Julia 朱莉娅';微分方程步长控制
我想用Julia中的Julia 朱莉娅';微分方程步长控制,julia,differential-equations,Julia,Differential Equations,我想用Julia中的微分方程解方程。对于某些初始值,我得到了错误: WARNING: dt <= dtmin. Aborting. If you would like to force continuation with dt=dtmin, set force_dtmin=true 我不知道该怎么做。以下是代码: using DifferentialEquations using Plots m = 1 l = 0.3 g = pi*pi function dbpen(du,u,pr
微分方程
解方程。对于某些初始值,我得到了错误:
WARNING: dt <= dtmin. Aborting. If you would like to force continuation with
dt=dtmin, set force_dtmin=true
我不知道该怎么做。以下是代码:
using DifferentialEquations
using Plots
m = 1
l = 0.3
g = pi*pi
function dbpen(du,u,pram,t)
th1 = u[1]
th2 = u[2]
thdot1 = du[1]
thdot2 = du[2]
p1 = u[3]
p2 = u[4]
du[1] = (6/(m*l^2))*(2*p1-3*p2*cos(th1-th2))/(16-9*(cos(th1-th2))^2)
du[2] = (6/(m*l^2))*(8*p2-3*p1*cos(th1-th2))/(16-9*(cos(th1-th2))^2)
du[3] = (-0.5*m*l^2)*(thdot1*thdot2*sin(th1-th2)+(3*g/l)*sin(th1))
du[4] = (-0.5*m*l^2)*(-thdot1*thdot2*sin(th1-th2)+(g/l)*sin(th2))
end
u0 = [0.051;0.0;0.0;0.0]
tspan = (0.0,100.0)
prob = ODEProblem(dbpen,u0,tspan)
sol = solve(prob)
plot(sol,vars=(0,1))
似乎你的颂歌很僵硬,默认情况下需要非常小的dt。 您可以切换到stiff ODE解算器或给出如下提示:
sol = solve(prob,alg_hints=[:stiff])
function dbpen(du,u,pram,t)
th1 = u[1]
th2 = u[2]
p1 = u[3]
p2 = u[4]
du[1] = (6/(m*l^2))*(2*p1-3*p2*cos(th1-th2))/(16-9*(cos(th1-th2))^2)
du[2] = (6/(m*l^2))*(8*p2-3*p1*cos(th1-th2))/(16-9*(cos(th1-th2))^2)
thdot1 = du[1]
thdot2 = du[2]
du[3] = (-0.5*m*l^2)*(thdot1*thdot2*sin(th1-th2)+(3*g/l)*sin(th1))
du[4] = (-0.5*m*l^2)*(-thdot1*thdot2*sin(th1-th2)+(g/l)*sin(th2))
end
参考:包文档中的ODE示例最近我更改了此警告,以明确告诉用户这很可能是模型的问题。如果您看到这一点,那么通常会有两个可能的问题:
thdot1 = du[1]
thdot2 = du[2]
为您提供可以无限小/无限大的虚拟值。原因是你应该覆盖它们!看起来你真正想做的是计算前两个导数项,然后在第二组导数项中使用它们。为此,您必须确保首先更新值!一个可能的代码如下所示:
sol = solve(prob,alg_hints=[:stiff])
function dbpen(du,u,pram,t)
th1 = u[1]
th2 = u[2]
p1 = u[3]
p2 = u[4]
du[1] = (6/(m*l^2))*(2*p1-3*p2*cos(th1-th2))/(16-9*(cos(th1-th2))^2)
du[2] = (6/(m*l^2))*(8*p2-3*p1*cos(th1-th2))/(16-9*(cos(th1-th2))^2)
thdot1 = du[1]
thdot2 = du[2]
du[3] = (-0.5*m*l^2)*(thdot1*thdot2*sin(th1-th2)+(3*g/l)*sin(th1))
du[4] = (-0.5*m*l^2)*(-thdot1*thdot2*sin(th1-th2)+(g/l)*sin(th2))
end
使: