Effacer les filtres
Effacer les filtres

What modifications do I need to make in the following codes for solving the boundary value problem similar to the Blasius equation using Shooting method with R-K 4 numerical analysis

4 vues (au cours des 30 derniers jours)
The equation is (1+2M*eta)f'''+ 2Mf"+ f*f"- (f')^2- K1*f'= 0 ; f' is df/d(eta) 'eta' is a similarity variable
where M- is a curvature parameter, M = 1
k1- is another parameter, k1= 0.1
boundary conditions are: eta=0, f(0)=0, f'(0)=1, f'(inf)=0, inf= 10
The following is the code for Blasius equation
% Fourth Runge Kutta for solving Blasius Equation
%-----------------------------------------------
%Blasius Equation : 2f"+ff'=0
%-----------------------------------------------
%Boundary Condition
%1. f'(0) = 0
%2. f'(inf) = 0
%3. f(0) = 0
%from shooting method
%4. f"(0) = 0.332
%-----------------------------------------------
%find: f,f" and f'''
%-----------------------------------------------
%Given
%eta = x
%f = y0,f'=y1,f"=y2,f'''= -(1/2)*y0*y2
%-----------------------------------------------
func1 = inline('y1','x','y0','y1','y2');
%-----------------------------------------------
func2 = inline('y2','x','y0','y1','y2');
%-----------------------------------------------
func3 = inline('-0.5*y0*y2','x','y0','y1','y2');
%-----------------------------------------------
%input: x = 0 , y0 = 0 , y1= 0
% y2 = 0.332 , total = 7 and h = 0.1
%-----------------------------------------------
x = input('\n Enter the value of x : ');
y0 = input( 'Enter the value of y0 : ');
y1 = input( 'Enter the value of y1 : ');
y2 = input( 'Enter the value of y2 : ');
total = input('Enter the value of total : ');
h = input( 'Enter the value of h : ');
fprintf('\n Solution with step size =%5.3f is:',h);
fprintf('\n x y0 y1 y2');
fprintf('\n%16.3f%16.3f%16.3f%16.3f',x,y0,y1,y2);
for i = 1:(total/h)
ak1y0 = func1(x,y0,y1,y2);
ak1y1 = func2(x,y0,y1,y2);
ak1y2 = func3(x,y0,y1,y2);
xx = x + h/2.;
yy0 = y0 + h*ak1y0/2.;
yy1 = y1 + h*ak1y1/2.;
yy2 = y2 + h*ak1y2/2.;
ak2y0 = func1(xx,yy0,yy1,yy2);
ak2y1 = func2(xx,yy0,yy1,yy2);
ak2y2 = func3(xx,yy0,yy1,yy2);
yy0 = y0 + h*ak2y0/2.;
yy1 = y1 + h*ak2y1/2.;
yy2 = y2 + h*ak2y2/2.;
ak3y0 = func1(xx,yy0,yy1,yy2);
ak3y1 = func2(xx,yy0,yy1,yy2);
ak3y2 = func3(xx,yy0,yy1,yy2);
all_x(i) = x;
all_y0(i) = y0;
all_y1(i) = y1;
all_y2(i) = y2;
xx = x + h;
yy0 = y0 + h*ak3y0;
yy1 = y1 + h*ak3y1;
yy2 = y2 + h*ak3y2;
ak4y0 = func1(xx,yy0,yy1,yy2);
ak4y1 = func2(xx,yy0,yy1,yy2);
ak4y2 = func3(xx,yy0,yy1,yy2);
y0 = y0 + (ak1y0 + 2.*ak2y0 + 2.*ak3y0 + ak4y0)*h/6.;
y1 = y1 + (ak1y1 + 2.*ak2y1 + 2.*ak3y1 + ak4y1)*h/6.;
y2 = y2 + (ak1y2 + 2.*ak2y2 + 2.*ak3y2 + ak4y2)*h/6.;
x = x + h;
fprintf('\n%16.3f%16.3f%16.3f%16.3f',x,y0,y1,y2);
end
plot(all_x, all_y0, 'k-', all_x, all_y1, 'b-', all_x, all_y2, 'g-')

Réponse acceptée

Torsten
Torsten le 8 Nov 2017
Use MATLAB's "bvp4c" instead of a code that you don't understand.
Best wishes
Torsten.

Plus de réponses (0)

Catégories

En savoir plus sur Call Python from MATLAB dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by