komputasi sistem fisis 3
TRANSCRIPT
8/19/2019 Komputasi Sistem Fisis 3
http://slidepdf.com/reader/full/komputasi-sistem-fisis-3 1/4
1
Homework 3
Name : Nadya Amalia
Student ID : 20213042
Subject : Computational Physical Systems (FI5005)
Lecturer : Dr.rer.nat. Linus Ampang Pasasa
FINITE-DIFFERENCEMETHOD
Use the finite-difference method to solve the problem
ݕ
ᇱᇱ =ݕ
+ݔ
(ݔ
− 4), 0 ≤ݔ
≤ 4
with (4) = 0ݕ = (0)ݕ and = 4 subintervals.
SOLUTION
Analytical Solution
The general solution of the second nonhomogeneous linear equation
ݕ
ᇱᇱ − ݕ (ݔ)ݎ =
can be expressed in the form
ݕ + ݕ =
The corresponding homogeneous equation ݕᇱᇱ − ݕ
= 0 has characteristic equation
ݎ
ଶ − 1 = 0
ݎ
ଶ = 1
ݎ
= ±1
Complementary solution
= ଵݕభ௫ + ଶ
మ௫
= ݕ ଵ
௫ + ଶି௫
The nonhomogeneous equation has ݎ(ݔ) = ݔ(ݔ − ଶݔ = (4 − . It is a degree 2 polinomial. Weݔ4
will let be a genetic quadratic polinomial: = ݔ
ଶ + ݔ + . It followsܥ ݔ 2 = + and
= 2 . Substitute them into the equation
(2
) − (ݔ
ଶ +ݔ
+ܥ
ଶݔ = ( − 4ݔ
−ݔ
ଶ −ݔ + (2 −
ܥ
ଶݔ = ( − ݔ4
The corresponding terms on both side should have the same coefficients, therefore, equating
the coefficients of like terms.
ݔ
ଶ : − = 1 = −1
ݔ : −
= −4 →
= 4
1 : 2 −ܥ = 0 ܥ = −2
Therefore, = ଶݔ− + ݔ4 − 2 and ݕ + ݕ = = ଵ௫ + ଶ
ି௫ − ݔ
ଶ + ݔ4 − 2.
8/19/2019 Komputasi Sistem Fisis 3
http://slidepdf.com/reader/full/komputasi-sistem-fisis-3 2/4
2
Boundary value conditions
ݕ
(4) = 0ݕ = (0)
For ݔ = 0 : ଵ + ଶ
ି − (0)ଶ + (4)(0) − 2 = 0
ଵ + ଶ = 2
For ݔ = 4 : ଵସ + ଶ
ିସ − (4)ଶ + (4)(4) − 2 = 0
ଵସ + ଶ
ିସ = 2
Substitute ଵ = 2 − ଶ
(2 − ଶ) ସ + ଶିସ = 2
− ଶସ + ଶ
ିସ = 2 − 2 ସ
− ଶ( ସ − ିସ) = 2 − 2 ସ
ଶ =2 − 2 ସ
−( ସ − ିସ)
We obtain ଶ = 1,96403 and ଵ = 0,03597.
Thus, ݕ = 0,03597 ௫ + 1,96403 ି௫ −ݔ
ଶ + ݔ4 − 2.
Numerical Solution
We denote the numerical solution at any point ݔ by ݕ
ݕ
ᇱᇱ
−ݎ = ݕ
The two boundary value conditions are = 0ݔ and . = 4ݔ
We also have = 4 and
ℎ =ݔ − ݔ
=4 − 0
4 = 1
Thus, we have five node points, they are , = 0ݔ ,ଵ = 1ݔ ,ଶ = 2ݔ ଷ = 3ݔ and ସ = 4ݔ .
