Constraint effect of enterprise productivity based on constrained form variational computing

Based on the analysis of related concepts of formal variational calculation, it can be seen that if {u_i (x),...,u_N (x)} is a set of independent functions of X, and A represents the polynomial algebra of ua(i)≡Diua u_a^{\left( i \right)} \equiv D^i u_a on complex domain C, then D = d/dz can be regarded as the total derivative operator, and it can be obtained as A^N = {f | f = (f₁,..., f_N), f_i ∈ A}

At this point, the vector fields corresponding to all f ∈ A^N are: ∂f=∑fa(i)∂/∂ua(i),fa(i)=Difa \partial _f = \sum f_a^{\left( i \right)} \partial /\partial u_a^{\left( i \right)} ,f_a^{\left( i \right)} = D^i f_a

In particular, under the condition that D_u = (Du_i) corresponds to ∂Du=∑astua(i+1)∂/∂ua(i)=d/df=D \partial _{Du} = \sum\limits_{a_s t} u_a^{\left( {i + 1} \right)} \partial /\partial u_a^{\left( i \right)} = d/df = D we can get: ∂jg=∑ fa(i)∂g/∂ua(i)=(dg(u+εf)/dε)ε=0[ ∂f,∂g ]=∂[ f,g ], \begin{array}{*{20}c} {\partial _j g = \sum f_a^{\left( i \right)} \partial g/\partial u_a^{\left( i \right)} = \left( {dg\left( {u + \varepsilon f} \right)/d\varepsilon } \right)_{\varepsilon = 0} } \hfill \\ {\left[ {\partial _f ,\partial _g } \right] = \partial _{\left[ {f,g} \right]} ,} \hfill \\ \end{array} Thus, represents the commutator [∂f, ∂_g], and [f, g]_a = ∂_fga − [∂_f, ∂_g] = ∂_{[f, g]}, ∂_gf_a.

Open Problem 8 . Combined with the above formula analysis, it can be seen that the vector field set Π = ∂f{|f ∈ A^N} is closed in the commutative suboperation and can form the Lie algebra. The definition is shown below, where W refers to linearity in relation to AI: q−theshape:C0(Π,A)=A,Cq(Π,A)={ω(a1,...,aq)|a1∈Π,ω(a1,...,aq)∈A}External Differential: d:Cq→Cq+1,(dω)(a1,...,aq+1)=∑i(df)a=af,f∈C0,(−1)i+1aiω(a1...∂i...aq+1)+∑i<f(−1)i+jω([ai,aj],a1,...,∂i,...,∂j,...,aq+1) \begin{gathered} \begin{array}{*{20}c} {q - theshape:C^0 \left( {\Pi ,A} \right) = A,} \hfill \\ {C^q \left( {\Pi ,A} \right) = \left\{ {\omega \left( {a_1 ,...,a_q } \right)\left| {a_1 \in \Pi ,\omega \left( {a_1 ,...,a_q } \right) \in A} \right.} \right\}} \hfill \\ \end{array} \hfill \\ External{\text{ }}Differential:\,\begin{array}{*{20}c} {d:C^q \to C^{q + 1} ,} \hfill \\ {\left( {d\omega } \right)\left( {a_1 ,...,a_{q + 1} } \right) = \sum\limits_i^{\left( {df} \right)a = af,f \in C^0 ,} \left( { - 1} \right)^{i + 1} a_i \omega \left( {a_1 ...\partial _i ...a_{q + 1} } \right)} \hfill \\ { + \sum\limits_{i < f} \left( { - 1} \right)^{i + j} \omega \left( {\left[ {a_i ,a_j } \right],a_1 ,...,\partial _i ,...,\partial _j ,...,a_{q + 1} } \right)} \hfill \\ \end{array} \hfill \\ \end{gathered} where A represents that the corresponding term does not appear.

Inner product: ia:Cq+1→Cq,iaf=0,f∈C0,(iaω)(a1...aq)=ω(a,a1,...,aq) \begin{array}{*{20}c} {i_a :C^{q + 1} \to C^q ,} \hfill \\ {i_a f = 0,f \in C^0 ,} \hfill \\ {\left( {i_a \omega } \right)\left( {a_1 ...a_q } \right) = \omega \left( {a,a_1 ,...,a_q } \right)} \hfill \\ \end{array}

Li derivative: L_a = i_ad + di_aQ^q (A) = {ω|ω ∈ C^q, ω (fa₁, a₂,...,a_q) = fω(a₁,...,a_q), f ∈ A

