BTCS Solution to the Heat Equation

Transcription

1 BTCS Solution to the Heat Equation ME 448/548 Notes Gerald Recktenwald Portland State University Department of Mechanical Engineering ME 448/548: BTCS Solution to the Heat Equation

2 Overview 1 Use nodal values of u k i and uk 1 i in the finite difference approximation to u/ t 2 Use nodal values of i 1, uk+1 i and i 1 in the finite difference approximation to 2 u/ x 2 3 Computational formula is implicit: all i must be solved simultaneously 4 Solution is more complex, but unconditionally stable 5 Truncation error is the same as FTCS ME 448/548: BTCS Solution to the Heat Equation page 1

3 Finite Difference Operators Choose the backward difference to evaluate the time derivative at t = t k u t = uk i uk 1 i + O( t) (1) tk,x i t Approximate the spatial derivative with the central difference operator and take all nodal values at time t k 2 u = uk i 1 2uk i + uk i+1 + O( x 2 ) (2) x 2 x 2 tk,x i ME 448/548: BTCS Solution to the Heat Equation page 2

4 BTCS Approximation to the Heat Equation Making these substitutions in the heat equation gives u k i uk 1 i t = α uk i 1 2uk i + uk i+1 x 2 + O( t) + O( x 2 ) (3) It is not possible to solve for u k i in terms of other known values at t k 1 Drop truncation error terms and shift the time step by one: k 1 k and k k + 1 i u k i t = α uk+1 i 1 2uk+1 i + i+1 (4) x 2 ME 448/548: BTCS Solution to the Heat Equation page 3

5 BTCS Computational Molecule t k+1 k k 1 BTCS scheme requires simultaneous calculation of u at all nodes on the k+1 mesh line Solution is known for these nodes t=0, k=1 i=1 i 1 i i+1 n x x=0 x=l ME 448/548: BTCS Solution to the Heat Equation page 4

6 BTCS Approximation to the Heat Equation Move all unknown nodal values in Equation (3) to the left hand side to get [ α ] [ 1 x 2 i 1 + t + 2α ] x 2 i + [ α ] x 2 i+1 = 1 t uk i (5) Nodal values at t k+1 are all on the left hand side, and the lone nodal value from t k is on the right hand side The terms in square brackets are the coefficients in a system of linear equations ME 448/548: BTCS Solution to the Heat Equation page 5

7 BTCS System of Equations The system of equations can be represented in matrix form as a 1 b c 2 a 2 b c 3 a 3 b c n x 1 a n x 1 b n x c n x a n x nx 1 nx = d 1 d 2 d 3 d n x 1 d n x (6) where the coefficients of the interior nodes (i = 2, 3,, N 1) are a i = (1/ t) + (2α/ x 2 ), b i = c i = α/ x 2, d i = (1/ t)u k i (7) ME 448/548: BTCS Solution to the Heat Equation page 6

8 BTCS System of Equations At each time step, and for an x-direction mesh, we must solve the nx nx system of equations A = d (8) where A is the coefficient matrix, is the column vector of unknown values at t k+1, and d is a set of values reflecting the values of u k i, boundary conditions, and source terms (if present) Fortunately (in 1D), matrix A is tridiagonal, which allows for a very efficient solution of Equation (8) ME 448/548: BTCS Solution to the Heat Equation page 7

9 Solving the BTCS System of Equations At each time step we need to solve A = d We could use a simplistic approach and use a standard Gaussian elimination routine However A is tridiagonal and substantial speed and memory savings can be had by exploiting that structure Furthermore, using LU factorization leads to even more savings by reducing the computational cost per time step ME 448/548: BTCS Solution to the Heat Equation page 8

