1 Introduction

At the heart of economic analysis is the concept of optimization under constraints, and the most common application is through dynamic optimization. There are in essence three major approaches, viz. that of dynamic programming, calculus of variations and its extension which is optimal control theory (see, for example, Chiang [1]).

Many of the models utilize the Lagrangian or its generalization, and the current value Hamiltonian whenever the integrand function contains a discount factor. These models range from the simplest to those of neoclassical economic growth [2, 3], the optimal firm-level investment [4] and human capital and earnings [5] models.

The Euler–Lagrange equations give necessary conditions for extremal values of the functional which has the Lagrangian as its integrand. This involves the time variable, the state variables and its derivatives up to some finite order. Many of the problems give rise to a Lagrangian which is of the first order. Thus, in this case one obtains a dynamical system of the second order. Further advances have led to the development of optimal control theory (see, for example, [1]) which in addition to the time and state variables also incorporates the control and costate variables. In the latter approach, Pontryagin’s maximum principle provides necessary conditions for the resolution of the continuous time optimal control problem which has a current value Hamiltonian, and therefore, a first-order system of dynamical ODEs is formulated for the control, state and costate variables.

There are several dynamic economic models which pervade the existing literature. Different approaches have been advocated to deal with these models. Predominantly qualitative theory and linear approximations around steady states (see [6–10]) have been applied to analyze these model. Numerical methods too have been utilized to solve the nonlinear dynamical systems at hand (see, for example, [11]). Notwithstanding analytical solutions as well as existence and uniqueness, theorems have proved beneficial in the analysis of such models (see, for example, [12–18]).

A separate strand of the literature has looked at the first integrals or conservation laws for differential equations (see, for example, [19–26]). Kara and Mahomed [20] developed the partial Lagrangian approach to constructing conservation laws for partial differential equations (PDEs). All different approaches to constructing conservation laws for PDEs are discussed by Naz et al. [21]. Gan and Qu [22] studied the approximate conservation laws of perturbed PDEs. Kara et al. [23] provided the methodology to establish the first integrals for ODEs. A review of different approaches to deriving first integrals for ODEs is provided by Naz et al. [24]. The approximate first integrals for nonlinear oscillators are discussed in [25, 26]. Dorodnitsyn and Kozlov [27] analyzed the invariance and first integrals of continuous and discrete Hamiltonian equations. In a recent study [28, 29], the authors proposed a partial Hamiltonian approach for many state, costate and control variables which takes into account the intrinsic and desirable current value Hamiltonian to unearth first integrals and reductions of the economic models under study. This approach provides a general and algorithmic procedure for the search for reductions and solutions of such type of nonlinear economic equations. The characterization of Hamiltonian symmetries and their first integrals is given in [30].

Significant contributions have been made in the analysis and solutions of nonlinear dynamical systems of economic models.

In the present study, we focus on a simple approach which yields first integrals, reductions and closed-form solutions (if there are sufficient operators) for dynamical systems of ODEs arising in economic growth theory. We utilize a partial Lagrangian or what we call a discount free Lagrangian framework for several state variables. This approach is applicable if one is able to construct a partial or discount free Lagrangian for the underlying dynamical system. We point out that this is not always possible as a system of first-order ODEs which is a consequence of optimal control theory may not in general be cast as a second-order system. Therefore, the method we invoke is applicable to an arbitrary system of ODEs provided that a partial or discount free Lagrangian exist. The utility of this method lies in the fact that the control variables, which occurs in the case of the partial Hamiltonian approach, do not arise in the unfolding analysis as the discount free Lagrangian is independent of these control variables from the outset. We apply this approach to a system of two ODEs which involves time, one state variable and its first derivative to show its effectiveness.

The outline of the paper is as follows. In Sect. 2, we provide the overviews of the partial Hamiltonian approach for what follows. The partial or discount free Lagrangian approach for dynamical systems is developed in Sect. 3. Herein we show how it naturally arises from the partial Hamiltonian by means of the partial Legendre transformation. Definitions and propositions are provided on the notion of discount factor free Lagrangians with the third one providing an algorithm on finding first integrals corresponding to this discount free Lagrangian in a simple manner. In Sect. 4, we provide examples using this approach of three fundamental Economic growth models. We compare our results with the partial Hamiltonian method to show its utility. Concluding remarks are finally given in Sect. 5.

