The Ambient Charge in Hyperbolic Duopoly and Triopoly: Static and Dynamic Analysis

Abstract

This paper presents a static and a dynamic model of an environmental policy of the ambient tax related to nonpoint source pollution that has many different sources. We apply the model to controllability by the ambient tax rate, showing that increasing the rate reduces the total level of the pollution. We also consider dynamic characteristics in the discrete time scales and numerically show the birth of complex dynamics via period-doubling bifurcation.

Appendices

Appendix 1

In this Appendix, we determine the sigh of the derivative of E ^c with respect to m. Equation (1.14) implies that the sign of the derivative is the same as the sign of aS ₁ − (1 + k + h)S ₂. It is repeated below as (1.19) for convenience,

$$\displaystyle \begin{aligned} sigh\left[ \frac{\partial E^{c}}{\partial m}\right] =sigh\left[ aS_{1}-(1+k+h)S_{2}\right]. {} \end{aligned} $$

(1.19)

In this appendix, we derive the conditions under which aS ₁ − (1 + k + h)S ₂ < 0. It is already shown that S ₁ is negative. Hence if S ₂ ≥ 0, then the sign is definitely negative. So we assume that S ₂ < 0.

We will now prove that the ambient charge is effective in controlling the NPS pollution. We can always number the firms so that

$$\displaystyle \begin{aligned} e_{1}\geq e_{2}\geq e_{3} \end{aligned}$$

or

$$\displaystyle \begin{aligned} 1\geq h\geq k. \end{aligned}$$

From the definition of the feasible set N, the nonnegative conditions of the Cournot outputs can be rewritten as

$$\displaystyle \begin{aligned} a\geq 1-k-h,\ a\geq h-k-1\ \text{and}\ a\geq k-h-1 \end{aligned}$$

from which the first condition is the only meaningful if k + h < 1 which is assumed first.⁵ Then we have

$$\displaystyle \begin{aligned} aS_{1}-(1+k+h)S_{2}=S_{1}(a-S) {} \end{aligned} $$

(1.20)

where

$$\displaystyle \begin{aligned} S=\frac{(1+k+h)S_{2}}{S_{1}}>0\ \text{as}\ S_{1}<0\ \text{and}\ S_{2}<0. \end{aligned}$$

If S < 1 − k − h, then the second factor of the right-hand side of (1.20) is positive,

$$\displaystyle \begin{aligned} a-S>a-\left( 1-k-h\right) \geq 0 \end{aligned}$$

where the last inequality is due to the non-negativity condition of $x_{1}^{c}\geq 0$. Since S ₁ < 0, (1.19) implies ∂E ^c∕∂m < 0. Thus our job is to show that S < 1 − k − h always.

Since S ₁ < 0, then this inequality can be rewritten as

$$\displaystyle \begin{aligned} (1+k+h)S_{2}>\left( 1-k-h\right) S_{1} \end{aligned}$$

or

$$\displaystyle \begin{aligned} -2\left[ 3\left( k^{3}+h^{3}\right) -4\left( k^{2}+h^{2}\right) +3\left( k+h\right) +kh\left( k-4+h\right) -2\right] >0. {} \end{aligned} $$

(1.21)

Since the condition S ₂ < 0 can be rewritten as

$$\displaystyle \begin{aligned} 2\left[ 2\left( k+h+kh\right) -(h^{2}+k^{2})\right] <2, \end{aligned}$$

the square bracketed terms of (1.21) is larger than

$$\displaystyle \begin{aligned} & 3\left( k^{3}+h^{3}\right) -4\left( k^{2}+h^{2}\right) +3\left( k+h\right) +kh\left( k-4+h\right)\\ &\quad -2\left[ 2\left( k+h+kh\right) -(h^{2}+k^{2})\right] \end{aligned} $$

which is, after arranging the terms,

$$\displaystyle \begin{aligned} 3\left( k^{3}+h^{3}\right) -\left( k^{2}+h^{2}\right) -\left( k+h\right) +k^{2}\left( h-1\right) +h^{2}\left( k-1\right) -8kh. \end{aligned}$$

