Last name Department Computer Name Legi No. Date Total. Always keep your Legi visible on the table.

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1 First name Last name Department Computer Name Legi No. Date Total Grade Fill in this cover sheet first. Always keep your Legi visible on the table. Keep your phones, tablets and computers turned off in your bag. Start each handwritten problem on a new sheet. Put your name on each sheet. Do not write with red/green/pencil. Write your solutions clearly and work carefully. Write all your solutions only in the folder results! Any other location will not be backed-up and will be discarded. Files in resources may be overridden at any time. Make sure to regularly save your solutions. Time spent on restroom breaks is considered examination time. Never turn off or log off from your computer!

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3 Final exam Numerical Methods for CSE R. Hiptmair, G. Alberti, F. Leonardi February 2, 2016 A CMake file is provided with the templates. To generate a Makefile for all problems, type cmake. in the folder ~/results. Compile your programs with make. For example, in order to compile and run the C++ code related to Problem 3, type cmake., then make (or make problem3 to compile only problem3.cpp) and finally./problem3. If you want to manually compile your code, use: 1 g++ s t d =c ++11 I / u s r / i n c l u d e / e i g e n 3 Wno d e p r e c a t e d d e c l a r a t i o n s f i l e n a m e. cpp o programname or 1 c l a n g ++ s t d =c ++11 I / u s r / i n c l u d e / e i g e n 3 Wno d e p r e c a t e d d e c l a r a t i o n s f i l e n a m e. cpp o programname The flag -Wno-deprecated-declarations suppresses some EIGEN warning. For each problem requiring C++ implementation, a template file named problemx.cpp is provided. For your own convenience, there is a marker TODO in the places where you are supposed to write your own code. 1

4 Problem 1 Complex roots (16 pts.) Given a complex number w = u + iv, u, v R with v 0, its root w = x + iy can be defined by x = ( u 2 + v 2 + u)/2, (1) y = ( u 2 + v 2 u)/2. (2) (1a) For what w C will the direct implementation of (1) and (2) be vulnerable to cancellation? Solution: As u/v or when v = 0 and u < 0, we have u 2 + v 2 u, and so cancellations may arise in u 2 + v 2 + u. Therefore, the implementation of x will be vulnerable to cancellation. Similarly, as u/v + or when v = 0 and u > 0, we have u 2 + v 2 u, and so cancellations may arise in u 2 + v 2 u. Therefore, the implementation of y will be vulnerable to cancellation. In conclusion, the implementation of these formulas will be vulnerable to cancellation when u is much bigger than v. (1b) Compute xy as an expression of u and v. Solution: An immediate calculation shows that xy = v/2. (1c) Implement a function 1 s t d : : complex <double > myroot ( s t d : : complex <double > w ) ; that computes the root of w as given by (1) and (2) without the risk of cancellation. You may use only real arithmetic: for instance, you may not apply the function sqrt with complex arguments. Test your implementation with w = i and with w = i. DEPENDS ON: subproblems (1a) and (1b). Solution: See implementation in problem1_solution.cpp and below. 1 # i n c l u d e < i o s t r e a m > 2 # i n c l u d e <complex > 3 # i n c l u d e <cmath > 4 5 //! \brief Compute complex root 2

5 6 //! \param[in] w complex number with non negative imaginary parts 7 //! \return the square root of w with non negative real and imaginary parts 8 s t d : : complex <double > myroot ( s t d : : complex <double > w ) { 9 double x, y ; 10 double u = w. r e a l ( ) ; 11 double v = w. imag ( ) ; // TODO: problem 1c: construct x and y as functions of u and v i f ( v ==0) return s q r t ( u ) ; i f ( u > 0) { 18 x = s q r t ( ( s q r t ( u*u+v*v ) +u ) / 2. ) ; 19 y = v / (2 *x ) ; 20 } e l s e { 21 y = s q r t ( ( s q r t ( u*u+v*v ) u ) / 2. ) ; 22 x = v / (2 *y ) ; 23 } return s t d : : complex <double > ( x, y ) ; 26 } // Test the implementation 29 i n t main ( ) { 30 s t d : : c o u t << " *** PROBLEM 1, t e s t i n g : " << s t d : : e n d l ; s t d : : complex <double > w(1 e 2 0, 5 ) ; 33 s t d : : c o u t << " The s q u a r e r o o t of " << w << " i s " << myroot (w) << s t d : : e n d l ; 34 s t d : : c o u t << " The c o r r e c t s q u a r e r o o t of " << w << " i s " << s q r t (w) << s t d : : e n d l << s t d : : e n d l ; w= s t d : : complex <double >( 5, 1 e 2 0 ) ; 37 s t d : : c o u t << " The s q u a r e r o o t of " << w << " i s " << myroot (w) << s t d : : e n d l ; 38 s t d : : c o u t << " The c o r r e c t s q u a r e r o o t of " << w << " i s " << s q r t (w) << s t d : : e n d l << s t d : : e n d l ; 39 } 3

