Irreversibility and the economics of forest conservation

Abstract

Regenerating forest on land used for non-forest economic activities can be difficult; this introduces some irreversibility in the process of deforestation. We analyze the effect of such irreversibility (reforestation cost) on the efficiency of forest conservation in a general model of optimal forest management where trees are classified in age classes and land has alternative economic use. Irreversibility may lead to a continuum of optimal steady states that differ in the area under forest cover; increase in irreversibility can only add steady states with smaller forest cover. High irreversibility discourages expansion of forests but at the same time, makes it optimal to conserve a minimal forested area in the long run; in particular, it is optimal to maintain a forested area above a critical size if the initial forest cover lies above it, while forests that are initially smaller than the critical level are optimally managed at constant size. We characterize the exact condition under which it is optimal to avoid total deforestation; the extent of irreversibility does not matter for this. Weak irreversibility may be associated with cyclical fluctuations in optimal forest cover; we characterize upper and lower bounds on the forest cover along an optimal path.

Appendices

Appendix A: Results without strict concavity of u(x)

In this section, we briefly outline modified versions of some of our results that hold when $A_{2}$ is replaced by the a weaker assumption on u:

$A^{\prime }_{2}$::

u is concave in ${\mathbb {R}}_+$.²⁴

In the absence of strict concavity of both u and w, we can no longer assert uniqueness of optimal program. The possibility of having non-unique optimal programs requires us to modify almost all the definitions and statements of this paper. Instead of doing this, we will illustrate the role of strict concavity for the monotonicity and convergence of the optimal path by continuing to assume that there is a unique optimal path from every initial state. Under this assumption, we describe how the results of Sects. 3 and 5 are modified when $A_{2}$ is replaced by $A_{2}^{\prime }$. (Results in other sections are not substantively modified.)

1.1 Alternative results of Sect. 3

In the absence of $A_2$, the definitions of $z_0$ and $z_{\delta }$ have to be adapted to the fact that the equations $\varDelta (x)=0$ and $\varDelta (x)=\delta$ may not define a unique solution:

$$\begin{aligned}&z_{0}=\min \{z\in [0,1]:\varDelta (z)=0\} \\&z_{\delta }= \max \{z\in [0,1] :\varDelta (z)=\delta \} \end{aligned}$$

Proposition 1 has a weaker version where 1. and 2. must be modified.

Proposition 1’

Under $A_1$, $A^{\prime }_2$, $A_3$ and $A_4$, the set of oss coincides with $\{{{\varvec{x}}}(z) / z\in [z_{\delta },z_0]\}$, i.e., the set of states with forest cover $z\in [z_{\delta }, z_0]$ where the forest area z is evenly divided among trees of age classes 1 through $\sigma .$ In particular,

• 1. If $\varDelta (0)<0,$ then $I(\delta )=\{0\}$ for all $\delta \ge 0$, i.e., regardless of the extent of irreversibility (cost of reforestation) the totally deforested state is the unique oss.

• 2. If $\varDelta (1)> \delta ,$ then $I(\delta )=\{1\}$, i.e., there is a unique oss and it is characterized by full forest cover.

• 3. If $\delta>0, \varDelta (0)>0$ and $\varDelta (1)<\delta ,$ then $z_{\delta }<z_{0}$ and there is a continuum of oss; the set of forest covers that can be sustained in an oss is a closed interval $[z_{\delta },z_{0}]$ with non-empty interior. In addition, $I(\delta )=\{{z}\in [0,1]:0\le \varDelta ({z})\le \delta \}$.

And, finally, the statement of Corollary 4 has to be modified (because certain properties are no longer equivalent).

Corollary 4’

• 1. An oss with positive forest cover exists if $\varDelta (0)>0$.

• 2. Every oss has positive forest cover if, and only if, $\varDelta (0)>\delta$. Equivalently, the total deforestation state, ${{{\varvec{x}}}}(0)$, is an oss if, and only if, $\varDelta (0)\le \delta$.

• 3. A steady state ${{{\varvec{x}}}}(1)$ with full forest cover is an oss if, and only if, $\varDelta (1)\ge 0$. A sufficient condition for uniqueness is $\varDelta (1)>\delta$.

1.2 Alternative results of Sect. 5

Section 5 characterizes necessary and sufficient conditions to have global weak conservation. When $A_{2}$ is replaced by $A_{2}^{\prime }$, the necessary condition is weaker.

Proposition 4’

Under $A_1$, $A^{\prime }_2$, $A_3$ and $A_4$

$$\begin{aligned} \varDelta (0)>0 \Rightarrow \text {There is global weak conservation} \Rightarrow \varDelta (0)\ge 0 \end{aligned}$$

Analogously, the equivalence in Proposition 5 is lost. Indeed, all that can be said is that

$$\begin{aligned} (2) \Rightarrow (1) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5) \Rightarrow \varDelta (0)\ge 0 . \end{aligned}$$

It is worth pointing out that all the results of Sect. 5 can be retrieved in their original form under an additional assumption:

• u is strictly concave in a neighborhood of $x=0$.

This additional assumption and $A_{2}^{\prime }$ are still weaker than $A_{2}$.

Appendix B: Proofs

1.1 Preliminary results: Euler conditions

Before presenting the proofs of the results stated in the main part of the paper, we state and prove some useful Euler inequalities that any optimal program must satisfy.

Lemma 6

Let $\{{{{\varvec{x}}}} _{t}\}_{t=0}^{\infty }$ be an optimal program.

If $\min _{j=1\dots a}\{y_{t+j}\}>0$ for some $t\ge 0$, then

$$\begin{aligned} \sum _{j=1}^{a}b^{j}w^{\prime }(y_{t+j})+\nu ^{\prime }_+(z_{t+1}-z_{t})\ge b^{a}f_{a}u^{\prime }(c_{t+a})+b^a\nu ^{\prime }_-(z_{t+a+1}-z_{t+a}). \end{aligned}$$

(11)

If $c_{a,t+a}>0$ for some t, then

$$\begin{aligned} \sum _{j=1}^{a}b^{j}w^{\prime }(y_{t+j})+\nu ^{\prime }_-(z_{t+1}-z_{t})\le b^{a}f_{a}u^{\prime }(c_{t+a})+b^a\nu ^{\prime }_+(z_{t+a+1}-z_{t+a}). \end{aligned}$$

(12)

Proof

We define the unitary vector ${{\varvec{e}}}_j \in {\mathbb {R}}^n$ such that $e_j=1$ and $e_k=0$ for all $k\ne j$. Consider an alternative program $\{ \widehat{{{{\varvec{x}}}}}_{s}\}_{s=0}^{\infty }$ such that

$$\begin{aligned} \begin{array}{rcl} \widehat{{{{\varvec{x}}}}}_{t+j}&{}=&{}{{\varvec{x}}}_{t+j}+\epsilon {{\varvec{e}}}_j \qquad \forall j=1,\dots ,a. \\ \widehat{{{{\varvec{x}}}}}_{s}&{}=&{}{{\varvec{x}}}_{s} \qquad \qquad \text{ else } \end{array} \end{aligned}$$

In words, whenever $\epsilon >0$ ($\epsilon <0$), the alternative program is one where the area under alternative use is reduced (increased) by $\epsilon$ at stage $t+1$ and replanted to yield more (less) young forest next period. This modification of the young forest area is allowed to grow undisturbed until age a at which point the consumption of forest of age a is modified. Of course, the area under alternative use can only be reduced if it is strictly positive along the periods involved, i.e., $\min _{j=1,\dots ,a}\{y_{t+j}\}>0$. On the other hand, to increase the land under alternative use along the stages $t+1,\dots ,t+a$, we decrease the harvest of age class a at stage $t+a$; hence, we need $c_{a,t+a}>0$. In conclusion, the alternative program is feasible for $-c_{a,t+a}<\epsilon < \min \{y_{t+1},y_{t+2},\dots ,y_{t+a}\}$.

