Introduction to PDEs and Numerical Methods Lecture 7. Solving linear systems

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1 Platzhalter für Bild, Bild auf Titelfolie hinter das Logo einsetzen Introduction to PDEs and Numerical Methods Lecture 7. Solving linear systems Dr. Noemi Friedman,

2 Reminder: Instationary heat equation what to solve? Stability checking from eigenvalue analysis: Method of lines Euler forward method find the eigenvalues (λ j ) and eigenvectors (v j ) of matrix A u n+ = I + ΔtA B u n u n+ = Bu n Euler backward method u n = I ΔtA B u n+ Theta method I θδta u n+ = I + θ ΔtA u n B u n+ = u n B θ u n+ = B 2θ u n Solve system of equations Gx = b solve for x B θ B 2θ Dr. Noemi Friedman PDE lecture Seite 2

3 Reminder: Stationary heat equation what to solve? Conclusion instationary heat equation with implicit FD System of linear equations: methods (Euler backward, Theta method) stationary heat equation Ax = b solve for x Where the A matrix is in general sparse, banded can get very large with refined spatial and temporal discretisation for D heat equation with three-point-stencils: tridiagonal for D heat equation with five-point-stencils: pentadiagonal for 2D heat equation: banded with sparse band Dr. Noemi Friedman PDE lecture Seite

4 Existance and uniquness of solutions to Ax=b Nullspace of A If the only element of the nullspace is the zero vector: trivial nullspace Unique solution: if and only if the nullspace is trivial Infinitive number of solutions: if nullspace is not trivial and b satisfies the compatibility condition (Fredholm alternative): Dr. Noemi Friedman PDE lecture Seite 4

5 How to solve? Direct methods If the matix A is a) Diagonal b) Lower triangular c) Upper triangular d) Orthogonal matrix A T = A e) Tridiagonal matrix Then to solve Ax = b is easy. a) a a a b) a 0 0 a 2 a 22 0 a a 2 a x x 2 x = x x 2 x = b b 2 b b b 2 b 2 x = b a x 2 = b 2 a 22 x = b a x = b a x 2 = f(x ) x = f(x, x 2 ) c) a a 2 a 0 a 22 a a x x 2 x = b b 2 b Backward substitution 2 x = f(x 2, x ) x 2 = f(x ) x = b a Forward substitution Dr. Noemi Friedman PDE lecture Seite 5

6 How to solve? Direct methods.) Solve system of equation directly Calculate inverse from Cramer-rule: x i = det(a i) det(a) Ax = b 2 n +! operations A = a a 2 a 2 a 22 a i a i2 a n a n2 a i a n a 2i a 2n a ii a in a ni a nn A i = a a 2 a 2 a 22 a i a i2 a n a n2 b i a n b 2i a 2n b ii a in b ni a nn Gauß-Jordan elimination Gauß elimination, LU decomposition (Cholesky decomp. If A is symmetric and pos.def., Thomas algorithm, if matrix is tridiagonal) Dr. Noemi Friedman PDE lecture Seite 6

7 Solve Ax=b Gauß-Jordan elimination Gauß-Jordan elimination for solving the system of equation: a x + a 2 x 2 + a x = b () a 2 x + a 22 x 2 + a 2 x = b 2 a x + a 2 x 2 + a x = b () Where a ii and b i are known x, x 2, x =? A = a a 2 a a 2 a 22 a 2 a a 2 a b = b b 2 b x = x x 2 x Ax = b Method: apply elementary transformation on matrix A (multiply both side of the equation with elementary tranformation matrices T i ) to get diagonal matrix. The elementary operations: Switch rows of A Multiply a row with a nonzero scalar Add to one row a scalar multiple of another Dr. Noemi Friedman PDE lecture Seite 7

8 Solve Ax=b Gauß-Jordan elimination I. Devide equation () by a * : x + a 2 a x 2 + a a x = b a () a 2 x + a 22 x 2 + a 2 x = b 2 a x + a 2 x 2 + a x = b () T 0 Ax =T 0 b a 2 a 0 0 a a a T 0 A = a 2 a 22 a 2 T 0 = T b = a a 2 a 0 0 * if a is nonzero, otherwise before this step rows should be switched such that first diagonal element of A is nonzero. If all elements in the first column of matrix A is zero, A= A and skip steps I and II Dr. Noemi Friedman PDE lecture Seite b a b 2 b

9 Solve Ax=b Gauß-Jordan elimination II. Add a 2 times equation () to equation and a times equation () to equation (): x + a 2 a x 2 + a a x = b a () 0 + a 22 a 2 a 2 a x 2 + a 2 a 2 a a x = b 2 a 2 b a 0 + a 2 a a 2 a x 2 + a a a a x = b a b a () T T 0 Ax = T T 0 b T = A T T 0 A = 0 0 a 2 0 a 0 a a 2 a a 2 a 22 a 2 = a a 2 a a 2 a a a 0 a 22 a 2 a 2 a a 2 a 2 a a 0 a 2 a a 2 a a a a a Dr. Noemi Friedman PDE lecture Seite 9

