komputasi sistem fisis 3

8/19/2019 Komputasi Sistem Fisis 3

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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.



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



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 ;



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

komputasi sistem fisis 3

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