As $\{{{{\varvec{x}}}}_{t}\}_{t=0}^{\infty }$ is optimal, the modification must not be gainful, hence:

$$\begin{aligned}&\sum _{j=1}^{\infty }b^{j-1}[u(c_j)+w(y_{j})-\nu (z_{j+1}-z_{j})] \\&\quad \ge \sum _{j=1}^{\infty }b^{j-1}[u(\widehat{c}_j)+ w( \widehat{y}_{j}) -\nu ({\widehat{z}}_{j+1}-{\widehat{z}}_{j})] \end{aligned}$$

which is equivalent to

$$\begin{aligned}&b^au(c_{t+a})+\sum _{j=1}^a b^jw(y_{t+j})-\nu (z_{t+1}-z_{t})-b^a\nu (z_{t+a+1}-z_{t+a}) ~ \ge \\&b^au(c_{t+a}+f_a\epsilon )+\sum _{j=1}^a b^jw(y_{t+j}-\epsilon ) - \nu (z_{t+1}+\epsilon -z_{t})\\&\qquad -\,b^a\nu (z_{t+a+1}-z_{t+a}-\epsilon ), \end{aligned}$$

and reordering we get

$$\begin{aligned}&\sum _{j=1}^a b^j[w(y_{t+j})-w(y_{t+j}-\epsilon )]-\nu (z_{t+1}-z_{t})+\nu (z_{t+1}-z_{t}+\epsilon )~ \\&\quad \ge b^a[u(c_{t+a}+f_a\epsilon )-u(c_{t+a})]+b^a\nu (z_{t+a+1}-z_{t+a}) \\&\qquad -\,b^a\nu (z_{t+a+1}-z_{t+a}-\epsilon ) \end{aligned}$$

If $\min _{j=1\dots a} \{y_{t+j}\}>0$ we can consider $\epsilon >0$. Dividing through by $\epsilon$ and taking the limit as $\epsilon \rightarrow 0^+,$ we obtain inequality (11). If $c_{a,t+a}>0$ we can consider $\epsilon <0$. Dividing through by $\epsilon <0$ and taking the limit as $\epsilon \rightarrow 0^-,$ we obtain inequality (12). $\square$

1.2 Main results

Lemma 2

If ${{\varvec{x}}}$ is an oss it must be of the form ${{\varvec{x}}}(z)$ for some value of $z\in [0,1]$.

Proof

We will prove that only the $\sigma$-age class can be harvested along an optimal stationary path. This implies that if the total forest area is z, the state must be ${{\varvec{x}}}(z)$.²⁵ We will assume that along a stationary path every harvested fraction of land is immediately replanted with trees to be harvested at the exact same age as that of their predecessors.²⁶

Assume that ${{\varvec{x}}}$ is an oss, i.e., the optimal path starting from it is constant ${{\varvec{x}}}_t={{\varvec{x}}}$. Of course, along this path, harvest is constant, let us denote the total consumption by C. To reach a contradiction, let us assume that there is some age $s\ne \sigma$ such that $c_{s}>0$. We consider an $\epsilon$ area of land that at $t=0$ is planted with trees that will be harvested at age s and modify the stationary path by harvesting this area exactly when the age of the trees is $\sigma$. This means that every $\sigma$ years, the harvest of the $\sigma$ age class is increased by $\epsilon$ and that every s years, the harvest of the s age class is reduced by $\epsilon$.

The marginal benefit of the perturbed path with respect to the constant one is

$$\begin{aligned} -\frac{b^{s}f_{s}}{1-b^{s}}u'(C)+\frac{b^{\sigma }f_{\sigma }}{1-b^{\sigma }} u'(C)>0 \end{aligned}$$

that is strictly positive due to $A_4$. In consequence, it is not optimal to harvest from $s\ne \sigma .$$\square$

Proposition 1

The set of oss coincides with $\{{{\varvec{x}}}(z) / z\in [z_{\delta },z_0]\}$, i.e., the set of states with forest cover $z\in [z_{\delta }, z_0]$ where the forest area z is evenly divided among trees of age classes 1 through $\sigma .$ In particular,

• 1. If $\varDelta (0)\le 0$, then $I(\delta )=\{0\}$ for all $\delta \ge 0$, i.e., regardless of the extent of irreversibility (cost of reforestation) the totally deforested state is the unique oss.

• 2. If $\varDelta (1)\ge \delta$, then $I(\delta )=\{1\}$, i.e., there is a unique oss and it is characterized by full forest cover.

• 3. If $\delta>0, \varDelta (0)>0$ and $\varDelta (1)<\delta ,$ then $z_{\delta }<z_{0}$ and there is a continuum of oss; the set of forest covers that can be sustained in an oss is a closed interval $[z_{\delta },z_{0}]$ with non-empty interior. In addition, $I(\delta )=\{{z}\in [0,1]:0\le \varDelta ({z} )\le \delta \}$.

Proof

The proof will be divided into two parts:

• First, we show: if ${{\varvec{x}}}$ is an oss $\Rightarrow {{\varvec{x}}}={{\varvec{x}}}(z)$ and $z\in [z_{\delta },z_0]$

• Second, we show the counterpart: if ${{\varvec{x}}}={{\varvec{x}}}(z)$ and $z\in [z_{\delta },z_0] \Rightarrow {{\varvec{x}}}$ is an oss

• First part.  Thanks to Lemma 2 we already know that ${{\varvec{x}}}$ is of the form ${{\varvec{x}}}(z)$, so we only need to show that $z\in [z_{\delta },z_0]$. We claim that

  $$\begin{aligned}&\hbox {If }z\in I\hbox { with }z<1\hbox { then }\varDelta (z)\le \delta \end{aligned}$$

  (13)
  $$\begin{aligned}&\hbox {If }z\in I\hbox { with }z>0\hbox { then }\varDelta (z)\ge 0. \end{aligned}$$

  (14)

  Postponing briefly the proof of (13) and (14), we use them to show that $z\in [z_{\delta },z_0]$. Observe that the definition of $z_0$ and $z_{\delta }$ can be stated as $z_q=\max \{0,\min \{1,\varDelta ^{-1}(q)\}\}$ for $q=0,\delta$.