We are given the data values = 0ݕ = (ݔ)ݕ and ସ = 0ݕ = (ସݔ)ݕ
By using the approximations
ݕ =1
ℎଶ(ାଵݕ − + ݕ2 (ିଵݕ
= ݕ1
ℎ
(ାଵݕ − (ݕ
with
ℎ = 1
in the differential equation we obtain
(ାଵݕ − + ݕ2 (ିଵݕ −
ݎ = ݕ
ିଵݕ − 2ݕ − + ݕ = ାଵݕ
ݎ
ିଵݕ − + ݕ3 ݎ = ାଵݕ
For = 1,,ଵ = 1ݔ
= 0ݕ : ݕ − 3
+ ଵݕ= ଶݕ
)ଵݔ ଵݔ − 4)
0 − + ଵݕ3 ݕ
ଶ = 1(1 − 4)
+ ଵݕ3− = ଶݕ −3
8/19/2019 Komputasi Sistem Fisis 3
http://slidepdf.com/reader/full/komputasi-sistem-fisis-3 3/4
3
For = 2, ଶ = 2ݔ : ଵݕ − 3+ ଶݕ
= ଷݕ)ଶݔ
ଶݔ − 4)
ଵݕ − + ଶݕ3 ଷ = 2(2ݕ − 4)
ଵݕ − 3+ ଶݕ
= ଷݕ −4
For = 3, ,ଷ = 3ݔ ସ = 0ݕ : ଶݕ − + ଷݕ3 )ଷݔ = ସݕଷݔ − 4)
ଶݕ − + ଷݕ3 0 = 3(3 − 4)
ଶݕ − 3= ଷݕ −3
We obtain the following system of equations
⎣
1 0 0 0 0
1 − 3 1 0 00 1 − 3 1 00 0 1 − 3 1
0 0 0 0 1 ⎦ ⎣
ݕ
0
ݕ
1
ݕ
2
ݕ
3
ݕ
4⎦
=
⎣
0
−3−4−3
0 ⎦
Or in a compact form as
=
.
MATLAB Source Code
clc, clear all
% NADYA AMALIA (20213042)
% COMPUTATIONAL PHYSICAL SYSTEMS
% DR.rer.nat. LINUS AMPANG PASASA
% FINITE-DIFFERENCE METHOD
% y"(x) - y(x) = r(x) , r(x)= x(x - 4)
% y(0)= alfa , y(l)= beta
% The definate of the global values
global alfa beta h l;
% Problem Data
alfa = 0;
beta = 0;
l = 4 ;
n = 4 ;
h = l/n;
% Making of the vector b
b = zeros(n,1);
b(1) = 0
b(n+1) = 0
for i = 2:n
b(i) = (i-1)*((i-1)-4)
end
% Making of the matrix A
A = zeros(n,n) ;
A(1,1) = 1
A(n+1,n+1) = 1
for i = 2:n
A(i,i-1) = 1
A(i,i) = -3
A(i,i+1) = 1
end
% The solution of the system
y = A \ b ;
8/19/2019 Komputasi Sistem Fisis 3
http://slidepdf.com/reader/full/komputasi-sistem-fisis-3 4/4
4
ym = linspace(0,4,n+1);
ym(1:n+1) = y;
% The analytical solution's presentment
% y(x) = 0.03597(e^x) + 1.96403(e^(-x)) - x^2 + 4x - 2
e = 2.7182818;
delta = linspace(0,4,100);
u = (0.03597.*(e.^delta)) + (1.96403.*(e.^(-delta))) - (delta.^2) +(4.*delta) - 2;
subplot(1,2,1);
p = plot(delta, u, '*');
title('Plotting of analytical solution of y"(x) - y(x) = x(x - 4)');
xlabel('x'); ylabel('y(x)');
grid on
set(p, 'Color', 'red')
% Presentment of the approximation
intervall = linspace(0,4,n+1);
subplot(1,2,2);
plot(intervall, ym, '-*') ;
title('Plotting of numerical solution of y"(x) - y(x) = x(x - 4) using
Finite-Difference Method');
xlabel('interval [0,4]'); ylabel('yi');
grid on;
From the calculation for numerical solution we obtain
=
⎣
0
1,85714
2,57143
1,85714
0 ⎦
Thus,
,(1) = 1,85714ݕ = ଵݕ
= ଶݕݕ
(2) = 2,57143, and
(3) = 1,85714ݕ = ଷݕ
0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3Plotting of analytical solution of y"(x) - y(x) = x(x - 4)
x
y ( x ) =
0 ,
0 3 5 9 7 e x
+
1 ,
9 6 4 0 3 e
( - x ) - x
2
+
4 x
- 2
0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3Plotting of numerical solution of y"(x) - y(x) = x(x - 4) using Finite-Difference Method
interval [0,4]
y i