6 Problem 2 Ellpack sparse matrix format (24 pts.) The following class (EllpackMat in file problem2.cpp) implements the Ellpack sparse matrix format for a generic matrix A R m,n : 1 using T r i p l e t = Eigen : : T r i p l e t <double >; 2 using T r i p l e t s = s t d : : v e c t o r < T r i p l e t >; 3 using V e c t o r = Eigen : : VectorXd ; 4 using i n d e x _ t = s t d : : s i z e _ t ; 5 6 c l a s s EllpackMat { 7 p u b l i c : 8 EllpackMat ( c o n s t T r i p l e t s & t r i p l e t s, i n d e x _ t m, i n d e x _ t n ) ; 9 10 double operator ( ) ( i n d e x _ t i, i n d e x _ t j ) c o n s t ; void mvmult ( c o n s t V e c tor &x, V ector &y ) c o n s t ; 13 void mtvmult ( c o n s t V ector &x, V ector &y ) c o n s t ; 14 p r i v a t e : 15 s t d : : v e c t o r <double > v a l ; 16 s t d : : v e c t o r < i n d e x _ t > c o l ; i n d e x _ t maxcols ; // Max nnz elements per row 19 i n d e x _ t m,n ; // Number of rows, number of columns 20 } ; Here, maxcols = max{#{(i, j) A i,j 0, j = 1,..., n} i = 1,..., m}. The access operator operator() is implemented as follows: 1 double EllpackMat : : operator ( ) ( i n d e x _ t i, i n d e x _ t j ) c o n s t { 2 a s s e r t (0 <= i && i < m && 0 <= j && j < n 3 && " Index o u t of bounds! " ) ; 4 5 f o r ( i n d e x _ t l = i * m a x c o l s ; l < ( i +1) *maxcols ; ++ l ) { 6 i f ( c o l [ l ] == j ) return v a l [ l ] ; 7 } 8 return 0. ; 4

7 9 } and is well defined for i from 0 to m-1 and for j from 0 to n-1. (2a) Implement an efficient constructor that builds an object of type EllpackMat from the data forming a matrix in triplet format, i.e. implement the constructor with signature 1 EllpackMat ( c o n s t T r i p l e t s & t r i p l e t s, i n d e x _ t m, i n d e x _ t n ) ; where triplets is a std::vector of EIGEN triplets. The arguments m and n represent the number of rows and of columns of the matrix A, respectively. The data in the triplets vector must be compatible with the matrix size provided to the constructor. No assumption is made on the ordering of the triplets. Values belonging to multiple occurrences of index pairs are to be summed up. However, in this sub-problem you may assume that duplicate index pairs do not occur in the triplets vector. HINT: An object of Eigen::Triplets<double> type is a structure representing a triplet (r, c, v). It provides the methods index_t row(), index_t col() and double value() which return r, c and v, respectively. HINT: As first thing, you will have to determine the quantity maxcols. Solution: See implementation in problem2_solution.cpp. (2b) Implement an efficient method 1 void mvmult ( c o n s t V e c t o r &x, V ector &y ) c o n s t ; that, given an input n-vector x and an output m-vector y as arguments, returns the matrixvector product A*x in y, where A is the matrix represented by *this. The implementation must deliver optimal complexity of O(nnz(A)). Solution: See implementation in problem2_solution.cpp. (2c) Implement an efficient method 1 void mtvmult ( c o n s t V e c t o r &x, V ector &y ) c o n s t ; that, given an input m-vector x and an output n-vector y as arguments, returns the matrixvector product A *x in y, where A is the transposed of the matrix represented by *this. The implementation must deliver optimal complexity of O(nnz(A)). Solution: See implementation in problem2_solution.cpp. 5

8 HINT: You can test your implementations of (2a), (2b) and (2c) using the code written in main(). A 3 6 EIGEN-Sparse matrix A (Eigen::SparseMatrix<double> A) is built from the triplets: {(1, 2, 4), (0, 0, 5), (1, 2, 6), (2, 5, 7), (0, 4, 8), (1, 3, 9), (2, 2, 10), (2, 1, 11), (1, 0, 12)}. An equivalent matrix E of Ellpack type is also built. We then use the vectors x = [4, 5, 6, 7, 8, 9], y = [1, 2, 3] and compute the product Ax and Ex. Finally we output the l 2 norm of the difference. We do the same for A y and E y. Solution: 1 # i n c l u d e < i o s t r e a m > 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 # i n c l u d e < E i g e n / S p a r s e > 7 8 using T r i p l e t = Eigen : : T r i p l e t <double >; 9 using 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 using V e c t o r = Eigen : : VectorXd ; 11 using i n d e x _ t = s t d : : p t r d i f f _ t ; //! \brief Class representing a sparse matrix in Ellpack format 14 c l a s s EllpackMat { 15 p u b l i c : EllpackMat ( c o n s t T r i p l e t s & t r i p l e t s, i n d e x _ t m, i n d e x _ t n ) ; double operator ( ) ( i n d e x _ t i, i n d e x _ t j ) c o n s t ; void mvmult ( c o n s t V ector &x, V ector &y ) c o n s t ; void mtvmult ( c o n s t V ector &x, V ector &y ) c o n s t ; 24 p r i v a t e : 25 s t d : : v e c t o r <double > v a l ; //!< Vector containing value corresponding to entry in col 26 s t d : : v e c t o r < i n d e x _ t > c o l ; //! Vector containing columns, partitioned into same-size rows 27 6

