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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?

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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?
  13 Kommentare
Torsten
Torsten am 5 Nov. 2014
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
Avan Al-Saffar am 5 Nov. 2014
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

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