VIEWS: 1 PAGES: 54 POSTED ON: 1/25/2012 Public Domain
Derivative Boundary Value Problems d2F F 2 f ( x , F, F ' ) Consider: dx 4 Derivative Boundary Value Problems d2F F 2 f ( x , F, F ' ) Consider: dx 4 with B.C.’s: F (0) F0 0 and F ( ) Fn 1 Derivative Boundary Value Problems d2F F 2 f ( x , F, F ' ) Consider: dx 4 with B.C.’s: F (0) F0 0 and F ( ) Fn 1 Note that in this case, we have derivative boundary conditions. Derivative Boundary Value Problems Let us solve this using finite differences. Using the finite difference approximations: Derivative Boundary Value Problems Let us solve this using finite differences. Using the finite difference approximations: d2F Fi 1 2 Fi Fi 1 Fi 2 2 dx x x h 4 i Derivative Boundary Value Problems Let us solve this using finite differences. Using the finite difference approximations: d2F Fi 1 2 Fi Fi 1 Fi 2 2 dx x x h 4 i Now if we attempt to set up the difference equations we obtain: Derivative Boundary Value Problems Our Finite Difference Analog is: Derivative Boundary Value Problems Our Finite Difference Analog is: h2 Fi 1 2 Fi Fi 1 0 4 Derivative Boundary Value Problems Our Finite Difference Analog is: h2 Fi 1 2 Fi Fi 1 0 4 Evaluating this equation at i = 1 through 3, we obtain: Derivative Boundary Value Problems For i = 1: F0 2 1 2 F F 0 4 4 1 2 Derivative Boundary Value Problems For i = 1: F0 2 1 2 F F 0 4 4 1 2 For i = 2 F 2 1 2 F F 0 1 4 4 2 3 Derivative Boundary Value Problems For i = 1: F0 2 1 2 F F 0 4 4 1 2 For i = 2 F 2 1 2 F F 0 1 4 4 2 3 For i = 3 F2 2 1 2 F F 0 4 4 3 4 Derivative Boundary Value Problems These equations become: F0 -1.84579F1 + F2 = 0 F1 - 1.84579 F2 + F3 = 0 F2 - 1.84579 F3 + F4 = 0 Notice that we now have 3 equations with 5 unknowns! But we also know: Derivative Boundary Value Problems These equations become: F0 -1.84579F1 + F2 = 0 F1 - 1.84579 F2 + F3 = 0 F2 - 1.84579 F3 + F4 = 0 Notice that we now have 3 equations with 5 unknowns! But we also know: Fn 1 Fn 1 Fn 2h Derivative Boundary Value Problems If we use our knowledge of the derivative at x0 and xn, i.e., F’0 = 0 and F’4 = 1: Derivative Boundary Value Problems If we use our knowledge of the derivative at x0 and xn, i.e., F’0 = 0 and F’4 = 1: F F1 F0 1 1 2 / 4 Derivative Boundary Value Problems If we use our knowledge of the derivative at x0 and xn, i.e., F’0 = 0 and F’4 = 1: F F1 F5 F3 F0 1 1 F4 1 2 / 4 2 / 4 Derivative Boundary Value Problems If we use our knowledge of the derivative at x0 and xn, i.e., F’0 = 0 and F’4 = 1: F F1 F5 F3 F0 1 1 F4 1 2 / 4 2 / 4 Notice, now we have two new equations, but we have also introduced two new unknowns! We need two more equations! Derivative Boundary Value Problems Two that we can use are : Derivative Boundary Value Problems Two that we can use are : 2 1 2 F F 0 For i = 0 F1 4 4 0 1 Derivative Boundary Value Problems Two that we can use are : 2 1 2 F F 0 For i = 0 F1 4 4 0 1 2 1 2 F F 0 For i = 4 F3 4 4 4 5 Derivative Boundary Value Problems Two that we can use are : 2 1 2 F F 0 For i = 0 F1 4 4 0 1 2 1 2 F F 0 For i = 4 F3 4 4 4 5 This gives us two more equations, and we have the same number of unknowns. Derivative Boundary Value Problems So now our equations become: F-1 + -1.84579F0 + F1 = 0 - F-1 + F1 = 0 -1.84579F1 + F2 = 0 F1 - 1.84579 F2 + F3 = 0 F2 - 1.84579 F3 + F4 = 0 -F3 + F5 = 1.5708 F3 - 1.84579 F4 + F5 = 0 Notice that we now have 7 eqns with 7 unkns! Derivative Boundary Value Problems Or: [A][F] = [b] where: Derivative Boundary Value Problems 1 0 1 0 0 0 0 1 a1 1 0 0 0 0 0 1 a1 1 0 0 0 A 0 0 1 a1 1 0 0 0 0 0 1 a1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 a1 1 Derivative Boundary Value Problems where -a1 = -1.84579 and : Derivative Boundary Value Problems 0 where 0 -a1 = -1.84579 and : 0 0 b 0 2 0 Derivative Boundary Value Problems 0 1.88253 where 0 2.03981 -a1 = -1.84579 and : 0 1.88253 0 b F= 1.43494 0 0.76606 2 0.02095 0 0.80474 Derivative Boundary Value Problems Comparing to exact solution of F = -2cos(x/2): 2 F j 0 x j 2. cos 2 2 4 2 0 2 4 x j Derivative Boundary Value xj Problems 2. cos 2 1.88253 Compared to 1.414 the exact 2.03981 solution: 2 1.88253 1.848 F= 1.43494 1.414 0.76606 0.765 0.02095 0 0.80474 0.765 Nonlinear Problems 2 d F dF 2 F dx e 0 Consider: x dx Nonlinear Problems 2 d F dF 2 F dx e 0 Consider: x dx with B.C.’s: F(0) F0 1 and F(1) Fn 1 Nonlinear Problems 2 d F dF 2 F dx e 0 Consider: x dx with B.C.’s: F(0) F0 1 and F(1) Fn 1 Note that in this case, we have a nonlinear D.E. (because of the second term) Nonlinear Problems Let us solve this using finite differences. Using the finite difference approximations: Nonlinear Problems Let us solve this using finite differences. Using the finite difference approximations: 2 d F Fi 1 2 Fi Fi 1 dx 2 x x h2 i Nonlinear Problems Let us solve this using finite differences. Using the finite difference approximations: 2 d F Fi 1 2 Fi Fi 1 dx 2 x x h2 i dF Fi 1 Fi 1 dx x xi 2h Nonlinear Problems Our Finite Difference Analog is: Nonlinear Problems Our Finite Difference Analog is: h 2 xi Fi 1 2 Fi Fi 1 ( Fi Fi 1 Fi Fi 1 ) h e 2 Nonlinear Problems Our Finite Difference Analog is: h 2 xi Fi 1 2 Fi Fi 1 ( Fi Fi 1 Fi Fi 1 ) h e 2 Evaluating this equation with h = 0.2 and thus at i = 1 through 4, we obtain: Nonlinear Problems For i = 1: 1 2 F F2 0.1( F F2 F F0 ) 0.04e 0.2 1 1 1 Nonlinear Problems For i = 1: 1 2 F F2 0.1( F F2 F F0 ) 0.04e 0.2 1 1 1 For i = 2 F 2 F2 F3 0.1( F2 F3 F2 F ) 0.04e 1 1 0. 4 Nonlinear Problems For i = 1: 1 2 F F2 0.1( F F2 F F0 ) 0.04e 0.2 1 1 1 For i = 2 F 2 F2 F3 0.1( F2 F3 F2 F ) 0.04e 1 1 0. 4 For i = 3 F2 2 F3 F4 0.1( F3 F4 F3 F2 ) 0.04e 0.6 Nonlinear Problems For i = 1: 1 2 F F2 0.1( F F2 F F0 ) 0.04e 0.2 1 1 1 For i = 2 F 2 F2 F3 0.1( F2 F3 F2 F ) 0.04e 1 1 0. 4 For i = 3 F2 2 F3 F4 0.1( F3 F4 F3 F2 ) 0.04e 0.6 For i = 4 F3 2 F4 1 0.1( F4 F4 F3 ) 0.04e 0.8 Nonlinear Problems Using MathCad to solve these four simultaneous non-linear equations, the results are : Nonlinear Problems Using MathCad to solve these four simultaneous non-linear equations, the results are : 1.000 0.49454 [F] = -0.01211 -0.45794 -0.79503 -1.000 Nonlinear Problems Graphically: 1 0.5 F 0 i 0.5 1 0 0.5 1 x i Two Dimensional Steady-State.. Consider: T T 2 2 0 x 2 y 2 Laplace Equation Two Dimensional Steady-State.. Analytical: 4To 1 y T 2n 1 sin( 2n 1) a n 0 b sinh( b 1)( 2n 1) cos ec ( 2n 1) a a Two Dimensional Steady-State.. Consider: System of grid-points (i,j) and its four immediate neighbors i,j+1 I-1,j i,j i+1,j Dx Dy i,j-1 Two Dimensional Steady-State.. T T 2 2 0 x 2 y 2 Difference equation: Ti 1, j 2Ti , j Ti 1, j Dx 2 Ti , j 1 2Ti , j Ti , j 1 0 Dy 2 Two Dimensional Steady-State.. If Dx = Dy: Ti 1, j Ti 1, j Ti , j 1 Ti , j 1 Ti , j 4 Two Dimensional Steady-State.. Example: What is the temperature at P 0 100 0 P= (100 + 0 + 0 + 0)/4 = 25 100 0 not very accurate…grid is too coarse! 100 0 0 Two Dimensional Steady-State.. Apply the equation to each interior node: 0 0 100 0 Ta = (Tb + 0 + 100 + Tc)/4 a b Tb = (0 + 0 + Ta + Td)/4 100 0 100 c d Tc = (Td + Ta + 0 + 100)/4 0 Td = (0 + Tb + Tc + 0)/4 100 0 0 0 Four equations --> four unknown! Two Dimensional Steady-State.. Gauss-Seidel Iteration: a. replace all “=“ by “<--” b. make starting guesses c. compute new approximations for Ta, Tb, Tc, Td d. repeat step c until converge.