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Are there some issues, if I run programs in MATLAB R2017a that I have created using MATLAB R2010a?

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naygarp
naygarp 2017-11-13
关闭： MATLAB Answer Bot 2021-8-20
I have used a program that I ran in MATLAB R2010a and later I ran the same code in MATLAB R2017a, I have some issues with the graphs. I am getting some different graphs in MATLAB R2017a
  2 个评论
naygarp
naygarp 2017-11-13
This is the Blasisus equation...Solving it using Shooting method with Runge-Kutta 4th order numerical technique, I got the code from MATLAB central website, the graphs obtained are wayward, I got a differnt sets of graphs using MATLAB R2010a.
%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-')

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回答（1 个）

Jan
Jan 2017-11-13
The graphic system has changed in R2014b substantially. The new system is called "HG2". There are many different possibility that this change influence the output. So please post the relevant code, the input data and screenshot of the unexpected output. Otherwise the question far to general to provide explicit help.
  1 个评论
naygarp
naygarp 2017-11-13
This is the Blasisus equation...Solving it using Shooting method with Runge-Kutta 4th order numerical technique, I got the code from MATLAB central website, the graphs obtained are wayward, I got a differnt sets of graphs using MATLAB R2010a.
%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-')

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