Question

. Using algebraic polynomial approximation, determine two parameter Galerkin solution for the governing equation: 1+k where, BCs: wo=0 at x=0; I dx² T qo varies linearly along x as- O at x=0 to 4o at x=L. Assume k=L=1 and qo= 3.

Answer 1

Galerkin methods are a class of methods for converting a continuous operator problem to a discrete problem. In order to understand the best approximation property of Galerkin’s method it is necessary to define a new norm and associated inner product in the space of functions.

the Galerkin Soluti method in which kin solution i It is a special case of the patrov- Galerkin in which the coordinate functions and weighted functions are some. In this method, the weight functions are chosen ato be the trial functions themselves. We usually use this mothed for developing finite element equations for problems. In Galeotin inethod, we set: Wi = N; list, 2, ...., n) The unknown coefficients in the approximal solution are-dotesmine! by setting the integral over D of the weighted residud to ye so For one-dinensional in the interval [a,b ], J WE R(x) dx = (Ni R(x) dx = 0, (ind, 2 .....,n). - For the Galeskne method, the linear system takes the form IN, LENI) dz: son, L (N2) dx .... SNL (Na) dx ] '(Nm) de sangenis) de finqlade ... from Lande INAL (N2) dx INAL (Ni) dx .. 1ML afdoc 'h Given: Bic's :- WCO) - 0 - and w(i)= 4 Assurne : We co+cix+62 22 (3 xc3 o. At Wzo, & zo in eqni o l nud . i O Coto 7602d At Wet, azt in ean. os 4= otc,.(2) + C2 (0) C3 (d)3 > cit Cz+C3=4*41:4-6263

From egn. Wz (4-C2-C37x +2 x2 +63 73 .: 426 -(236-(3x+ (22? +(3*3 = 4x + cz (1-x)+ (362-> differentiate eqın. © w-sitx: dw : 4+C2 (22-1) + 6(322-1) [iem m-xm-'] . . 82 - Cz (2) + 6x cj dx2 thens, using ean. @ find out the solutions; given, for the governing equation 4th and so othes way! ; 3 4 w XL1 X2 exi - 2+1 - $itt . U s Intes polation Linear intes . 0(x) 2 di Liti- + ZinZI Boundary condition: 0.= 0; and $420 Apply Galeoxin's method: weighting a x-xa-l Wizli -1:-) foo xi-l Exsxi li+l - for li LIL Xiti 542+1 - Xi Integration by parts: I dwi . = Wi wi dx L

Viti - du Cati wi da = 0 da Jx - die