10 Solve Ax=b Gauß-Jordan elimination III. Devide equation by a 22 * T 2 T T 0 Ax = T 2 T T 0 b T 2 T T 0 A = a a 2 a 0 a 2 a 22 0 a 2 a where a = 0 or T 2 = a IV. Add a 2 times equation to equation () and a 2 times equation to equation () (transformation T ) T T 2 T T 0 Ax = T T 2 T T 0 b A T T 2 T T 0 A = a 0 a 0 a a T = a a 2 *if a 22 is zero, 2nd and rd equations should be switched before step III. If all elements in the second column is zero, skip steps III and IV. with A = A Dr. Noemi Friedman PDE lecture Seite 0

11 Solve Ax=b Gauß-Jordan elimination V. Devide equation () by a * (transformation T 4 ) T 4 T T 2 T T 0 Ax =T 4 A = T 4 T T 2 T T 0 b T 4 A = a 0 a 0 a 22 a where a and a 22 = 0 or T 4 = a VI. Add a times equation () to equation () and a 2 times equation () to equation () (transformation T 5 ) T 5 T 4 T T 2 T T 0 Ax = T 5 T 4 T T 2 T T 0 b A T 5 T 4 A = a a *if a is zero skip steps V and VI. T 5 = 0 a a 2 If in A there are zero diagonal elements, matrix A is singular. Otherwise A is the unit matrix, and T 5 T 4 T T 2 T T 0 is the inverse of matrix A ** ** if rows had to be switched in between, the corresponding tranformations has to be inserted in between the T i transformations Dr. Noemi Friedman PDE lecture Seite

12 Solve Ax=b Gauß-Jordan elimination Example Solve with Gauß-Jordan elimination 2x x 2 + 2x = 6 () 4x x 2 + 6x = 20 6x 7x 2 + x = () x, x 2, x =? A = b = 6 20 x = x x 2 x Ax = b I () () devide equation by () () T 0 = Dr. Noemi Friedman PDE lecture Seite 2

13 Solve Ax=b Gauß-Jordan elimination Example II () add -4 times () () add -6 times () () () T = IV. (Second diagonal is, step III. is not needed) () () add 0.5 times add 4 times () () T 2 = V () () Devide by () () T = /0 VI () () add -2 times () add -2 times () () () T 4 = Dr. Noemi Friedman PDE lecture Seite

14 Solve Ax=b Gauß-Jordan elimination x = () x 2 = 2 x = () And the inverse of matrix A: A = T 4 T T 2 T T 0 T 0 = T = T 2 = T = /0 T 4 = A = T 4 T T 2 T T 0 = Note: results can be calculated without the transformation matrices, but they are necessary for the inverse calculation Dr. Noemi Friedman PDE lecture Seite 4

15 Solve Ax=b Gauß elimination, inverse calculation Inverse of A can be calculated also from A Unit matrix Example () () devide equation by () () add -4 times () add -6 times () Dr. Noemi Friedman PDE lecture Seite 5

16 Solve Ax=b Gauß elimination, inverse calculation Example () () add 0.5 times add 4 times () () Devide by () () add -2 times () add -2 times () A = Dr. Noemi Friedman PDE lecture Seite 6

17 Solve Ax=b Gauß elimination Method: same as in Gauß-Jordan elimination, but A is transformed to upper triangular matrix through the elementary operations. Method is presented through the previous example. Example 2x x 2 + 2x = 6 () 4x x 2 + 6x = 20 6x 7x 2 + x = () x, x 2, x =? A = b = 6 20 x = x x 2 x Ax = b Dr. Noemi Friedman PDE lecture Seite 7

18 Solve Ax=b Gauß elimination Example () () I. devide equation by 2 II III. IV () add -4 times () () add -6 times () () () () () add 4 times Devide by () () () () () () () () T 0 = T = 2 T 2 = T = Dr. Noemi Friedman PDE lecture Seite /0

19 Solve Ax=b Gauß elimination, LU decomposition Example x 0.5x 2 + x = () x 2 + 2x = x = () If equation Ax=b has to be solved for several b: Using back-substitution, each x i can be solved for () x = x 2 = 2x = 2 = 2 () x = + 0.5x 2 x = = T 0 - T - T 2 - T - T T 2 T T 0 A = A = LU L: lower triangular matrix U: upper triangular matrix.) Calculate L and U 2.) Solve Ax=LUx=b in two steps: solve y from Ly = b where y = Ux (equation can be solved with forward substitution) solve x from Ux= y (equation can be solved with backward substitution) Note: LU decomposition can be calculated from different algorithms Dr. Noemi Friedman PDE lecture Seite 9