9 28 i n d e x _ t maxcols ; // Max elements per row 29 i n d e x _ t m,n ; // Number of rows, number of columns 30 } ; //! \brief Construct a matrix Ellpack from triplet format 33 //! \param[in] triplets vector of Eigen::Triplets 34 //! \param[in] m rows of matrix 35 //! \param[in] n columns of matrix 36 //! \param[in] maxcols max. number of nonzeros of the matrix 37 EllpackMat : : EllpackMat ( c o n s t T r i p l e t s & t r i p l e t s, i n d e x _ t m, i n d e x _ t n ) 38 : m(m), n ( n ) { 39 // TODO: problem2a: implement constructor building matrix from Eigen Triplet format // Find maxcols, counters denotes number of non-empty columns for each row (assuming all triplets uniquely 42 // define one entry) 43 s t d : : v e c t o r < unsigned i n t > c o u n t e r s (m) ; 44 // Fill counter with 0 45 s t d : : f i l l ( c o u n t e r s. b e g i n ( ), c o u n t e r s. end ( ), 0) ; 46 // maxcols starts with no entry for each row 47 maxcols = 0 ; 48 // Loop over each triplet and increment corresponding counter, update maximum if needed 49 f o r ( c o n s t T r i p l e t& t r : t r i p l e t s ) { 50 i f ( ++ c o u n t e r s [ t r. row ( ) ] > maxcols ) maxcols = c o u n t e r s [ t r. row ( ) ] ; 51 } 52 s t d : : c o u t << " Maxcols : " << maxcols << s t d : : e n d l ; // Reserve space 55 c o l. r e s i z e ( m*maxcols, 1) ; 56 v a l. r e s i z e ( m*maxcols,0 ) ; // Loop over each triplet, then find a column where to put the value 59 f o r ( c o n s t T r i p l e t& t r : t r i p l e t s ) { 60 a s s e r t (0 <= t r. row ( ) && t r. row ( ) < m && 0 <= t r. c o l ( ) && t r. c o l ( ) < n && 61 " Index o u t of bounds! " ) ; 7

10 62 63 i n d e x _ t l ; 64 f o r ( l = t r. row ( ) *maxcols ; l < ( t r. row ( ) +1) *maxcols ; ++ l ) { 65 // If value must be added to existing one 66 i f ( c o l [ l ] == t r. c o l ( ) ) { 67 v a l [ l ] += t r. v a l u e ( ) ; 68 break ; 69 // If value is new (find the next empty place) 70 } e l s e i f ( c o l [ l ] == 1 ) { 71 c o l [ l ] = t r. c o l ( ) ; 72 v a l [ l ] = t r. v a l u e ( ) ; 73 break ; 74 } 75 } 76 a s s e r t ( l < ( t r. row ( ) +1) *maxcols && 77 "You d i d n o t r e s e r v e enough columns! " ) ; 78 } 79 } //! \brief Retrieve value of entry at index (i,j) 82 //! \param[in] i row index i 83 //! \param[in] j column index j 84 //! \return value at (i,j) 85 double EllpackMat : : operator ( ) ( i n d e x _ t i, i n d e x _ t j ) c o n s t { 86 a s s e r t (0 <= i && i < m && 0 <= j && j < n && " Index o u t of bounds! " ) ; f o r ( i n d e x _ t l = i * m a x c o l s ; l < ( i +1) *maxcols ; ++ l ) { 89 i f ( c o l [ l ] == j ) return v a l [ l ] ; 90 } 91 return 0. ; 92 } //! \brief Computes (*this)*x 95 //! \param[in] x vector x for mat-vec mult 96 //! \param[out] y result y = (*this)*x 97 void EllpackMat : : mvmult ( c o n s t V ector &x, V ector &y ) c o n s t { 98 a s s e r t ( x. s i z e ( ) == n && " I n c o m p a t i b l e v e c t o r x s i z e! " ) ; 99 a s s e r t ( y. s i z e ( ) == m && " I n c o m p a t i b l e v e c t o r y s i z e! " ) ; 100 8

11 101 // TODO: problem 2b: implement operation y = this*x f o r ( i n d e x _ t i = 0 ; i < m; ++ i ) { 104 y ( i ) = 0 ; 105 f o r ( i n d e x _ t l = i * m a x c o l s ; l < ( i +1) *maxcols ; ++ l ) { 106 i f ( c o l [ l ] == 1 ) break ; 107 y ( i ) += x ( c o l [ l ] ) * v a l [ l ] ; 108 } 109 } 110 } //! \brief Computes (*this) *x, A is transposed 113 //! \param[in] x vector x for mat-vec mult 114 //! \param[out] y result y = (*this) *x 115 void EllpackMat : : mtvmult ( c o n s t V ector &x, V ector &y ) c o n s t { 116 a s s e r t ( x. s i z e ( ) == m && " I n c o m p a t i b l e v e c t o r x s i z e! " ) ; 117 a s s e r t ( y. s i z e ( ) == n && " I n c o m p a t i b l e v e c t o r y s i z e! " ) ; // TODO: problem 2c: implement operation y = this^t*x y = V e c t o r : : Zero ( n ) ; 122 f o r ( i n d e x _ t i = 0 ; i < m; ++ i ) { 123 f o r ( i n d e x _ t l = i * m a x c o l s ; l < ( i +1) *maxcols ; ++ l ) { 124 i f ( c o l [ l ] == 1 ) break ; 125 y ( c o l [ l ] ) += x ( i ) * v a l [ l ] ; 126 } 127 } 128 } i n t main ( i n t, c h a r * * ) { 131 // Vector of triplets 132 T r i p l e t s t r i p l e t s ; // Data 135 unsigned i n t m = 3, n = 6 ; 136 unsigned i n t n t r i p l e t s = 9 ; // Reserve space for triplets 139 t r i p l e t s. r e s e r v e ( n t r i p l e t s ) ; // Fill in some triplet 9

