How can i plot this graph?

for x=0:0.5:5;

if 0<x<2

y=((1+0.5*(1-cos(2*pi*x/5))).^-0.5);

else x>2;

y=1;

end

end

plot(x,y)

Answer 1

[Ameer Hamza]

You can use this vectorized version which is usually faster in MATLAB as compared to for loop.

x = 0:0.1:5;
y = (x<2)*((1+0.5*(1-cos(2*pi*x/5))).^-0.5) + (x>=2)*1;
plot(x,y)

[Farooq Hussain]

Thank you very much for your dear Ameer ....Will this vectorized version work correctly if i use (x<2)*((1+0.5*(1-cos(2*pi*x/5))).^-0.5) + (x>=2)*1; by using RK method? As i use this expression but, i don not get the required graph..Any suggestion please

[Ameer Hamza]

Can you paste your code here?

[Farooq Hussain]

%%% Range-Kutta method %%%%%

L==8;

CM=-1;

a=0;

h=0.5;

n=20;

x=a:h:n;

%%% Functions %%%

%%% A(x) =(1).*((x < 0)) + ((1-0.5*CM*(1-cos(2*pi*x/L))).^-0.5).*((0<= x) & (x <=L))+(1).*((x >L));

f1=@(x,u) u* ((1).*((x < 0)) +((1+0.5*(1-cos(2*pi*x/L))).^-0.5).*((0<= x) & (x <=L))+(1).*((x >L)));

%%% Initial Conditions %%%

u(1)=01;

for i=1:length(x)-1;

m1=h*f1(x(i), u(i));

m2=h*f1(x(i)+0.5*h, u(i)+0.5*m1);

m3=h*f1(x(i)+0.5*h, u(i)+0.5*m2);

m4=h*f1(x(i)+h, u(i)+m3);

u(i+1)=u(i)+(1/6)*(m1+2*m2+2*m3+m4);

end

plot(x,u,'k')

[Farooq Hussain]

my code does not read the value of A(x)...this is the issue

[Ameer Hamza]

Variables a and L are not defined in your code. Also can you show me the mathematical equation you are trying to implement.

[Farooq Hussain]

can i take your email please...so that i can email you the pdf of the paper i have reviewed...i manged to do all coding and plot graphs.....every thing is clear to me but piece-wise function A(x) is not clear to me...have been trying by different methods.....but there is something i am missing....Thanking in anticpation

[Farooq Hussain]

here is the link of that paper:

https://authors.library.caltech.edu/174/1/WNG177.pdf

[Ameer Hamza]

It is not practical for me to understand the complete paper. Can you write down the logic to generate the matrix A, and I will try to suggest how it can be written in MATLAB.

[Farooq Hussain]

simply how can i incorporate or use A(x) by using this RK code...

A(x) in MATLAB code...I had successfully used A(x) in Mathematica.... I will send you my complete code after simplifying..so that you have no problem in understanding this...

Finally, thanks for your cooperation and time..I really appreciate this.

[Ameer Hamza]

I haven't used Mathematica in a while, so cannot remember correctly. In MATLAB, you don't need to explicitly write an index on the left-hand side if you are initializing a variable. You can just write

A = (1).*((x < 0)) + ((1-0.5*CM*(1-cos(2*pi*x/L))).^-0.5).*((0<= x) & (x <=L))+(1).*((x >L));

[Farooq Hussain]

Assalam o alaikum!

Dear Ameer Hamaz...Hope you are doing well...Please replace the value of A(x)

in my code u*A(x)......when

L=10;

a=0;

h=0.5;

n=20;

x=a:h:n;

f1=@(x,u) u * A(x);

%%% Initial Conditions %%%

u(1)=1;

for i=1:length(x)-1;

m1=h*f1(x(i), u(i));

m2=h*f1(x(i)+0.5*h, u(i)+0.5*m1);

m3=h*f1(x(i)+0.5*h, u(i)+0.5*m2);

m4=h*f1(x(i)+h, u(i)+m3);

u(i+1)=u(i)+(1/6)*(m1+2*m2+2*m3+m4);

end

plot(x,u,'k')

[Ameer Hamza]

You can define it like this

A = @(x) (0<x & x<L).*(1+0.5*(1-cos(2*pi*x/L))-1/2) + (L<x).*1;
f1=@(x,u) u * A(x);

Answer 2

[Thiago Henrique Gomes Lobato]

Making y=... substitute the whole array. Try something like this:

x=0:0.5:5;
y = zeros(size(x));
for idx = 1:length(x)
    if 0<x(idx) && x(idx)<2
        y(idx)=((1+0.5*(1-cos(2*pi*x(idx)/5))).^-0.5);
    else x(idx)>2;
        y(idx)=1;
    end   
end
 plot(x,y)