2 Preliminaries

Let t be the independent variable which is usually time and \((q,p)=(q^1,\ldots ,q^n, p_1,\ldots ,p_n)\) the phase space coordinates. In the applications to equations of economics, \(q^1,\ldots ,q^n\) are the state variables and \(p_1,\ldots ,p_n\) the costate variables. We provide here the overview of partial Hamiltonian approach. The following definitions and results are adapted from [1, 27–29].

Definition 1

The Euler operator \({\delta }/{\delta q^i}\) and the variational operator \({\delta }/{\delta p_i}\) are defined as

$$\begin{aligned} \frac{\delta }{\delta q^i}=\frac{\partial }{\partial q^i}-D\frac{\partial }{\partial \dot{q}^i},\quad i=1,2, \ldots , n, \end{aligned}$$
(1)

and

$$\begin{aligned} \frac{\delta }{\delta p_i}=\frac{\partial }{\partial p_i}-D\frac{\partial }{\partial \dot{p}_i},\quad i=1,2, \ldots , n, \end{aligned}$$
(2)

where

$$\begin{aligned} D=\frac{\partial }{\partial t}+\dot{q}^i\frac{\partial }{\partial q^i}+\dot{p}_i\frac{\partial }{\partial p_i}+\cdots \end{aligned}$$
(3)

is the total derivative operator with respect to the time t. The summation convention applies for repeated indices here and in the sequel.

The variables t, q, p are independent and connected by the differential relations

$$\begin{aligned} \dot{p}_i=D(p_i),\; \dot{q}^i=D(q^i),\quad i=1,2,\ldots ,n. \end{aligned}$$
(4)

In economic analysis, the optimal control problem is stated as (see, for example, [1])

$$\begin{aligned} \text {Maximize}\; {\mathcal {F}}= & {} \int _0^TF(t,q,c)\;\mathrm{e}^{-\rho t}\;\hbox {d}t\nonumber \\ \text {subject to}\; \dot{q}^i= & {} f^i(t,q,c),\quad i=1,\ldots ,n, \end{aligned}$$
(5)

where c is the control vector \(c=(c_1,\ldots ,c_m)\), \(m\le n\) or \(m>n\) as well as appropriate boundary conditions imposed. The integrand contains the discount factor \(\mathrm{e}^{-\rho t}\). We can have minimization problems as well.

Definition 2

The present value Hamiltonian is defined as

$$\begin{aligned} \bar{H}=F(t,q,c)\mathrm{e}^{-\rho t}+\bar{p}_if^i(t,q,c). \end{aligned}$$
(6)

The current value Hamiltonian H(t, q, p, c) (see, for example, [1]) is

$$\begin{aligned} H=F(t,q,c)+p_if^i(t,q,c), \end{aligned}$$
(7)

where the following relations hold, viz.

$$\begin{aligned} p_i=\bar{p}_i\mathrm{e}^{\rho t},\; H=\bar{H}\mathrm{e}^{\rho t}, q^i=\bar{q}^i, c_i=\bar{c}_i. \end{aligned}$$
(8)

This current value Hamiltonian is independent of the discount factor and deemed more desirable in economic analysis [1].

The maximum principle gives necessary conditions for optimal control for \(\bar{H}\)

$$\begin{aligned} \frac{\partial \bar{H}}{\partial \bar{c}_i}= & {} 0,\; \frac{\partial ^2 \bar{H}}{\partial \bar{c}_i{^2}}<0\nonumber \\ \bar{ \dot{q}}^i= & {} \frac{\partial \bar{H}}{\partial \bar{p}_i},\nonumber \\ \bar{\dot{p}}^i= & {} -\frac{\partial \bar{H}}{\partial \bar{q}^i},\quad i=1,\ldots ,n. \end{aligned}$$
(9)

and these conditions in terms of the current value Hamiltonian H defined in (7) satisfy the system

