Model Solution to equations 1. f$_{x}$ ∗ = F t − 1 ( φ − 1 u φ$_{z}$ ) . 2. f$_{x}$ ∗ = F t − 1 ( φ$_{−}$$_{u}$φ$_{z}$ ) . 3. Under Assumption A f$_{x}$ ∗ = F t − $^{1}$(exp ∫ ζ ζ$_{0}$ d ∑ k =1$^{κ}$ $_{k}$( ξ ) dξ ) , where $^{κ}$k solves κ $_{k}$φ$_{z}$ − [( φ$_{z}$ ) ′ k − ε$_{k}$ ] = 0; f$_{u}$ = F t − $^{1}$( φ − 1 x ∗ ε ) . Under Assumption B f$_{u}$ = F t − $^{1}$(exp ∫ ζ ζ$_{0}$ d ∑ k =1$^{κ}$ $_{k}$( ξ ) dξ ); $^{κ}$k solves κ $_{k}$φ$_{z}$ − ε$_{k}$ = 0; f$_{x}$ ∗ = F t − $^{1}$( φ − 1 u ε ) . 4 f$_{x}$ ∗ , f$_{u}$ obtained similarly to those in 3.; φ$_{u}$$_{x}$ = φ − 1 x ∗ φ$_{x}$. 4a. f$_{u}$$_{x}$, f$_{u}$ obtained similarly to φ$_{x}$ ∗ , φ$_{u}$ in 3.; φ$_{x}$ ∗ = φ − 1 u x φ$_{x}$. 5. Three steps: 1. (a) Get F t ( gf$_{x}$ ∗ ) , φ$_{u}$ similarly to φ$_{x}$ ∗ , φ$_{u}$ in model 3 (under Assumption A use F t ( gf$_{x}$ ∗ )(0)); 2. Obtain φ$_{x}$ ∗ = φ − 1 u φ$_{z}$ ; 3. Get g = [ F t − $^{1}$( φ$_{x}$ ∗ )] − 1 F t − $^{1}$( F t ( gf$_{x}$ ∗ )) . 6. φ$_{−}$$_{u}$ = φ − 1 z φ$_{x}$ and g = F t − $^{1}$( φ − 1 x φ$_{z}$ε ) . 7. φ$_{x}$ ∗ , F t ( g )obtained similarly to φ$_{x}$ ∗ , φ$_{u}$ in 3 (under Assumption A use F t ( g )(0)) . 45