Thomas Algorithm for Tridiagonal Matrix

Transcription

1 P a g e 1 Thomas Algorithm for Tridiagonal Matrix Special Matrices Some matrices have a particular structure that can be exploited to develop efficient solution schemes. Two of those such systems are banded and symmetric matrices. A banded matrix is a square matrix that has all elements equal to zero, with the exception of a band centered on the main diagonal. Banded systems are frequently encountered in the solution of differential equations. In addition, other numerical methods such as cubic splines involve the solution of banded systems. The dimensions of a banded system can be quantified by two parameters: the bandwidth BW and the halfbandwidth HBW (see figure below). These two values are related by BW = 2HBW + 1. In general, then, a banded system is one for which a ij = 0 if i j > HBW. For a triadiagonal matrix BW=3, HBW=1. For a pentadiagonal matrix BW=5, and HBW=2. Next efficient elimination methods are described for both. Heat Transfer Problem As an example to illustrate the solution mechanism for banded matrices, specifically for tridiagonal matrices, we are going to use a simple problem from the heat transfer field. Assume that the extremes of a cylindrical metal rod are maintained at different temperatures, 0 C (left-hand extreme) and 1 C (right-hand extreme) and we wish to determine the steady state temperature at interior nodes. To do this, we divide the rod into 11 small imaginary cylinders, the temperatures of each cylinder is represented by the geometric center, named interior nodes as shown in the figure below: 0 o C 1 o C x 11 nodes

2 P a g e 2 At steady state, the heat transfer equation takes the form: 2 u x 2 = 0 0 x 1 BC s: u = 0 x = 0 u = 1 x = 1 where u is the temperature, and x is the spatial one dimensional coordinate. Expand the second derivative with central difference representation of error order θ(h 2 ): Rearrange and simplify u(i + 1) 2u(i) + u(i 1) x 2 = 0 i = 2, 3, 4, ix 1 u(i 1) 2u(i) + u(i + 1) = 0 i = 2, 3, 4, ix 1 Choose the number of points to be solved ix = 11 (i.e., the number of nodes) and expand the above equation such as the resulting index for u are i=1, 2, 3, ix (within the range of your problem discretization). u(1) 2u(2) + u(3) = 0 u(2) 2u(3) + u(4) = 0 u(3) 2u(4) + u(5) = 0 u(4) 2u(5) + u(6) = 0 u(5) 2u(6) + u(7) = 0 u(6) 2u(7) + u(8) = 0 u(7) 2u(8) + u(9) = 0 u(8) 2u(9) + u(10) = 0 u(9) 2u(10) + u(11) = 0 The range of this discretization is 1 i ix for later coding with matlab Name the coefficients of these equations p, q, r and the right-hand side vector s: p(1)u(1) + q(1)u(2) + r(1)u(3) = s(1) p(2)u(2) + q(2)u(3) + r(2)u(4) = s(2) p(3)u(3) + q(3)u(4) + r(3)u(5) = s(3) p(4)u(4) + q(4)u(5) + r(4)u(6) = s(4) p(5)u(5) + q(5)u(6) + r(5)u(7) = s(5) p(6)u(6) + q(6)u(7) + r(6)u(8) = s(6) p(7)u(7) + q(7)u(8) + r(7)u(9) = s(7) p(8)u(8) + q(8)u(9) + r(8)u(10) = s(8) p(9)u(9) + q(9)u(10) + r(9)u(11) = s(9)

3 P a g e 3 The matrix form yields a quasi tridiagonal matrix: The empty cells within the matrix coefficients matrix are actually filled with zeros. Please notice that that the above matrix is not square; we need to make it square. Thomas algorithm consists of two steps, direct sweep and inverse sweep (Con el permiso de Thomas, this method is just Gaussian elimination applied to a tridiagonal matrix avoiding the job on zeros entries).

