The result is not what I want it is completely different ,,so Is there anyone could help me to use another routines to get the result for this code please?

This code what I am doing,, My problem is in plot(4),The result is not what I want it is completely different ,,so Is there anyone could help me to use another routines to get the result for this code please?

function Runfisher

A=1;

b=3;

x0=[12 5];

dt=0.01;

tspan=0:dt:1;

[t,x]=ode45(@fisher,tspan,x0,[],b);

figure(1)

plot(t,x)

figure(2)

plot(x(:,1),x(:,2),'-')

grid

dt= 0.095;

a1 = ( 15*x(1) - b*x(1).*x(2) );

b1 = ( -5*x(2) + 0.5*x(1).*x(2) );

I = A.*(dt.^2).*sum.*( ( ( ( (a1.*(15-b*x(2))+b1.*(.5*x(2))).*a1 ) + ( (a1.*(-b*x(1))+ b1.*(-5+.5*x(1))).*b1) ).^2)./(( (a1.^2) + (b1.^2) ).^2) )

ball=2:1:10 ;

n = length(ball) ;

Iall = zeros(1,n) ;

for i = 1 : n

      [t,x]=ode45(@fisher,(0:0.0001:1),x0,[],ball(i)) ;

      figure(3)

      plot(x(:,1),x(:,2))

      hold on  

      dt= 0.095;

      a1 =  ( 15*x(1) - ball(i).*x(1).*x(2) );

      b1 = ( -5*x(2) + 0.5*x(1).*x(2) );

      I = A.*(dt.^2).*sum.*( ( ( ( (a1.*(15-ball(i).*x(2))+b1.*(.5*x(2))).*a1 ) + ( (a1.*(-ball(i).*x(1))+ b1.*(-5+.5*x(1))).*b1) ).^2)./(( (a1.^2) + (b1.^2) ).^2) ) ;

      Iall(i) = I 

end

figure(4)

plot(ball,Iall,'r-')

1;

% function dxdt = fisher(t,x,b)

% dxdt=zeros(2,1);

% dxdt(1) = 15 * x(1) - b * x(1).* x(2);

% dxdt(2) = -5 * x(2) + .5 * x(1).* x(2);

% end

The right plot for figure(4) should increase until (ball =3 ) then it will deceasing until the last value. But what I am getting is just increasing line . I do not what is the reason so could anyone help me please?

[Torsten]

Questions:

1. What is the value of the variable "sum" in the calculation of I ?

2. x is an (nx2) matrix where n is the number of output times from ODE45.

So what are x(1) and x(2) in the calculation of I ?

3. Why is the function "fisher" commented out ?

Best wishes

Torsten.

[Avan Al-Saffar]

Many thanks for your concern.

According to your questions, I noticed that may be I am not using the correct routine so please I will answer your quires and if you can help me to solve my problem:

1- the value of the '' sum '' is from 1 to n ( the number of output from ODE45), This mean that I need to get the result of summation .

2- May be I could not understand your question well but My systems contains two equations with two variables x(1) and x(2) which they represent the Predator-Prey variables and they are depending on time .

3- I just referred it out here to informed you what is the system that I am calling in my ODE45

[Torsten]

ad 1) "sum" is a MATLAB operator that sums elements of arrays or matrices.

So "sum.*" has to be replaced simply by "sum".

ad 2)

I guess "I" should contain an approximation of an integral over time.

Thus you will have to replace

x(1) by x(:,1),

x(2) by x(:,2),

a1 = ( 15*x(1) - ball(i).*x(1).*x(2) ); by a1(:) = ( 15*x(:,1) - ball(i)*x(:,1).*x(:,2) );

and

b1 = ( -5*x(2) + 0.5*x(1).*x(2) ); by b1(:) = ( -5*x(:,2) + 0.5*x(:,1).*x(:,2) );

Of course I don't know whether the function to be integrated is correct and whether the approximation of the integral by your sum is correct.

Best wishes

Torsten.

[Avan Al-Saffar]

Many thanks for you Torsten.

I did what you mentioned but I received this error message :

In an assignment A(:) = B, the number of elements in A and B must be the same.

