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Assignment
Chapter 5 problems
5.06 and 5.
Problem 5.
Find the solution of differential equation d^2 ɸ/dx^2 +ɸ+x= Subject to boundary conditions ɸ (0) = ɸ (1) = 0, using the collection method with x=1\4 and x=3\4 as the collection points.
Solution:
As there are only two boundary conditions and two collection points so we selected a ɸ(x) which has 4 constants so a cubic model is selected.
ɸ(x) = a 0 + a 1 x + a 2 x^2 + a 3 x^3
by applying boundary conditions:
ɸ(0)= 0
or a 0 = 0
and ɸ(1) = 0 so 0 = a 1 +a 2 +a 3 or a 1 = -a 2 -a 3
putting the values
ɸ(x) = (-a 2 -a 3 )x + a 2 x^2 +a 3 x^3
Differentiating the above equation
dɸ/dx = (-a 2 -a 3 ) + 2a 2 x +3a 3 x^2
d^2 ɸ/dx^2 = 2a 2 +6a 3 x
Substituting in original equation gives
R(x)=2a 2 +6a 3 x+(-a 2 -a 3 ) + 2a 2 x +3a 3 x^2 +x
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R(x)=(2-x-x^2 )a 2 +(5x+x^3 )a 3 +x
At collection points the residue is zero so
R(1/4)= 1.8125a 2 +1.2656a 3 +0.25=
R(3/4)= 1.8125a 2 +4.17187a 3 +0.75=
Solving the two equations
a 3 = -0.17204 a 2 = -0.0178 a 1 = 0.
Final equation becomes:
ɸ(x) = 0.18948x – 0.0178x^2 – 0.17204x^3
Problem 5.
Find the solution of problem 5.25 using collocation with the trial solution ɸ(x) = c 1 x(1-x) + c 2 x^2 (1-x) Assume two sub domains as x = 0 to ¼ and x= ¾ to 1
Solution:
In sub domain collocation method ʃRdx = 0 so from problem number R(x) = -2c 1 + 2c 2 (1-3x) + 400x^2 Integrating from 0 to ¼ ʃR(x) = -2c 1 (0.25)+ 2c 2 (0.25-3/2(0.25)^2 )+400/3(0.25)^3 =-0.5c 1 +0.3125c 2 +2.0833 = 0
Integrating from ¾ to 1 ʃR(x) = -0.5c 1 – 0.81c 2 + 77.0833 = 0
Solving
C 1 = 45.962 c 2 = 66.
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