$$\begin{aligned} \frac{\partial {H}}{\partial {c}_i}= & {} 0,\; \frac{\partial ^2 {H}}{\partial {c}_i{^2}}<0\nonumber \\ \dot{q}^i= & {} \frac{\partial H}{\partial p_i}, \nonumber \\ \dot{p}^i= & {} -\frac{\partial H}{\partial q^i}+\varGamma _i,\quad i=1,\ldots ,n, \end{aligned}$$
(10)

where we take \(\varGamma _i\) as a nonzero function of \(t, p_i, q^i\) in general. Note that each \(\varGamma _i\) is mostly taken as a linear function of \(p_i\) in economic applications in the earlier literature of optimal control theory in which the functional maximized contains the discount factor \(\mathrm{e}^{-\rho t}\) in the integrand [1] and this is known as constant time preferences in economic growth theory. For endogenous time preferences, \(\varGamma _i\) is taken more generally. The sufficiency conditions relate the \(p_i\) and \(c_i\). In correspondence to mechanics, they deal with non-conservative forces if the \(\varGamma _i\) are nonzero.

We have that the operator X given by

$$\begin{aligned} X=\xi (t,q,p)\frac{\partial }{\partial t}+ \eta ^i (t, q,p)\frac{\partial }{\partial q^i} + \zeta _i (t,q,p) \frac{\partial }{\partial p_i} \end{aligned}$$
(11)

is a generator of point symmetry of the current value Hamiltonian system of ODEs (10) if the following determining system involving the coefficient functions \(\xi \), \(\eta ^i\) and \(\zeta _i\), viz.

$$\begin{aligned} \dot{\eta }^i-\dot{q}^i\dot{\xi }-X\left( \frac{\partial H}{\partial p_i}\right)= & {} 0,\nonumber \\ \dot{\zeta }_i-\dot{p}_i\dot{\xi }+X\left( \frac{\partial H}{\partial q^i}-\varGamma _i\right)= & {} 0,\quad i=1,\ldots ,n \end{aligned}$$
(12)

identically holds on the system (10).

Definition 3

(see [28]). An operator X of the form (11) is said to be a partial Hamiltonian operator corresponding to a current value Hamiltonian H(t, q, p, c) which satisfies (7), if there exists a function B(t, q, p, c) such that

$$\begin{aligned}&\zeta _i \frac{\partial H}{\partial p_i}+p_i D(\eta ^i)-X(H) -H D(\xi )\nonumber \\&\quad =D(B) +\left( \eta ^i -\xi \frac{\partial H}{\partial p_i}\right) (-\varGamma _i) \end{aligned}$$
(13)

holds.

Note that it is important to point out that if H is a present value Hamiltonian, then Eq. (12) becomes the usual determining equation for symmetries of the Hamiltonian action since \(\varGamma _i=0\) in this case. Also in this case (13) with \(\varGamma _i=0\) (see [27]), the Hamiltonian action with gauge term is invariant with respect to the group generated by operator X of (11).

The following theorem is essential for the construction of first integrals for the system (10).

Theorem 1

(see [28]). The first integral corresponding to the system (10) associated with a partial Hamiltonian operator X of the current value Hamiltonian H(t, q, p, c) is determined from

$$\begin{aligned} I= p_i \eta ^i -\xi H-B, \end{aligned}$$
(14)

where B(t, p, q, c) is a gauge-like function.

If \(\varGamma _i=0\) and \(B=B(t,p,q)\), then this formula (14) is valid as well for an invariant Hamiltonian action up to divergence [27].

3 A partial Lagrangian approach for dynamical systems

We provide the connection between a partial Hamiltonian and what we term a partial or discount free Lagrangian when it exists. The dynamic optimization problems can be treated by the calculus of variation techniques and the optimal control theory. In the definition given below, we define the optimal control problem stated in (5) in terms of the calculus of variation problem.