12 142 t r i p l e t s. push_back ( T r i p l e t (1, 2, 4 ) ) ; 143 t r i p l e t s. push_back ( T r i p l e t (0, 0, 5 ) ) ; 144 t r i p l e t s. push_back ( T r i p l e t (1, 2, 6 ) ) ; 145 t r i p l e t s. push_back ( T r i p l e t (2, 5, 7 ) ) ; 146 t r i p l e t s. push_back ( T r i p l e t (0, 4, 8 ) ) ; 147 t r i p l e t s. push_back ( T r i p l e t (1, 3, 9 ) ) ; 148 t r i p l e t s. push_back ( T r i p l e t (2, 2, 1 0 ) ) ; 149 t r i p l e t s. push_back ( T r i p l e t (2, 1, 1 1 ) ) ; 150 t r i p l e t s. push_back ( T r i p l e t (1, 0, 1 2 ) ) ; // Build matrix (Eigen Sparse) 153 Eigen : : S p a r s e M a t r i x <double > S ( m,n ) ; 154 S. 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 ( ) ) ; // Build Ellpack matrix 157 EllpackMat E ( t r i p l e t s, m, n ) ; //// PROBLEM 2 TEST 160 s t d : : c o u t << " *** PROBLEM 2, t e s t i n g : " << s t d : : e n d l ; 161 s t d : : c o u t << " T e s t of y = A^ t * x " << s t d : : e n d l ; 162 Eigen : : VectorXd x ( 6 ) ; 163 x << 4, 5, 6, 7, 8, 9 ; 164 Eigen : : VectorXd Ex = Eigen : : VectorXd : : Zero (m) ; 165 E. mvmult ( x, Ex ) ; 166 s t d : : c o u t << " S p a r s e S*x =" << s t d : : e n d l << S*x << s t d : : e n d l ; 167 s t d : : c o u t << " E l l p a c k E*x =" << s t d : : e n d l << Ex << s t d : : e n d l ; 168 s t d : : c o u t << " T e s t of x = A*y " << s t d : : e n d l ; 169 Eigen : : VectorXd y ( 3 ) ; 170 y << 1, 2, 3 ; 171 Eigen : : VectorXd Ey = Eigen : : VectorXd : : Zero ( n ) ; 172 E. mtvmult ( y, Ey ) ; 173 s t d : : c o u t << " S p a r s e S^T*y =" << s t d : : e n d l << S. t r a n s p o s e ( ) *y << s t d : : e n d l ; 174 s t d : : c o u t << " E l l p a c k E^T*y =" << s t d : : e n d l << Ey << s t d : : e n d l ; 175 } 10

13 Problem 3 Piecewise cubic Hermite interpolation (24 pts.) We are concerned with piecewise cubic Hermite interpolation of given data points (t j, y j ) R 2, j = 0,..., n, n N. Assume that the nodes are ordered, namely t j < t j+1 for every j = 0,..., n 1. The slopes c j = s (t j ), j = 0,..., n of the interpolant s at the points t j are computed from the data according to the formula: 1 for j = 0, c j = minmod( j, j+1 ) for j = 1,..., n 1, n for j = n, (3) where j = y j y j 1 t j t j 1 and 0 if sgn(x) sgn(y), minmod(x, y) = min( x, y ) sgn(x) otherwise, (4) with the sign function given by 1 if x < 0, sgn(x) = 0 if x = 0, 1 if x > 0. (5) (3a) Implement a C++ function: 1 double minmod ( double x, double y ) ; for the computation of the function minmod as defined in (4). Solution: See implementation in problem3_solution.cpp. (3b) Implement a C++ function 1 using V e c t o r = Eigen : : VectorXd ; 2 V ector d e l t a ( c o n s t V e c t o r & t, c o n s t V ector & y ) ; for the computation of the quantities j, j = 1,..., n, from the data (t j, y j ). The arguments t and y pass the t j and y j, j = 0,..., n. The return vector contains the computed values 1,..., n. Solution: See implementation in problem3_solution.cpp. 11

14 (3c) The template file problem3.cpp contains the declaration of a class mmpchi, reproduced here: 1 //! \brief Class representing a piecewise cubic Hermite interpolant s with minmod-reconstructed slopes 2 c l a s s mmpchi { 3 p u b l i c : 4 //! \brief Construct the slopes c j of the interpolant, from the data (t j, y j ). 5 //! \param[in] t Vector of nodes t j, j = 0,..., n 6 //! \param[in] y Vector of values y j, j = 0,..., n 7 mmpchi( c o n s t V e c t o r & t, c o n s t V ector &y ) ; 8 9 //! \brief Evaluate *this interpolant at the points defined by x and store the result in v 10 //! \param[in] x Vector of points x i, i = 0,..., m 11 //! \return v Vector of values s(x i ), i = 0,..., m 12 V e c t o r e v a l ( c o n s t V ector &x ) c o n s t ; p r i v a t e : 15 c o n s t V e c t o r t, y ; //!< Vectors containing nodes and values 16 V e c t o r c ; //!< Vector containing slopes for the interpolant 17 } ; This class implements a piecewise cubic Hermite interpolant using the minmod slope reconstruction (3). Some class member functions still have to be implemented. Implement an efficient constructor 1 mmpchi : : mmpchi( c o n s t V ector & t, c o n s t V ector &y ) ; The function mmpchi is the constructor of the class, in which you should compute the slopes c j according to (3) and store the value in the corresponding class variable. Solution: See implementation in problem3_solution.cpp. (3d) Implement an efficient method: 1 V ector mmpchi : : e v a l ( c o n s t V ector &x ) c o n s t ; The function eval evaluates the interpolant defined by the class at the points in x and returns the values in a vector of the same length as x. No assumption is made on the ordering of x. 12

