;2 us ing V ec to r = Eigen : : VectorXd ;3

4 void GSIt ( c o n s t Ma t r ix & A, c o n s t V ec to r & b , V ec to r & x ,double r t o l ) ;

solving the linear system Ax = b using the iterative scheme (2). To that end, apply theiterative scheme to an initial guess x(0) provided in x. The approximated solution givenby the final iterate is then stored in x as an output.

Use a correction based termination criterion with relative tolerance rtol using the Eu-clidean vector norm.

Solution: See implementation in problem1_solution.cpp.

(1d) Test your implementation (use n = 9 and rtol = 10e-8) with the linear sys-tem given by

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

3 1 0 . . . 02 ⋱ ⋱ ⋱ ⋮

0 ⋱ ⋱ ⋱ 0⋮ ⋱ ⋱ ⋱ 10 . . . 0 2 3

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

∈ Rn,n, b =

⎡⎢⎢⎢⎢⎢⎣

1⋮

1

⎤⎥⎥⎥⎥⎥⎦

∈ Rn. (10)

Output the l2-norm of the residual of the approximated solution. Use b as your initialguess.


(1e) Using the same matrix A and the same r.h.s. vector b as above (1d), in Table 1 wehave tabulated the quantity ∥x(k) −A−1b∥2 for k = 1, . . . ,20.

Describe qualitatively and quantitatively the convergence of the iterative scheme with re-spect to the number of iterations k.

Solution: Asymptotic linear convergence can be expected after an initial phase where theerror decreases much faster. The quotient (rate of convergence) ∣x(k) − x∗∣/∣x(k−1) − x∗∣approaches 0.679048.

1 # i n c l u d e 2

3 # i n c l u d e < Eigen /Dense >

3

k ∥x(k) −A−1b∥2 k ∥x(k) −A−1b∥21 0.172631 11 0.003415632 0.0623049 12 0.002342553 0.0464605 13 0.001601734 0.0360226 14 0.001092885 0.0272568 15 0.0007445956 0.0200715 16 0.0005067847 0.0144525 17 0.0003446828 0.0102313 18 0.0002343169 0.00715417 19 0.00015923410 0.00495865 20 0.000108185

Table 1: Euclidean norms of error vectors for 20 iterates

4

5 us ing Ma t r ix = Eigen : : MatrixXd ;6 us ing V ec to r = Eigen : : VectorXd ;7

8 //! \brief Use symmetric Gauss-Seidel iterations toapproximate the solution of the system Ax = b

9 //! \param[in] A system matrix to be decomposed, must beinvertible (L + D + U)

10 //! \param[in] b r.h.s. vector11 //! \param[in,out] x initial guess as input and last value of

iteration as output12 //! \param[in] rtol relative tolerance for termination criteria13 void GSIt ( c o n s t Ma t r ix & A, c o n s t V ec to r & b , V ec to r & x ,

double r t o l ) {14

15 // Extract strictly triangular part16 auto U = Eigen : : T r i a n g u l a r V i ew < M a t r i x ,

Eigen : : S t r i c t l y U p p e r >(A) ;17 auto L = Eigen : : T r i a ng u l a r V i e w < M a t r i x ,

Eigen : : S t r i c t l y L o w e r >(A) ;18

19 // Extract non-strictly triangular parts (we only requireL+D and U+D)

20 auto UpD = Eigen : : T r i a n g u l a r V i ew < M a t r i x , Eigen : : Upper >(A) ;21 auto LpD = Eigen : : T r i an g u l a rV i e w < M a t r i x , Eigen : : Lower >(A) ;22

4

23 // Temporary storage24 V ec to r temp ( x . s i z e ( ) ) ;25 Vec to r* xo ld = &x ;26 Vec to r* xnew = &temp ;27

28 // Counter for iterations29 unsigned i n t k = 0 ;30 // Error of this and previous iteration31 double e r r ;32 // double errold = 0.;33 do {34 // Apply iteration scheme35 *xnew = UpD . s o l v e ( b ) − UpD . s o l v e ( L*LpD . s o l v e ( b −

U**xold ) ) ;36 //Compute error37 e r r = ( *xo ld − *xnew ) . norm ( ) ;38 // Output error and rate for problem 1e39 // std::cout << k++ << "\t& " << (A*(*xnew) -

b).norm() << "\t& " << (*xold - *xnew).norm() / errold <<"\t\\\\" << std::endl;

40

41 // Proceed to next step42 s t d : : swap ( x o l d , xnew ) ;43 // errold = err;44 } whi le ( e r r > r t o l * ( *xnew ) . norm ( ) ) ; // Termination

criteria when tolerance reached45

46 x = *xnew ; // Output was last value of iterations47

48 re turn ;49 }50

51 i n t main ( i n t , ch a r * * ) {52 // Build matrix and r.h.s.53 unsigned i n t n = 9 ;54

55 Ma t r ix A = Ma t r i x : : Zero ( n , n ) ;56 f o r ( unsigned i n t i = 0 ; i < n ; ++ i ) {57 i f ( i > 0 ) A( i , i −1) = 2 ;58 A( i , i ) = 3 ;59 i f ( i < n−1) A( i , i +1) = 1 ;

5

60 }61 V ec to r b = V ec to r : : C o n s t a n t ( n , 1 ) ;62

63 //// PROBLEM 1d64 s t d : : c o u t << " *** PROBLEM 1d : " << s t d : : e n d l ;65

66 V ec to r x = b ;67 GSIt ( A, b , x , 10e −8) ;68

69 double r e s i d u a l = ( A*x − b ) . norm ( ) ;70

71 s t d : : c o u t << " R e s i d u a l = " << r e s i d u a l << s t d : : e n d l ;72 }

Problem 2 Efficient sparse solver (24 pts.)Fix n ∈ N, n ≥ 2, and let 1 ≤ i0, j0 ≤ n, i0 > j0 and ci ∈ R for i = 1, . . . , n− 1. We consider an × n linear system of equations Ax = b, where A ∈ Rn,n is given by

(A)i,j =

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

1 if i = j,ci if i = j − 1,1 if i = i0, j = j0,0 otherwise,

for 1 ≤ i, j ≤ n.