Definition 4

The optimal control problem stated in (5) can be re-cast as the following calculus of variations problem

$$\begin{aligned} \mathrm{Maximize}\; {\mathcal {F}}=\int _0^TF(t,q,\dot{q})\;\mathrm{e}^{-\rho t}\;\hbox {d}t \end{aligned}$$
(15)

under the conditions of invertibility given by \(p_i=g_i(t,q^i,\dot{q}^i)\) and the relationship between the control vector c and costate vector p which arises from the sufficient conditions of optimal control from the maximum principle yielding \(c_i=h_i(p_i)\). Also appropriate boundary conditions are imposed.

Next, we define the partial Legendre transformation in the more general setting with \(p_i={\partial L}/{\partial \dot{q}^i}\) which relates the current value Hamiltonian and what we define as a partial or discount free Lagrangian when it exists.

Proposition 1

(Partial Legendre transformation) If H is the current value Hamiltonian, then under conditions of invertibility given by \(p_i=g_i(t,q^i,\dot{q}^i)\) and the relationship between the control vector c and costate vector p which arises from the sufficient conditions of optimal control from the maximum principle yielding \(c_i=h_i(p_i)\), there exists the partial or discount free Lagrangian \(L(t,q^i,\dot{q}^i)\) satisfying

$$\begin{aligned} L\left( t,q^i,\dot{q}^i\right) =p_i\dot{q}^i-H\left( t,q^i,p_i,c_i\right) . \end{aligned}$$
(16)

The relation (16) is called a partial Legendre transformation.

Proof

The Legendre transformation [31] connects a standard Lagrangian \(\bar{L}\) and canonical Hamiltonian \(\bar{H}\) as

$$\begin{aligned} \bar{L}\left( t,q^i,\dot{q}^i\right) =\bar{p}_i\dot{q}^i-\bar{H}\left( t,q^i,\bar{p}_i,c_i\right) . \end{aligned}$$
(17)

The coordinate transformation \( p_i=\bar{p}_i\mathrm{e}^{\rho t},\; H=\bar{H}\mathrm{e}^{\rho t}, q^i=\bar{q}^i, c_i=\bar{c}_i,\;L=\bar{L}\mathrm{e}^{\rho t},\) directly transforms (17) to the partial Legendre transformation (16). We term L as a partial or discount free Lagrangian and H is the current value Hamiltonian. \(\square \)

Proposition 2

(Partial Euler–Lagrange equations) If H is the current value Hamiltonian, then under conditions of invertibility given by \(p_i=g_i(t,q^i,\dot{q}^i)\) and the relationship between the control vector c and costate vector p which arises from the sufficient conditions of optimal control from the maximum principle yielding \(c_i=h_i(p_i)\), there exists the discount factor independent Lagrangian \(L(t,q^i,\dot{q}^i)\) or discount free Lagrangian via the partial Legendre transformation (16) which satisfies

$$\begin{aligned} \frac{\delta L}{\delta q^i}=-\varGamma _i. \end{aligned}$$
(18)

We call Eq. (19) the partial Euler–Lagrange equations motivated by Kara and Mahomed [20] and Kara et al. [23].

Proof

Indeed, the action of the variational operator \({\delta }/{\delta p_i}\) on the partial Legendre transformation \(p_i\dot{q}^i-H\) (16) after use of (10) yields \({\delta L}/{\delta p_i}=0\). Further, if one acts with the Euler operator \({\delta }/{\delta q^i}\) on the partial Legendre transformation (16), then one straightforwardly obtains \(-{\partial H}/{\partial q^i}-\dot{p}_i\). The latter is equal to \(-\varGamma _i\) as a consequence of the second set of (10). We precisely arrive at

$$\begin{aligned} \frac{\delta L}{\delta q^i}=-\varGamma _i \end{aligned}$$
(19)

provided one can invert for the costate vector \(p_i=g_i(t,q^i,\dot{q}^i)\) and moreover, one requires that the control variables \(c_i\) be determined explicitly in terms of \(t,q^i\) and \(\dot{q}^i\) as H in the partial Legendre transformation (16) has explicit c in it. Note that there is a relation between the variables c and p due to the sufficient conditions for optimal control for the maximum principle. This completes the proof. \(\square \)