20 Stationary and instationary heat equation how to solve?.) Solve system of equation directly Ax = b Calculate inverse from Cramer-rule: x i = det(a i) det(a) 2 n +! operations Gauß-Jordan elimination Gauß elimination, LU decomposition (Cholesky decomp. If G is symmetric and pos.def., Thomas algorithm, if matrix is tridiagonal) A = L U 2n 2 operations Ax = L Ux y Ly = b Ux = y = b forward substitution back substitution Dr. Noemi Friedman PDE lecture Seite 20

21 Stationary and instationary heat equation how to solve? Direct solve A = LU LU factorisation with Gauß method a a 2 a 2 a 22 = l 0 l 2 l 22 u u 2 0 u 22 a = l u g 2 = l u 2 a 2 = l 2 u g 22 = l 2 u 2 + l 22 u 22 4 equations 6 unknowns 0 l 2 u u 2 0 u 22 a = u a 2 = u 2 0 l 2 a a 2 0 u 22 a 2 = l 2 a a 22 = l 2 a 2 + u 22 0 a 2 a a a 2 0 a 22 l 2 a Dr. Noemi Friedman PDE lecture Seite 2

22 Stationary and instationary heat equation how to solve? Direct solve Gauß factorization A = LU in general n 2 equations n 2 l + n unknowns ii = General algorithm for i = k + n l ik = a ik (k) a kk (k) for j = k +.. n a ij (k+) = a ij (k) l ik a kj (k) ~ 2n operations But even if the matrix is nonsingular the elements g kk (k) (pivot elements) can be zero Pivoting (flip rows or columns) can be also important for reducing roundoff errors g kk (k) won t be zero if the matrix is positivive definit or if it is diagonally dominant Dr. Noemi Friedman PDE lecture Seite 22

23 Stationary and instationary heat equation how to solve? Direct solve If A is positive definit+symmetric Tridiagonal system: A = A = LU = HH T a c e 2 a 2 c 2 e n a n L = LU factorisation with Cholesky decomposition Thomas algorithm β 2 β n U = ~ n operations (half of the Gauß method) α c α 2 c 2 α n α = a β i = e i α α i = a i β i c i i = 2 n What happens with the roundoff errors in A = LU = A + δa A Ax = b A + δa x = b + δb x x x λ max λ min δb b O(n) operations λ max λ min = K(A) Dr. Noemi Friedman PDE lecture Seite 2

24 Iterative methods illustrative example the Jacobi method x y z = x + y = 0 x + 4y + z = 2 y + 4z = 0 x = (0 y) 4 y = (2 x z) 4 z = (0 y) 4 x (0) y (0) z (0) : = x () y () z () : = The true solution: x () = 4 0 y 0 = 2.5 y () = 4 2 x 0 z 0 = z () = 4 0 y 0 = 2.5 x y z : = x y z = x () y () z () : = x (k+) = 4 0 y k = 2.5 y (k+) = 4 2 x k z k = z (k+) = 4 0 y k = 2.5 x i (k+) = a ii b i j i a ij x j (k) Dr. Noemi Friedman PDE lecture Seite 24

25 Iterative methods illustrative example the Gauß-Seidel method x y z = x + y = 0 x + 4y + z = 2 y + 4z = 0 x = (0 y) 4 y = (2 x z) 4 z = (0 y) 4 x (0) y (0) z (0) : = x () y () z () : = The true solution: x () = 4 0 y 0 = 2.5 y () = 4 2 x z 0 = 2.25 z () = 4 0 y =.975 x y z = x i (k+) = a ii x (k+) = 4 0 y k = 2.5 y (k+) = 4 2 x k+ z k = z (k+) = 4 0 y k+ = 2.5 (k) b i a ij x j(k+) a ij x j j<i j>i Dr. Noemi Friedman PDE lecture Seite 25

26 Iterative methods illustrative example the Jacobi method A = a a 2 a 2 a 22 a i a i2 a n a n2 a i a n a 2i a 2n a ii a in a ni a nn = D + E + F x i (k+) = a ii b i j i a ij x j (k) a a 22 a i a n a 2i a 2n a ii a nn x k+ x 2 k+ x i k+ x n k+ = b b 2 b i b n 0 a 2 a 2 0 a i a i2 a n a n2 a i a n a 2i a 2n 0 a in a ni 0 x k x 2 k x i k x n k D E + F x (k+) = D b E + F x (k) Dr. Noemi Friedman PDE lecture Seite 26

27 Iterative methods illustrative example the Jacobi method A = a a 2 a 2 a 22 a i a i2 a n a n2 a i a n a 2i a 2n a ii a in a ni a nn = D + E + F x i (k+) = a ii j<i b i a ij x j (k+) + aii x i (k+) = j<i a ij x j(k+) j>i bi j>i a ij x j (k) a ij x j (k) Ex (k+) + Dx (k+) = b Fx (k) E + D x (k+) = b Fx (k) x (k+) = E + D b Fx (k) Dr. Noemi Friedman PDE lecture Seite 27

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