Solving differential equations in Matlab - In-Vitro Dissolution

问题

I am trying to solve a similar problem to this one: Solving Differential Equations in Matlab

However, this time the scenario is not injection of a drug into the subcutaneous tissue and its subsequent dissolution, but a more simple situation where the suspension is allowed to dissolve in a dissolution bath of volume 900 ml.

function dydt=odefcnNY_v12(t,y,D,Cs,rho,r0,N,V)
dydt=zeros(2,1);
dydt(1)=(-D*Cs)/(rho*r0^2)*(1-y(2))*y(1)/(1e-6+y(1)^2); % dr*/dt
dydt(2)=(D*4*pi*N*r0*(1-y(2))*y(1))/V; %dC*/dt
end

i.e. the absorption term from the previous question is removed:

Absorption term: Af*y(2)

The compound is also different, so the MW, Cs and r0 are different, and the experimental setup is also different so W and V are now changed. To allow for these changes, the ode113 call changes to this:

MW=336.43; % molecular weight
D=9.916e-5*(MW^-0.4569)*60/600000 %m2/s - [D(cm2/min)=9.916e-5*(MW^-0.4569)*60], divide by 600,000 to convert to m2/s 
rho=1300; %kg/m3
r0=9.75e-8; %m dv50
Cs=0.032; %kg/m3 
V=0.0009;%m3 900 mL dissolution bath
W=18e-6; %kg 18mg
N=W/((4/3)*pi*r0^3*rho); % particle number
tspan=[0 200*3600]; %s in 200 hours
y0=[1 0];
[t,y]=ode113(@(t,y) odefcnNY_v12(t,y,D,Cs,rho,r0,N,V), tspan, y0);
plot(t/3600,y(:,1),'-o') %plot time in hr, and r*
xlabel('time, hr')
ylabel('r*, (rp/r0)')
legend('DCU')
title ('r*');
plot(t/3600,y(:,1)*r0*1e6); %plot r in microns
xlabel('time, hr');
ylabel('r, microns');
legend('DCU');
title('r');
plot(t/3600,y(:,2),'-') %plot time in hr, and C*
xlabel('time, hr')
ylabel('C* (C/Cs)')
legend('DCU')
title('C*');

The current problem is that this code has been running for 3 hours, and still not complete. What is different now to the previous question in the link above, that is making it take so long?

Thanks

回答1:

I can not really reproduce your problem. I use the "standard" python modules numpy and scipy, copied the block of parameters,

MW=336.43; # molecular weight
D=9.916e-5*(MW**-0.4569)*60/600000 #m2/s - [D(cm2/min)=9.916e-5*(MW^-0.4569)*60], divide by 600,000 to convert to m2/s 
rho=1300.; #kg/m3
r0=9.75e-8; #m dv50
Cs=0.032; #kg/m3 
V=0.0009;#m3 900 mL dissolution bath
W=18e-6; #kg 18mg
N=W/((4./3)*pi*r0**3*rho); # particle number
Af = 0; # bath is isolated

used the same ODE function like in the previous post (remember Af=0)

def odefcnNY(t,y,D,Cs,rho,r0,N,V,Af):
    r,C = y;
    drdt = (-D*Cs)/(rho*r0**2)*(1-C) * r/(1e-10+r**2); # dr*/dt
    dCdt = (D*4*pi*N*r0*(1-C)*r-(Af*C))/V;            # dC*/dt
    return [ drdt, dCdt ];

and solved the ODE

tspan=[0, 1.0]; #1 sec
#tspan=[0, 200*3600]; #s in 200 hours
y0=[1.0, 0.0];
method="Radau"

sol=solve_ivp(lambda t,y: odefcnNY(t,y,D,Cs,rho,r0,N,V,Af), tspan, y0, method=method, atol=1e-8, rtol=1e-11);

t = sol.t; r,C = sol.y;
print(sol.message)
print("C*=",C[-1])

This works in a snap, using 235 steps for the first second and 6 further steps to cover the constant behavior in the remaining time of the 200 hours.

I can only mess it up by increasing the tolerances to unreasonably large values like 1e-4, and only if the epsilon used in the mollification is 1e-12. Then the hard turn when the radius reaches zero is too hard, the step size controller falls into a loop. This is more an error of the crude implementation of the step size controller, that should not be the case in the Matlab routines.

来源：https://stackoverflow.com/questions/59101142/solving-differential-equations-in-matlab-in-vitro-dissolution

标签

matlab

ode