FEA | Galerkin Method | Subdomain Method | Petro - Galerkin Method | Least Square method |

Solve the following differential equation using Galerkin Method. (d^2 y)/(dx^2 )+3x dy/dx-6y=0, 0 x 1. Boundary conditions are y(0)=1,y^' (1)=0.1, Find y(0.2) and compare with exact solution. Solution: Weighted Residual method According to WRM , ∫_Ω▒〖w_i R.dx〗= 0 Where w_i= Error in function Let y = y_app and,y_app≠ y_exact ∴(d^2 y)/(dx^2 )+3x dy/dx-6y≠ 0 = R ……………………………………………………………………………….. (A) Where,R = Residue The approximate solution for differential equation can be written as, y_ap = c_1 + c_2 x + c_3 x^2 + c_4 x^3+ …… + c_(n+1) x^n Consider the equation of y_ap to the term x^((order of given equation+1)) x^(2+1) = x^3 y_ap = c_1 + c_2 x + c_3 x^2 + c_4 x^3 ………………………………………………………………….(1) 〖 y'〗_ap=C_2 x+〖3C〗_3 x+C_4 x^2 …………………………………………………………………. (2) Apply boundary condition one by one. From 1^st boundary condition, y(0)=1, We get 1=C_1 From 2^nd boundary condition, y^' (1)=0.1, We get0.1=C_2+〖2C〗_3+〖3C〗_4 ∴C_2=0.1-〖2C〗_3-〖3C〗_4 Rewrite the approximate solution as y_ap=1+(0.1-〖2C〗_3-〖3C〗_4 )x+C_2 x^2+C_3 x^3 =1+0.1x+C_3 (x^2-2x)+C_4 (x^3-3x) Finding the residue given by the differential equation (dy_ap)/dx=0.1+C_3 (2x-2)+C_4 (〖3x〗^2-3) (d^2 y_ap)/dx^2 =〖2c〗_3+〖6xC〗_4 Therefore the residue becomes R=(d^2 y_ap)/dx^2 +3x dy/dx-6y =〖2C〗_3+〖6xC〗_4+3x[0.1+C_3 (2x-2)+C_4 (〖3x〗^2-3) ]-6[1+0.1x+C_3 (x^2-2x)+C_4 (x^3-3x) ] =C_3 [2+3x(2x-2)+12x-〖6x〗^2 ]+C_3 [6x+3x(〖3x〗^2-3)-18x-〖6x〗^3 ]+0.3x-6-0.6x =C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3-6 The weighted integral from ∫_0^1▒w_i R dx=0 i.e.∫_0^1▒w_i [C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3x-6]dx=0 Methods: (I) Galerkin Method w_i=Coefficients of constant (C_i) in y. w_1=(x^2-2x) and w_2=(x^3-3x) Putting i=1,we get ∫_0^1▒〖(x^2-2x)[C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3-6]dx=0〗 i.e.-3.833 C_3-6.95 C_4+4.125=0 i.e.3.833 C_3-6.95 C_4+4.125=0……….(i) Similarly,putting i=2,we get ∫_0^1▒〖(x^3-3x)[C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3-6]dx=0〗 i.e.-7.3 C_3-13.371 C_4+7.74=0 i.e.7.3 C_3-13.371 C_4+7.74=0……….(ii) Solving equations (i) and (ii),we get C_3=2.64 and C_4=-0.862 Therefore,y=1+0.1x+2.64(x^2-2x)—0.862(x^3-3x) ⟹y(0.2)=0.58 II) Petrov-Galerkin method w_i can be any functions of x. So we can assume w_1=x and w_2=x^2 Putting i=1,we get ∫_0^1▒〖x[C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3x-6]dx=0〗 i.e.3 C_3+5.6 C_4-3.1=0 ………(i) Similarly,puttingi=2,we get ∫_0^1▒〖x^2 [C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3x-6]dx=0〗 i.e.2.17 C_3+4.25 C_4-2.075=0 ………(ii) Solving equations (i) and (ii),we get C_3=2.6 and C_4=-0.84 Therefore y=1+0.1x+2.6(x^2-2x)—0.84(x^3-3x) ⟹y(0.2=0.581) III) Subdomain method In this method w_i=1 but as we are having two parameters we divide the domain into two parts. Let us divide the domain in two equal parts. Hence we have ∫_0^0.5▒〖[C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3x-6]dx=0〗 i.e.1.75 C_3+1.922 C_4-3.0375=0………(i) And ∫_0.5^1▒〖[C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3x-6]dx=0〗 i.e.3.25 C_3+6.328 C_4-3.1125=0………(ii) Solving equations (i) and (ii),we get C_3=2.745 and C_4=-0.9193 Therefore,y=1+0.1+2.745(x^2-2x)-0.9193(x^3-3x) ⟹y(0.2)=0.576 IV) Least square method In this method,weight fuction is given by w_i=∂R/〖∂C〗_i For i=1,w_1=∂R/〖∂C〗_3 =(6x+2) ∴∫_0^1▒〖(6x+2)[C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3x-6]dx=0〗 i.e.28 C_3+50.1 C_4-30.90=0 ………(i) Similarly for i=2,w_2=∂R/〖∂C〗_4 =(〖3x〗^3+15x) ∴∫_0^1▒〖(〖3x〗^3+15x)[C_3 (6x+2)+C_4 (〖3x〗^3+15x)-0.3x-6]dx=0〗 i.e.50.1 C_3+94.286 C_4-51.18=0 ………(ii) Solving equations (i) and (ii),we get C_3=2.766 and C_3=-0.965 Therefore,y=1+0.1 x+2.766(x^2-2x)-0.965(x^3-3x) ⟹y(0.2)=0.5955