如何修复TypeError:在setindex中!在DifferentialEquations.jl中

时间:2019-01-29 13:21:46

标签: julia typeerror ode differential-equations

最近,我开始使用Julia的(v1.0.3)DifferentialEquations.jl软件包。我尝试解决一个简单的ODE系统,该系统的结构与我的真实模型相同,但更小。 该示例根据我使用的求解器来解决或引发错误。考虑一下这个MWE,它是CSTR中连续/平行反应的化学工程模型:

using DifferentialEquations
using Plots

# Modeling a consecutive / parallel reaction in a CSTR
# A --> 2B --> C, C --> 2B, B --> D
# PETERSEN-Matrix
#   No.     A       B       C       D       Rate
#   1      -1       2                       k1*A
#   2              -2       1               k2*B*B
#   3               2      -1               k3*C
#   4              -1               1       k4*B

function fpr(dx, x, params, t)
    k_1, k_2, k_3, k_4, q_in, V_liq, A_in, B_in, C_in, D_in = params
    # Rate equations
    rate = Array{Float64}(undef, 4)
    rate[1] = k_1*x[1]
    rate[2] = k_2*x[2]*x[2]
    rate[3] = k_3*x[3]
    rate[4] = k_4*x[2]

    dx[1] = -rate[1] + q_in/V_liq*(A_in - x[1])
    dx[2] = 2*rate[1] - 2*rate[2] + 2*rate[3] - rate[4] + q_in/V_liq*(B_in - x[2])
    dx[3] = rate[2] - rate[3] + q_in/V_liq*(C_in - x[3])
    dx[4] = rate[4] + q_in/V_liq*(D_in - x[4])
end 

u0 = [1.5, 0.1, 0, 0]
params = [1.0, 1.5, 0.75, 0.15, 3, 15, 0.5, 0, 0, 0]
tspan = (0.0, 15.0)
prob = ODEProblem(fpr, u0, tspan, params)
sol = solve(prob)
plot(sol)

这很好用。 但是,如果选择其他求解器,例如Rosenbrock23()Rodas4(),则ODE无法解决,并且出现以下错误:

ERROR: LoadError: TypeError: in setindex!, in typeassert, expected Float64,
got ForwardDiff.Dual{Nothing,Float64,4}

我不会在此处粘贴整个堆栈跟踪,因为它很长,但是您可以通过将sol = solve(prob)更改为sol = solve(prob, Rosenbrock23())来轻松地重现该堆栈跟踪。在我看来,当求解器尝试导出Jacobian时会发生错误,但是我不知道为什么。以及为什么默认求解器可以工作,而其他求解器却不能工作?

请,有人可以告诉我为什么会发生此错误以及如何解决该错误吗?

1 个答案:

答案 0 :(得分:2)

自动微分的工作原理是通过函数传递Dual类型,而不是通常使用的浮点数。因此出现问题是因为您将内部值rate固定为类型Vector{Float64}(请参见第三点herethis advice)。幸运的是,这很容易解决(恕我直言,甚至更好看):

julia> function fpr(dx, x, params, t)
           k_1, k_2, k_3, k_4, q_in, V_liq, A_in, B_in, C_in, D_in = params
           # Rate equations
           # should actually be rate = [k_1*x[1], k_2*x[2]*x[2], k_3*x[3], k_4*x[2]], as per @LutzL's comment
           rate = [k_1*x[1], k_2*x[2], k_3*x[3], k_4*x[2]]

           dx[1] = -rate[1] + q_in/V_liq*(A_in - x[1])
           dx[2] = 2*rate[1] - 2*rate[2] + 2*rate[3] - rate[4] + q_in/V_liq*(B_in - x[2])
           dx[3] = rate[2] - rate[3] + q_in/V_liq*(C_in - x[3])
           dx[4] = rate[4] + q_in/V_liq*(D_in - x[4])
       end

Rosenbrock23Rodas4都适用。

或者,您可以使用Rosenbrock23(autodiff=false)(我认为它将使用有限差分代替)关闭AD,或提供雅可比行列式。