  • 1. if $z=1$ we know thanks to (14) that $\varDelta (1)\ge 0$. Since $\varDelta (z)$ is a decreasing function, this condition implies $1\le \varDelta ^{-1}(0)$. By definition $z_0=1$ and evidently $z\in [z_{\delta },z_0]$.

  • 2. if $z=0$, (13) implies that $\varDelta (0)\le \delta$. We then have that $0\ge \varDelta ^{-1}(\delta )$. By definition $z_{\delta }=0$ and evidently $z\in [z_{\delta },z_0]$.

  • 3. We deal now with the more general case, where $0<z<1$. Using both (13) and (14), we get $0\le \varDelta (z)\le \delta$ and the monotonicity of $\varDelta$ yields $\varDelta ^{-1}(0)\ge z \ge \varDelta ^{-1}(\delta )$. Combining these inequalities, we get

    $$\begin{aligned} z\le \min \{1,\varDelta ^{-1}(0)\}\Rightarrow & {} z\le z_0 \\ z\ge \min \{0,\varDelta ^{-1}(\delta )\}\Rightarrow & {} z\ge z_{\delta } \end{aligned}$$

  This completes the proof of the first part. We present now the rather technical proof of (13) and (14).

  We define the unit vector ${{\varvec{e}}}_j \in {\mathbb {R}}^n$ such that $e_j=1$ and $e_k=0$ for all $k\ne j$ and $t(\sigma )$ stands for the remainder of the integer division of t by $\sigma$. We define a program alternative to the constant one starting at ${{\varvec{x}}}(z)$, $\{\widehat{{{{\varvec{x}}}}}_{s}\}_{s=0}^{\infty }$ such that

  $$\begin{aligned} \widehat{{{{\varvec{x}}}}}_{0}={{\varvec{x}}}_{0} \qquad \text {and}\qquad \widehat{{{{\varvec{x}}}}}_{t}={{\varvec{x}}}_{t}+\epsilon {{\varvec{e}}}_{t(\sigma )}~\forall t\ge 1. \end{aligned}$$

  In words, whenever $\epsilon >0$ ($\epsilon <0$), the alternative program is one where the area under alternative use is reduced (increased) by $\epsilon$ at stage $t=0$ and replanted to yield more (less) young forest next period. There is no further reallocation of land; and at every period, only the $\sigma$-age class is clearcut leaving the other classes untouched. This policy generates a $\sigma$-periodic path from $t=1$ onward.

  Of course, the area under alternative use can only be reduced if it is strictly positive, i.e., $1-z>0 \Rightarrow z<1$. On the other hand, to increase the land under alternative use, we decrease the area of one of the age classes; hence, we need $z>0$.

  As the constant path is optimal, the modification cannot be gainful. Proceeding analogously to the proof of Lemma 6 we obtain,

  $$\begin{aligned}&\frac{b^{\sigma }}{1-b^{\sigma }}\Big [u\Big (\frac{f_{\sigma }}{\sigma }z\Big )- u\Big (f_{\sigma }\Big (\frac{z}{\sigma }+\epsilon \Big )\Big )\Big ] \\&\qquad +\frac{b}{1-b}[w(y)-w(y-\epsilon )]+\nu (\epsilon )-\nu (0)\ge 0. \end{aligned}$$

  If $z<1$ we can consider $\epsilon >0$. Dividing through by $\epsilon$ and taking the limit as $\epsilon \rightarrow 0^+,$ we obtain (13):

  $$\begin{aligned} -\frac{b^{\sigma } f_{\sigma }}{1-b^{\sigma }}u'\Big (\frac{f_{\sigma }}{\sigma }z\Big )+\frac{b}{1-b}w'(y)+\delta \ge 0\Rightarrow & {} \delta \ge \varDelta (z) \end{aligned}$$

  If $z>0$ we can consider $\epsilon <0$. This time, dividing through by $\epsilon$ involves changing the sense of the inequality. Taking the limit as $\epsilon \rightarrow 0^-,$ we obtain (14):

  $$\begin{aligned} -\frac{b^{\sigma } f_{\sigma }}{1-b^{\sigma }}u'\Big (\frac{f_{\sigma }}{\sigma }z\Big )+\frac{b}{1-b}w'(y)\le 0\Rightarrow & {} 0\le \varDelta (z). \end{aligned}$$

• Second part.  Considering the definition of $z_0$ and $z_{\delta }$, we have the following equivalences

  • (a) $[z_{\delta },z_0]=\{0\}$ if, and only if, $\varDelta (0)\le 0$.

  • (b) $[z_{\delta },z_0]=\{1\}$ if, and only if, $\varDelta (1)\ge \delta$.

  • (c) $[z_{\delta },z_0]=\{{z}\in [0,1]:0\le \varDelta ({z})\le \delta \}$ if, and only if, $\delta>0, \varDelta (0)>0$ and $\varDelta (1)<\delta$.

  We will use these equivalences to prove the second part of the Proposition. In particular, we claim that

  $$\begin{aligned} (a)&\text{ If } \varDelta (0)\le 0\hbox { then }{{\varvec{x}}}(0)\hbox { is a oss} \end{aligned}$$

  (15)
  $$\begin{aligned} (b)&\text{ If } \varDelta (1)\ge \delta \hbox { then }{{\varvec{x}}}(1)\hbox { is a oss} \end{aligned}$$

  (16)
  $$\begin{aligned} (c)&\text{ If } \varDelta (z)\in [0,\delta ]\hbox { then }{{\varvec{x}}}(z)\hbox { is a oss (including the cases }z=0\hbox { and }z=1)\nonumber \\ \end{aligned}$$

  (17)

  These statements can be intuitively interpreted by recalling that $\varDelta (z)$ represents the marginal benefit of dedicating more land to forestry when the current state is ${{\varvec{x}}}(z)$. Indeed, $\varDelta (0)\le 0$ implies that, even when all the land is allocated to the alternative use, there is no incentive to dedicate land to forestry because the marginal benefit of such decision is negative. In other words, it means that any state ${{\varvec{x}}}(z)$ with $z>0$ cannot be sustainable because the marginal benefit of transferring land to the alternative use would be positive. In consequence, ${{\varvec{x}}}(0)$ is the only eventual oss. The inequality $\varDelta (1)\ge \delta$ implies that the marginal benefit of transferring land from alternative use to forestry is higher than the intrinsic marginal reforestation cost ($\delta$). This is so, even when all the land is already allocated to forestry, and the marginal benefit of this transference is at its minimum, implying that no state ${{\varvec{x}}}(z)$ with less than full forest cover can be a oss. Finally, in case (c) where $\varDelta (z)\in [0,\delta ]$ we see that the marginal benefit of transferring land to forestry is not enough to cover the intrinsic marginal reforestation costs and that the marginal benefit of relocating land to the alternative use is, at best, non-positive. Hence, ${{\varvec{x}}}(z)$ is a promising candidate to oss.

