在sympy中用隐函数求解KKT方程

如何解决在sympy中用隐函数求解KKT方程

我是 SymPy 的新手，想解决以下优化问题的 KKT 方程：

\begin{对齐*} &\max_{L_A,L_C,X_a,X_{na},s,u} U(L_A,X_{na})=(\hat{\imath}-L_C)(X_a-S_a)^{ \alpha_a}(X_{na}-S_{na})^{\alpha_{na}}(I-\hat{\imath}-L_A)^{\alpha_{A}}(\hat{\imath}- L_C)^{\alpha_C}\ st\quad& p_a X_a + p_{na} X_{na} + p_A (I-\hat{\imath}-L_A)+p_C (\hat{\imath}-L_C) = Y = p_a f(L,\theta,K)-p_A L_A - p_C L_C+p_A(I-\hat{\imath})+p_C(\hat{\imath}) \end{对齐*}

约束中有一个隐式函数（我的意思是 $f(L,K)$ 的实际形式是未知的）。由假设，该函数是正单调凹函数，即 $$\frac{\partial f}{\partial L}>0 \quad \frac{\partial f}{\partial \theta}>0 \quad \frac{\partial^2 f}{\partial L^2 }0.$$

为了使事情复杂化，函数 $f(L,K)$ 的输入 $L$ 有一个明确的形式，因为它涉及六个选择变量中的四个：

$$L(L_C,L_A,u,s) = L_A - s u L_C + s v(u) L_C + (1-s) v_0 L_C$$

其中 v(u) 又是正单调和凹面。

下面是我在 SymPy 中实现它的尝试。我在求解阶段（最后）遇到一个错误，我认为这是 SymPy 不知道如何取隐函数 ($f(\cdot)$) 与另一个函数 ($L(\ cdot)$) 本身定义。

我如何告诉 SymPy $f(\cdot)$ 应该是正单调和凹面？

任何关于如何进行的想法将不胜感激。

    # import stuff
from sympy.interactive import printing
printing.init_printing(use_latex=True)
from sympy import Function
from sympy.solvers import solve
import sympy as sp

# define the maximization variables
X1,X2,X3,X4,LC,LA,u = sp.var('X_1,X_2,X_3,X_4,u',positive=True);

# import exogenous variables

a1 = sp.symbols('alpha_1') # alphas
a2 = sp.symbols('alpha_2')
a3 = sp.symbols('alpha_3')
a4 = sp.symbols('alpha_4')

S1 = sp.symbols('S_1') # Subsistence levels
S2 = sp.symbols('S_2')
S3 = sp.symbols('S_3')
S4 = sp.symbols('S_4')

v0 = sp.symbols('v_0') # raw skill level

i = sp.symbols('\hat{\imath}') # children in the household - time endowment
I = sp.symbols('I') # household size and total time endowment

p1 = sp.symbols('p_1') # price vector
p2 = sp.symbols('p_2')
p3 = sp.symbols('p_3')
p4 = sp.symbols('p_4')

climate = sp.symbols('theta') # climate variable
capital = sp.symbols('K') # quasi fixed land and (non-human) capital

lam = sp.symbols('\lambda') # shadow prices (lagrange multiplier)

# technology of skill enhancement
v = Function('v')(u)

# labour equivalence function (cases 1 and 2)
L1 = Function('L_1')(u,LC)
L1 = LA-s * u * LC + s * v * LC + (1 - s) * v0 * LC
L2 = Function('L_2')(u,LC)
L2 = v * (LA/u)+v0*(LC-LA/u)

# define production function
f1 = Function('f')(L1,capital,climate) # positive monotonic and convex in L1,and climate
f2 = Function('f')(L2,climate)

# utility function (cases 1 and 2)
U1 = Function ('U_1')(X1,LC)
U2 = Function ('U_2')(X1,LC)
U1 = (i-LC) * (X1-S1)**a1 *(X2-S2)**a2 * (I-i-LA-S3)**a3 * (i-LC-S4)**a4 
U2 =          (X1-S1)**a1 *(X2-S2)**a2 * (I-i-LA-S3)**a3 * (i-LC-S4)**a4

# main constraints
#G1 = Function('G_1')(X1,f1)
G1 = p1*X1 + p2*X2 + p3*I - p3*i -p3*LA + p4*i - p4*LC - p1*f1 + p3*LA + p4*LC - p3*I-p3*i - p4*i
#G1 = Function('G_1')(X1,f2)
G2 = p1*X1 + p2*X2 + p3*I - p3*i -p3*LA + p4*i - p4*LC - p1*f2 + p3*LA + p4*LC - p3*I-p3*i - p4*i

# Lagrangians
Lagrange1 = U1 - lam * G1 # case 1
Lagrange2 = U2 - lam * G1 # case 2
Lagrange3 = U1 - lam * G2 # case 3
Lagrange4 = U2 - lam * G2 # case 4

# Calculate KKT-conditions
grad1 = [sp.diff(Lagrange1,c) for c in [X1,u]] # gradient of Lagrangian w.r.t choice variables
KKT1 = grad1 + [G1]

grad2 = [sp.diff(Lagrange2,u]] # gradient of Lagrangian w.r.t choice variables
KKT2 = grad2 + [G1]

grad3 = [sp.diff(Lagrange3,u]] # gradient of Lagrangian w.r.t choice variables
KKT3 = grad3 + [G2]

grad4 = [sp.diff(Lagrange4,u]] # gradient of Lagrangian w.r.t choice variables
KKT4 = grad4 + [G2]

### Here the thing breaks down - perhaps because the convexity of f1 is not specified.
########################################################################################################
candidates1 = sp.solve(KKT1,[X1,lam],dict=True) # solve the KKT equations #
candidates2 = sp.solve(KKT2,dict=True) # solve the KKT equations #
candidates3 = sp.solve(KKT3,dict=True) # solve the KKT equations #
candidates4 = sp.solve(KKT4,dict=True) # solve the KKT equations #
########################################################################################################


# return candidates for optima
print(candidates)
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