15 HINT: An efficient implementation may sort x first. To that end, you can use std::sort, in combination with the data() and size() method of the class Vector. Solution: The implementation computes the value of the interpolant using local Hermite polynomial evaluation. See implementation in problem3_solution.cpp. HINT: You can test your implementation using the code written in the main() function of problem3.cpp. The main() function will perform a convergence study applied to the function f(t) = 1, t [ 5, 5]. (6) 1 + t2 (3e) We use the piecewise cubic Hermite interpolation introduced above for function approximation. Consider interpolation error for the functions 1 f 1 (t) =, t [ 5, 5], (7) 1 + t2 1 f 2 (t) =, t [ 5, 5], (8) 1 + t and sequences of n equidistant interpolation nodes ( 5 = t 0 <... < t n 1 = 5) for n = 2 2, 2 3,..., In the following tables, we measure the interpolation error in the L -norm on [ 5, 5], which is approximated by sampling on a grid of M = 100, 000 nodes on [ 5, 5]. One table represents the error for f 1, the other for f 2. Assign each table to the corresponding function and explain the reasons for your choice. Determine the empirical rate of convergence for each table. Table A Table B n error n error e e e

16 Solution: The experimental rate of convergence is polynomial and limited to secondorder (O(h 2 )) in the case of a regular function (i.e. f 1 ) (table A), and limited to firstorder (O(h)) in case of irregular function (i.e. f 2 ) (table B). See the implementation in problem3_solution.cpp. 1 # i n c l u d e < i o s t r e a m > 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 using V e c t o r = Eigen : : VectorXd ; 8 using M a t r i x = Eigen : : MatrixXd ; 9 10 //! \brief Function for the evaluation of minmod(x,y) 11 double minmod ( double x, double y ) { 12 //// TODO: problem 3a, implement this function return x*y <= 0.? 0. : ( x >= 0.? s t d : : min ( x, y ) : s t d : : max ( x, y ) ) ; 15 } //! \brief Function for the computation of the values j 18 //! \param[in] t Vector of nodes t j, j = 0,..., n 19 //! \param[in] y Vector of values y j, j = 0,..., n 20 //! \return Vector ( 1,..., n ) relative to (t, y) data 21 V ector d e l t a ( c o n s t V e c t o r & t, c o n s t V ector & y ) { 22 a s s e r t ( t. s i z e ( ) == y. s i z e ( ) ) ; 23 a s s e r t ( t. s i z e ( ) > 1) ; //// TODO: problem 3b, implement this function V e c t o r d l t ( t. s i z e ( ) 1) ; f o r ( unsigned i n t j = 0 ; j < t. s i z e ( ) 1; ++ j ) { 30 d l t ( j ) = ( y ( j +1) y ( j ) ) / ( t ( j +1) t ( j ) ) ; 31 } return d l t ; 34 } 14

17 35 36 //! \brief Class representing a pecevise cubic Hermite interpolant s with minmod-reconstructed slopes 37 c l a s s mmpchi { 38 p u b l i c : 39 //! \brief Construct the slopes c j of the interpolant, from the data (t j, y j ). 40 //! \param[in] t Vector of nodes t j, j = 0,..., n 41 //! \param[in] y Vector of values y j, j = 0,..., n 42 mmpchi( c o n s t V e c t o r & t, c o n s t V ector &y ) ; //! \brief Evaluate interpolant defined by (*this) at the points defined by x and store the result in v 45 //! \param[in] x Vector of points x i, i = 0,..., m 46 //! \return v Vector of values s(x i ), i = 0,..., m 47 V e c t o r e v a l ( c o n s t V ector &x ) c o n s t ; 48 p r i v a t e : 49 c o n s t V e c t o r t, y ; //!< Vectors containing nodes and values 50 V e c t o r c ; //!< Vector containing slopes for the interpolant 51 } ; mmpchi : : mmpchi( c o n s t V ector & t, c o n s t V ector &y ) 54 : t ( t ), y ( y ), c ( t. s i z e ( ) ) { a s s e r t ( t. s i z e ( ) == y. s i z e ( ) ) ; 57 a s s e r t ( t. s i z e ( ) > 1) ; //// TODO: problem 3c, implement this function unsigned i n t n = t. s i z e ( ) 1; V e c t o r D = d e l t a ( t, y ) ; // Reconstruct slopes using minmod 66 c ( 0 ) = D( 0 ) ; 67 f o r ( unsigned i n t i = 1 ; i < n ; ++ i ) { c ( i ) = minmod (D( i ), D( i 1) ) ; 70 } 71 c ( n ) = D( n 1) ; 72 } 15

