Need the backward trajectories of ode plot

I have a system of differential equations which I can solve by Euler's method. The following code gives a plot of a trajectory that starts from x(1)=0.7; y(1)=0.11; and depicts its evolution in forwarding time. But I need a trajectory that starts from x(1)=0.7; y(1)=0.11; and evolved in backward time. That mean what will be the plot if t tends to -infinity. Please correct my code so that I can get backword evolution of trajectories:

clear
alpha=.5;gamma=1; delta=0.3; L=.4; beta=1.778; 
x(1)=0.7;
y(1)=0.11;
t(1)=0;
for i=1:50000
    t(i+1)=t(i)+.01;
    x(i+1)=x(i)+.01*[x(i)*((1-x(i))*(x(i)/L-1)-beta*y(i)/(x(i)+alpha))];
    y(i+1)=y(i)+.01*[beta*x(i)*y(i)/(x(i)+alpha)-gamma*y(i)-delta*y(i)^2];
end
plot(x,y, 'b')
axis([.4 1 0 .22])

Answer 1

[Jan]

Do you ask for changing the line:

t(i+1) = t(i) + 0.01;

to

t(i+1) = t(i) - 0.01;

?

[Atom]

I want to know the behevour of the trajectories as t tends to -infinity (in stead of t tends to +infinity). Your code t(i+1) = t(i) - 0.01; gives information in backward time. But both the codes give the same plot

clear
alpha=.5;gamma=1; delta=0.3; L=.4; beta=1.778; 
x(1)=0.7;
y(1)=0.11;
t(1)=0;
for i=1:50000
    t(i+1)=t(i)-.01;
    x(i+1)=x(i)+.01*[x(i)*((1-x(i))*(x(i)/L-1)-beta*y(i)/(x(i)+alpha))];
    y(i+1)=y(i)+.01*[beta*x(i)*y(i)/(x(i)+alpha)-gamma*y(i)-delta*y(i)^2];
end
plot(x,y, 'r')
axis([.4 1 0 .22])
hold on
alpha=.5;gamma=1; delta=0.3; L=.4; beta=1.778; 
x(1)=0.7;
y(1)=0.11;
t(1)=0;
for i=1:50000
    t(i+1)=t(i)+.01;
    x(i+1)=x(i)+.01*[x(i)*((1-x(i))*(x(i)/L-1)-beta*y(i)/(x(i)+alpha))];
    y(i+1)=y(i)+.01*[beta*x(i)*y(i)/(x(i)+alpha)-gamma*y(i)-delta*y(i)^2];
end
plot(x,y, 'b')
axis([.4 1 0 .22])

[Jan]

The formulas for x and y do not depend on t at all. So why do you assume, that there is any behaviour for t->anything?

I guess, that you want to solve the equations [x(i+1), y(i+1)] = f(x(i), y(i)) to [x(i), y(i)] = g(x(i+1), y(i+1)). This seems to be trivial in this case, but start with i=50000 and go back to 1 then.

[Atom]

Do you mean this?

alpha=.5;gamma=1; delta=0.3; L=.4; beta=1.778; 
  x(50000)=0.7;
  y(50000)=0.11;
  t(50000)=0;
  for i=50000:1
      t(i+1)=t(i)-.01;
      x(i+1)=x(i)+.01*[x(i)*((1-x(i))*(x(i)/L-1)-beta*y(i)/(x(i)+alpha))];
      y(i+1)=y(i)+.01*[beta*x(i)*y(i)/(x(i)+alpha)-gamma*y(i)-delta*y(i)^2];
  end
  plot(x,y, 'b')
  axis([.4 1 0 .22])

But this does not servers the purpose. Please help me to solve the issue. I am very sorry for the inconvenience for the repeated request.

[Jan]

"for i=50000:1" does not enter the loop at all. You need the stepsize of -1.

I cannot test it currently, but let me guess:

for i = 50000:-1:2
  t(i-1) = t(i) - 0.01;
  x(i-1) = x(i) + 0.01*[x(i)*((1-x(i))*(x(i)/L-1)-beta*y(i)/(x(i)+alp ha))];
  y(i-1) = y(i) + 0.01*[beta*x(i)*y(i)/(x(i)+alpha)-gamma*y(i)-delta*y(i)^2];
end