Julia Can'；我似乎无法在朱莉娅身上取得一致。你能给我一些建议吗？

我试图用Julia解决经济学中的一个生命周期问题，但我在NLsolve上遇到了麻烦。该模型归结为试图求解两个方程组，以找到每个工作周期的最佳休闲时间和资本存量。退休后的经济主体将休闲设置为1，我只需要解一个关于资本的非线性方程。这部分很好用。这是解决两个方程组，似乎要崩溃

一般来说，我对朱莉娅/编程相当陌生，所以任何建议都会非常有用。此外，对于本准则各方面的建议/要点/建议，我们将不胜感激。模型从最后一个时间段向后求解

我的尝试

using Parameters
using Roots
using Plots
using NLsolve
using ForwardDiff

Model = @with_kw (α = 0.66,
                  δ = 0.02,
                  τ = 0.015,
                  β = 1/1.01,
                  T = 70,
                  Ret = 40,
                  );

function du_c(c, l, η=2, γ=2)
    if c>0 && l>0
        return (c+1e-6)^(-η) * l^((1-η)*γ)
    else
        return Inf
    end
end

function du_l(c, l, η=2, γ=2)
    if l>0 && c>0
        return γ * (c+1e-6)^(1-η) * l^(γ*(1-η)-1)
    else
        return Inf
    end
end

function create_euler_work(x, y, m, k, l, r, w, t)
    # x = todays capital, y = leisure
    @unpack α, β, τ, δ, T, Ret = m
    c_1 = x*(1+r) + (1-τ)*w*(1-y) - k[t+1]
    c_2 = k[t+1]*(1+r) + (1-τ)*w*(1-l[t+1]) - k[t+2]
    return du_c(c_1,y) - β*(1+r)*du_c(c_2,l[t+1])
end

function create_euler_retire(x, m, k, r, b, t)
    # Holds at time periods Ret onwards 
    @unpack α, β, τ, δ, T, Ret = m
    c_1 = x*(1+r) + b - k[t+1]
    c_2 = k[t+1]*(1+r) + b - k[t+2]
    return du_c(c_1,1) - β*(1+r)*du_c(c_2,1)
end

function create_euler_lyw(x, y, m, k, r, w, b, t)
    # x = todays capital, y = leisure
    @unpack α, β, τ, δ, T, Ret = m
    c_1 = x*(1+r) + (1-τ)*w*(1-y) - k[t+1]
    c_2 = k[t+1]*(1+r) + b - k[t+2]
    return du_c(c_1,y) - β*(1+r)*du_c(c_2,1)
end


function create_foc(x, y, m, k, r, w, t)
    # x = todays capital, l= leisure
    @unpack α, β, τ, δ, T = m
    c = x*(1+r) + (1-τ)*w*(1-y) - k[t+1]
    return du_l(c,y) - (1-τ)*w*du_c(c,y)
end


function life_cycle(m, guess, r, w, b, initial)
    @unpack α, β, τ, δ, T, Ret = m
    k = zeros(T+1);
    l = zeros(T);
    k[T] = guess

    println("Period t = $(T+1) Retirment, k = $(k[T+1]), l.0 = NA")
    println("Period t = $T Retirment, k = $(k[T]), l = 1.0")
    ########################## Retirment ################################

    for t in T-1:-1:Ret+1 
        euler(x) = create_euler_retire(x, m, k, r, b, t)
        k[t] = find_zero(euler, (0,100))
        l[t] = 1
        println("Period t = $t Retirment, k = $(k[t]), l = $(l[t])")
    end 
    ###################### Retirement Year  #############################
    for t in Ret:Ret
        euler(x,y) = create_euler_lyw(x, y, m, k, r, w, b, t)
        foc(x,y) = create_foc(x, y, m, k, r, w, t)

        function f!(F, x)
            F[1] = euler(x[1], x[2])
            F[2] = foc(x[1], x[2])
        end

        res = nlsolve(f!, [5; 0.7], autodiff = :forward)
        k[t] = res.zero[1]
        l[t] = res.zero[2]
        println("Period t = $t Working, k = $(k[t]), l = $(l[t])")
    end 
    ############################ Working ###############################
    for t in Ret-1:-1:1
        euler(x,y) = create_euler_work(x, y, m, k, l, r, w, t)
        foc(x,y) = create_foc(x, y, m, k, r, w, t)

        function f!(F, x)
            F[1] = euler(x[1], x[2])
            F[2] = foc(x[1], x[2])
        end

        res = nlsolve(f!, [5; 0.7], autodiff = :forward)
        k[t] = res.zero[1]
        l[t] = res.zero[2]
        println("Period t = $t Working, k = $(k[t]), l = $(l[t])")
    end
    #####################################################################
    return k[1] - initial, k, l
end


m = Model();

residual, k, l = life_cycle(m, 0.3, 0.03, 1.0, 0.0, 0.0)

代码似乎在第35周期中断，错误是“在非线性系统解析过程中，对以下方程的评估导致了一个非有限数：[1,2]”，但在第37周期，解决方案似乎变得怪异