18 73 74 V ector mmpchi : : e v a l ( c o n s t V ector &x ) c o n s t { 75 V e c t o r v ( x. s i z e ( ) ) ; //// TODO: problem 3d, implement this function 78 V e c t o r xcopy = x ; 79 s t d : : s o r t ( xcopy. d a t a ( ), xcopy. d a t a ( ) +xcopy. s i z e ( ) ) ; unsigned i n t i = 0 ; 82 f o r ( unsigned i n t j = 0 ; j < t. s i z e ( ) 1; ++ j ) { 83 double t 1 = t ( j ) ; 84 double t 2 = t ( j +1) ; 85 double y1 = y ( j ) ; double h = t 2 t 1 ; 88 double a1 = y ( j +1) y1 ; 89 double a2 = a1 h*c ( j ) ; 90 double a3 = h*c ( j +1) a1 a2 ; 91 while ( i < xcopy. s i z e ( ) && xcopy ( i ) <= t 2 ) { 92 double t x = ( xcopy ( i ) t 1 ) / h ; 93 v ( i ) = y1 + ( a1 +( a2+ a 3 * t x ) * ( tx 1. ) ) * t x ; 94 i ++; 95 } 96 } return v ; 99 } i n t main ( i n t, c h a r * * ) { 102 //// PROBLEM 3 TEST auto f = [ ] ( double t ) > double { return 1. / ( 1. + t * t ) ; } ; 105 //auto f = [] (double t) -> double { return 1. / (1. + std::abs(t)); }; // Interval 108 c o n s t double a = 5., b = 5 ; 109 // Number of sampling nodes 110 c o n s t unsigned i n t M = 10000;

19 112 // Sampling nodes and values 113 c o n s t V e c t o r x = V ector : : LinSpaced ( M,a,b ) ; 114 V e c t o r v_ex (M) ; 115 // Fill in exact values of f 116 f o r ( unsigned i n t i = 0 ; i <M; ++ i ) { 117 v_ex ( i ) = f ( x ( i ) ) ; 118 } 119 // Output table with error w.r.t number of nodes 120 s t d : : c o u t << " n " << " \ t " << "L^ i n f e r r. " << s t d : : e n d l ; 121 f o r ( unsigned i n t n = 4 ; n <= 2048; n=n < <1) { 122 V e c t o r t = V e c t o r : : LinSpaced ( n, a, b ) ; 123 V e c t o r y ( n ) ; 124 f o r ( unsigned i n t i = 0 ; i <n ; ++ i ) { 125 y ( i ) = f ( t ( i ) ) ; 126 } mmpchi s ( t, y ) ; 129 c o n s t V e c t o r v = s. e v a l ( x ) ; 130 s t d : : c o u t << n << " \ t \ t " << ( v v_ex ). lpnorm < Eigen : : I n f i n i t y > ( ) << s t d : : e n d l ; 131 } } Problem 4 Quadrature by transformation (16 pts.) For f C([0, 2]) define I(f) = t f(t) dt. t (4a) For n N, derive a family of 2n-point quadrature formulas Q n (f) = 2n j=1 w n j f(c n j ), f C([0, 2]) for which Q n (f) converges to I(f) exponentially as n, if f has an analytic extension to a complex neighborhood of [0, 2]. Solution: We wish to transform the integral in I(f) into an integral of a smooth function 17

20 over a bounded interval. In order to this, write 1 2 t I(f) = f(t) dt + 0 t t f(t) dt. t Performing the substitutions s = t in the first integral and s = 2 t in the second integral we obtain I(f) = s2 f(s 2 ) ds s 2 2 s 2 f(2 s2 ) ds = f 1 (s) ds f 2 (s) ds where f 1 (s) = 2 s 2 f(s 2 ) and f 2 (s) = s 2 2 s 2 f(2 s2 ). With the change of variables t = 2s 1 we finally obtain I(f) = 1 1 f 1 ((t + 1)/2) ds + Applying Gauss quadrature to these integrals, we define Q n (f) = n j=1 1 1 n f 2 ((t + 1)/2) ds. (9) w j n f 1 ((ξj n + 1)/2) + w j n f 2 ((ξj n + 1)/2), where ξj n and w j n are nodes and weights of the Gauss quadrature of order n defined on [ 1, 1]. (Alternatively, it is also possible to use the Clenshaw-Curtis quadrature rule.) By construction, we have that Q n (f) I(f) exponentially as n if f has an analytic extension to a neighborhood of [1, 2], since this implies that the integrands in (9) have an analytic extension to a neighborhood of [ 1, 1]. Define now the nodes c n ((ξj n j = + 1)/2)2 if j = 1,..., n, 2 ((ξj n n + 1)/2)2 if j = n + 1,..., 2n. Therefore, the above expression can be written as j=1 Q n (f) = n j=1 where the weights are given by w j n 2 c n j f(c n j ) + n j=1 w n j 2 c n j+n c n j+n f(c n j+n) = 2n j=1 w n j f(c n j ), w wj n j n 2 c n j if j = 1,..., n, = w j n n 2 c n j c if j = n + 1,..., 2n. n j 18