At this point, Q-shaped infinity is equivalent to 0 and can be regarded as infinity ~ 0. Assuming ω₁ ∈ Q^q and ω = L_Dω₁, then we can get Q0q={ω|ω∈Qq,ω∼0} Q_0^q = \left\{ {\omega \left| {\omega \in Q^q ,\omega \sim {\text{0}}} \right.} \right\} and Q˜q=Qq/Q0q \tilde Q^q = Q^q /Q_0^q . And when q is equal to 0, you get QC0=DA={ Df|f∈A } Q_C^0 = DA = \left\{ {Df\left| {f \in A} \right.} \right\} and Q˜0=A/DA \tilde Q^0 = A/DA . The middle element is called functional Q˜0 \tilde Q^0 , and its element has Dhdef__∫fdx Dh\underline {\underline {def} } \int fdx , which can be written as (f,g)≡∫∑afagadx,f,g∈AN \left( {f,g} \right) \equiv \int \sum\limits_a f_a g_a dx,f,g \in A^N •

The usual form of the element in Q’(A) is: ω=∑aiωaidua(i)∼ω^=∑aξadua,ξa=∑i(−D)iωa*i \omega = \sum\limits_{ai} \omega _a^i du_a^{\left( i \right)} \sim \hat \omega = \sum\limits_a \xi _a du_a ,\xi _a = \sum\limits_i \left( { - D} \right)^i \omega _{a*}^i

If w = df, then we can get df=∑ai(∂f/∂ua(i))dua(i)∼∑a(∂f/∂uα)dua df = \sum\nolimits_{ai} (\partial f/\partial u_a^{(i)} )du_a^{(i)} \sim \sum\nolimits_a (\partial f/\partial u_\alpha )du_a

And we can specify δf/δua=∑i(−1)i(∂f/∂ua(i)) \delta f/\delta u_a = \sum\limits_i \left( { - 1} \right)^i \left( {\partial f/\partial u_a^{\left( i \right)} } \right) , which is also called the variational derivative of F with respect to Uα. According to f ~ g, δf/δu ⇒ δf/δu can be calculated, so ∂f/∂u ⇒ ∂f/∂u can be defined. Similarly, df^=df˜ d\hat f = d\tilde f can be obtained by combining f ~ g → df ~ dg and defining d˜f=∫∑a(δf/δua)dua \tilde df = \int \sum\limits_a \left( {\delta f/\delta u_a } \right)du_a . The final derivation is given as ∂h∫fdx=∫(∑a(δf/δua)ha)dx=(δf/δu,h) \partial _h \int fdx = \int \left( {\sum\limits_a \left( {\delta f/\delta u_a } \right)h_a } \right)dx = \left( {\delta f/\delta u,h} \right)

By constructing the equivalence between the Nash equilibrium model and the problem of a variational inequality, we can get K={(T,E,l,θ,ξ)|T∈R+mn,E∈R+m,l∈R+mδ,θ∈R+m,ξ∈R+m,hikEi≤lik;Gi(Ei)≤g(θi,Ti)} K = \left\{ {\left( {T,E,l,\theta ,\xi } \right)\left| {T \in R_ + ^{mn} ,E \in R_ + ^m ,l \in R_ + ^{m\delta } ,\theta \in R_ + ^m ,\xi \in R_ + ^m ,h_{ik} E_i \leqslant l_{ik} ;G_i \left( {E_i } \right) \leqslant g\left( {\theta _i ,T_i } \right)} \right.} \right\}

And let u = (u₁, u₂,...,u_m), then K → Rm belongs to a concave mapping that is continuously differentiable, and let (T^*, E^*, l^*, θ^*, ξ^*) ∈ K, then let all i ∈ {1, 2,...,m}, then (T_i, E_i, l_i, θ_i, ξ_i) ∈ K_i must satisfy the following conditions:

Equivalent to specifying (T^*, E^*, l^*, θ^* , ξ^* ) ∈ K, let the variational inequality as shown below hold, and then we get: ∑i=1m∑j=1n〈∂Tij∂fi(Ti*,θi*,ξi*)−Pj(T*)−∑g=1n∂Tij∂Pg(T*)Tig*,Tij−Tij*〉+∑i=1m∂Ei∂Gi(Ei*),Ei−Ei*〉+∑i=1m∑k=1δ〈P′k,lik−lik*〉+∑i=1m〈∂θi∂fi(Ti*,θi*,ξi*)+1,θi−θi*〉+∑i=1m〈∂ξi∂fi(Ti*,θi*,ξi*)+1,ξi−ξi*〉≥0∀(T,E,I,θ,ξ)∈K \begin{array}{*{20}c} {\sum\limits_{i = 1}^m \sum\limits_{j = 1}^n \left\langle {{\partial T_{ij} }^{\partial f_i \left( {T_i^* ,\theta i^ ,\xi _i^ } \right)} - P_j \left( {T^* } \right) - } \right.} \hfill \\ {\left. {\left. {\sum\limits_{g = 1}^n \begin{array}{*{20}c} {\partial P_g \left( {T^* } \right)} \\ {\partial T_{ij} } \\ \end{array} T_{ig}^ ,T_{ij} - T_{ij}^ } \right\rangle + \sum\limits_{i = 1}^m \begin{array}{*{20}c} {\partial G_i \left( {E_i^* } \right)} \\ {\partial E_i } \\ \end{array} ,E_i - E_i^* } \right\rangle + } \hfill \\ {\sum\limits_{i = 1}^m \sum\limits_{k = 1}^\delta \left\langle {P'_k ,l_{ik} - l_{ik}^ } \right\rangle + \sum\limits_{i = 1}^m \left\langle {{\partial \theta i }^{\partial f_i \left( {T_i^ ,\theta i^* ,\xi i^ } \right)} + 1,\theta i - \theta i^ } \right\rangle + } \hfill \\ {\sum\limits_{i = 1}^m \left\langle {{\partial \xi i }^{\partial f_i \left( {T_i^ ,\theta _i^ ,\xi i^* } \right)} + 1,\xi i - \xi _i^* } \right\rangle \geqslant 0} \hfill \\ {\forall \left( {T,E,I,\theta ,\xi } \right) \in K} \hfill \\ \end{array}

Also, clarify the following requirements for K={(T,E,l,θ,ξ)|T∈R+mn,E∈R+m,l∈R+mδ,θ∈R+m,‖T‖≤t0,‖E‖≤f0,‖l‖≤l0,‖θ‖≤θ0,‖ξ‖≤ξ0hikEi≤lik;Gi(Ei)≤g(θi,Ti)} \begin{array}{*{20}c} {K = \left\{ {\left( {T,E,l,\theta ,\xi } \right)} \right.\left| {T \in R_ + ^{mn} ,E \in R_ + ^m ,l \in R_ + ^{m\delta } ,\theta \in R_ + ^m ,} \right.} \hfill \\ {\left\| T \right\| \leqslant t_0 ,\left\| E \right\| \leqslant f_0 ,\left\| l \right\| \leqslant l_0 ,\left\| \theta \right\| \leqslant \theta _0 ,\left\| \xi \right\| \leqslant \xi _0 } \hfill \\ {h_{ik} E_i \leqslant l_{ik} ;G_i \left( {E_i } \right) \leqslant g\left. {\left( {\theta _i ,T_i } \right)} \right\}} \hfill \\ \end{array}

Then, in u = (u₁, u₂,...,u_m), K→ Rm belongs to the concave mapping of the first continuous partial derivative. G_i with respect to E_i has continuous rows and is convex, and G with respect to (θ_i, ξ_i) also has this characteristic. Therefore, it is proved that the inequality obtained by Theorem 1 above has a Nash equilibrium point. In this process, the marginal cost will increase with the increase of the quantity of products, scientific research funds and the cost of factor allocation efficiency, and the marginal pollution cost of all enterprises will also increase with the increase of the amount of pollution discharged. The regression model designed in this paper is as follows: tei=a0+a1fc1+a3X3+∂......................model1tei=a0+a1fc1+a2int1i+a3X3+∂.......model2tei=a0+a1fc1+a2int2i+a3X3+∂......model3tei=a0+a1fc1+a2int3i+a3X3+∂......model4tei=a0+a1fc1+a2int4i+a3X3+∂......model5 \begin{array}{*{20}c} {te_i = a_0 + a_1 fc_1 + a_3 X_3 + \partial ......................model1} \hfill \\ {te_i = a_0 + a_1 fc_1 + a_2 \operatorname{int} 1_i + a_3 X_3 + \partial .......model2} \hfill \\ {te_i = a_0 + a_1 fc_1 + a_2 \operatorname{int} 2_i + a_3 X_3 + \partial ......model3} \hfill \\ {te_i = a_0 + a_1 fc_1 + a_2 \operatorname{int} 3_i + a_3 X_3 + \partial ......model4} \hfill \\ {te_i = a_0 + a_1 fc_1 + a_2 \operatorname{int} 4_i + a_3 X_3 + \partial ......model5} \hfill \\ \end{array}

Model 1 is mainly used to study the effect of financing constraints on the production efficiency, model 2 is mainly used to study the effect of R&D financing constraints on the production efficiency, model 2 is a measure of financing constraints on investment – cash flow sensitivity of the influence on the production efficiency, model 4 is used to evaluate the financing constraint in the act of export influence on production efficiency and model 5 is used to analyse the impact of financing constraints on productivity in employee compensation in Pingliang. Meanwhile, the corresponding control variable set formula is as follows:

(1) Xi=y1sizei+y2alri+y3deni+y4subi+y5taxi X_i = y_1 size_i + y_2 alr_i + y_3 den_i + y_4 sub_i + y_5 tax_i

Among them, a₁, a₂, a₃ and y₁ to y₅, respectively, represent the influence levels of various control variables on production efficiency. For example, a₁ refers to the impact level of financing constraints on the production efficiency of enterprises in R&D expenditure. If this value is positive, it means that financing constraints will have a negative impact on production efficiency, which requires the market to relax constraints. On the contrary, if the value is negative, the impact of financing constraints on production efficiency is positive. In this case, it is necessary to strengthen the practice of financing constraints so as to improve the actual production efficiency of enterprises. Table 2 shows the indicators and measurement methods to be defined in this study. Table 2

Specific indicators and measurement methods