Error in Runsystem (line 41)

    a1(:) = ( 15*x(:,1) - ball(i)*x(:,1).*x(:,2) );

[Torsten]

I don't know why, but the setting of a1, b1 and I appear twice in your code (at the beginning and in the for-loop). Furthermore, since tspan changes, the dimensions of a1 and b1 will be different at both positions of your code. So the easiest way is to start your code from the line

ball=2:1:10;

and delete the upper part.

Best wishes

Torsten.

[Avan Al-Saffar]

They appear twice because for the first time I am fixing the value of (b) and later I am changing the value of (b) from 2 to 10. I will try your suggestion.

Many thanks Regards

[Avan Al-Saffar]

I still have the same plot even I applied your suggestion.

[Torsten]

Could you show the code you are using now ?

Best wishes

Torsten.

[Avan Al-Saffar]

function Runsystemvaryb

A=1;

x0=[12 5];

ball=2:1:10 ;

n = length(ball) ;

Iall = zeros(1,n) ;

for i = 1 : n

    [t,x]=ode45(@system,(0:0.0001:1),x0,[],ball(i)) ;

    figure(1)

    plot(x(:,1),x(:,2))

    hold on    

    dt= 0.055;

    % a1 =  ( 15*x(1) - ball(i).*x(1).*x(2) );

 % a1(:) = ( 15*x(:,1) - ball(i).*x(:,1).*x(:,2) );

    % b1 = ( -5*x(2) + 0.5*x(1).*x(2) );

    % b1(:) = ( -5*x(:,2) + 0.5*x(:,1).*x(:,2) );

    % I = A.*(dt.^2).*sum( ( ( ( (a1.*(15-ball(i)*x(2))+b1.*(.5*x(2))).*a1 ) + ( (a1.*(-ball(i)*x(1))+ b1.*(-5+.5*x(1))).*b1) ).^2)./(( (a1.^2) + (b1.^2) ).^2) )

    I = A.*(dt.^2).*sum( ( ( ( (( 15*x(:,1) - ball(i).*x(:,1).*x(:,2) ).*(15-ball(i).*x(:,2))+( -5*x(:,2) + 0.5*x(:,1).*x(:,2) ).*(.5*x(:,2))).*( 15*x(:,1) - ball(i).*x(:,1).*x(:,2) ) ) + ( (( 15*x(:,1) - ball(i).*x(:,1).*x(:,2) ).*(-ball(i).*x(:,1))+ ( -5*x(:,2) + 0.5*x(:,1).*x(:,2) ).*(-5+.5*x(:,1))).*( -5*x(:,2) + 0.5*x(:,1).*x(:,2) )) ).^2)./(( (( 15*x(:,1) - ball(i).*x(:,1).*x(:,2) ).^2) + (( -5*x(:,2) + 0.5*x(:,1).*x(:,2) ).^2) ).^2) ) 

    Iall(i) = I./10000

end

figure(2)

plot(ball,Iall,'r-')

1;

% function dxdt = system(t,x,b)

% dxdt=zeros(2,1);

% dxdt(1) = 15 * x(1) - b * x(1).* x(2);

% dxdt(2) = -5 * x(2) + .5 * x(1).* x(2);

% end

[Torsten]

I get the attached output for Iall.

Best wishes

Torsten.

[Torsten]

With an analogue Fortran-program, I get the following output for Iall:

ball Iall

   2.00000000000000        1.65285396632246 

   3.00000000000000        2.03520968157525

   4.00000000000000        3.46914314988618

   5.00000000000000        6.11508746765178     

   6.00000000000000        9.82670128843249   

   7.00000000000000        14.5325924533958   

   8.00000000000000        20.1308960907625    

   9.00000000000000        25.9469750868376   

   10.0000000000000        31.4212066754488

Best wishes

Torsten.

[Torsten]

Sorry, the numbers on the right should be one order of magnitude smaller (I divided by 1000, not by 10000).

Best wishes

Torsten.

[Avan Al-Saffar]

Dear Torsten

Yes, it is still increasing but what I am looking for is increasing curve until b=3 which will give us the higher value and then will decreasing until the last value which is b=10.

Many thanks for your help . I believe that I should re-understand what I am doing again and if you have any idea please post it to me here.

Regards Avan