Equation (1.21) is now written as

$$\displaystyle \begin{aligned} 2\left[ k^{2}\left( 1-h\right) +h^{2}\left( 1-k\right) -3\left( k^{3}+h^{3}\right) +\left( k^{2}+h^{2}\right) +\left( k+h\right) +8kh\right] . {}\end{aligned} $$

(1.22)

However 1 − k > h and 1 − h > k, (1.22) is larger than

$$\displaystyle \begin{aligned} \begin{array}{c} 2\left[ -2\left( k^{3}+h^{3}\right) +\left( k^{2}+h^{2}\right) +\left( k+h\right) +8kh\right] \\ \\ =2\left[ k^{2}(1-k)+k\left( 1-k^{2}\right) +h^{2}(1-h)+h\left( 1-h^{2}\right) +8kh\right] >0. \end{array} \end{aligned}$$

Hence S < 1 − k − h holds if k + h < 1. If k + h ≥ 1, then

$$\displaystyle \begin{aligned} S_{2}=2(h+k)^{2}-(h-k)^{2}-1\geq 0 \end{aligned}$$

since h ≤ 1 and k ≤ 1 are assumed. Hence ∂E ^c∕∂m is always negative.

Appendix 2

In this Appendix, we examine the signs of the derivatives in (1.16) and show that the ambient charge could have perverse effect on the individual level of emission.

Lemma 1

$\dfrac {\partial x_{1}^{c}}{\partial m}\geq 0$ is possible if a ≥ a ₂ and x ₁ ≤ k + h ≤ x ₂ where

$$\displaystyle \begin{aligned} a_{2}=2(5+2\sqrt{6}),\ \end{aligned}$$

$$\displaystyle \begin{aligned} x_{1}=\frac{a-\sqrt{a^{2}-20a+4}}{2} \end{aligned}$$

and

$$\displaystyle \begin{aligned} x_{2}=\frac{a+\sqrt{a^{2}-20a+4}}{2}. \end{aligned}$$

Proof

A new variable x = k + h is introduced. Then the square bracketed terms of the first equation of (1.16) can be rewritten as

$$\displaystyle \begin{aligned} f(x)=-x^{2}+ax+(1-5a) \end{aligned}$$

implying that

$$\displaystyle \begin{aligned} sigh\left[ \frac{\partial x_{1}^{c}}{\partial m}\right] =sigh\left[ f(x) \right]. \end{aligned}$$

Solving f ^′(x) = 0 presents x _m = a∕2 and the corresponding maximum value, f(x _m) = D∕4 with D = a ² − 20a + 4. It is checked that D = 0 is attained at the following two points,

$$\displaystyle \begin{aligned} a_{1}=2\left( 5-2\sqrt{6}\right) \simeq 0.202\ \text{and}\ a_{2}=2\left( 5+2 \sqrt{6}\right) \simeq 19.798 \end{aligned}$$

indicating that D ≥ 0 for a ≤ a ₁ or a ≥ a ₂ and D < 0 for a ₁ < a < a ₂. Solving f(x) = 0 yields two solutions,

$$\displaystyle \begin{aligned} x_{1}=\frac{a-\sqrt{D}}{2}\ \text{and}\ x_{2}=\frac{a+\sqrt{D}}{2}. \end{aligned}$$

Taking account of $f(0)\gtreqless 0$ depending on $a\lesseqgtr 1/5,\$we then identify four cases according to the value of a, 0 < a < 1∕5, 1∕5 ≤ a ≤ a ₁, a ₁ < a < a ₂ and a ≥ a ₂.

• (i) In case of 0 < a < 1∕5, we have x ₁ < 0 and x ₂ > 0. Hence, f(x) ≥ 0 for 0 ≤ x ≤ x ₂ and f(x) < 0 for x > x ₂. Returning to the definition of x, x ≤ x ₂ can be rewritten as

  $$\displaystyle \begin{aligned} k+h\leq \frac{a+\sqrt{D}}{2}. \end{aligned}$$

  On the other hand, the non-negativity condition of $x_{1}^{c}$ is given by

  $$\displaystyle \begin{aligned} k+h\geq 1-a. \end{aligned}$$

  Since it can be confirmed that

  $$\displaystyle \begin{aligned} (1-a)-\frac{a+\sqrt{D}}{2}>0\ \text{for}\ a<0.2, \end{aligned}$$

  the condition x ≤ x ₂ does not satisfy the non-negativity so that we can eliminate it for further consideration. Hence we have f(x) < 0 and therefore $\partial x_{1}^{c}/\partial m<0$ for x > x ₂.

• (ii) In case of 1∕5 ≤ a ≤ a ₁, we have 0 ≤ x ₁ < x ₂, leading to the following,

  $$\displaystyle \begin{aligned} \begin{array}{l} (\text{ii-a})\ f(x)\leq 0\ \text{for}\ 0\leq x\leq x_{1},\medskip \\ (\text{ii-b})\ f(x)\geq 0\ \text{for}\ x_{1}\leq x\leq x_{2},\medskip \\ (\text{ii-c})\ f(x)<0\ \text{for}\ x>x_{2}. \end{array} \end{aligned}$$

  Under the condition of 0.2 ≤ a ≤ a ₁, we can confirm that

  $$\displaystyle \begin{aligned} 1-a>x_{2}>x_{1}. \end{aligned}$$

  Hence x satisfying 0 ≤ x ≤ x ₂ violates the non-negativity condition. Therefore we can have only the last case.

• (iii) In case of a ₁ < a ≤ a ₂, D ≤ 0 where the equality holds only when a = a ₂. Hence f(x) < 0, so $\partial x_{1}^{c}/\partial m<0$ for x ₁ < x < x ₂.

• (iv) In case of a > a ₂, we have 0 < x ₁ < x ₂ and then

  $$\displaystyle \begin{aligned} \begin{array}{l} (\text{iv-a})\ f(x)\leq 0\ \text{for}\ 0\leq x\leq x_{1},\medskip \\ (\text{iv-b})\ f(x)\geq 0\ \text{for}\ x_{1}\leq x\leq x_{2},\medskip \\ (\text{iv-c})\ f(x)<0\ \text{for}\ x>x_{2}. \end{array} \end{aligned}$$

Summarizing the results, we have

$$\displaystyle \begin{aligned} \dfrac{\partial x_{1}^{c}}{\partial m}\geq 0\ \text{is}\ \text{possible}\ \text{if}\ a\geq a_{2}\ \text{and}\ x_{1}\leq k+h\leq x_{2}. \end{aligned}$$

This completes the proof. ■

We now proceed to the second equation of (1.16).

Lemma 2

$\dfrac {\partial x_{2}^{c}}{\partial m}\geq 0$ is possible if a ≥ 1 and k ≥ 5h − 1.

Proof

We denote by g ₁(h, k) the square bracketed terms of the second equation of (1.16),

$$\displaystyle \begin{aligned} g_{1}(h,k)=a(1+k-5h)-\left( 1+k-h\right) (1+h+k) \end{aligned}$$

where

$$\displaystyle \begin{aligned} sigh\left[ \frac{\partial x_{2}^{c}}{\partial m}\right] =sigh\left[ g_{1}(h,k)\right]. \end{aligned}$$

The non-negativity condition of $x_{2}^{c}$ is a ≥ h − k − 1. We identify the following three cases, (i) h − k − 1 ≥ 0, (ii) h − k − 1 < 0 with 1 + k − 5h ≤ 0, and (iii) 1 + k − 5h > 0 under which h − k − 1 < 0 always holds.

• (i) Since − (1 + k − h) ≥ 0 is assumed, multiplying both sides of the non-negativity condition by 1 + h + k presents

  $$\displaystyle \begin{aligned} a\left( 1+k+h\right) \geq -(1+k-h)\left( 1+h+k\right) \geq 0. \end{aligned}$$

  The first inequality implies that the bracketed term is less than or equal to

  $$\displaystyle \begin{aligned} a(1+k-5h)-a\left( 1+k+h\right) =2a(1+k-2h)<0 \end{aligned}$$

  where the last inequality is due to the assumption h − k − 1 ≥ 0 or 1 + k − h ≤ 0. Therefore g ₁(h, k) < 0 for any a ≥ 0 if h − k − 1 ≥ 0 .

• (ii) Now suppose that h − k − 1 < 0 with which the non-negativity condition is always satisfied. The second term of the bracketed terms is negative. So if 1 + k − 5h ≤ 0, then g ₁(h, k) < 0 for any a ≥ 0.

• (iii) Assume finally that 1 + k − 5h > 0. Then g ₁(h, k) is a linear function of a with a positive slope. The non-negativity condition for the equilibrium outputs are satisfied if

  $$\displaystyle \begin{aligned} a\geq \max \left\{ h-1-k;\ 1-h-k;\ k-1-h\right\} =a^{\ast }. \end{aligned}$$

The value of g ₁(h, k) is zero if

$$\displaystyle \begin{aligned} a=\Omega (h,k)=\frac{\left( 1+k-h\right) (1+h+k)}{1+k-5h}. \end{aligned}$$

We can next prove that Ω(h, k) > 1, which can be written as

$$\displaystyle \begin{aligned} \left( 1+k-h\right) (1+h+k)-\left( 1+k-5h\right) >0. \end{aligned}$$

The simplified left-hand side satisfies

$$\displaystyle \begin{aligned} \begin{array}{l} 1+k+h+k+kh+k^{2}-h-h^{2}-kh-1-k+5h\medskip \\ =k+k^{2}+5h-h^{2}\medskip \\ =k(k+1)+5h-h^{2}\medskip \\ >\left( 5h-1\right) 5h+5h-h^{2}\medskip \\ =24h^{2}>0. \end{array}\end{aligned} $$

Therefore g ₁(h, k) > 0 with positive equilibrium output value if

$$\displaystyle \begin{aligned} a>\max \left\{ a^{\ast },\ \Omega (h,k)\right\}\end{aligned} $$

which completes the proof. ■

Notice that the sign of a − Ω(h, k) > 1 is the same as that of

$$\displaystyle \begin{aligned} \begin{array}{l} a+ak-5ah-\left( 1+k-h+k+kh+k^{2}-h-h^{2}-kh\right) \\ =-k^{2}+k(a-2)+\left( a-5ah-1+h^{2}\right) \end{array} \end{aligned}$$

which is a concave parabola with roots

$$\displaystyle \begin{aligned} k_{1}=f_{1}(h)=\frac{1}{2}\left[ -2+a-\sqrt{a^{2}-20ah+4h^{2}}\right] \end{aligned}$$

and

$$\displaystyle \begin{aligned} k_{2}=f_{2}(h)=\frac{1}{2}\left[ -2+a+\sqrt{a^{2}-20ah+4h^{2}}\right] \end{aligned}$$

In Fig. 1.3a, four blue curves $\left (\mathrm {i.e.,} \ k_{2}=f_{2}(h)\right )$ and three red curves (i.e., k ₁ = f ₁(h)) are illustrated for a = 2, 3, 4, 5 where the curves shift outward as a increases. Notice that a >  Ω(h, k) holds in each gray region, that is, g ₁(h, k) > 0.

The derivative $\partial x_{3}^{c}/\partial m$ can be obtained from $\partial x_{2}^{c}/\partial m$ by interchanging h and k, so from Lemma 2, we have the following result.

Lemma 3

$\dfrac {\partial x_{3}^{c}}{\partial m}\geq 0$ is possible if a > 1 and h > 5k − 1.

As in the previous case, we can have the roots of equation a − Ω(k, h) as

$$\displaystyle \begin{aligned} h_{1}=f_{1}(k)=\frac{1}{2}\left[ -2+a-\sqrt{a^{2}-20ak+4k^{2}}\right] \end{aligned}$$

and

$$\displaystyle \begin{aligned} h_{2}=f_{2}(k)=\frac{1}{2}\left[ -2+a+\sqrt{a^{2}-20ak+4k^{2}}\right] \end{aligned}$$

where are illustrated in Fig. 1.3b.

These lemmas lead to Theorem 3.