21 (4b) Given n > 0, determine the maximal d N such that I(p) = Q n (p) for every polynomial p P d 1. DEPENDS ON: subproblem (4a). Solution: The above expression is never exact for polynomials f, because of the presence of the square roots in (9). In other words, d = 0, for which P 1 =. Problem 5 Mono-implicit Runge-Kutta single step method (28 pts.) A so-called s-stage mono-implicit Runge-Kutta single step method (MIRK) for the autonomous ODE ẏ = f(y), f D R d R d, is defined as: i 1 g i = (1 v i )y 0 + v i y 1 + h d i,j f(g j ), i = 1,... s, y 1 = y 0 + h b j f(g j ) s j=1 for suitable coefficients v i, b i, d i,j R. (5a) Single step methods defined by (10) belong to the class of implicit Runge-Kutta methods. Write down the corresponding Butcher scheme in terms of the coefficients v i, b i, d i,j. Solution: Let us define d i,j = 0 for j i. Replacing y 1 in the equation for g i, one obtains: s i 1 s g i = (1 v i )y 0 + v i (y 0 + h b j f(g j )) + h d i,j f(g j ) = y 0 + h (v i b j + d i,j )f(g j ), y 1 = y 0 + h b j f(g j ) s j=1 j=1 = a i,j Set k i = f(g i ). Then, k i = f(y 0 + h s j=1 (v i b j + d i,j ) k j ) and Therefore, the Butcher scheme becomes: j=1 j=1 s j=1 (10) (11) y 1 = y 0 + h b j k j (12) j=1 c D + vb b, where (D) i,j = d i,j, b = [b 1,..., b s ], v = [v 1,..., v s ] and c i = s j=1 a i,j. 19

22 (5b) Compute the stability function of a MIRK scheme defined as in (10) for s = 2. Solution: We apply the MIRK scheme to the ODE y = λy. Denote by g = [g 1, g 2 ] and z = hλ, then: where 1 = [1,..., 1]. Consider Then whence g = (1 v)y 0 + vy 1 + zdg (I zd)g = (1 v)y 0 + vy 1 y 1 = y 0 + zb g = y 0 + zb (I zd) 1 ((1 v)y 0 + vy 1 ). (1 zb (I zd) 1 v)y 1 = (1 + zb (I zd) 1 (1 v))y 0, S(z) = 1 + zb (I zd) 1 (1 v). 1 zb (I zd) 1 v Alternative: use formula from Thm, from the course together with the Butcher scheme derived in the previous sub-problem. (5c) Now, we consider the special case of a scalar ODE (d = 1) and s = 2. Abbreviating z = [g 1, g 2, y 1 ], rewrite (10) as a non-linear system of equations in the form F(z) = 0 for an explicitly specified suitable function F R 3 R 3. Solution: F( g 1 g 2 y 1 g 1 (1 v 1 )y 0 v 1 y 1 ) = g 2 (1 v 2 )y 0 v 2 y 1 hd 2,1 f(g 1 ) y 1 y 0 h 2 j=1 b j f(g j ) (5d) Find the Jacobian DF(z) of the function F (in terms of d i,j, v i, b i and the derivative of f) from the previous sub-problem. DEPENDS ON: subproblem (5c). Solution: DF( g 1 g 2 y v 1 ) = I 3 hd 2,1 f (g 1 ) 0 v 2. hb 1 f (g 1 ) hb 2 f (g 2 ) 0 In the next steps, we will implement the MIRK scheme for d = 1, s = 2. 20

23 (5e) Implement a function 1 using V e c t o r = Eigen : : VectorXd ; 2 template < c l a s s F u n c, c l a s s Jac > 3 void n e w t o n 2 s t e p s ( c o n s t Func & F, c o n s t J a c & DF, 4 V e c t o r & z ) ; that approximates the solution of F(z) = 0 by performing two steps of the Newton method applied to F(z) = 0. Use z to pass the initial guess and to return the approximated solution. Objects of type Func and Jac have to supply suitable evaluation operators operator(). Solution: See implementation in problem5_solution.cpp. (5f) Consider the particular MIRK scheme given by the coefficients: v 1 = 1, v 2 = , d 21 = , b 1 = 37 82, b 2 = (13) Using the function newton2steps, implement a function 1 template < c l a s s F u n c, c l a s s Jac > 2 double MIRKstep ( c o n s t Func & f, c o n s t J a c & d f, 3 double y 0, double h ) ; that realizes one step of the MIRK scheme defined by (10) for s = 2 and a scalar ODE, that is, d = 1. The right hand side function f and its Jacobian are passed through f and df. The solution of the nonlinear system arising from (10) is approximated using two Newton steps, namely by using the function newton2steps. The initial guess has to be chosen appropriately! DEPENDS ON: subproblems (5c) and (5d). Solution: See implementation in problem5_solution.cpp. (5g) Implement a function 1 template < c l a s s F u n c, c l a s s Jac > 2 double MIRKsolve ( c o n s t Func & f, c o n s t J a c & d f, 3 double y 0, double T, unsigned i n t N) ; for the solution of a scalar ODE up to time T, using N equidistant steps of the monoimplicit Runge-Kutta single step method defined by (13). The initial value is passed in y0. 21

