我正在网上阅读这篇文章,其中有人提到使用“if statements”和“abs()”函数可能会对 MATLAB 的可变步长 ODE 求解器(如 ODE45)产生负面影响。根据 OP,它会显着影响时间步长(需要太低的时间步长),并在最终积分微分方程时给出较差的结果。我想知道这是否是真的,如果是,为什么。此外,如何在不求助于固定步骤求解器的情况下缓解这个问题。我在下面给出了一个示例代码来说明我的意思:
function [Z,Y] = sauters(We,Re,rhos,nu_G,Uinj,Dinj,theta,ts,SMDs0,Uzs0,...
Uts0,Vzs0,zspan,K)
Y0 = [SMDs0;Uzs0;Uts0;Vzs0]; %Initial Conditions
options = odeset('RelTol',1e-7,'AbsTol',1e-7); %Tolerance Levels
[Z,Y] = ode45(@func,zspan,Y0,options);
function DY = func(z,y)
DY = zeros(4,1);
%Calculate Local Droplet Reynolds Numbers
Rez = y(1)*abs(y(2)-y(4))*Dinj*Uinj/nu_G;
Ret = y(1)*abs(y(3))*Dinj*Uinj/nu_G;
%Calculate Droplet Drag Coefficient
Cdz = dragcof(Rez);
Cdt = dragcof(Ret);
%Calculate Total Relative Velocity and Droplet Reynolds Number
Utot = sqrt((y(2)-y(4))^2 + y(3)^2);
Red = y(1)*abs(Utot)*Dinj*Uinj/nu_G;
%Calculate Derivatives
%SMD
if(Red > 1)
DY(1) = -(We/8)*rhos*y(1)*(Utot*Utot/y(2))*(Cdz*(y(2)-y(4)) + ...
Cdt*y(3)) + (We/6)*y(1)*y(1)*(y(2)*DY(2) + y(3)*DY(3)) + ...
(We/Re)*K*(Red^0.5)*Utot*Utot/y(2);
elseif(Red < 1)
DY(1) = -(We/8)*rhos*y(1)*(Utot*Utot/y(2))*(Cdz*(y(2)-y(4)) + ...
Cdt*y(3)) + (We/6)*y(1)*y(1)*(y(2)*DY(2) + y(3)*DY(3)) + ...
(We/Re)*K*(Red)*Utot*Utot/y(2);
end
%Axial Droplet Velocity
DY(2) = -(3/4)*rhos*(Cdz/y(1))*Utot*(1 - y(4)/y(2));
%Tangential Droplet Velocity
DY(3) = -(3/4)*rhos*(Cdt/y(1))*Utot*(y(3)/y(2));
%Axial Gas Velocity
DY(4) = (3/8)*((ts - ts^2)/(z^2))*(cos(theta)/(tan(theta)^2))*...
(Cdz/y(1))*(Utot/y(4))*(1 - y(4)/y(2)) - y(4)/z;
end
end
其中函数“dragcof”由以下给出:
function Cd = dragcof(Re)
if(Re <= 0.01)
Cd = (0.1875) + (24.0/Re);
elseif(Re > 0.01 && Re <= 260.0)
Cd = (24.0/Re)*(1.0 + 0.1315*Re^(0.32 - 0.05*log10(Re)));
else
Cd = (24.0/Re)*(1.0 + 0.1935*Re^0.6305);
end
end