Remark 1

We have an alternative proof of Proposition 2. The standard Lagrangian for the calculus of variations problem as stated in (15) is \(\bar{L}=F(t,q^i,\dot{q}^i)\;\mathrm{e}^{-\rho t}\). If one acts with the Euler operator \({\delta }/{\delta q^i}=0\) on \(\bar{L}\) and using the fact that \({\delta \bar{L}}/{\delta q^i}=0\), one can easily arrive at

$$\begin{aligned} \frac{\delta F}{\delta q^i}=-\rho \frac{\partial F}{\partial \dot{q}^i}. \end{aligned}$$
(20)

Setting \(L=-F\), \(p_i= \frac{\partial L}{\partial \dot{q}^i}\) and \(\varGamma _i=\rho p_i\) Eq. (20) provides the partial Euler–Lagrange equations (19). Then, \(L=-F\) is the partial or discount free Lagrangian.

The relation (19) to be valid, we must have that L exists as a function of the time, the state variables and their derivatives only. In this case, we say that L is a partial Lagrangian corresponding to the partial Hamiltonian H with the equivalence being provided by the partial Legendre transformation (16).

In the same way that the current value Hamiltonian (which is the partial Hamiltonian) which is free of the discount factor is desirable, the partial or discount free Lagrangian that is defined by means of (19) also displays the same advantages. It is independent of time. We call it a discount free Lagrangian.

We therefore have the following important result for discount factor free Lagrangians that arise in economics.

Proposition 3

Suppose that the Lagrangian L which is a discount free Lagrangian corresponding to the current value Hamiltonian H satisfies the partial Euler–Lagrange equation (19) and X as given by

$$\begin{aligned} X=\xi \;{{\partial }\over {\partial t}}+\eta ^i\;{{\partial }\over {\partial q^i}}+\zeta _{t}^i\;{{\partial }\over {\partial \dot{q}^i}}, \end{aligned}$$
(21)

where \(\zeta _t^i=D \eta ^i-\dot{q}^iD \xi \), is a partial Noether operator corresponding to L. Then, X satisfies

$$\begin{aligned} X(L)+LD(\xi )=D(B)+(\eta ^i -\xi \dot{q}^i)(-\varGamma _i),\nonumber \\ i=1,\ldots , n \end{aligned}$$
(22)

and the corresponding first integral associated with L is obtained by the explicit formula

$$\begin{aligned} I=\xi L+\left( \eta ^i -\xi \dot{q}^i\right) \frac{\partial L}{\partial \dot{q}^i}-B, \end{aligned}$$
(23)

for a suitable gauge term \(B(t,q^i,\dot{q}^i)\).

Proof

The partial Hamiltonian operator determining Eq. (13) with the help of partial Legendre transformation (16) yields

$$\begin{aligned}&\zeta _i{\dot{q}}^i+p_iD\eta ^i-\zeta _i{\dot{q}}^i-p_i\zeta _t^i+XL\nonumber \\&\quad -\,p_i{\dot{q}}^iD\xi +LD\xi =DB+\left( \eta ^i-\xi {\dot{q}}^i\right) (-\varGamma _i)\nonumber \\ \end{aligned}$$
(24)

which easily results in (22) after replacing \(\zeta ^i_t\) with \(D\eta ^i-{\dot{q}}^iD\xi \). For the first integral I, Eq. (14) with the aid of partial Legendre transformation (16) yields

$$\begin{aligned} I= \eta ^i \frac{\partial L}{\partial \dot{q}^i} -\xi \left( \dot{q}^i \frac{\partial L}{\partial \dot{q}^i}-L\right) -B, \end{aligned}$$
(25)

and this is (23). Note that we have \(p_i=g_i(t,q,\dot{q})\) in view of Definition 4 and thus \(\varGamma _i\) and B are in terms of the variables \(t,q^i,\dot{q}^i\). This completes our proof. \(\square \)

The approach presented here is termed as a “partial or discount free Lagrangian approach for dynamical systems.” This provides a direct methodology to construct first integrals for first-order dynamical or Hamiltonian systems arising from optimal control problems. These types of systems arise naturally in economics. The partial or discount free Lagrangian is always of first order. The partial Lagrangian approach [20, 23] can have first-order and higher-order partial Lagrangians. Moreover, it works for even-ordered differential equations or systems of differential equations. One can apply it to odd-ordered differential equations or systems of differential equations by increasing the order of differential equation to an even order.

Remark 2

The partial Hamiltonian approach developed in [28] and the partial Lagrangian approach presented here both are applicable provided the dynamic optimization problem can be stated in optimal control as well as in the calculus of variations form. This depends on invertibility conditions of the control variables. We cannot apply the partial Lagrangian approach if the problem cannot be stated in the calculus of variations form, but we can solve those dynamic optimization problems by utilizing the more generalized partial Hamiltonian approach.

We discuss two typical examples of control problems in economics. One can be expressed as a problem in the calculus of variations in a simple way, and in the second, it is quite cumbersome and the problem is better treated in its original formulation by using a Hamiltonian.

1. Maximize

$$\begin{aligned} \int _0^{\infty } \left[ \alpha q-\beta q^2- \alpha u^2- \gamma u\right] e^{-rt}\hbox {d}t \end{aligned}$$
(26)

subject to

$$\begin{aligned} \dot{q}=u, \end{aligned}$$
(27)

where \(\alpha , \beta , \gamma \) are positive constants, r a constant discount factor, q(t) the state variable, and u(t) is the control variable. It is easy to observe that we can state this problem equivalently as an infinite-horizon problem in the calculus of variations as

Maximize

$$\begin{aligned} \int _0^{\infty } \left[ \alpha q-\beta q^2- \alpha \dot{q}^2- \gamma \dot{q}\right] e^{-rt}\hbox {d}t \end{aligned}$$
(28)

by simply expressing u in terms of \(\dot{q}\) through Eq. (27). This problem can be further simplified by eliminating the discount factor using the present formulation of a discount free Lagrangian. Here both the partial Hamiltonian and partial Lagrangian approaches are applicable.

2. In the model of Lucas–Uzawa [14, 15], the representative agents utility function is defined as

$$\begin{aligned} \mathrm{Max}_{c,u} \quad \int _0^{\infty }\frac{c^{1-\sigma }-1}{1-\sigma } e^{-\rho t}, \; \sigma \not =1 \end{aligned}$$
(29)

subject to the constraints of physical and human capital:

$$\begin{aligned} \dot{k}(t)= & {} \gamma k^\beta (uh)^{1-\beta } -\pi k-c, \; k_0=k(0), \nonumber \\ \dot{h}(t)= & {} \delta (1-u)h,\; h_0=h(0), \end{aligned}$$
(30)

where \(1/{\sigma }\) is the constant elasticity of intertemporal substitution, \(\rho >0\) is the discount factor, \(\beta \) is the elasticity of output with respect to physical capital, \(\gamma >0\) is the technological levels in the goods sector, \(\delta >0\) is the technological levels in the education sector, k is the level of physical capital, h is the level of human capital, c is per capita consumption, and u is the fraction of labor allocated to the production of physical capital. Here we see that it is quite cumbersome to express c and u in terms of the state variables and their derivatives. Thus, we cannot easily cast this problem in the calculus of variations as its description is unduly complicated. Here we can utilize the partial Hamiltonian approach to finding the first integrals and closed-form solution.

We invoke our approach in the next section and demonstrate their usefulness in a simple way.

4 Applications to economics

In this section, we explain our method via three economic growth models: capitalists decision model of endogenous growth, the Ramsey growth model and a one-sector economic growth model with logarithmic utility preferences. First we solve a simple capitalists decision model of endogenous growth to demonstrate how our approach works. Then, we solve the Ramsey growth model and a one-sector economic growth model with logarithmic utility preferences. Moreover, we compare our results for the Ramsey model derived in [28] by the partial Hamiltonian approach.