(2a) Efficiently construct the matrix A defined above as an EIGEN matrix of typeEigen::SparseMatrix<double> using an intermediate triplet format. To that end,implement a function

1 us ing V ec to r = Eigen : : VectorXd ;2 us ing Ma t r ix = Eigen : : S p a r s e M a t r i x <double >;3

4 Ma t r ix bu i ldA ( c o n s t V ec to r & c , unsigned i n t i 0 , unsigned i n tj 0 ) ;

whose, given the coefficients ci in c and the indices i0 and j0, returns the sparse matrix A.


(2b) Implement a function

6

1 V ec to r solveLSE ( c o n s t V ec to r & c , c o n s t V ec to r & b ,2 unsigned i n t i 0 , unsigned i n t j 0 ) ;

that computes the solution x ∈ Rn of the system Ax = b with optimal asymptotic com-plexity O(n).

HINT: If (A)i0,j0 were zero, A would be a banded upper triangular matrix and the taskwould be easy. Regard A as a small modification of such a matrix.

Solution: We can either employ a rank-1 modification technique using the Sherman-Morrison-Woodbury formula, or use an elementary Gaussian elimination. See implemen-tation in problem2_solution.cpp.

1 # i n c l u d e 2

3 # i n c l u d e < v e c t o r >4

5 # i n c l u d e < E i g e n / S p a r s e >6 # i n c l u d e <Eigen/SparseLU >7

8 us ing T r i p l e t = Eigen : : T r i p l e t <double >;9 us ing T r i p l e t s = s t d : : v e c t o r < T r i p l e t > ;

10

11 us ing V ec to r = Eigen : : VectorXd ;12 us ing Ma t r ix = Eigen : : S p a r s e M a t r i x <double >;13

14 //! \brief Efficiently construct the sparse matrix A given c,i_0 and j_0

15 //! \param[in] c contains entries c_i for matrix A16 //! \param[in] i0 row index i_017 //! \param[in] j0 column index j_018 //! \return Sparse matrix A19 Ma t r ix bu i ldA ( c o n s t V ec to r & c , unsigned i n t i 0 , unsigned i n t

j 0 ) {20 a s s e r t ( i 0 > j 0 ) ;21

22 unsigned i n t n = c . s i z e ( ) + 1 ;23 Ma t r ix A( n , n ) ;24 T r i p l e t s t r i p l e t s ;25

26 unsigned i n t n t r i p l e t s = 2*n ;

7

27

28 // Reserve space29 t r i p l e t s . r e s e r v e ( n t r i p l e t s ) ;30

31 // Build triplets vector32 f o r ( unsigned i n t i = 0 ; i < n ; ++ i ) {33 t r i p l e t s . push_back ( T r i p l e t ( i , i , 1 ) ) ;34 i f ( i < n−1) t r i p l e t s . push_back ( T r i p l e t ( i , i +1 , c [ i ] ) ) ;35 }36 t r i p l e t s . push_back ( T r i p l e t ( i 0 , j 0 , 1 ) ) ;37

38 // Construct sparse matrix from its triplets39

40 A. s e t F r o m T r i p l e t s ( t r i p l e t s . b e g i n ( ) , t r i p l e t s . end ( ) ) ;41 re turn A;42 }43

44 //! \brief Solve system Ax = b with optimal complexity O(n)45 //! \param[in] c Entries for matrix A46 //! \param[in] b r.h.s. vector47 //! \param[in] i0 index48 //! \param[in] j0 index49 //! \return Solution x, s.t. Ax = b50 V ec to r solveLSE ( c o n s t V ec to r & c , c o n s t V ec to r & b , unsigned

i n t i 0 , unsigned i n t j 0 ) {51 a s s e r t ( c . s i z e ( ) == b . s i z e ( ) −1 && " S i z e mismatch ! " ) ;52 a s s e r t ( i 0 > j 0 ) ;53

54 //// PROBLEM 2b55

56 // Allocate solution vector57 V ec to r r e t ( b . s i z e ( ) ) ;58

59 unsigned i n t n = b . s i z e ( ) ;60

61 T r i p l e t s t r i p l e t s ;62

63 unsigned i n t n t r i p l e t s = 2*n ;64

65 // Reserve space66 t r i p l e t s . r e s e r v e ( n t r i p l e t s ) ;

8

67

68 // Build triplets vector69 f o r ( unsigned i n t i = 0 ; i < n ; ++ i ) {70 t r i p l e t s . push_back ( T r i p l e t ( i , i , 1 ) ) ;71 i f ( i < n−1) t r i p l e t s . push_back ( T r i p l e t ( i , i +1 , c [ i ] ) ) ;72 }73

74 // Construct sparse matrix from its triplets75 Ma t r ix A( n , n ) ;76 A. s e t F r o m T r i p l e t s ( t r i p l e t s . b e g i n ( ) , t r i p l e t s . end ( ) ) ;77

78 // Vectors u and v79 V ec to r u = V ec to r : : Zero ( n ) ;80 Eigen : : MatrixXd v = Eigen : : MatrixXd : : Zero (1 , n ) ;81 u ( i 0 ) = 1 ;82 v ( 0 , j 0 ) = 1 ;83

84 // Apply SMW formula85 V ec to r Ainv_b = A. t r i a n g u l a r V i e w < Eigen : : Upper > ( ) . s o l v e ( b

) ;86 r e t = Ainv_b − A. t r i a n g u l a r V i e w < Eigen : : Upper > ( ) . s o l v e ( u *

( v * Ainv_b ) ( 0 ) ) / ( 1 . + ( v *A. t r i a n g u l a r V i e w < Eigen : : Upper > ( ) . s o l v e ( u ) ) ( 0 ) ) ;

87

88 re turn r e t ;89 }90

91 i n t main ( i n t , ch a r * * ) {92 // Setup data for problem93 unsigned i n t n = 150 ;94

95 unsigned i n t i 0 = 5 , j 0 = 4 ;96

97 V ec to r b = V ec to r : : Random ( n ) ; // Random vector for b98 V ec to r c = V ec to r : : Random ( n−1) ; // Random vector for c99

100 //// PROBLEM 2a101 s t d : : c o u t << " *** PROBLEM 2 a : " << s t d : : e n d l ;102

103 Ma t r ix A = bui ldA ( c , i 0 , j 0 ) ;104

9

105 // Solve sparse system using sparse LU and our own routine106 A. makeCompressed ( ) ;107 Eigen : : SparseLU < Matr ix > s p l u ;108 s p l u . a n a l y z e P a t t e r n (A) ;109 s p l u . f a c t o r i z e (A) ;110

111 s t d : : c o u t << " E r r o r : " << s t d : : e n d l << (solveLSE ( c , b , i 0 , j 0 ) − s p l u . s o l v e ( b ) ) . norm ( ) <<s t d : : e n d l ;

112 }

Problem 3 Not-a-knot cubic spline interpolation (24 pts.)We are given an interval [a, b] ⊂ R and a knot set

M = {a = t0 < t1 < ⋅ ⋅ ⋅ < tn−1 < tn = b}

for some n ∈ N ∖ {0}. Let S3,M denote the space of cubic spline functions on M, anddefine

S3,M = {s ∈ S3,M ∶ s′′′ is continuous at t1 and at tn−1}.

(3a) Derive a linear system of equations whose solution yields the slopes s′(tj), j =

0, . . . , n, of s ∈ S3,M satisfying the interpolation conditions s(tj) = yj for given yj ∈ R,j = 0, . . . , n.

Solution: We know that (see (3.5.9) of the lecture notes), for general spline interpolants,the slopes cj ∶= s′(tj) satisfy

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

b0 a1 b1 0 ⋯ 00 b1 a2 b2 ⋱ 0⋮ ⋱ ⋱ ⋱ ⋮

⋮ bn−3 an−2 bn−2 00 ⋯ 0 bn−2 an−1 bn−1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎣

c0⋮

cn

⎤⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

3 (y1−y0h21

+y2−y1h22

)

⋮

3 (yn−1−yn−2h2n−1

+yn−yn−1h2n

)

⎤⎥⎥⎥⎥⎥⎥⎥⎦

,

wherehi = ti − ti−1, ai =

2

hi+

2

hi+1and bi =

1

hi+1.

This is an underdetermined linear system with n + 1 unknowns and n − 1 equations. Thetwo missing equations determining the splines in S3,M will be obtained by imposing thecontinuity conditions on s′′′ at t1 and at tn−1.

10

By differentiating three times the expression in (3.5.5) of the lecture notes for s∣[tj−1,tj], weobtain

s′′′∣[tj−1,tj]

(t) = 6h−3j (2yj−1 − 2yj + hjcj−1 + hjcj).

Therefore, imposing s′′′∣[t0,t1]

(t1) = s′′′∣[t1,t2](t1) yields

h−21 c0 + (h−21 − h−22 )c1 − h−22 c2 = 2(h−32 (y1 − y2) + h

−31 (y1 − y0)).

Similarly, imposing the continuity condition in tn−1 yields

h−2n−1cn−2 + (h−2n−1 − h−2n )cn−1 − h

−2n cn = 2(h−3n (yn−1 − yn) + h

−3n−1(yn−1 − yn−2)).

Finally, adding these two equations to the initial system gives

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

b20 b20 − b21 −b21 0 ⋯ 0

b0 a1 b1 0 ⋯ 00 b1 a2 b2 ⋱ 0⋮ ⋱ ⋱ ⋱ ⋮

⋮ bn−3 an−2 bn−2 00 ⋯ 0 bn−2 an−1 bn−10 ⋯ 0 b2n−2 b2n−2 − b

2n−1 −b2n−1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎣

c0⋮

cn

⎤⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

2 (y1−y0h31

+y1−y2h32

)

3 (y1−y0h21

+y2−y1h22

)

⋮

3 (yn−1−yn−2h2n−1

+yn−yn−1h2n

)

2 (yn−1−yn−2h3n−1

+yn−1−ynh3n

)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

as desired.

(3b) The provided C++ function (in problem3.cpp)

1 us ing V ec to r = Eigen : : VectorXd ;2

3 void n a t s p l i n e s l o p e s ( c o n s t V ec to r & t , c o n s t V ec to r &y , V ec to r&c )

computes the slopes cj ∶= s′(tj), j = 0, . . . , n, of the natural cubic spline interpolants ∈ S3,M through the data points (tj, yj), j = 0, . . . , n.