4 P a g e 4 DIRECT SWEEP (forward pass on Gaussian Elimination) First step Eliminate u(1) and u(11) (i.e, first and last) from the matrix of coefficients in the first and last equations, by passing them to the other side of the equal sign. This elimination is possible as u(1) and u(11) are known temperatures. For the first equation: p(1)u(1) + q(1)u(2) + r(1)u(3) = s(1) q(1)u(2) + r(1)u(3) = s(1) - p(1)u(1) and for the last equation: p(9)u(9) + q(9)u(10) + r(9)u(11) = s(9) p(9)u(9) + q(9)u(10) = s(9) - r(9)u(11) which square and simplifies the original matrix to a tridiagonal matrix. A tridiagonal system that is, one with a bandwidth of 3 can be expressed generally as columns rows q(1) r(1) u(2) s(1) - p(1)u(1) 2 p(2) q(2) r(2) u(3) s(2) 3 p(3) q(3) r(3) u(4) s(3) 4 u(5) s(4) 5... = u(6) = s(5) 6... u(7) s(6) 7... u(8) s(7) 8 u(9) s(8) 9 p(9) q(9) u(10) s(9) - r(9)u(11) Rename: s(1) = s(1) - p(1)u(1) s(9) = s(9) r(9)u(11) Every time you solve a problem similar to this one, you must develop a square matrix as the above. This is the MATRIX representation of your problem

5 P a g e 5 This is a clean matrix equation suitable for a solution representing your problem. columns rows q(1) r(1) u(2) s(1) 2 p(2) q(2) r(2) u(3) s(2) 3 p(3) q(3) r(3) u(4) s(3) 4 u(5) s(4) 5... u(6) = s(5) 6... u(7) s(6) 7... u(8) s(7) 8 u(9) s(8) 9 p(9) q(9) u(10) s(9) The following procedure is equivalent to the steps of Gauss elimination by using the banded matrix notation which prevents from working with the zero elements. The first equation: q(1)u(2) + r(1)u(3) = s(1) Eq-1 Normalize the first equation, dividing it by q1 (pivot element). In the following equations we have simplified the notation, e.g. q(1) becomes q1, etc. u2 + r1 s1 u3 = q1 q1 rename: u2 + r 1u3 = s 1 Eq-2, where r 1 = r1 q1 and s 1 = s1 q1 Eq-3 Multiplied the first equation by p2 and subtract it from the second equation: p2u2 + q2u3 + r2u4 = s2 Eq-4 (p2u2 + p2 r 1u3 = p2ŝ1) Multiply Eq-2 by p2 Eq-5 (q2 p2r 1)u3 + r2u4 = s2 p2s 1 Eq-6 Notice that the above procedure works for eliminating u2. Normalize E-6 dividing by q2 p2r 1 (i.e., the coefficient of u3 u3 + r2 q2 p2r 1 u4 = s2 p2s 1 q2 p2r 1 Eq-7 Rename Eq-7: u3 + r 2u4 = s 2 Eq-8 where r 2 = r2 q2 p2r 1 and s 2 = s2 p2s 1 q2 p2r 1 Eq-9 We repeat the procedure with the 3 rd equation and eliminate u3 using the previous equation (Eq-8):

6 P a g e 6 p3u3 + q3u4 + r3u5 = s3 Eq-10 (p3u3 + p3r 2u4 = p3s 2) Multiply Eq-8 by p3 Eq-11 (q3 p3r 2)u4 + r3u5 = s3 p3s 2 Eq-12 Normalize Eq-12 dividing by q3 p3r 2 u4 + r3 q3 p3r 2 u5 = s3 p3s 2 q3 p3r 2 Eq-13 Rename: u4 + r 3u5 = s 3 Eq-14 where r 3 = r3 q3 p3r 2 and s 3 = s3 p3s 2 q3 p3r 2 Eqs-15 By looking at Eqs-9 and Eqs-15 we derived the recurrence formulas: r i = r i q i p i r Eq-16 and s i i 1 = si p i s i 1 qi p i r i 1 Eq-17 For the first equation in the coefficient Matrix where p1 is not used in the left hand side, it is enough to set p1=0 for both (Eq-16 and Eq-17) recurrence formulas to yield an expression for the r1 and s1: r 1 = r 1 q 1 Eq-3a and s 1 = s 1 q1 Eq-3b NOTE: This doesn t mean that p(1)=0 for the right-hand side vector INVERSE SWEEP Here we compute from the last equation backwards to the first. Start with the last equation: p9u9 + q9u10 = s9 r9u11 Eq-18 Since ix =11, the above equation can be expressed also as: p(ix-2)u(ix-2) + q(ix-2)u(ix-1) = s(ix-2) Eq-19 Write the second to last equation by looking the pattern of either Eq.-2 or Eq. 8: u(ix 2) + r (ix 3)u(ix 1) = s (ix 3) Eq-20 Multiply Eq-20 by p(ix-2) and subtract it from Eq-19:

7 P a g e 7 Last equation p(ix 2)u(ix 2) + q(ix 2)u(ix 1) = s(ix 2) Second last (p(ix 2)u(ix 2) + p(ix 2)r (ix 3)u(ix 1) = p(ix 2)s (ix 3)) [q(ix 2) p(ix 2)r (ix 3)]u(ix 1) = s(ix 2) p(ix 2)s (ix 3) Eq-20 Normalize Eq-20 u(ix 1) = s(ix 2) p(ix 2)s (ix 3) q(ix 2) p(ix 2)r (ix 3) Eq-21 Rename the right-hand side of Eq-21: s (ix 2) = s(ix 2) p(ix 2)s (ix 3) q(ix 2) p(ix 2)r (ix 3) Eq-22 Then Eq-21 becomes u(ix 1) = s (ix 2) Eq-23 As expected, the definition of s is the same as in the previous direct sweep step. For the current example ix=11, therefore ux(ix-1) = u(10), and Eq-21 can be easily computed: u(10) = s(9) p(9)s (8) q(9) p(9)r (8) Eq-24 Observe that all constants in the RHS of Eq-24 are known values. Now the following equation from bottom up (second to last equation): u(ix 2) + r (ix 3)u(ix 1) = s (ix 3) Eq-20 Solve by u(ix-2): u(ix 2) = s (ix 3) r (ix 3)u(ix 1) Eq-25 Again observe that you already know every variable and parameter on the RHS. Generalize u(i) = s (i 1) r (i 1)u(i + 1) Eq-26 Please become aware of that the last computed u(i) will be, e.g., u(2) = s (1) r (1)u(3) The formulas to apply then For the last equation

8 P a g e 8 u(ix 1) = s (ix 2) Eq-23 For the next ones, from bottom up: u(i) = s (i 1) r (i 1)u(i + 1) Eq-26 And so on. For the last equation to be solved we have: u(2) = s (1) r (1)u(3) Summarizing the equations: DIRECT SWEEP r i = r i q i p i r Eq-16 and s i i 1 = si p i s i 1 qi p i r i 1 Eq-17 In Eq-16, indices run as i=2, 3, 4,, ix-3 (note that r(9) = 0 is in the other side of the equation and r(9) =0 does not exist). Eq-17 applies for i=2, 3,, ix-2 For the first equation where i=1 and p(1) is in the other side of the equation, the next two equations apply: r 1 = r 1 q 1 Eq-3a and s 1 = s 1 q1 Eq-3b INVERSE SWEEP The formulas to apply then For the last equation u(ix 1) = s (ix 2) For the next ones up u(i) = s (i 1) r (i 1)u(i + 1) Eq-23 Eq-26 Therefore, for this step you compute the unknowns in the order u(ix-1), u(9), u(8),, u(2) the last unknown to be computed, i.e., i=2: u(2) = s (1) r (1)u(3)

9 P a g e 9 For the heat transfer problem-example and for developing MATLAB code u(i 1) 2u(i) + u(i + 1) = 0 i = 2, 3, 4, ix 1 For values of the variables and coefficients: u(1), u(2), u(3),, u(11) p(1), p(2), p(3),, p(9) q(1), q(2), q(3),, q(9) r(1), r(2), r(3),.,r(9) s(1), s(2), s(3),., s(9) u(1), u(ix) p(1), p(ix-2) q(1), q(ix-2) r(1),.r(ix-2) s(1), s(ix-2) Actual values: u(1)=0.0 u(ix)=u(11)=1 p(i)=1.0 q(i)=-2.0 r(i)=1.0 BC BC i=1,2,.. ix-2 i=1,2, ix-2 i=1,2, ix-2 The values of s(1) and s(9) were renamed, here we go back to the original definition: s(1)=s(1)-p(1)u(1)=0-(1)(0)=0 % Note that p(1) is not zero for the right-hand side vector.. s(9)=s(9)-r(9)u(11)=0-(1)(1)=-1 % Note that r(9) is not zero for the right-hand side vector In other words, for the current problem: s(i) =0 i=2, ix-3 (i.e., ix-3=8); all are zeros except for s(9) s(9)=s(ix-2)=-1 After identifying p(i), q(i), r(i), s(i) and u(i) we start computing r (i), and s (i)

