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Julia programming

Now let us turn back to our Volterra-Lotka problem, with these Uber-methods. We assume we still have f1 and f2 and their parameters defined from before. The Heun's version yields:

@time for n=1:npts-1
    y1[n+1]=y1[n]+(h/2)*(f1((n-1)*h,y1[n],y2[n])+f1(n*h,y1[n]+h*f1((n-1)*h,y1[n],y2[n]),y2[n]))
    y2[n+1]=y2[n]+(h/2)*(f2((n-1)*h,y1[n],y2[n])+f2(n*h,y1[n],y2[n]+h*f2((n-1)*h,y1[n],y2[n])))
end

The Volterra-Lotka model reads:

Euler's method in this case reads:

$$\begin{align} y_{1,n+1}&=y_{1,n}+hf_1(t_n,y_{1,n},y_{2,n})\\ y_{2,n+1}&=y_{2,n}+hf_2(t_n,y_{1,n},y_{2,n}) \end{align}$$

The functions are implemented as:

function f1(t,y1, y2)
    α*y1-β*y1*y2
end
function f2(t,y1, y2)
    δ*y1*y2-γ*y2
end

npts=10000
h=.001
 
y1=0.0*collect(1:npts);
y2=0.0*collect(1:npts);
 
y1[1]=1;
y2[1]=.1;
 
α=2/3;
β=4/3;
γ=1;
δ=1;
 
@time for i=1:npts-1
    y1[i+1]=y1[i]+h*f1((i-1)*h,y1[i],y2[i])
    y2[i+1]=y2[i]+h*f2((i-1)*h,y1[i],y2[i])
    end

plot([[y1[i] for i=1:npts], [y2[i] for i=1:npts]])

plot([(y1[i], y2[i]) for i=1:npts])

http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx

http://calculuslab.deltacollege.edu/ODE/7-C-3/7-C-3-h.html

Backward Euler method, or Implicit Euler method,

— • 21:51, 15 March 2021