Based on natsplineslopes, implement a corresponding function (in problem3.cpp)

1 void n o t k n o t s l o p e s ( c o n s t V ec to r & t , c o n s t V ec to r &y , V ec to r&c )

that computes the slopes cj ∶= s′(tj), j = 0, . . . , n, of s ∈ S3,M satisfying s(tj) = yj forgiven yj ∈ R, j = 0, . . . , n.

Solution: See the implementation in problem3_solution.cpp.

11

1 # i n c l u d e < v e c t o r >2 # i n c l u d e 3 # i n c l u d e < Eigen /Dense >4 # i n c l u d e < E i g e n / S p a r s e >5 # i n c l u d e < c a s s e r t >6

7 us ing V ec to r = Eigen : : VectorXd ;8 us ing Ma t r ix = Eigen : : MatrixXd ;9

10 //! \brief Computes the slopes for natural cubic splineinterpolation

11 //! \param[in] t Vector of nodes12 //! \param[in] y Vector of values at the nodes13 //! \param[out] c Vector of slopes at the nodes14 void n a t s p l i n e s l o p e s ( c o n s t V ec to r & t , c o n s t V ec to r &y , V ec to r

&c ) {15 // Size check16 a s s e r t ( ( t . s i z e ( ) == y . s i z e ( ) ) && " E r r o r : mismatched

s i z e o f t and y ! " ) ;17

18 // n+1 is the number of conditions (t goes from t_0 to t_n)19 i n t n = t . s i z e ( ) − 1 ;20

21 V ec to r h ( n ) ;22 // Vector containing increments (from the right)23 f o r ( i n t i = 0 ; i < n ; ++ i ) {24 h ( i ) = t ( i +1) − t ( i ) ;25 // Check that t is sorted26 a s s e r t ( ( h ( i ) > 0 ) && " E r r o r : a r r a y t must be

s o r t e d ! " ) ;27 }28

29 // System matrix and rhs as in (3.5.9), we remove firstand last row (known by natural contition)

30 Eigen : : S p a r s e M a t r i x <double > A( n+1 , n +1) ;31 V ec to r b ( n +1) ;32

33 // WARNING: sparse reserve space34 A. r e s e r v e ( 3 ) ;35

12

36 // Fill in natural conditions (3.5.10) for matrix37 A. c o e f f R e f (0 , 0 ) = 2 . / h ( 0 ) ;38 A. c o e f f R e f (0 , 1 ) = 1 . / h ( 0 ) ;39 A. c o e f f R e f ( n ,n −1) = 1 . / h ( n−1) ;40 A. c o e f f R e f ( n , n ) = 2 . / h ( n−1) ;41

42 // Reuse computation for rhs43 double bo ld = ( y ( 1 ) − y ( 0 ) ) / ( h ( 0 ) *h ( 0 ) ) ;44 b ( 0 ) = 3 . *bo ld ; // Fill in natural conditions (3.5.10)45 // Fill matrix A and rhs b46 f o r ( i n t i = 1 ; i < n ; ++ i ) {47 // Fill in a48 A. c o e f f R e f ( i , i −1) = 1 . / h ( i −1) ;49 A. c o e f f R e f ( i , i ) = 2 . / h ( i −1) + 2 . / h ( i ) ;50 A. c o e f f R e f ( i , i +1) = 1 . / h ( i ) ;51

52 // Reuse computation for rhs b53 double bnew = ( y ( i +1) − y ( i ) ) / ( h ( i ) *h ( i ) ) ;54 b ( i ) = 3 . * ( bnew + bo ld ) ;55 bo ld = bnew ;56 }57 b ( n ) = 3 . *bo ld ; // Fill in natural conditions (3.5.10)58 // Compress the matrix59 A. makeCompressed ( ) ;60 s t d : : c o u t << A;61

62 // Factorize A and solve system A*c(1:end) = b63 Eigen : : SparseLU < Eigen : : S p a r s e M a t r i x <double >> l u ;64 l u . compute (A) ;65 c = l u . s o l v e ( b ) ;66 }67

68 //! \brief Computes the slopes for not-a-knot cubic splineinterpolation

69 //! \param[in] t Vector of nodes70 //! \param[in] y Vector of values at the nodes71 //! \param[out] c Vector of slopes at the nodes72 void n o t k n o t s p l i n e s ( c o n s t V ec to r & t , c o n s t V ec to r &y , V ec to r

&c ) {73 // Size check74 a s s e r t ( ( t . s i z e ( ) == y . s i z e ( ) ) && " E r r o r : mismatched

13

s i z e o f t and y ! " ) ;75

76 // n+1 is the number of conditions (t goes from t_0 to t_n)77 i n t n = t . s i z e ( ) − 1 ;78

79 V ec to r h ( n ) ;80 // Vector containing increments (from the right)81 f o r ( i n t i = 0 ; i < n ; ++ i ) {82 h ( i ) = t ( i +1) − t ( i ) ;83 // Check that t is sorted84 a s s e r t ( ( h ( i ) > 0 ) && " E r r o r : a r r a y t must be

s o r t e d ! " ) ;85 }86

87 // System matrix and rhs as in (3.5.9), adding the tworemaining conditions

88 Eigen : : S p a r s e M a t r i x <double > A( n+1 , n +1) ;89 V ec to r b ( n +1) ;90

91 // WARNING: sparse reserve space92 A. r e s e r v e ( 3 ) ;93

94 // Fill in additional equations95 A. c o e f f R e f (0 , 0 ) = 1 . / ( h ( 0 ) *h ( 0 ) ) ;96 A. c o e f f R e f (0 , 1 ) = 1 . / ( h ( 0 ) *h ( 0 ) ) − 1 . / ( h ( 1 ) *h ( 1 ) ) ;97 A. c o e f f R e f (0 , 2 ) = − 1 . / ( h ( 1 ) *h ( 1 ) ) ;98 A. c o e f f R e f ( n ,n −2) = 1 . / ( h ( n−2) *h ( n−2) ) ;99 A. c o e f f R e f ( n ,n −1) = 1 . / ( h ( n−2) *h ( n−2) ) −