10 P a g e 10 r (1) = r(1) q(1) r (i) = r(i) [q(i) p(i)r (i 1) i=1 i=2,3, ix-3 s (1) = s(1) q(1) = 0 2 = 0 i=1 s (i) = (s(i) p(i)s (i 1)) (q(i) p(i)r (i 1)) i=2,3,..,ix-2 Now keep going with the INVERSE SWEEP to find u(i) s At the end of your computations you should be able to summarize your results in a Table: i p(i) q(i) r(i) s(i) u(i) in which not all the cells are filled with values. Please as an assignment fill the table above for the heat transfer problem, this practice is critical for your success in a test. EXACT SOLUTION We have chosen a problem that has an exact (and simple) solution, just to show the solution mechanism for the numerical method. The analytical (exact) solution is then 2 u x 2 = 0 0 x 1 x ( u ) = 0 x K 1 u x = K 1 ==> u = K 1 x + K 2 Apply boundary conditions to find the values of K 1 and K 2:

11 P a g e 11 u=0 x=0 K 2=0 u=1 x=1 K 1=1 Solution u = x In an actual exam, the evaluation of the Thomas Algorithm may involve. (1) Write down the system of equations in standarized form (2) Write down the system in Matrix form (square matrix) (3) Identify the p(i), q(i), r(i), and s(i) (4) Compute the r (i), and s (i) parameters (5) Compute the u(i) s (6) Fill a table like i p(i) q(i) r(i) s(i) u(i) ix-2

12 Excel computation File: Tridiagonal Algorithm You start computing here and go down TRIDIAGONAL ALGORITHM DIRECT SWEEP (Computation start with r^(1), s^(1) up to r^(9) and s^(9)) i p q r s ř ŝ u BC's The values of s(1) and s(9) could be tricky as they were both renamed to find recurrence formulas: s(1)=s(1)-p(1)u(1)=0-(1)(0)=0 s(9)=s(9)-r(9)u(11)=0-(1)(1)=-1 P a g e 12 s(1) and s(9) in the RHS both are zero, but not the new s(1) and s(9) Boundary conditions are known values 11 1 BC's first step second step third step INVERSE SWEEP (Computation goes from u(10) up to u(2) i p q r s ř ŝ u BC's BC's You start computing here and go up MATLAB, to the RESCUE Following are several ways of creating tridiagonal matrices. You need to create a tridiagonal matrix to use standard methods of solving a matrix system with matlab, such as x=a\b % Tridiagonal Maker % file: tridiamaker.m % TM [tridiagonal matrix] clc, clear, close % Option 1: n = 5; p = -1;

13 P a g e 13 q = 2; r = -1; TM=full(gallery('tridiag',n,p,q,r)) fprintf('\n\n'); % Option 2 n = 5 ; nones = ones(n, 1) ; TM = diag(2 * nones, 0) - diag(nones(1:n-1), -1) - diag(nones(1:n-1), 1) fprintf('\n\n'); % Option 3 n=5; TM=toeplitz([2-1 zeros(1, n-2)], [2-1 zeros(1, n-2)]) OUTPUT TM = Tridiagonal solver routine for MATLAB If you want to be loyal to Thomas algorithm, here you have a shared code: function u = tridiag(a,b,c,r,n) ; % Function tridiag: % Inverts a tridiagonal system whose lower, main and upper diagonals % are respectively given by the vectors a, b and c. r is the right hand % side, and N is the size of the system. The result is placed in u. % (Adapted from Numerical Recipes, Press et al. 1992) if (b(1)==0) fprintf(1,'reorder the equations for the tridiagonal solver...') pause

14 P a g e 14 end beta = b(1) ; u(1) = r(1)/beta ; % Start the decomposition and forward substitution for j = 2:N gamma(j) = c(j 1)/beta ; beta = b(j) a(j)*gamma(j) ; if (beta==0) fprintf(1,'the tridiagonal solver failed...') pause end u(j) = (r(j) a(j)*u(j 1))/beta ; end % Perform the back substitution for j = 1:(N 1) k = N j ; u(k) = u(k) gamma(k+1)*u(k+1) ; end REFERENCES