Saddle-node bifurcations in an optimal growth model with preferences for wealth habit

Appendix

We prove Proposition 2 and Proposition 3 after providing some lemmas.

Lemma 1 Let k be an optimal path from k₀>0. Then, there cannot be an integer T such that γk_t=k_t+1 for all t≥T.

Proof: Let k be an optimal path from k₀>0. Assume that there exists such T. Since k_t→0, under the Assumption 3 there exists an integer T′≥T such that βf′(kT′+1)>1. The positivity of the optimal consumption implies that k_t+1<f(k_t) for all t so that there exists ε>0 small enough such that

γkt<(1+ε)kt+1≤f(kt),∀t≥T′.

Define k˜ as k˜t=kt for t=1, …, T′ and k˜t=(1+ε)kt for t≥T′+1. It is feasible as we have:

γk˜t=k˜t+1 and k˜t+1=(1+ε)kt+1≤f(kt)<f(k˜t), for t≥T′+1.

Next, we show that k˜ dominates k for some ε small enough. Define Δ(ε)=U(k˜)−U(k). By setting f^(k)=f(k)−γk, we have:

Δ(ε)=βT′[u(f^(kT′)−γεkT′)−u(f^(kT′))]+βT′ηw(γεkT′)+βT′+1[u(f˜((1+ε)kT′+1))−u(f˜(kT′+1)]+∑τ>T′+1+∞βτ[u(f˜((1+ε)kτ))−u(f˜(kτ)]

Since the second and the last terms are positive and u, f are concave and differentiable we get

Δ(ε)βT′>[u(f^(kT′)−γεkT′)−u(f^(kT′))]+β[u(f^((1+ε)kT′+1))−u(f^(kT′+1)].

As ε→0, we have:

Δ(ε)βT′>γkT′[−u′(f^(kT′))+βu′(f^(kT′+1))f^′(kT′+1)]>0,

which contradicts with the optimality of k.              ■

Lemma 2 Let k be an optimal path from k₀>0. Then, there exists an integer T such that γk_t<k_t+1 for all t≥T.

Proof: Let k be an optimal path from k₀>0. Assume on the contrary that for any T there exists T′≥T such that γkT′−1=kT′. Note that T′ can be chosen so that γkT′<kT′+1 and βf′(kT′)>1 by Lemma 1 and by the fact that k_t→0.

The positivity of the optimal consumption implies that kT′<f(kT′−1) for all t so that there is ε>0 small enough to verify that

kT′+ε<f(kT′−1) and γ(kT′+ε)<kT′+1.

Let k˜ be a feasible sequence defined as

k˜t=kt for t≠T′ and k˜T′=kT′+ε.

Let us define Δ:ℝ+→ℝ+ by

Δ(ε)=u(f(kT′−1)−kT′−ε)+ηw(ε)+βu(f(kT′+ε)−kT′+1)+βηw(kT′+1−γ(kT′+ε)).

Differentiating Δ(ε) with respect to ε and evaluating at ε=0, we obtain that

Δ′(0)>−u′(f(kT′−1)−kT′−ε)+βu′(f(kT′+ε)−kT′+1)f′(kT′)≥0,

contradicting with the fact that Δ(ε) must have a maximum at ε=0.           ■

Lemma 3 Let k be an optimal path from k₀>0. Then, there exists an integer T such that

ηw′ (kt+1−γkt)−βηγw′ (kt+2−γkt+1)≥0, ∀t≥T.

Proof: Let k be an optimal path from k₀>0. By Assumption 3, k_t∈[0, A(k₀)] for all t and by Proposition 2-(ii), k is monotonic. Therefore, k must converge to some k^ss. Assume on the contrary that for any integer T there exists τ≥T such that

ηw′ (kτ+1−γkτ)−βηγw′ (kτ+2−γkτ+1)<0.

As τ→+∞, we have

η(1−βγ)w′ (δkSS)>0,

which contradicts with the continuity of w′.                 ■