4.1 Capitalists decision model of endogenous growth

Consider the one-sector capitalists decision model of endogenous growth presented in [32]. The decision problem faced by capitalists is

$$\begin{aligned} \mathrm{Max}_c \quad \int _0^{\infty } e^{-\rho t}\frac{c^{1-\theta }-1}{1-\theta }\hbox {d}t ,\; \theta >0,\;\theta \not =1 \end{aligned}$$
(31)

subject to

$$\begin{aligned} c(t)+i(t)= & {} [1-\tau _k]r k(t)+\tau _k \delta k(t), \end{aligned}$$
(32)
$$\begin{aligned} \dot{k}= & {} i(t)-\delta k(t), \end{aligned}$$
(33)

where \(\rho \) is the constant rate of time preference, \({1}/{\theta }\) is the intertemporal elasticity of substitution, c(t) is the consumption per worker, i(t) is investment, k(t) is the physical capital per worker, r is rental rate of capital to competitive firms, \(\tau _k\) is tax rate for gross rental income, and \(\delta \) is the depreciation rate of capital. Equations (32) and (33) yield

$$\begin{aligned} \dot{k} =(1-\tau _k)(r-\delta )k-c. \end{aligned}$$
(34)

The initial and transversality conditions are of the following form:

$$\begin{aligned} c(0)=c_0, \;k(0)=k_0, \end{aligned}$$
(35)

and

$$\begin{aligned} \lim _{t\rightarrow \infty } e^{-\rho t} c(t)^{-\theta } k(t)=0 . \end{aligned}$$
(36)

The analogue problem in the calculus of variations can be formulated by expressing c in terms of k and \(\dot{k}\) from Eq. (34) and substituting this into Eq. (31). The optimal control problem (31) and (34) in the calculus of variations takes the following form:

$$\begin{aligned}&\mathrm{Max}_k \quad \int _0^{\infty } e^{-\rho t}\frac{\left[ -\dot{k} +(1-\tau _k)(r-\delta )k\right] ^{1-\theta }-1}{1-\theta }\hbox {d}t,\nonumber \\&\quad k(0)=k_0. \end{aligned}$$
(37)

We utilize the partial Lagrangian approach to deriving first integrals for this problem as the partial Lagrangian is time independent, i.e., independent of the discount factor. The desirable partial Lagrangian for our model is

$$\begin{aligned} L= \frac{\left[ -\dot{k} +(1-\tau _k)(r-\delta )k\right] ^{1-\theta }}{\theta -1} +\frac{1}{{1-\theta }}. \end{aligned}$$
(38)

It is worthy to re-iterate here that the partial Lagrangian is autonomous. The partial Euler–Lagrange equation by Eq. (19) is

$$\begin{aligned} \frac{\delta L}{\delta k}=-\rho \left[ -\dot{k} +(1-\tau _k)(r-\delta )k\right] ^{-\theta }. \end{aligned}$$
(39)

The partial Noether operators determining Eq. (22) in terms variables t and k results in

$$\begin{aligned} XL+LD_t(\xi )=D_t(B)+(\eta -\xi \dot{k})\frac{\delta L}{\delta k}, \end{aligned}$$
(40)

where X is the first-order prolonged operator given by

$$\begin{aligned} X= & {} \xi (t,k) \frac{\partial }{\partial t}+\eta (t,k) \frac{\partial }{\partial k}\nonumber \\&+\,\left[ \eta _t+\dot{k}(\eta _k-\xi _t)-\dot{k}^2 \xi _k\right] \frac{\partial }{\partial \dot{k}}. \end{aligned}$$
(41)

Equation (40) for the partial Lagrangian (38) results in

$$\begin{aligned}&-(1-\tau _k)(r-\delta )\eta +\eta _t+\dot{k}(\eta _k-\xi _t)-\dot{k}^2 \xi _k\nonumber \\&\qquad +\,\frac{1}{\theta -1}[-\dot{k}+(1-\tau _k)(r-\delta )k][\xi _t+\dot{k} \xi _k] \nonumber \\&\qquad +\,\frac{1}{1-\theta }(\xi _t+\dot{k} \xi _k)[-\dot{k} +(1-\tau _k)(r-\delta )k]^{\theta }\nonumber \\&\quad =(B_t+\dot{k} B_k)\left[ -\dot{k} +(1-\tau _k)(r-\delta )k\right] ^{\theta }\nonumber \\&\qquad +\,(\eta -\xi \dot{k})(-\rho ), \end{aligned}$$
(42)

