Solution of Falkner-Skan Equation by Finite Differences

SOLUTION OF FALKNER-SKAN EQUATION BY FINITE DIFFERENCES F'' - U = 0 F' - 5F = 0 F(0) = 0 F'(0) = 1 U(0) = 1 DISCRETE SYSTEM (TRAPEZOIDAL FORMULA) F_{i+1} - F_{i} - h/2(U_{i+1} + U_{i}) = R_{F_{i}} = 0 U_{i+1} - U_{i} - h/2(S_{i+1} + S_{i}) = R_{U_{i}} = 0 S_{i+1} - S_{i} + (1 + βu)/2 * h/2(F_{i+1}S_{i+1} + F_{i}S_{i}) + βua(1 - 1/2(U_{i+1}^2 + U_{i}^2)) = R_{S_{i}} = 0 BCs: F_{1} = R_{BC_{1}} = 0 U_{1} = R_{BC_{2}} = 0 U_{N-1} = R_{BC_{3}} = 0 NEWTON-RAPHSON SYSTEM FOR SF, SU, SS SF_{i} = ∂R_{F_{i}}/∂F_{i} SU_{i} = ∂R_{U_{i}}/∂U_{i} SS_{i} = ∂R_{S_{i}}/∂S_{i} + (1 + βu)/2 * h/2(F_{i+1} + F_{i}) SOLUTION OF FALKNER-SKAN EQUATION BY FINITE DIFFERENCES cont'd 2/4 βu is a global variable which in general appears in each equation, and thus disrupts the block-tridiagonal structure of the Jacobian matrix. This problem is sidestepped by writing the Newton system   as: Title: APPLICATION OF NEWTON-RAPHSON METHOD TO DISCRETE FALINER-SKAN EQUATIONS Subheadings: NON-LINEAR SYSTEM TO BE SOLVED SOLUTION BY ITERATION Equations: A series of nonlinear equations, primarily involving variables like F, u, S, β, and subscripts like i, n, and k. These equations are written in a standard mathematical notation, using symbols like =, +, -, *, and parentheses. Labels: R_q, R_u, R_s, R_β, R_oc, R_ai, R_Bϵ, and R_θ are labels for different equations or residual functions. F, u, S, β, and other variables are labeled as unknowns or parameters. Formatting: The text is divided into multiple columns and rows, with equations and labels aligned within specific sections. Some equations are enclosed in brackets or boxes for emphasis. Introduction: The standard Newton System has the form: [SF SU SS ...][S] = [R] [Ra][U] = [Rs] [R][V] = [Rv] ... [Rn][W] = [Rw] Note that the residual R is generally a function of all variables, which spoils the diagonal structure of the coefficient matrix, making its solution awkward. Solution Process: Step 1: The last equation (bottom matrix row) is put aside, and the S column is placed on the right-hand side. This is now easily solved. [SF SU SS ...][S] = [R] [Ra][U] = [Rs] [R][V] = [Rv] ... [Rn-1][W] = [Rw-1] This results in a block matrix structure. Step 2: It is still necessary to find S so that SF, SU, SS, ... can be completely determined. This is done by using the last equation which was put aside: [Rn][S] = [Rw] By substituting the solution from Step 1, the final equation can be solved for S.