10 LU Factorization Given the square n n matrix A, and n 1 column vectors x and b, the canonical system of equations is written Ax = b (9) The LU factorization of a matrix A involves finding the lower triangular matrix L and the upper triangular matrix U such that The factorization alone does not solve Ax = b A = LU (10) Gaussian elimination only transforms an augmented coefficient matrix to triangular form It is the backward substitution phase that obtains the solution Similarly the factorization of A into L and U sets up the solution Ax = b via two triangular solves ME 448/548: BTCS Solution to the Heat Equation page 9

11 LU Factorization Since A = LU, the system Ax = b is equivalent to (LU)x = b (11) Matrix multiplication is associative, so regroup the left hand side (LU)x = b L(Ux) = b Let y = Ux, so that Equation (11) becomes Ly = b Given y, we then have the system Ux = y, which is easily solved for x with a backward substitution ME 448/548: BTCS Solution to the Heat Equation page 10

12 Solving Ax = b via LU Factorization Put the pieces together to obtain an algorithm for solving Ax = b Algorithm 1 Factor A into L and U Solve Ly = b for y Solve Ux = y for x Solve Ax = b with LU factorization forward substitution backward substitution The last two steps, solve Ly = b and solve Ux = y, are efficient because L and U are triangular matrices ME 448/548: BTCS Solution to the Heat Equation page 11

13 LU Factorization for triangular systems Store the diagonals of A as three vectors, a, b and c a 1 b 1 c 2 a 2 b 2 c n 1 a n 1 b n 1 c n a n x 1 x 2 x n 1 x n = d 1 d 2 d n 1 d n The L and U matrix factors of the tridiagonal coefficient matrix have the form 1 f 1 b 1 e 2 1 f 2 b 2 L =, U = e n 1 1 f n 1 b n 1 e n 1 f n ME 448/548: BTCS Solution to the Heat Equation page 12

14 LU Factorization for triangular systems To find formulas for e i and f i, multiply the L and U factors, and set the result equal to A to get e i f i 1 = c i, e i b i 1 + f i = a i, b i = b i Solve the first and second equationsfor e i and f i e i = c i /f i 1, f i = a i e i b i 1 which apply for i = 2,, nmultiplying the first row of L with the first column of U gives f 1 = a 1 ME 448/548: BTCS Solution to the Heat Equation page 13

15 LU Factorization for triangular systems The preceding formulas are directly translated into Matlab code f(1) = a(1); for i=2:n e(i) = c(i)/f(i-1); f(i) = a(i) - e(i)*b(i-1); end Given e and f vectors, the solution to the system is y(1) = d(1); % Forward substitution: solve L*y = d for i=2:n y(i) = d(i) - e(i)*y(i-1); end x(n) = y(n)/f(n); % Backward substitution: solve U*x = y for i=n-1:-1:1 x(i) = ( y(i) - b(i)*y(i+1) )/f(i); end ME 448/548: BTCS Solution to the Heat Equation page 14

16 demobtcs Code % --- Assign physical and mesh parameters alfa = 01; L = 1; tmax = 2; % Diffusion coefficient, domain length and max time dx = L/(nx-1); dt = tmax/(nt-1); % --- Coefficients of the tridiagonal system b = (-alfa/dx^2)*ones(nx,1); % Super diagonal: coefficients of u(i+1) c = b; % Subdiagonal: coefficients of u(i-1) a = (1/dt)*ones(nx,1) - (b+c); % Main Diagonal: coefficients of u(i) a(1) = 1; b(1) = 0; % Fix coefficients of boundary nodes a(end) = 1; c(end) = 0; [e,f] = tridiaglu(a,b,c); % Save LU factorization % --- Assign IC and save BC values in ub IC creates u vector x = linspace(0,l,nx) ; u = sin(pi*x/l); ub = [0 0]; % --- Loop over time steps for k=2:nt d = [ub(1); u(2:nx-1)/dt; ub(2)]; % Update RHS, preserve BC u = tridiaglusolve(e,f,b,d); % Solve the system end ME 448/548: BTCS Solution to the Heat Equation page 15