Problem 1
a)
Given second order equation
𝜕2𝑦
= 𝑦 + sinx
𝜕𝑥 2
From Central Difference formula, we know that
𝜕2𝑦
𝑦𝑖−1 − 2𝑦𝑖 + 𝑦𝑖+1
=
𝜕𝑥 2
ℎ2
Substitute this in the given ODE,
𝑦𝑖−1 − 2𝑦𝑖 + 𝑦𝑖+1
= 𝑦𝑖 + sin (𝑥𝑖 )
ℎ2
By solving, we get
𝑦𝑖−1 − (2 + ℎ2 )𝑦𝑖 + 𝑦𝑖+1 = ℎ2 sin (𝑥𝑖 )
This is then ODE discretized into a finite difference formula with i = 1,2,3 ……….
(n-1).
b)
We have the value of h as 1
Iterate the discretized ODE by varying the ‘i’ values from 1 to n-1
If we do that, we get
For i = 1:
−3𝑦1 + 𝑦2 = sin(𝑥1 )
For i = 2:
𝑦1 − 3𝑦 + 𝑦3 = sin (𝑥2 )
For i = 3:
𝑦2 − 3𝑦3 + 𝑦4 = sin (𝑥3 )
For i = 4:
𝑦3 − 3𝑦4 + 𝑦5 = sin (𝑥4 )
If we compile these equations into Matrix form, we get,
−3
1
0
0
0
0
0
0
[ 0
1
−3
1
0
0
0
0
0
0
0
0 0 0
1
0 0 0
−3
1 0 0
1
−3 1
0
0
1 −3 1
0
0
1 −3
0
0 0 1
0
0 0 0
0
0 0 0
𝑦2
0 0 0
𝑦2
0 0 0
𝑦3
0 0 0
𝑦4
0 0 0
𝑦5 =
0 0 0
𝑦6
0 0 0
−3 1
0 𝑦7
1 −3 1 𝑦8
0
1 −3] [𝑦9 ]
sin(𝑥1 ) − 𝑦0
sin(𝑥2 )
sin(𝑥3 )
sin(𝑥4 )
sin(𝑥5 )
sin(𝑥6 )
sin(𝑥7 )
sin(𝑥8 )
[sin(𝑥9 ) − 𝑦𝑛−1 ]
From the matrix, upper and lower diagonal elements are 1 and main diagonal
elements are 3.
init = bvpinit(linspace(0,20),[0.2,0.2]);
result = bvp4c(@ODE, @BC, init);
plot(result.x, result.y(1,:))
title('Boundary Value Problem')
xlabel('distance')
ylabel('Height of cable')
function dydx = ODE(x,yv)
dydx = [yv(2); 0.041*sqrt(1+(yv(2))^2)];
end
function sol = BC(y_a, y_b)
BC_a = 10;
BC_b = 15;
sol = [y_a(1) - BC_a; y_b(1) - BC_b];
end
Published with MATLAB® R2019b
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