1 . / ( h ( n−1) *h ( n−1) ) ;100 A. c o e f f R e f ( n , n ) = − 1 . / ( h ( n−1) *h ( n−1) ) ;101

102 // Reuse computation for rhs103 double bo ld = ( y ( 1 ) − y ( 0 ) ) / ( h ( 0 ) *h ( 0 ) ) ;104 b ( 0 ) = 2 . * ( ( y ( 1 )−y ( 0 ) ) / ( h ( 0 ) *h ( 0 ) *h ( 0 ) ) +

( y ( 1 )−y ( 2 ) ) / ( h ( 1 ) *h ( 1 ) *h ( 1 ) ) ) ;105 // Fill matrix A and rhs b106 f o r ( i n t i = 1 ; i < n ; ++ i ) {107 // Fill in a108 A. c o e f f R e f ( i , i −1) = 1 . / h ( i −1) ;109 A. c o e f f R e f ( i , i ) = 2 . / h ( i −1) + 2 . / h ( i ) ;110 A. c o e f f R e f ( i , i +1) = 1 . / h ( i ) ;

14

111

112 // Reuse computation for rhs b113 double bnew = ( y ( i +1) − y ( i ) ) / ( h ( i ) *h ( i ) ) ;114 b ( i ) = 3 . * ( bnew + bo ld ) ;115 bo ld = bnew ;116 }117 b ( n ) = 2 . * ( ( y ( n−1)−y ( n−2) ) / ( h ( n−2) *h ( n−2) *h ( n−2) ) +

( y ( n−1)−y ( n ) ) / ( h ( n−1) *h ( n−1) *h ( n−1) ) ) ;118 // Compress the matrix119 A. makeCompressed ( ) ;120 s t d : : c o u t << A;121

122 // Factorize A and solve system A*c(1:end) = b123 Eigen : : SparseLU < Eigen : : S p a r s e M a t r i x <double >> l u ;124 l u . compute (A) ;125 c = l u . s o l v e ( b ) ;126 }127

128 i n t main ( ) {129 s t d : : c o u t << " Noth ing t o do ! " << s t d : : e n d l ;130 }

Problem 4 On the Gauss points (24 pts.)Given f ∈ C0([0,1]), the integral

g(x) ∶= ∫1

0e∣x−y∣f(y)dy

defines a function g ∈ C0([0,1]). Throughout this problem, we approximate the integralby means of n-point Gauss quadrature, which yields a function gn(x).

(4a) Let {ξnj }nj=1 be the nodes for the Gauss quadrature on the interval [0,1] and write

wnj for the corresponding weights. Assume that the nodes are ordered, namely ξnj < ξnj+1for every j = 1, . . . , n − 1. We can write

(gn(ξnl ))

nl=1 =M (f(ξnj ))

nj=1,

for a suitable matrix M ∈ Rn,n. Give a formula for the entries of M.

Solution: Applying Gauss quadrature to the expression for g(ξnl ) we obtain gn(ξnl ) =

∑nj=1w

nj e

∣ξnl −ξnj ∣f(ξnj ), whence Mlj = wnj e

∣ξnl −ξnj ∣ for l, j = 1, . . . , n.

15

(4b) Implement a function (in problem4.cpp)


3 template <typename Func t ion >4 V ec to r comp_g_gaussp t s ( F u n c t i o n f , unsigned i n t n )

that computes the n-vector (gn(ξnl ))nl=1 with optimal complexityO(n) (excluding the com-

putation of the Gauss nodes and weights), where f is an object with an evaluation operatordouble operator()(double x), e.g. a lambda function, that represents the func-tion f .

HINT: You may use the provided function (in problem4.cpp)

1 void g a u s s r u l e ( i n t n , Ve c t o r & w, Ve c t o r & x i )

that computes the weights w and the ordered nodes xi relative to the n-point Gauss quadra-ture on the interval [−1,1].

Solution: In order to write a function with optimal complexity, we need to use the struc-ture of the matrix M. Write

gn(ξnl ) =

n

∑j=1

wnj e∣ξnl −ξ

nj ∣f(ξnj ) =

l−1

∑j=1

wnj eξnl e−ξ

nj f(ξnj ) +

n

∑j=l

wnj e−ξnl eξ

nj f(ξnj ).

Setting pnl = ∑l−1j=1w

nj e

−ξnj f(ξnj ) and qnl = ∑nj=lw

nj e

ξnj f(ξnj ), this identity can be rewrittenas

gn(ξnl ) = e

ξnl pnl + e−ξnl qnl .

This expression for l = 1, . . . , n can be computed withO(n) operations. It remains to com-pute pnl and qnl in O(n) operations. This can be done by using these recursive formulas,that immediately follow from the definition: for every l = 1, . . . , n − 1 we have

pn1 = 0, qnn = wnne

ξnnf(ξnn),

pnl+1 = pnl +w

nl e

−ξnl f(ξnl ), qnl = qnl+1 +w

nl e

ξnl f(ξnl ).

See the implementation in problem4_solution.cpp.

(4c) Test your implementation by computing g(ξ2111) (ξ2111 = 1/2) for f(y) = e−∣0.5−y∣.What result do you expect?

Solution: For reasons of symmetry, we have ξ2111 = 1/2, and so g21(ξ2111) should be equalto 1 for f(y) = e−∣0.5−y∣, since Gauss quadrature is exact for constant functions.