  We will use this intuitive interpretation to define the dual variables of two auxiliary variables accounting for reforestation ($r_t$) and deforestation ($d_t$). The new variables $r_t$ and $d_t$ are both nonnegative and satisfy $z_{t+1}=z_t+r_t-d_t$ for all t. By introducing them, we avoid dealing with the non-differentiable function $\nu$. Indeed, We can substitute $\nu (z_{t+1}-z_{t})$ by the term ${\tilde{\nu }}(r_t)$ where ${\tilde{\nu }}(r)$ is a smooth and concave function that coincides with $\nu (r)$ for nonnegative values of r. We have then $\tilde{\nu }'(0)=\delta$.

  Eliminating the variables $c_{a,t}$ and $c_t$ from the optimization problem (6), we get

  $$\begin{aligned} \left\{ \begin{array}{ll} \text{ maximize } &{} \sum _{t=0}^{\infty }b^{t}[u(\sum _{a=1}^{n-1}f_a(x_{a,t}-x_{a+1,t+1})+f_{n}x_{n,t}) +w(y_{t}) -\tilde{\nu }(r_t)] \\ \text{ subject } \text{ to } &{} \sum _{a=1}^n x_{n,t}+y_t\le 1 \qquad \forall t\\ &{} x_{a,t}-x_{a+1,t+1}\ge 0 \qquad a=1,\dots ,n-1,~~\forall t\\ &{}y_{t+1}=y_t-r_t+d_t \qquad \forall t\\ &{} x_{n,t},y_t,r_t,d_t\ge 0 \qquad ~~~ \forall t \end{array} \right. \end{aligned}$$

  (18)

  And the associated Lagrangian is:

  $$\begin{aligned} {\mathcal {L}}= & {} \sum _{t=0}^{\infty }b^{t}\left[ u\left( \sum _{a=1}^{n-1}f_a(x_{a,t}-x_{a+1,t+1})+f_{n}x_{n,t}\right) +w(y_{t})\right. \\&\left. -\,\tilde{\nu }(r_t)\right] + \theta _t(1-\sum _a x_{a,t}-y_t)\\&+\sum _{a=1}^{n-1}\rho _{a,t}(x_{a,t}-x_{a+1,t+1}) + \mu _t(y_{t+1}-y_t+r_t-d_t)+\lambda ^x_t x_{n,t}\\&+\,\lambda ^y_t y_t + \lambda ^r_t r_t +\lambda ^d_t d_t. \end{aligned}$$

  The partial derivatives of ${\mathcal {L}}$ evaluated along the constant path are:

  $$\begin{aligned} {\mathcal {L}}_{x_{1,t}}= & {} b^tf_1u'\Big (\frac{f_{\sigma }}{\sigma }{\hat{z}}\Big )-\theta _t+\rho _{1,t} \\ {\mathcal {L}}_{x_{a,t}}= & {} (b^tf_a-b^{t-1}f_{a-1})u'\Big (\frac{f_{\sigma }}{\sigma }{\hat{z}}\Big )-\theta _t+\rho _{a,t}-\rho _{a-1,t-1} \qquad \forall a=2,\dots ,n-1 \\ {\mathcal {L}}_{x_{n,t}}= & {} (b^tf_n-b^{t-1}f_{n-1})u'\Big (\frac{f_{\sigma }}{\sigma }{\hat{z}}\Big )-\theta _t-\rho _{n-1,t-1}+\lambda ^x_t \\ {\mathcal {L}}_{y_{t}}= & {} b^tw'({\hat{y}})-\theta _t+\mu _{t-1}-\mu _t+\lambda ^y_t \\ {\mathcal {L}}_{r_{t}}= & {} -b^t\delta +\mu _t+\lambda ^r_t \\ {\mathcal {L}}_{d_{t}}= & {} -\mu _t+\lambda ^d_t \end{aligned}$$

  To prove that ${{\varvec{x}}}(z)$ is a oss, we need to show that the constant path is a solution of the optimization problem (18) when starting from it.

  We first deal with (17). To show that the stationary solution is optimal, we propose the following set of $\ell _1$ multipliers where $\tau _a$ stands for $\frac{b^a}{1-b^a}$:

  $$\begin{aligned} \left\{ \begin{array}{rcl} \theta _t&{}=&{}b^t\frac{f_{\sigma } \tau _{\sigma }}{\tau _1}u'\Big (\frac{f_{\sigma }}{\sigma }{\hat{z}}\Big )\\ \rho _{a,t}&{}=&{} \frac{b^t}{\tau _a}(\tau _{\sigma }f_{\sigma }-\tau _a f_a)u'\Big (\frac{f_{\sigma }}{\sigma }{\hat{z}}\Big )\\ \mu _{t}&{}=&{} {b^t}\varDelta ({\hat{z}})\\ \lambda ^r_t&{}=&{}b^t[\delta -\varDelta ({\hat{z}})]\\ \lambda ^d_t&{}=&{}b^t\varDelta ({\hat{z}})\\ \lambda ^x_t&{}=&{}b^t\frac{f_{\sigma } \tau _{\sigma }}{\tau _n}u'\Big (\frac{f_{\sigma }}{\sigma }{\hat{z}}\Big )\\ \lambda ^y_t&{}=&{}0 \end{array} \right. \end{aligned}$$

  Due to the definition of $\sigma$ (see $A_4$), complementarity slackness and the non-negativity of all the multipliers are satisfied.²⁷ A straightforward computation shows that $\nabla {\mathcal {L}}= 0$ so the proposed path is a stationary point of ${\mathcal {L}}$. As the optimization problem (6) is strictly concave, optimality follows.

  The proof of (16) is analogous. In this case, we propose the following set of $\ell _1$ nonnegative multipliers,

  $$\begin{aligned} \left\{ \begin{array}{rcl} \theta _t&{}=&{}b^t\frac{f_{\sigma } \tau _{\sigma }}{\tau _1}u'(f_{\sigma }/\sigma )\\ \rho _{a,t}&{}=&{} \frac{b^t}{\tau _a}(\tau _{\sigma }f_{\sigma }-\tau _a f_a)u'(f_{\sigma }/\sigma )\\ \lambda ^d_t&{}=&{}b^t\delta \\ \lambda ^x_t&{}=&{}b^t\frac{f_{\sigma } \tau _{\sigma }}{\tau _n}u'(f_{\sigma }/\sigma )\\ \lambda ^y_t&{}=&{}b^t\tau _1\varDelta (1)\\ \mu _{t}&{}=&{} \lambda ^r_t ~=~0 \end{array} \right. \end{aligned}$$

  and the verification follows the same lines of the previous case.

Finally, we deal with (15). From the results in Piazza and Roy (2015), we know that if $\varDelta (0)\le 0$ then ${{\varvec{x}}}(0)$ is an oss in the absence of reforestation costs. The stationary path remaining at the total deforestation state will not see its total discounted utility modified when deforestation costs are introduced. It is evident that no other path will yield higher total discounted utility with the introduction of reforestation costs. In consequence, the total deforestation state continues to be an oss with reforestation costs and (15) follows. $\square$

Lemma 4

Suppose that $\varDelta (0)\le \delta .$ Let $\{x_{t}\}$ be an optimal path and let $\{z_{t}\}$ be the path of area under forest cover associated with it. Then, $z_{t}\ge z_{t+1}$ for all t, i.e., the area under forest cover is always non-increasing along an optimal program. If, further, $\varDelta (0)>0,$ then

$$\begin{aligned} z_{t}\ge \min \{{\underline{z}},z_{o}\}\text { for all }t. \end{aligned}$$

Proof

To prove that $\varDelta (0)\le \delta \Rightarrow z_t$ is non-increasing, we proceed by contradiction. Assume that there exists at least one optimal path, such that the forest cover increases at least once. Without loss of generality, we assume that the increase takes place at $t=0$.

If $z_o<z_1$, then $x_{1,1}>0$ and there is a such that $c_{a,a}>0$. We apply then (12)

$$\begin{aligned} 0\ge & {} \sum _{j=1}^{a}b^jw'(y_{j})+\underbrace{\nu '(z_1-z_0)}_{>\delta } -b^af_au'(c_{a})-b^a\nu '_+(z_{a+1}-z_{a}) \\> & {} \frac{b(1-b^a)}{1-b}w'(1)-b^af_au'(0)+\delta -b^a\nu '_+(z_{a+1}-z_{a}) \\= & {} (1-b^a)\left[ \frac{b}{1-b}w'(1)-\frac{ b^af_a}{1-b^a}u'(0) \right] +\delta -b^a\nu '_+(z_{a+1}-z_{a}) \\\ge & {} -(1-b^a)\varDelta (0)+\delta - b^a\nu '_+(z_{a+1}-z_{a}) \end{aligned}$$

We have different scenarios depending on the change of the forest cover at $t=a$. For instance, if $z_a>z_{a+1}$ we have $\nu '(z_{a+1}-z_{a})=0$ and if $z_a=z_{a+1}$ we have $\nu '_+(z_{a+1}-z_{a})=\delta$. In both cases, we conclude $0>-(1-b^a)\varDelta (0)+\delta - b^a\nu '_+(z_{a+1}-z_{a})\ge -(1-b^a)\varDelta (0)+(1-b^a)\delta$. This implies that $\varDelta (0)>\delta$, contradicting $\varDelta (0)\le \delta$ and proving the first statement of the lemma.

The case where $z_{a}<z_{a+1}$ is the difficult one, because we may have $\nu '(z_{a+1}-z_{a})>\delta$. Before going further, we recall the deduction of the Euler inequality (12). It was based on a perturbation where the area under alternative use is increased by $\epsilon$ at stage $t+1$ and less young forest is replanted. This modification of the young forest area is allowed to grow undisturbed until age a at which point the consumption of forest of age a is modified. To compensate for the perturbation, the area under alternative use can only be increased by decreasing harvest of age class a at stage $t=a$; hence, we need $c_{a,a}>0$. But if $z_a<z_{a+1}$, then $x_{a+1,1}>0$ and there exists $a'$ such that $c_{a+a',a'}>0$. We modify the perturbation above keeping the extra $\epsilon$ of land allocated to the alternative use for another $a'$ stages. Observe that the term $\nu '_+(z_{a+1}-z_{a})$ disappears because land is not returned to forest use.

We then have

$$\begin{aligned} 0\ge & {} \sum _{j=1}^{a+a'} b^jw'(y_{j}) -b^af_au'(c_{a})-b^{a+a'}f_{a'}u'(c_{a+a'})\\&+\,\nu '(z_{1}-z_{0})-b^{a+a'}\nu '_+(z_{a+a'+1}-z_{a+a'})\\> & {} (1-b^a)\left[ \frac{b}{1-b}w'(1)-\frac{b^{a}f_{a}}{1-b^{a}}u'(0)\right] \\&+\, b^{a}(1-b^{a'})\left[ \frac{b}{1-b}w'(1)- \frac{b^{a'}f_{a'}}{1-b^{a'}}u'(0)\right] +\delta -b^{a+a'}\nu '_+(z_{a+1}-z_{a})\\> & {} -\varDelta (0)(1-b^{a+a'})+\delta -b^{a+a'}\nu '_+(z_{a+a'+1}-z_{a+a'}) \end{aligned}$$

If $z_{a+a'}\ge z_{a+a'+1}$, the proof is complete because $\nu '_+(z_{a+a'+1}-z_{a+a'})\le \delta$ and again $\varDelta (0)>\delta$ is reached.

If not, we repeat the process. Let us introduce some notation for the last part of the proof. We denote $a_1=a$ and $a_2=a'$ and $s_j=\sum _{i\le j}a_i$. If $z_{s_j}\le z_{s_j+1}$ then $x_{s_j+1,1}>0$ and there exists $a_{j+1}$ such that $c_{s_{j}+a_{j+1},a_{j+1}}>0$. We modify the perturbation above keeping the extra $\epsilon$ of land allocated to the alternative use for another $a_{j+1}$ stages. We repeat the process until we reach some stage such that $z_{s_j}>z_{s_j+1}$.

If such stage is not reached, then the process is repeated infinitely to get

$$\begin{aligned} 0\ge & {} \lim _{J\rightarrow \infty }\sum _{j=1}^{J}b^{s_{j-1}}(1-b^{a_j})\left[ \frac{b}{1-b}w'(1)-\frac{b^{a_j}f_{a_j}}{1-b^{a_j}}u'(0)\right] \\&+\,\nu '(z_{1}-z_{0})-\lim _{J\rightarrow \infty } b^{s_{J}}\nu '_+(z_{s_{J}+1}-z_{s_{J}})\\> & {} -\lim _{J\rightarrow \infty }\sum _{j=1}^{J}b^{s_{j-1}}(1-b^{a_j})\varDelta (0) +\delta -\lim _{J\rightarrow \infty } b^{s_{J}}\nu '_+(z_{s_{J}+1}-z_{s_{J}})\\= & {} -\varDelta (0)\big (1-\lim _{J\rightarrow \infty }b^{s_J}\big ) + \delta -\lim _{J\rightarrow \infty } b^{s_{J}}\nu '_+(z_{s_{J}+1}-z_{s_{J}})\\= & {} -\varDelta (0)+\delta \end{aligned}$$

a contradiction. The proof follows.

We proceed now to prove that $0<\varDelta (0)\le \delta \Rightarrow z_{t}\ge \min \{{\underline{z}},z_{o}\}\text { for all }t$. Given the monotonicity of $z_t$, there are only two possibilities: either (i) $z_t=z_o$ for all t or (ii) there exists t such that $z_{t-1}>z_{t}$. The proposition follows trivially in (i).