24 Solution: See implementation in problem5_solution.cpp. (5h) Apply your implementation to the IVP ẏ = 1 + y 2, y(0) = 0 (14) on [0, 1]. The file problem5.cpp contains a partial template for this sub-problem. The exact solution of (14) is y ex (t) = tan t. Compute the solution y n at T = 1 with a sequence of uniform temporal meshes with n = 4,..., 512 intervals. Compute and output the error y n (1) y ex (1) and determine the rate of convergence of the scheme. DEPENDS ON: subproblems (5c), (5d), (5e), (5f) and (5g). Solution: The experimental rate of convergence of the methods is O(h 2 ). See implementation in problem5_solution.cpp. Solution: 1 # i n c l u d e < i o s t r e a m > 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 using V e c t o r = Eigen : : VectorXd ; 8 using M a t r i x = Eigen : : MatrixXd ; 9 10 //! \brief Perform 2 steps of newton method applied to F and its jacobian DF 11 //! \tparam Func type for function F 12 //! \tparam Jac type for jacobian DF of F 13 //! \param[in] F function F, for which F(z) = 0 is needed 14 //! \param[in] DF Jacobian DF of the function F 15 //! \param[in,out] z initial guess and final approximation for F(z) = 0 16 template < c l a s s F u n c, c l a s s Jac > 17 void n e w t o n 2 s t e p s ( c o n s t Func & F, c o n s t J a c & DF, Vector & z ) { 18 // TODO: problem 5e: two newton steps V e c t o r znew = z DF( z ). l u ( ). s o l v e ( F ( z ) ) ; 21 z = znew DF( znew ). l u ( ). s o l v e ( F ( znew ) ) ; 22 } 23 22

25 24 //! \brief Perform a single step of the MIRK scheme applied to the scalar ODE y = f(y) 25 //! \tparam Func type for function f 26 //! \tparam Jac type for jacobian df of f 27 //! \param[in] f function f, as in y = f(y) 28 //! \param[in] df Jacobian df of the function f 29 //! \param[in] y0 previous value 30 //! \param[in] h step-size 31 //! \return value y1 at next step 32 template < c l a s s F u n c, c l a s s Jac > 33 double MIRKstep ( c o n s t Func & f, c o n s t J a c & d f, double y 0, double h ) { 34 c o n s t double v1 = 1 ; 35 c o n s t double v2 = /2025. ; 36 c o n s t double d21 = 164. /2025. ; 37 c o n s t double b1 = 3 7. / 8 2. ; 38 c o n s t double b2 = 4 5. / 8 2. ; // TODO: problem 5f: implement MIRK step auto F = [& f, &y 0, &v 1, &v 2, &d 2 1, &b 1, &b 2, &h ] ( c o n s t V e c t o r & z ) > Vector { 43 V e c t o r r e t ( 3 ) ; 44 r e t << z ( 0 ) (1 v1 ) *y0 v1*z ( 2 ), 45 z ( 1 ) (1 v2 ) *y0 v2*z ( 2 ) h*d21*f ( z ( 0 ) ), 46 z ( 2 ) y0 h* ( b1*f ( z ( 0 ) ) + b2*f ( z ( 1 ) ) ) ; 47 return r e t ; 48 } ; 49 auto DF = [& d f, &v 1, &v 2, &d 2 1, &b 1, &b 2, &h ] ( c o n s t V e c t o r & z ) > M atrix { 50 M a t r i x M(3, 3 ) ; 51 M << 0, 0, v 1, 52 h*d21*df ( z ( 0 ) ), 0, v 2, 53 h*b1*df ( z ( 0 ) ), h * b 2 * d f ( z ( 1 ) ), 0 ; 54 return M a trix : : I d e n t i t y (3, 3 ) M; 55 } ; 56 V e c t o r z ( 3 ) ; 57 z << 0, 0, y 0 ; // FIXME: initial data 58 n e w t o n 2 s t e p s ( F, DF, z ) ; return z ( 2 ) ; 23

26 61 } //! \brief Solve an ODE y = f(y) using MIRK scheme on equidistant steps 64 //! \tparam Func type for function f 65 //! \tparam Jac type for jacobian df of f 66 //! \param[in] f function f, as in y = f(y) 67 //! \param[in] df Jacobian df of the function f 68 //! \param[in] y0 initial value 69 //! \param[in] T final time 70 //! \param[in] N number of steps 71 //! \return value approximating y(t) 72 template < c l a s s F u n c, c l a s s Jac > 73 double MIRKsolve ( c o n s t Func & f, c o n s t J a c & d f, double y 0, double T, unsigned i n t N) { 74 // TODO: problem 5g: implement MIRK solver c o n s t double h = T / N; 77 double y n e x t = y0 ; 78 f o r ( unsigned i n t i = 0 ; i < N; ++ i ) { 79 y n e x t = MIRKstep ( f, d f, y n e x t, h ) ; 80 } 81 return y n e x t ; 82 } i n t main ( i n t, c h a r * * ) { auto f = [ ] ( double y ) > double { return 1 + y*y ; } ; 87 auto df = [ ] ( double y ) > double { return 2*y ; } ; c o n s t double y0 = 0. ; 90 c o n s t double T = 1. ; c o n s t double yex = t a n ( 1 ) ; //// PROBLEM 5h TEST 95 s t d : : c o u t << " *** PROBLEM 5h : " << s t d : : e n d l ; 96 // TODO: problem 5h: solve IVP y = f(y) up to T s t d : : c o u t << "N" << " \ t " << " yend " << " \ t " << " e r r " << s t d : : e n d l ; 24

27 99 f o r ( unsigned i n t N = 4 ; N < 512; N=N< <1) { 100 double yend = MIRKsolve ( f, d f, y 0, T, N) ; 101 double e r r = s t d : : abs ( yex yend ) ; 102 s t d : : c o u t << N << " \ t " << yend << " \ t " << e r r << s t d : : e n d l ; 103 } 104 } 25