16

1 # i n c l u d e < v e c t o r >2 # i n c l u d e 3 # i n c l u d e < Eigen /Dense >4


7 //! \brief Golub-Welsh implementation 5.3.358 //! \param[in] n number of Gauss nodes9 //! \param[out] w weights for interval [-1,1]

10 //! \param[out] xi ordered nodes for interval [-1,1]11 void g a u s s r u l e ( i n t n , Eigen : : VectorXd & w, Eigen : : VectorXd &

x i ) {12 w. r e s i z e ( n ) ;13 x i . r e s i z e ( n ) ;14 i f ( n == 0 ) {15 x i ( 0 ) = 0 ;16 w( 0 ) = 2 ;17 } e l s e {18 V ec to r b ( n−1) ;19 Eigen : : MatrixXd J = Eigen : : MatrixXd : : Zero ( n , n ) ;20

21 f o r ( i n t i = 1 ; i < n ; ++ i ) {22 double d = ( i ) / s q r t ( 4 . * i * i − 1 . ) ;23 J ( i , i −1) = d ;24 J ( i −1 , i ) = d ;25 }26

27 Eigen : : E i g e n S o l v e r < Eigen : : MatrixXd > e i g ( J ) ;28

29 x i = e i g . e i g e n v a l u e s ( ) . r e a l ( ) ;30 w = 2 * e i g . e i g e n v e c t o r s ( ) . r e a l ( )31 . topRows <1 >()32 . c w i s e P r o d u c t (33 e i g . e i g e n v e c t o r s ( )34 . r e a l ( )35 . topRows <1 >()36 ) ;37 }38

39 s t d : : v e c t o r < s t d : : p a i r < d o u b l e , d o u b l e >> P ;

17

40 P . r e s e r v e ( n ) ;41 f o r ( unsigned i n t i = 0 ; i < n ; ++ i ) {42 P . push_back ( s t d : : p a i r < d o u b l e , d o u b l e >( x i ( i ) ,w ( i ) ) ) ;43 }44 s t d : : s o r t ( P . b e g i n ( ) , P . end ( ) ) ;45 f o r ( unsigned i n t i = 0 ; i < n ; ++ i ) {46 x i ( i ) = s t d : : ge t <0 >(P [ i ] ) ;47 w( i ) = s t d : : ge t <1 >(P [ i ] ) ;48 }49 }50

51 //! \brief Compute the function g in the Gauss nodes52 //! \param[in] f object with an evaluation operator (e.g. a

lambda function) representing the function f53 //! \param[in] n number of nodes54 //! \param[out] Eigen::VectorXd containing the function g

calculated in the Gauss nodes.55 template <typename Func t ion >56 Eigen : : VectorXd comp_g_gaussp t s ( F u n c t i o n f , unsigned i n t n ) {57

58 V ec to r g ( n ) ;59 V ec to r w( n ) , x i ( n ) , p ( n ) , q ( n ) , e x p x i ( n ) , f x i ( n ) ;60 g a u s s r u l e ( n , w, x i ) ; // Compute Gauss nodes and weights

relative to [-1,1]61 w = w/2 . ; // Rescale the weights to [0,1]62 x i = ( x i + Eigen : : MatrixXd : : Ones ( n , 1 ) ) / 2 . ; // Rescale the

nodes to [0,1]63

64 f o r ( i n t l =0 ; l <n ; l ++) {65 f x i ( l ) = f ( x i ( l ) ) ;66 e x p x i ( l ) = exp ( x i ( l ) ) ;67 }68

69 // Initialize the vectors p and q70 p ( 0 ) = 0 ;71 q ( n−1) = w( n−1) * e x p x i ( n−1) * f x i ( n−1) ;72

73 // Fill in the vectors p and q (O(n) complexity)74 f o r ( i n t l =0 ; l <n−1; l ++) {75 p ( l +1) = p ( l ) + w( l ) / e x p x i ( l ) * f x i ( l ) ;76 q ( n−2− l ) = q ( n−1− l ) + w( n−2− l ) * ex px i ( n−2− l ) * f x i ( n−2− l ) ;

18

77 }78

79 // Finally, constructing the output (O(n) complexity)80

81 g = ( p . a r r a y ( ) * e x p x i . a r r a y ( ) + q . a r r a y ( ) /e x p x i . a r r a y ( ) ) . m a t r i x ( ) ;

82 re turn g ;83 }84

85 // Test the implementation by calculating g(xi^21_10) forf(y)=exp(-|0.5-y|).

86 i n t main ( ) {87 i n t n = 2 1 ;88 auto f = [ ] ( double y ) { re turn exp (− s t d : : abs ( .5 − y ) ) ; } ;89 Eigen : : VectorXd g = comp_g_gaussp t s ( f , n ) ;90 s t d : : c o u t << " g ( x i ^ " << n << " _ " << ( n +1) / 2 << " ) = " <<

g ( ( n +1) /2 −1) << s t d : : e n d l ;91 }

Problem 5 Construction of an evolution operator (28 pts.)Let Ψh define the discrete evolution of an order p Runge-Kutta single step method for theautonomous ODE y = f(y), f ∶D ⊆ Rd → Rd. We define a new evolution operator:

Ψh ∶=1

1 − 2p(Ψh − 2p ⋅ (Ψh/2 ○Ψh/2)) , (11)

where ○ denotes the composition of mappings.

(5a) Let Ψh be the evolution operator of the explicit Euler method. Give the explicitformula for Ψh.