In (ii), let T be any stage where the forest cover decreases, i.e., $z_{T-1}>z_T$. We know that $z_t\le z_T<z_{T-1}\le 1$ for all $t>T$; hence, $z_{T-1+j}<1$ for all $j=1,\dots ,\sigma$ and then $\min _{j=1,\dots ,\sigma }\{y_{T-1+j}\}>0$. We apply (11) to get

$$\begin{aligned} 0\le & {} \sum _{j=1}^{{\sigma }}b^jw'(y_{T-1+j})+\nu '_+(z_{T}-z_{T-1}) -b^{\sigma }u'(c_{T-1+{\sigma }})-b^{\sigma }\nu '_-(z_{T+{\sigma }}-z_{T-1+{\sigma }}) \\= & {} \sum _{j=1}^{{\sigma }}b^jw'(y_{T-1+j}) -b^{\sigma }u'(c_{T-1+{\sigma }}) \\\le & {} \sum _{j=1}^{{\sigma }}b^jw'(y_{T-1+j})-b^{\sigma }u'(f_m(1-y_{T-1+{\sigma }})) \\\le & {} \frac{b(1-b^{\sigma })}{1-b}w'(1-z_T)-b^{\sigma }u'(f_m(z_{T})) \end{aligned}$$

Implying that $g(z_T)\le 0$ and, in consequence, $z_T\ge {\underline{z}}$. $\square$

Proposition 4

There is global weak conservation if, and only if, $\varDelta (0)>0$.

Proof

It is sufficient to show that

$$\begin{aligned} \varDelta (0)\le 0\Leftrightarrow \hbox {Global weak conservation does not hold.} \end{aligned}$$

If $\varDelta (0)\le 0$, the results in Piazza and Roy (2015) assure that every optimal path is characterized by immediate deforestation in the absence of reforestation costs. Evidently, when reforestation costs are introduced, any path characterized by immediate deforestation will not change its total benefit, while the total benefit of the other paths will either remain constant or decrease. Hence, if $\varDelta (0)\le 0$ the optimal paths are the same with and without reforestation costs. And they are all characterized by immediate and eventual total deforestation.

If there is no global weak conservation, there exists at least one optimal path, $\{x_t\}$, starting from a positive forest cover state characterized by eventual total deforestation.

Along $\{x_t\}$, we know that $z_t\rightarrow 0$. This convergence can be finite or asymptotic, i.e., either there exists T such that $z_T>0$ and $z_t=0$ for all $t>T$ or there exists a subsequence $\{t_k\}_{k\in I\!\!N}$ such that $z_{t_k-1}>z_{t_k}$.

In the first case, we apply (11) for $t=T$ and $a=\sigma$ to get,

$$\begin{aligned} 0\le \frac{b(1-b^{\sigma })}{1-b}w'(1)-b^{\sigma }f_{\sigma }u'(0) = -(1-b^{\sigma })\varDelta (0) \Rightarrow \varDelta (0)\le 0. \end{aligned}$$

In the second case, we evaluate (11) along the subsequence ${{\varvec{x}}}_{t_k}$ and $a=\sigma$,

$$\begin{aligned} 0\le & {} \sum _{j=1}^{\sigma }b^j w'(y_{t_k+j})-b^{\sigma }f_{\sigma }u'(c_{t_k+\sigma })- b^{\sigma }\nu '_-(z_{t_k+\sigma +1}-z_{t_k+\sigma })\\\le & {} \sum _{j=1}^{\sigma }b^j w'(y_{t_k+j})-b^{\sigma }f_{\sigma }u'(c_{t_k+\sigma }) \end{aligned}$$

and letting $k\rightarrow \infty$ we get $0\le -(1-b^{\sigma })\varDelta (0) \Rightarrow \varDelta (0)\le 0$. $\square$

Proposition 5

The following are equivalent

• 1. There is global weak conservation

• 2. $\varDelta (0)>0$

• 3. There exists at least one oss with strictly positive forest cover

• 4. There exists at least one optimal path where eventual total deforestation does not occur

• 5. There exists at least one optimal path where immediate total deforestation does not occur

Proof

The equivalence 1.$\Leftrightarrow$ 2. is exactly Proposition 4 and Corollary 4 delivers 2.$\Leftrightarrow$ 3. To complete the proof, we show that 3.$\Rightarrow$ 4.$\Rightarrow$ 5.$\Rightarrow$ 2., which are all fairly straightforward.

Given a positive forest cover oss, ${{\varvec{x}}}({\hat{x}})$, the constant path starting from it is not characterized by eventual total deforestation, proving 3.$\Rightarrow$ 4.

4.$\Rightarrow$ 5. follows trivially. Indeed, any path not characterized by eventual total deforestation is not characterized by immediate total deforestation, either.

It is only left to show that 5.$\Rightarrow$ 2., and we proceed by contradiction. If $\varDelta (0)\le 0$, we know thanks to Piazza and Roy (2015), that every optimal program is characterized by immediate total deforestation in the absence of reforestation costs. Following the same reasoning of the proof of Proposition 4, we conclude that all the optimal paths are characterized by immediate deforestation in the presence of such costs, contradicting 5. $\square$

Proposition 6

If $\{x_{t}\}$ is an optimal program and $\{z_{t}\}$ is its associated path of area under forest cover, then for every $t\ge 0,$

$$\begin{aligned} \max _{j=1,\dots ,\sigma }\{z_{t+j}\}\ge \min \{{\underline{z}},z_{o}\} \end{aligned}$$

Proof

We first note that if $\max _{j=1,\dots ,a}\{z_{t+j}\}=1$, then (7) holds trivially. From now on, we assume that the maximum is less than one which allows us to use (11) for $a=\sigma$.

We proceed by induction on t. First consider $t=0$. If $z_1\ge z_o$, the inequality follows trivially. If not, we have $\nu '_+(z_1-z_o)=0$ and (11) implies,

$$\begin{aligned} 0\le & {} \sum _{j=1}^{\sigma } b^j w'(1-z_{j})-b^{\sigma }f_{\sigma }u'(c_{{\sigma }})-b^{\sigma }\nu '_-(z_{{\sigma }+1}-z_{{\sigma }})\\< & {} \frac{b(1-b^{\sigma })}{1-b}w'(1-\max _{j=1,\dots ,{\sigma }}\{z_{j}\}) -b^{\sigma }f_{\sigma }u'(f_m\max _{j=1,\dots ,{\sigma }}\{z_{j}\}) \end{aligned}$$

Considering the definition of ${\underline{z}}$, we get that $\max _{j=1,\dots ,{\sigma }}\{z_{j}\}>{\underline{z}}$, and hence, (7) holds for $t=0$.

We now assume that (7) holds for $t=k-1$ and prove that it is true for $t=k$.

If $z_{k+1}<z_{k}$, we apply (11) at $t=k$ and follow the same steps of the case $t=0$ to conclude.

