Solve the following differential Equation using Galerkin Method.

$\hspace{2cm} \frac{d^2 \phi}{dx^2} + cos \hspace{0.1cm}\phi \hspace{0.1cm}x = 0 \hspace{2cm} 0 \lt x \lt 1.$

Boundary Conditions are : $\hspace{0.2cm} \phi(0) = 0 \hspace{0.2cm} \phi$(1) = 0

Find $\phi (0.25), \phi\hspace{0.2cm} (0.5) and\hspace{0.2cm} \phi\hspace{0.2cm} (0.75)$ Compare your answer with exact solution

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

Marks: 10M

Year: Dec2016

let the approximate solution,

$\phi = c_o + \in ci sin \frac{\pi}{2}(2i-1)x$

Boundary condition 1: $\phi (o)=0$

$\therefore c_o=0$

$\therefore \phi = \in c_i sin \frac{\pi}{2}(2i-1) x.$

=$c_i sin \frac{\pi x}{2}+ c_2 sin \frac{3 \pi x}{2}+ c_3 sin \frac{5 x}{2}$

Boundary condition 2 : $\phi$ (1)=0

$\therefore 0 = c_1-c_2+c_3$

$\therefore c_2 =c_1+c_3$

$\phi =c_1\bigg[sin \frac{\pi x}{2}+ sin \frac{3 \pi x}{2}\bigg]+c_3 \bigg[sin \frac{3 \pi x}{2}+sin \frac{5 \pi x}{2}\bigg]$

$\phi ' =c_1\frac{\pi}{2}\bigg[cos\frac{\pi x}{2}+3cos \frac{3 \pi x}{2}\bigg]+c_3\frac{\pi}{2}\bigg[3cos \frac{3 \pi x}{2}+5 cos \frac{5 \pi x}{2}\bigg]$

$\phi '' = -c_1 \frac{\pi^2}{4}\bigg[sin \frac{\pi x}{2}+9 sin \frac{3 \pi x}{2}\bigg]-c_3\frac{\pi^2}{4}\bigg[9sin \frac{3\pi x}{2}+25 sin \frac{5 \pi x}{2}\bigg]$

$\therefore$ Residue, R,

R = $\phi''$-cos $\pi x$

=$\frac{\pi^2}{4}\bigg[c_1\bigg[sin \frac{\pi x}{2}+9 sin\frac{3 \pi x}{2}\bigg]+c_3\bigg[9sin \frac{3 \pi x}{2}+25 sin \frac{5x \pi}{2}\bigg]-cos\hspace{0.2cm} \pi x$

$W_i$= coefficients of ci in u .

for i = 1 $\Rightarrow W_1= sin \frac{\pi}{2}x+ sin \frac{3 \pi}{2}x.$

for i = 2 $\Rightarrow W_2= sin \frac{3 \pi}{2}x + sin \frac{5\pi}{2}x.$

for i = 1

$\int\limits_o^1 w_1 R dx=0$

$\frac{\pi^2}{4} \int\limits_0^1 \bigg[sin \frac{\pi}{2}x+sin\frac{3 \pi}{2}x\bigg]\bigg[c_1\bigg(sin \frac{\pi}{2}x+9 sin \frac{3\pi}{3}x\bigg)$

$+ c_3\bigg(9 sin \frac{3\pi x}{2}+25sin \frac{3pi x}{2}\bigg)-cos \pi x\bigg] dx=c$

$\therefore c_1 \int\limits_0^1\bigg(sin\frac{\pi}{2}+sin \frac{3 \pi}{2}x\bigg) \bigg(sin\frac{\pi}{2}x+9 sin \frac{3\pi}{2}x\bigg)dx$

$c_3\int\limits_0^1\bigg(sin\frac{\pi}{2}x+sin\frac{3\pi}{2}x\bigg)\bigg(9sin\frac{3\pi}{2}x+25 sin \frac{5\pi}{2}x)$

$- \int\limits_0^1\bigg(sin\frac{\pi}{2}x+sin\frac{3\pi}{2}x\bigg)cos \hspace{0.2cm}\pi x dx \frac{4}{\pi^2=0}$

$5 c_1+2.84c_3-0.069=0$

$5 c_1+2.84 c_3 = 0.069 \hspace{2cm}------(1)$

Exact solution :

$\frac{d^2\phi}{dx^2}+cos \pi x=0$

$\frac{d^2 \phi}{dx^2}=-cos \pi x$

$\frac{d \phi}{dx}=\frac{sin \pi x}{\pi}+c^1$

$\phi=\frac{cos \pi x}{\pi^2}+c_1x+c_2$

for x = 0 : $\phi$ = 0

$0 = \frac{1}{\pi^2}+c_2 \hspace{0.6cm} \therefore c_2=-\frac{1}{\pi ^2}$

$\therefore \phi = \frac{cos \pi x}{\pi ^2}-\frac{1}{\pi^2}+c_1x$

for x = 1 : $\phi$ = 0

$0 = -\frac{1}{\pi ^2}-\frac{1}{\pi^2}+c_1 \hspace{0.6cm} \therefore c_1=+\frac{2}{\pi^2}$

$\therefore u =\frac{cos \pi x}{\pi^2}+\frac{2x}{\pi^2}-\frac{1}{\pi^2}$

u (0.25) = 0.02098 u (0.5) = 0 u (0.75) = -0.02098

for i= 2

$\int\limits_0^1 w_2R dx=0$

$\frac{\pi^2}{4} \int\limits_0^1\bigg(sin\frac{3 \pi x}{2}+ sin \frac{5 \pi x}{2}\bigg)\bigg[c_1\bigg(sin\frac{\pi x}{2}+9 sin \frac{3 \pi x}{2}\bigg)+c_3\bigg(9sin\frac{3 \pi x}{2}+25sin\frac{5 \pi x}{2}\bigg)-cos \pi x \bigg] -dx=c$

$c_1 \int\limits_0^1\bigg(sin\frac{3 \pi x}{2}+sin \frac{5 \pi x}{2}\bigg)\bigg(sin \frac{\pi x}{2} +9 sin\frac{3 \pi x}{2}\bigg)dx$

$+c_3\int\limits_0^1\bigg(sin\frac{3 \pi x}{2}+ sin \frac{5 \pi x}{2}\bigg)\bigg(9 sin \frac{3 \pi x}{2}+25 sin \frac{5 \pi x}{2}\bigg) dx$

$- \int\limits_0^1\bigg(sin\frac{3 \pi x}{2}+sin \frac{5 \pi x}{2}\bigg)cos \pi x dx.\frac{4}{\pi^2}=0$

3.9 $c_1+14.75 c_3=0.0122\hspace{2cm}------(2)$

from (1)& (2)

$c_1$=0.0069 & $c_3$=0.0122.

$\therefore \phi = 0.0069 \bigg(sin\frac{\pi x}{2}+sin\frac{3 \pi x}{2}\bigg)+0.0122\bigg(sin\frac{3\pi x}{2}+sin\frac{5\pi x}{2}\bigg)$

u (0.25) = 0.03156

u(0.5) = 9.7581$\times10^{-3}$ = 0.0097581

u (0.75) = -5.6033 $\times10^{-3}=-0.0056$