Solution: Writing the forward Euler method as Ψh(y0) ∶= y0 + hf(y0) (and p = 1), weobtain:

Ψh(y0) =1

1 − 2p(y0 + hf(y0) − 2p ⋅ (y0 +

h

2f(y0) +

h

2f(y0 +

h

2f(y0)))

= y0 − (h(1 − 1)f(y0) − hf(y0 +h

2f(y0)))

= y0 + hf(y0 +h

2f(y0)).

19

Remark 1. This is the evolution operator of the explicit trapezoidal method, a 2-stageexplicit Runge-Kutta method.

(5b) Implement a C++ function (in problem5.cpp)


3 t empla te < c l a s s Opera to r >4 V ec to r p s i t i l d e ( c o n s t O p e r a t o r &P s i , unsigned i n t p ,5 double h , c o n s t V ec to r &y0 ) ;

that returns Ψhy0 when given the underlying Ψ.

Objects of type Operator (here and in the following) must provide an evaluation opera-tor with signature:

1 V ec to r operator ( ) ( double h , c o n s t V ec to r &y ) ;

providing the evaluation of Ψh(y). A suitable C++ lambda function satisfies this require-ment.


(5c) Implement a C++ function (in problem5.cpp)

1 t empla te < c l a s s Opera to r >2 s t d : : v e c t o r < Vector > o d e i n t e q u i ( c o n s t O p e r a t o r &P s i ,3 double T, c o n s t V ec to r &y 0 , unsigned i n t N) ;

for the approximation of the solution of the initial value problem y = f(y),y(0) = y0. Thefunction uses a single step method defined by the discrete evolution operator Ψ (given asPsi) on N equidistant timesteps with final time T > 0. The function returns the approxi-mated value at each step (including y0: y0, y1, . . . ) in a std::vector<Vector>.

HINT: You can use std::vector<Vector>::push_back to add elements to theend of a std::vector<Vector>.


(5d) In the case of the IVP

y = 1 + y2, y(0) = 0, (12)

20

with exact solution yex(t) = tan(t), determine empirically (using odeintequi) theorder of the single step method induced by Ψh, when Ψ is the discrete evolution op-erator for the explicit Euler method. Monitor the error ∣yN(1) − yex(1)∣ at final timeT = 1, where yN is the solution approximated through Ψh using N uniforms steps withN = 2q, q = 2, . . . ,12.

Solution: The experimental convergence order is 2 (O(h2)), the order of convergence offorward Euler is p = 1. See implementation in problem5_solution.cpp.

(5e) In general, the method defined by Ψh has order p + 1. Thus, it can be used foradaptive timestep control and prediction.

Complete the implementation of a function (found in problem5.cpp)1 t empla te < c l a s s Opera to r >2 s t d : : v e c t o r < Vector > o d e i n t s s c t r l ( c o n s t O p e r a t o r &P s i ,3 double T, c o n s t V ec to r &y 0 ,4 double h 0 , unsigned i n t p ,5 double r e l t o l , double a b s t o l ,6 double hmin ) ;

for the approximation of the solution of the IVP by means of adaptive timestepping basedon Ψ and Ψ, where Ψ is passed through the argument Psi. Step rejection and stepsizecorrection and prediction has to be employed. The argument T supplies the final time, y0the initial state, h0 an initial stepsize, p the order to the discrete evolution Ψ, reltoland abstol the respective tolerances, and hmin a minimal stepsize that will triggerpremature termination. Compute the solution obtained using this function applied to theIVP (12) up to time T = 1 with the following data: h0 = 1/100, reltol = 10e-8,abstol = 10e-8, hmin = 10e-6. Output the approximated solution at time T = 1.

Solution: See implementation in problem5_solution.cpp.1 # i n c l u d e 2

3 # i n c l u d e < v e c t o r >4

5 # i n c l u d e < Eigen /Dense >6

7 us ing V ec to r = Eigen : : VectorXd ;8 us ing Ma t r ix = Eigen : : MatrixXd ;9

10 //! \brief Evolves the vector y0 using the evolution operator\tilde{\Psi} with step-size y0

21

11 //! \tparam Operator type for evolution operator (e.g. lambdafunction type)

12 //! \param[in] Psi original evolution operator, must haveoperator(double, const Vector&)

13 //! \param[in] p parameter p for construction of Psi tilde14 //! \param[in] h step-size15 //! \param[in] y0 previous step16 //! \return Evolved step Ψhy017 t empla te < c l a s s Opera to r >18 V ec to r p s i t i l d e ( c o n s t O p e r a t o r& P s i , unsigned i n t p , double h ,

c o n s t V ec to r & y0 ) {19 re turn ( P s i ( h , y 0 ) − (2 < <( p ) ) * P s i ( h / 2 . , P s i ( h / 2 . , y 0 ) ) ) /

( 1 . − (2 < <( p ) ) ) ;20 }21

22 //! \brief Evolves the vector y0 using the evolution operatorfrom time 0 to T using equidistant steps



25 //! \param[in] T final time26 //! \param[in] y0 initial data27 //! \param[in] N number of steps28 //! \return Vector of all steps y_0, y_1, ...29 t empla te < c l a s s Opera to r >30 s t d : : v e c t o r < Vector > o d e i n t e q u i ( c o n s t O p e r a t o r& P s i , double T,

c o n s t V ec to r &y 0 , i n t N) {31 double h = T / N;32 double t = 0 . ;33 s t d : : v e c t o r < Vector > Y;34 Y. r e s e r v e (N+1) ;35 Y. push_back ( y0 ) ;36 V ec to r y = y0 ;37