If $z_{k+1}\ge z_{k}$, then

$$\begin{aligned} \max _{j=1,\dots ,{\sigma }}\{z_{k-1+j}\}= \max _{j=2,\dots ,{\sigma }}\{z_{k-1+j}\} \le \max _{j=2,\dots ,{\sigma }+1}\{z_{k-1+j}\}= \max _{j=1,\dots ,{\sigma }}\{z_{k+j}\}. \end{aligned}$$

In consequence, $\min \{{\underline{z}},z_o\}\le \max _{j=1,\dots ,{\sigma }}\{z_{k-1+j}\} \le \max _{j=1,\dots ,{\sigma }}\{z_{k+j}\}$ where the first inequality is (7) for $t=k-1$. The proof follows by induction on t. $\square$

Proposition 7

If $\{x_{t}\}$ is an optimal program and $\{z_{t}\}$ is its associated path of area under forest cover, then

$$\begin{aligned} \lim \inf _{t}z_{t}<{\bar{z}}(0). \end{aligned}$$

Proof

To obtain a contradiction, suppose that there is T such that the optimal program satisfies $z_t\ge {\bar{z}}(0)$ for all $t\ge T$. Take $t\ge T$ with $x_{t+1,1}>0$.²⁸ As a consequence, there exists a such that $c_{t+a,a}>0$, using (12) we get

$$\begin{aligned} 0\ge & {} \sum _{j=1}^a b^jw'(1-z_{t+j})-b^af_au'(c_{t+a})+\underbrace{\nu '_-(z_{t+1}-z_{t})}_{\ge 0}-b^a\nu '_+(z_{t+a+1}-z_{t+a})\\> & {} \frac{b(1-b^a)}{1-b}w'(1-{\bar{z}}(0))-b^af_au'(0)-b^a\nu '_+(z_{t+a+1}-z_{t+a})\\\ge & {} (1-b^a)\left[ \frac{b}{1-b}w'(1-{\bar{z}}(0))-\frac{b^{a}f_{a}}{1-b^{a}}u'(0)\right] -b^a\nu '_+(z_{t+a+1}-z_{t+a})\\\ge & {} (1-b^a)\left[ \frac{b}{1-b}w'(1-{\bar{z}}(0))-\frac{b^{\sigma }f_{\sigma }}{1-b^{\sigma }}u'(0)\right] -b^a\nu '_+(z_{t+a+1}-z_{t+a}) \end{aligned}$$

The term between brackets is zero due to (8). If $z_{t+a}>z_{t+a+1}$ then $\nu '_+(z_{t+a+1}-z_{t+a})=0$ implying that the last term is zero and a contradiction is reached.

The case where $z_{t+a}\le z_{t+a+1}$ is the difficult one. The proof is analogous to that of Lemma 4. Indeed, if $z_{t+a}\le z_{t+a+1}$, then $x_{t+a+1,1}>0$ and there exists $a'$ such that $c_{t+a+a',a'}>0$. We consider then a perturbation where the area under alternative use is increased by $\epsilon$ at stage $t+1$ and the fraction of land is only returned to the forestry use after $a+a'$ time steps. This causes a reduction in $\epsilon$ of the consumption of age class a at stage $t+a$ and of age class $a'$ at stage $t+a+a'$.

We get then

$$\begin{aligned} 0\ge & {} \sum _{j=1}^{a+a'} b^jw'(1-z_{t+j}) -b^af_au'(c_{t+a})-b^{a+a'}f_{a'}u'(c_{t+a+a'})\\&+\,\nu '_-(z_{t+1}-z_{t})-b^{a+a'}\nu '_+(z_{t+a+a'+1}-z_{t+a+a'})\\> & {} (1-b^a)\left[ \frac{b}{1-b}w'(1-{\bar{z}}(0))-\frac{b^{\sigma }f_{\sigma }}{1-b^{\sigma }}u'(0)\right] \\&+\, b^{a}(1-b^{a'})\left[ \frac{b}{1-b}w'(1-{\bar{z}}(0))- \frac{b^{\sigma }f_{\sigma }}{1-b^{\sigma }}u'(0)\right] -b^{a+a'}\nu '_+(z_{t+a+1}-z_{t+a})\\= & {} -b^{a+a'}\nu '_+(z_{t+a+a'+1}-z_{t+a+a'}) \end{aligned}$$

If $z_{t+a+a'}>z_{t+a+a'+1}$, the proof is complete because $\nu '_+(z_{t+a+a'+1}-z_{t+a+a'})=0$ and again a contradiction is reached.

If not, we repeat the process until we reach some stage such that the forest cover is decreased when the perturbed fraction of land is returned to the forestry use, i.e., until we reach some stage such that $z_{t+s_j}>z_{t+s_j+1}$.²⁹

If such stage is not reached, then the process is repeated infinitely to get

$$\begin{aligned} 0\ge & {} \lim _{J\rightarrow \infty }\sum _{j=1}^{J}b^{s_{j-1}}[\sum _{i=1}^{a_j}b^{i}w'(1-z_{t+s_j+i}) -b^{a_j}f_{a_j}u'(c_{s_{j}})] +\nu '_-(z_{t+1}-z_{t})\\&-\lim _{J\rightarrow \infty } b^{s_{J}}\nu '_+(z_{t+s_{J}+1}-z_{t+s_{J}})\\> & {} \lim _{J\rightarrow \infty }\sum _{j=1}^{J}b^{s_{j-1}}(1-b^{a_j})\left[ \frac{b}{1-b}w'({\bar{z}}(0)) \right. \\&\left. -\frac{b^{\sigma }f_{\sigma }}{1-b^{\sigma }}u'(0)\right] -\lim _{J\rightarrow \infty } b^{s_{J+1}}\nu '_+(z_{t+s_{J}+1}-z_{t+s_{J}})\\= & {} -\lim _{J\rightarrow \infty } b^{s_{J}}\nu '_+(z_{t+s_{J+1}+1}-z_{t+s_{J+1}})=0. \end{aligned}$$

a contradiction. The proof follows. $\square$

Proposition 8

If $\{x_{t}\}$ is an optimal program, $\{z_{t}\}$ is its associated path of area under forest cover and there is T such that $z_{T}<z_{T+1}$ (forest cover increases in period T), then

$$\begin{aligned} \min _{j=1,\dots ,n}\{z_{t+j}\}\le {\bar{z}}(\delta )\text { for all }t\ge T. \end{aligned}$$

Proof

If $z_T<z_{T+1}$, we know that $x_{T+1,1}>0$. The trees planted at T must be harvested between $T+1$ and $T+n$. Hence, there exists $a\in [1,n]$ such that $c_{T+a,a}>0$ and using (12)

$$\begin{aligned} 0\ge & {} \sum _{j=1}^a b^j w'(1-z_{T+j})+\underbrace{\nu '_-(z_{T+1}-z_{T})}_{\ge \delta } -b^af_au'(c_{T+a})-b^a\nu '_+(z_{T+a+1}-z_{T+a})\\> & {} \frac{b(1-b^a)}{1-b}w'\big (1-\min _{j=1,\dots ,a}\{z_{T+j}\}\big ) -b^af_au'(0)+\delta -b^a\nu '_+(z_{T+a+1}-z_{T+a}) \end{aligned}$$

We have different scenarios depending on the change in the forest cover at $t=T+a$. For instance, if $z_{T+a}>z_{T+a+1}$ we have $\nu '(z_{T+a+1}-z_{T+a})=0$ and if $z_{T+a}=z_{T+a+1}$ we have $\nu '_+(z_{T+a+1}-z_{T+a})=\delta$. In both cases, we have

$$\begin{aligned} 0\ge & {} \sum _{j=1}^a b^j w'(1-z_{T+j})+\underbrace{\nu '_-(z_{T+1}-z_{T})}_{\ge \delta } -b^af_au'(c_{T+a})-b^a\nu '_+(z_{T+a+1}-z_{T+a})\\> & {} (1-b^a)\left[ \frac{b}{1-b}w'(1-\min _{j=1,\dots , a} \{1-z_{T+j}\})-\frac{b^{a}f_{a}}{1-b^{a}} u'(0)+\delta \right] \\= & {} (1-b^a)\left[ \frac{b}{1-b}w'(1-\min _{j=1,\dots , n} \{1-z_{T+j}\})-\frac{b^{\sigma }f_{\sigma }}{1-b^{\sigma }} u'(0)+\delta \right] \end{aligned}$$

Considering the definition of ${\bar{z}}(\delta )$, we get $\min _{j=1,\dots ,n}\{z_{T+j}\}<{\bar{z}}(\delta )$, and hence, (9) holds for $t=T$. The case where $z_{T+a}<z_{T+a+1}$ is more difficult, but the proof follows the same lines as those of Lemma 4 and Proposition 7, details are left to the reader.

