问题
I’m building a function in Octave that can solve N coupled ordinary differential equation of the type:
dx/dt = F(x,y,…,z,t)
dy/dt = G(x,y,…,z,t)
dz/dt = H(x,y,…,z,t)
With any of these three methods (Euler, Heun and Runge-Kutta-4).
The following code correspond to the function:
function sol = coupled_ode(E, dfuns, steps, a, b, ini, method)
range = b-a;
h=range/steps;
rows = (range/h)+1;
columns = size(dfuns)(2)+1;
sol= zeros(abs(rows),columns);
heun=zeros(1,columns-1);
for i=1:abs(rows)
if i==1
sol(i,1)=a;
else
sol(i,1)=sol(i-1,1)+h;
end
for j=2:columns
if i==1
sol(i,j)=ini(j-1);
else
if strcmp("euler",method)
sol(i,j)=sol(i-1,j)+h*dfuns{j-1}(E, sol(i-1,1:end));
elseif strcmp("heun",method)
heun(j-1)=sol(i-1,j)+h*dfuns{j-1}(E, sol(i-1,1:end));
elseif strcmp("rk4",method)
k1=h*dfuns{j-1}(E, [sol(i-1,1), sol(i-1,2:end)]);
k2=h*dfuns{j-1}(E, [sol(i-1,1)+(0.5*h), sol(i-1,2:end)+(0.5*h*k1)]);
k3=h*dfuns{j-1}(E, [sol(i-1,1)+(0.5*h), sol(i-1,2:end)+(0.5*h*k2)]);
k4=h*dfuns{j-1}(E, [sol(i-1,1)+h, sol(i-1,2:end)+(h*k3)]);
sol(i,j)=sol(i-1,j)+((1/6)*(k1+(2*k2)+(2*k3)+k4));
end
end
end
if strcmp("heun",method)
if i~=1
for k=2:columns
sol(i,k)=sol(i-1,k)+(h/2)*((dfuns{k-1}(E, sol(i-1,1:end)))+(dfuns{k-1}(E, [sol(i,1),heun])));
end
end
end
end
end
When I use the function for a single ordinary differential equation, the RK4 method is the best as expected, but when I ran the code for a couple system of differential equation, RK4 is the worst, I've been checking and checking and I don't know what I am doing wrong.
The following code is an example of how to call the function
F{1} = @(e, y) 0.6*y(3);
F{2} = @(e, y) -0.6*y(3)+0.001407*y(4)*y(3);
F{3} = @(e, y) -0.001407*y(4)*y(3);
steps = 24;
sol1 = coupled_ode(0,F,steps,0,24,[0 5 995],"euler");
sol2 = coupled_ode(0,F,steps,0,24,[0 5 995],"heun");
sol3 = coupled_ode(0,F,steps,0,24,[0 5 995],"rk4");
plot(sol1(:,1),sol1(:,4),sol2(:,1),sol2(:,4),sol3(:,1),sol3(:,4));
legend("Euler", "Heun", "RK4");
回答1:
Careful: there's a few too many h's in the RK4 formulæ:
k2 = h*dfuns{ [...] +(0.5*h*k1)]);
k3 = h*dfuns{ [...] +(0.5*h*k2]);
should be
k2 = h*dfuns{ [...] +(0.5*k1)]);
k3 = h*dfuns{ [...] +(0.5*k2]);
(last h's removed).
However, this makes no difference for the example that you provided, since h=1 there.
But other than that little bug, I don't think you're actually doing anything wrong.
If I plot the solution generated by the more advanced, adaptive 4ᵗʰ/5ᵗʰ order RK implemented in ode45:
F{1} = @(e,y) +0.6*y(3);
F{2} = @(e,y) -0.6*y(3) + 0.001407*y(4)*y(3);
F{3} = @(e,y) -0.001407*y(4)*y(3);
tend = 24;
steps = 24;
y0 = [0 5 995];
plotN = 2;
sol1 = coupled_ode(0,F, steps, 0,tend, y0, 'euler');
sol2 = coupled_ode(0,F, steps, 0,tend, y0, 'heun');
sol3 = coupled_ode(0,F, steps, 0,tend, y0, 'rk4');
figure(1), clf, hold on
plot(sol1(:,1), sol1(:,plotN+1),...
sol2(:,1), sol2(:,plotN+1),...
sol3(:,1), sol3(:,plotN+1));
% New solution, generated by ODE45
opts = odeset('AbsTol', 1e-12, 'RelTol', 1e-12);
fcn = @(t,y) [F{1}(0,[0; y])
F{2}(0,[0; y])
F{3}(0,[0; y])];
[t,solN] = ode45(fcn, [0 tend], y0, opts);
plot(t, solN(:,plotN))
legend('Euler', 'Heun', 'RK4', 'ODE45');
xlabel('t');
Then we have something more believable to compare to.
Now, plain-and-simple RK4 indeed performs terribly for this isolated case:
However, if I simply flip the signs of the last term in the last two functions:
% ±
F{2} = @(e,y) +0.6*y(3) - 0.001407*y(4)*y(3);
F{3} = @(e,y) +0.001407*y(4)*y(3);
Then we get this:
The main reason RK4 performs badly for your case is because of the step size. The adaptive RK4/5 (with a tolerance set to 1 instead of 1e-12 as above) produces an average δt = 0.15. This means that basic error analysis has indicated that for this particular problem, h = 0.15 is the largest step you can take without introducing unacceptable error.
But you were taking h = 1, which then indeed gives a large accumulated error.
The fact that Heun and Euler perform so well for your case is, well, just plain luck, as demonstrated by the sign inversion example above.
Welcome to the world of numerical mathematics - there never is 1 method that's best for all problems under all circumstances :)
回答2:
Apart from the error described in the older answer, there is indeed a fundamental methodological error in the implementation. First, the implementation is correct for scalar order-one differential equations. But the moment you try to use it on a coupled system, the de-coupled treatment of the stages in the Runge-Kutta method (note that Heun is just a copy of the Euler step) reduces them to an order-one method.
Specifically, starting in
k2=h*dfuns{j-1}(E, [sol(i-1,1)+(0.5*h), sol(i-1,2:end)+(0.5*h*k1)]);
the addition of 0.5*k1 to sol(i-1,2:end) means to add the vector of slopes of the first stage, not to add the same slope value to all components of the position vector.
Taking this into account results in the change to the implementation
function sol = coupled_ode(E, dfuns, steps, a, b, ini, method)
range = b-a;
h=range/steps;
rows = steps+1;
columns = size(dfuns)(2)+1;
sol= zeros(rows,columns);
k = ones(4,columns);
sol(1,1)=a;
sol(1,2:end)=ini(1:end);
for i=2:abs(rows)
sol(i,1)=sol(i-1,1)+h;
if strcmp("euler",method)
for j=2:columns
sol(i,j)=sol(i-1,j)+h*dfuns{j-1}(E, sol(i-1,1:end));
end
elseif strcmp("heun",method)
for j=2:columns
k(1,j) = h*dfuns{j-1}(E, sol(i-1,1:end));
end
for j=2:columns
sol(i,j)=sol(i-1,j)+h*dfuns{j-1}(E, sol(i-1,1:end)+k(1,1:end));
end
elseif strcmp("rk4",method)
for j=2:columns
k(1,j)=h*dfuns{j-1}(E, sol(i-1,:));
end
for j=2:columns
k(2,j)=h*dfuns{j-1}(E, sol(i-1,:)+0.5*k(1,:));
end
for j=2:columns
k(3,j)=h*dfuns{j-1}(E, sol(i-1,:)+0.5*k(2,:));
end
for j=2:columns
k(4,j)=h*dfuns{j-1}(E, sol(i-1,:)+k(3,:));
end
sol(i,2:end)=sol(i-1,2:end)+(1/6)*(k(1,2:end)+(2*k(2,2:end))+(2*k(3,2:end))+k(4,2:end));
end
end
end
As can be seen, the loop over the vector components is recurring frequently. One can hide this by using a full vectorization using a vector-valued function for the right side of the coupled ODE system.
The plot for the second component of the solution with these changes gives the much more reasonable plot for step size 1
and with a subdivision into 120 intervals for step size 0.2
where the graph for RK4 did not change much while the other two moved towards it from below and above.
来源：https://stackoverflow.com/questions/46654283/runge-kutta-for-coupled-odes