38 whi le ( t < T ) {39 y = P s i ( h,Y . back ( ) ) ;40

41 Y. push_back ( y ) ;42 t += s t d : : min ( T− t , h ) ;

22

43 }44

45 re turn Y;46 }47

48 //! \brief Evolves the vector y0 using the evolution operatorfrom time 0 to T using adaptive error control



51 //! \param[in] T final time52 //! \param[in] y0 initial data53 //! \param[in] h0 initial step size54 //! \param[in] p parameter p for construction of Psi tilde55 //! \param[in] reltol relative tolerance for error control56 //! \param[in] abstol absolute tolerance for error control57 //! \param[in] p parameter p for construction of Psi tilde58 //! \param[in] hmin minimal step size59 //! \return Vector of all steps y_0, y_1, ...60 t empla te < c l a s s Opera to r >61 s t d : : v e c t o r < Vector > o d e i n t s s c t r l ( c o n s t O p e r a t o r& P s i , double

T, c o n s t V ec to r &y 0 , double h 0 ,62 unsigned i n t p , double

r e l t o l , double a b s t o l ,double hmin ) {

63 // Time tracker64 double t = 0 . ;65 // Step size tracker66 double h = h0 ;67 // Vector of snapshots68 s t d : : v e c t o r < Vector > Y;69 // Save initial data70 Y. push_back ( y0 ) ;71 // Start with y0 as value72 V ec to r y = y0 ;73

74 // Kill if final time reached or min step size reached75 whi le ( t < T && h > hmin ) {76 // Compute with two methods77 V ec to r y_h igh = p s i t i l d e ( P s i , p , h , y ) ;

23

78 V ec to r y_low = P s i ( h , y ) ;79 // Estimate error80 double e r r _ e s t = ( y_h igh − y_low ) . norm ( ) ;81

82 // Compute tolerance (min of reltol and abstol)83 double t o l = s t d : : max ( r e l t o l * y _ h i g h . norm ( ) , a b s t o l ) ;84

85 // Accept step if true (error smaller than tol)86 i f ( e r r _ e s t < t o l ) {87 t += h ;88

89 y = y_h igh ;90 Y. push_back ( y_h igh ) ;91

92 }93 // Predict step-size94 h *= s t d : : max ( . 5 , s t d : : min ( 2 . , s t d : : pow ( t o l / e r r _ e s t ,

1 . / ( p + 1 . ) ) ) ) ;95 // Force time if needed96 h = s t d : : min ( T− t , h ) ;97 }98 i f ( t < T ) s t d : : c e r r << " F a i l e d t o r e a c h f i n a l t ime ! " <<

s t d : : e n d l ;99

100 re turn Y;101 }102

103 i n t main ( i n t , ch a r * * ) {104

105 auto f = [ ] ( c o n s t V ec to r &y ) −> Ve c t o r { re turnV ec to r : : Ones ( 1 ) + y*y ; } ;

106 V ec to r y0 = V ec to r : : Zero ( 1 ) ;107 double T = 1 . ;108 auto y_ex = [ ] ( double t ) −> Ve c t o r { Ve c t o r y ( 1 ) ; y <<

t a n ( t ) ; re turn y ; } ;109

110 unsigned p = 1 ;111

112 auto P s i = [& f ] ( double h , c o n s t V ec to r & y0 ) −> Ve c t o r {re turn y0 + h*f ( y0 ) ; } ;

113 auto P s i T i l d e = [& P s i , &f , &p ] ( double h , c o n s t V ec to r &

24

y0 ) −> Ve c t o r { re turn p s i t i l d e ( P s i , p , h , y0 ) ; } ;114

115 //// PROBLEM 5d116 s t d : : c o u t << " *** PROBLEM 5d : " << s t d : : e n d l ;117

118 s t d : : c o u t << " E r r o r t a b l e f o r e q u i d i s t a n t s t e p s : " <<s t d : : e n d l ;

119 s t d : : c o u t << "N" << " \ t " << " E r r o r " << s t d : : e n d l ;120 f o r ( i n t N = 4 ; N < 4096 ; N=N<<1 ) {121 s t d : : v e c t o r < Vector > Y = o d e i n t e q u i ( P s i T i l d e , T , y 0 , N) ;122 double e r r = (Y. back ( ) − y_ex ( 1 ) ) . norm ( ) ;123 s t d : : c o u t << N << " \ t " << e r r << s t d : : e n d l ;124 }125

126 //// PROBLEM 5e127 s t d : : c o u t << " *** PROBLEM 5 e : " << s t d : : e n d l ;128

129 double h0 = 1 . /100 . ;130 s t d : : v e c t o r < Vector > Y;131 double e r r = 0 ;132 Y = o d e i n t s s c t r l ( P s i , T , y 0 , h 0 , p , 10e−8 , 10e−8 , 10e −6) ;133 e r r = (Y. back ( ) − y_ex ( 1 ) ) . norm ( ) ;134 s t d : : c o u t << " A d a p t i v e e r r o r c o n t r o l r e s u l t s : " <<

s t d : : e n d l ;135 s t d : : c o u t << " E r r o r " << " \ t " << "No . S t e p s " << " \ t " <<

" y ( 1 ) " << " \ t " << " y ( ex ) " << s t d : : e n d l ;136 s t d : : c o u t << e r r << " \ t " << Y. s i z e ( ) << " \ t " << Y. back ( )

<< " \ t " << y_ex ( 1 ) << s t d : : e n d l ;137 }

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First name Grade - Seminar for Applied Mathematicsgrsam/HS16/NumCSE/old_exams/final_exam... · 1 Vector solveLSE ( const Vector & c, const Vector & b, 2 unsignedint i 0 , unsignedint

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