We now assume that (9) holds for $t=k-1$ and prove that it is true for $t=k$.

If $z_{k}\le z_{k+1}$, we apply (12) at $t=k$ and follow the same steps of the case $t=T$.

If $z_{k}>z_{k+1}$, then

$$\begin{aligned} \min _{j=1,\dots ,n}\{z_{k+j}\} = \min _{j=0,\dots ,n}\{z_{k+j}\} \le \min _{j=0,\dots ,n-1}\{z_{k+j}\}= \min _{j=1,\dots ,n}\{z_{k-1+j}\}\le {\bar{z}}(\delta ) \end{aligned}$$

where the last inequality is (9) for $t=k-1$. The proof follows by induction on t. $\square$

Lemma 5

The solution to (10), $\{z_t\}$, satisfies the following:

• 1. If the initial forest cover $z_o$ is larger or equal to $z^s$, the static optimal forest size, then the optimal path is $z_t=z^s$ for all $t \ge 1$.

• 2. If the initial forest cover $z_o$ is below $z^s$, the optimal path is non-decreasing, bounded above by $z^s$ and converges to a value in the interval $[z_o,z^s]$

Proof

• 1. The policy $z_t=z^s$ for all $t\ge 1$ yields the maximum benefit when $\nu (\cdot ) = 0$, and the reforestation costs do not affect its total benefit.

• 2. We prove first that $z_t\le z^s$ for all t by contradiction. Assume that it is not true that $z_t\le z^s$ for all t and take T to be lowest value of t such that $z_T>z^s$. From the paragraph above, we know that $z_t=z^s$ for all $t>T$.³⁰ We propose the following alternative path

  $$\begin{aligned} \hat{z}_t=\left\{ \begin{array}{ll} z^s &{} \text { if }t=T \\ z_t &{}\text { if }t\ne T, \end{array} \right. \end{aligned}$$

  i.e., the forest cover at stage T is smaller along the alternative path than in the original one. The two paths coincide in every other time period.

  The modification only affects the benefit at two stages: $t=T-1$ and $t=T$, and we claim that it yields a strictly larger benefit. Indeed, at $t=T-1$ we have

  $$\begin{aligned}&u(f_1 z_{T-1})+w(1-z_{T-1})-\nu (z_{T}-z_{T-1}) \\&\quad < u(f_1 z_{T-1})+w(1-z_{T-1})-\nu (z^s-z_{T-1}) \end{aligned}$$

  where we are using that $z_T>\hat{z}_T=z^s\ge z_{T-1}$ which is equivalent to $z_{T}-z_{T-1}> z^s-z_{T-1}\ge 0$ and implies $\nu (z_{T}-z_{T-1})> \nu (z^s-z_{T-1})$. And, at $t=T$ we have

  $$\begin{aligned} u(f_1 z_{T})+w(1-z_{T})-\nu (z^s-z_{T}) < u(f_1 z^s)+w(1-z^s)-\nu (z^s-z^s) \end{aligned}$$

  where we are using that $z_{T+1}=\hat{z}_{T+1}= z^s$. Indeed, this last fact implies that $\nu (z^s-z_T)=\nu (z^s-z^s)=0$, and, on the other hand, we have $z^s$ is the unique maximum of $u(f_1z)+w(1-z)$ which gives $u(f_1 z_{T})+w(1-z_{T})<u(f_1 z^s)+w(1-z^s)$. We have shown that the alternative path provides a strictly larger total benefit, reaching a contradiction.

  We now show that the sequence is non-decreasing, again by contradiction. Assume that there is T such that $z_T<z_{T-1}$. We propose the following alternative path

  $$\begin{aligned} \hat{z}_t=\left\{ \begin{array}{ll} z_{T-1} &{} \text { if }t=T \\ z_t &{}\text { if }t\ne T, \end{array} \right. \end{aligned}$$

  i.e., forest cover in stage T is larger along the alternative path than in the original one and the two paths coincide in every other time stage. The total benefit at stage $t=T-1$ is not modified because no reforestation costs are incurred along the two paths. However, at stage $t=T$ the benefit along $\{{\hat{z}}_t\}$ is greater. Indeed, given that $u(f_1z)+w(1-z)$ is concave and maximized at $z^s$ and the fact that $z_{T}<z_{T-1}\le z^s$ we know that $u(f_1z_T)+w(1-z_T) < u(f_1z_{T-1})+w(1-z_{T-1}).$ Regarding the reforestation costs, we have that $z_{T-1}>z_T$ implies $\nu (z_{T+1}-z_{T}) \ge \nu (z_{T+1}-z_{T-1})$.³¹ Thus,

  $$\begin{aligned}&u(f_1 z_{T})+w(1-z_{T})-\nu (z_{T+1}-z_{T}) \\&\quad < u(f_1 z_{T-1})+w(1-z_{T-1})-\nu (z_{T+1}-z_{T-1}). \end{aligned}$$

  that is, $\{{\hat{z}}_t\}$ yields a strictly larger benefit in stage T. The rest of the stage utilities remain unchanged. This yields a contradiction.

$\square$

Proposition 9

Assume $n=1$, then the following are equivalent

• 1. there is global strong conservation

• 2. there is global weak conservation

• 3. $\varDelta (0)>0$.

Proof

The proof of this proposition is as follows. By definition $1\Rightarrow 2$. We prove next that $2\Rightarrow 3$ and $3\Rightarrow 1$.

The function $\varDelta (z)$ is simply $\varDelta (z)=\frac{b}{1-b}[f_1 u'(f_1 z)-w'(1-z)]$. Observe that $z^s=0$ if and only if $\varDelta (0)\le 0$. We have then that total deforestation is optimal from every initial state if $\varDelta (0)\le 0$. Or, equivalently, that if there is a path such that total deforestation is not optimal, then $\varDelta (0)>0$. This is equivalent to $2\Rightarrow 3$.

If $\varDelta (0)>0$, then $z^s>0$. Lemma 5 implies that the forest cover remains always above $\min \{z_o,z^s\}$, which in turn implies that there is global strong conservation. $\square$