问题
I am trying to find a way to make this code faster.
Nagumo1 is the function that calculates the value of the two derivatives at time t.
function x = nagumo(t, y, f)
Iapp = f(t);
e = 0.1;
F = 2/(1+exp(-5*y(1)));
n0 = 0;
x = zeros(2, 1);
z(1) = y(1) - (y(1).^3)/3 - y(2).^2 + Iapp; %z(1) = dV/dt
z(2) = e.*(F + n0 - y(2)); %z(2) = dn/dt
x = [z(1);z(2)];
end
It is a system of differential equations that represents a largely simplified model of neuron. V represents a difference of electric potential, n represents the number of K+/Na+ canals and Iapp is the electric current applied to the neuron. The time variable (t) is measured in msec.
I want to use the Euler explicit method, with a variable step size, to numerically resolve the differential equation system and graphe the solution.
function x = EulerExplicit1(V0, n0, tspan, Iapp)
format long;
erreura = 10^-3;
erreurr = 10^-6;
h = 0.1;
to =tspan(1, 1) + h;
temps = tspan(1, 1);
tf = tspan(1, 2);
y = zeros(1,2);
yt1 = zeros(1, 2);
yt2 = zeros(1, 2);
y = [V0, n0];
z = y;
i = 1;
s = zeros(1, 2);
st1 = zeros(1, 2);
while temps<tf
s = nagumo1(to+i*h, y, Iapp);
y = y + h*s;
yt1 = y + (h/2)*s;
st1 = nagumo1(to+(i*h+h/2), yt1, Iapp);
yt2 = yt1 + (h/2)*st1;
if abs(yt2-y)>(erreura+erreurr*abs(y))
test = 0;
elseif h<0.4
h = h*2;
test = 0;
end
while test == 0
if abs(yt2-y)>(erreura+erreurr*abs(y)) & h>0.01
h = h/2;
s = nagumo1(to+i*h, y, Iapp);
y = y + h*s;
yt1 = y + (h/2)*s;
st1 = nagumo1(to+i*h+h/2, yt1, Iapp);
yt2 = yt1 + (h/2)*st1;
else
test = 1;
end
end
z = [ z ; y ];
temps = [temps; temps(i)+h];
i = i+1;
end
x = zeros(size(z));
x = z;
disp('Nombre d iterations:');
disp(i);
plot(temps, x(:, 1:end), 'y');
grid;
end
I haven't included any comments, because I think it is clear. I just want to maintain the adaptable step h and make the code faster. Ideally I would like to find a way to initialize z and temps(time), but when I try to do that then I have a problem plotting my solution. Note that when erreura(absolute error) and erreurr(relative error) are greater than 10^-6 my solution varies a lot in comparison to ode45 solution (which i consider to be accurate).
Any ideas?
P.S. if you want to test use values varying between -2, 2 for V, 0,1, 1 for n, 0.1, 1 for Iapp (defined a function handle @(t)).
回答1:
Before trying to speed up the interpreted code, you should care to get a correct solution at all. That there is still something amiss is visible in the time computations to+i*h that are only valid for a fixed step size. I'll explain the adaptive method from first principles.
The error estimation by Richardson extrapolation
uses the approximation that the numerical solution at time t computed with step size h relates to the exact solution in first order as
y(h;t)=y_exact(t) + C*t*h + O(t*h²)
gives that the advancement in one and two steps of half size has the errors
y(h;h) = y_exact(h) + C*h² + O(h³)
y(h/2;h) = y_exact(h)+C*h²/2 + O(h³)
and thus
y(h;h)-y(h/2;h) = C*h²/2 + O(h³)
is an estimator of the local error at step size h/2. We know that the local errors in first order add to the global error (in a better approximation there is some compounding with the Lipschitz constant as "annual" interest rate). Thus in the reverse direction we desire to get that the local error is a h sized part of the global error. Divide all local error quantities by h to get values that directly compare to the global error.
The adaptive step size controller
now tries to keep that local error estimate local_err = norm(y(h;h)-y(h/2;h))/h = norm(C)*h/2 inside some corridor [tol/100, tol] where ´tol´ stands for the desired global error. The ideal step size from the current data is thus computed as
tol = norm(C)*h_ideal/2 = local_err*h_ideal/h
<==>
h_ideal = tol / local_err * h
In the algorithm one would compute these integration steps and error estimates and then accept the steps and advance the computation if inside the tolerance bounds, then adapt the step size by above formula go to the next iteration of the loop. Instead of using the computed ideal step size one could also modify the step size by constant factors in the direction of the ideal step size. In general this will only increase the number of rejected steps to still reach the ideal step size.
To avoid oscillations and too abrupt changes in the tried and used step sizes, introduce some kind of moving average, dampen the change factor in direction 1 like in
a = tol / (1e-12+local_err);
h = 0.25*(3+a^0.8)*h ;
In code this could look like
while t < t_final
if t+1.1*h > t_final
h = t_final - t
force_advance = True
end
s1 = f(t,y)
s05 = f(t+0.5*h, y+0.5*h*s1)
s2 = 0.5*(s1+s05)
localerr = norm(s2-s1)
tol = abstol+norm(y)*reltol
if force_advance | (0.01*tol < localerr & localerr < tol)
y = y + h*s2
t = t + h
sol_y(end+1)=y
sol_t(end+1)=t
force_advance = False
end
a = tol / (1e-19+localerr) )
h = 0.25*(3+a^0.8)*h ;
if h < h_min
h = h_min
force_advance = True
end
if h > h_max
h = h_max
force_advance = True
end
end
The practical application of this method gives the following plot.
In the top the solution curves are depicted. One sees a higher density at the curved or rapidly changing parts and a lower density where the solution curve is more straight. In the lower part the error against the solution of lowest tolerance is displayed. The difference is scaled by the tolerance of the solution, so that all share the same scale. As one can see, the output traces the tolerance demanded at input closely.
回答2:
Look at those lines:
z = [ z ; y ];
temps = [temps; temps(i)+h];
These are really slow, and I do understand you cannot pre-allocate when using a variable step size. Assuming the size of the data you use is considerable I'd suggest you work with traditional files. A replacement for z would be:
fp = fopen('z_file.dat','wb');
...
fwrite(fp,y,'double');
...
fclose(fp);
fp = fopen('z_file.dat','r');
z = fread(fp,length_of_z,'double');
fclose(fp);
Where you need to know the amount of stored data (length_of_z which I guess is your "i"). Also, this will only speed up things if the amount of data is rather larger.
来源：https://stackoverflow.com/questions/49256309/matlab-euler-explicit-ode-solver-with-adaptable-step-is-there-a-way-to-make-cod