Solve following differential equation $\frac{d^2y}{dx^2} - 10x^2 = 5 ; \hspace{0.6cm} 0 \leq x \leq 1$

BCs : y(0) = y(1) = 0. using Rayieigh-Ritz method, mapped over entire domain using one parameter method.

Mumbai University > Mechanical Engineering > Sem 6 > Finite Element Analysis

Marks: 12M

Year: Dec 2015

$y = C_0 + C-1x + C-2 x^2$

$y(0) = 0 \hspace{2cm} \therefore 0 = C_0 + 0 + 0\\ \hspace{2.5cm} \therefore C_0 = 0\\ \hspace{2cm} y = C_1x + C_2 x^2$

$y(1) = 0 \hspace{1.5cm} \therefore 0 = C_1 + C_2\\ \hspace{2.5cm} C_1 = - C_2\\ \hspace{1cm} \therefore y = C_1x - C_1 x^2\\ \hspace{1cm} \therefore y = C_1(x - x^2$

$R = y'' - 10x^2 - 5\\ \int\limits_0^1 W_iR dx = 0\\ \int\limits_0^1 w \Big(\frac{d^2y}{dx^2} - 10x^2 - 5 \Big)dx = 0\\ \int\limits_0^1 w \frac{d^2y}{dx^2} dx - 10 \int\limits_0^1 Wx^2 dx - 5 \int\limits_0^1 w dx = 0\\ w = x - x^2\\ \frac{dw}{dx} = 1 - 2x$

$w \frac{dy}{dx} |_0^1 - \int\limits_0^1 \frac{dw}{dx}.\frac{dy}{dx}dx - 10 \int\limits_0^1 wx^2dx - 5 \int\limits_0^1 w dx = 0 \hspace{2cm} ----- (1)$

$y = C_1(x - x^2)$

$\frac{dy}{dx} = C_1(1 - 2x)$

$\bigg[ w \frac{dy}{dx} \bigg]_0^1 = 0 - 0 = 0$

Equation 1:

$- \int\limits_0^1 (1 - 2x)C_1(1 - 2x)dx - 10 \int\limits_0^1 (x - x^2)x^2 dx - 5 \int\limits_0^1 (x - x^2)dx = 0\\ - 0.3334 - 0.5 - 0.8333 = 0\\ \therefore C_1 = -4\\ \therefore y = -4(x - x^2)$