where B(t, k) is the gauge term. Separation of Eq. (42) with respect to powers of \(\dot{k}\) provides

$$\begin{aligned}&\left[ -\dot{k} +(1-\tau _k)(r-\delta )k\right] ^{\theta }: B_t=\frac{1}{1-\theta }\xi _t,\; \xi _k=B_k, \end{aligned}$$
(43)
$$\begin{aligned}&\quad \dot{k}^2:\xi _k=0, \end{aligned}$$
(44)
$$\begin{aligned}&\quad \dot{k}: \eta _k-\xi _t\frac{\theta }{\theta -1}-\rho \xi =0, \end{aligned}$$
(45)
$$\begin{aligned}&\quad \dot{k}^0: -(1-\tau _k)(r-\delta )\eta +\eta _t\nonumber \\&\qquad +\,\frac{1}{\theta -1}(1-\tau _k)(r-\delta )k \xi _t+\eta \rho =0. \end{aligned}$$
(46)

Equations (43)–(45) yield

$$\begin{aligned} \xi= & {} a_1(t),\; \eta =\left[ -\frac{\theta }{1-\theta } \dot{a}_1+\rho a_1\right] k+a_2(t), \nonumber \\&B=\frac{1}{1-\theta }a_1(t). \end{aligned}$$
(47)

Equation (46) with \(\xi , \eta , B\) from (47) gives rise to

$$\begin{aligned}&k: \theta \ddot{a}_1-\bigg [(1-\tau _k)(r-\delta )(1-\theta ) -\rho (1-\theta )-\theta (n-\rho )\bigg ]\dot{a}_1\nonumber \\&\qquad -\,\rho (1-\tau _k)(r-\delta )(\theta -1) +\rho ^2(\theta -1)a_1=0, \end{aligned}$$
(48)
$$\begin{aligned}&\quad k^0:\dot{a}_2-1-\tau _k)(r-\delta )a_2+\rho a_2=0. \end{aligned}$$
(49)

Equations (47)–(49) lead to the following partial Noether operators and gauge terms:

$$\begin{aligned} \xi= & {} c_1e^{-\rho t}+c_2e^{\frac{\rho (1-\theta )-(1-\tau _k)(r-\delta )(1-\theta )}{\theta }t}, \nonumber \\ \eta= & {} \frac{\rho }{1-\theta }c_1ke^{-\rho t}\nonumber \\&+\,(1-\tau _k)(r-\delta )c_2 ke^{\frac{\rho (1-\theta )-(1-\tau _k)(r-\delta )(1-\theta )}{\theta }t}\nonumber \\&+\,c_3e^{((1-\tau _k)(r-\delta )-\rho )t},\nonumber \\ B= & {} \frac{1}{1-\theta }\left[ c_1e^{-\rho t}+c_2e^{\frac{\rho (1-\theta ) -(1-\tau _k)(r-\delta )(1-\theta )}{\theta }t}\right] . \end{aligned}$$
(50)

The following first integrals corresponding to the operators and gauge terms given in (50) are established from Eq. (23):

$$\begin{aligned} I_1= & {} e^{-\rho t}[-\dot{k}+(1-\tau _k)(r-\delta )k]^{-\theta } k\nonumber \\&\times \,\left[ \frac{\rho +(\theta -1)(1-\tau _k)(r-\delta )}{(1-\theta )}\right] \nonumber \\&-\,\frac{\theta }{1-\theta }\left[ -\dot{k}+ (1-\tau _k)(r-\delta )k\right] ^{1-\theta }e^{-\rho t},\nonumber \\ I_2= & {} \frac{\theta }{\theta -1}\left[ -\dot{k}+ (1-\tau _k)(r-\delta ) k\right] ^{1-\theta }e^{\frac{\rho (1-\theta ) -(1-\tau _k)(r-\delta )(1-\theta )}{\theta }t}\nonumber \\ I_3= & {} [-\dot{k}+(1-\tau _k)(r-\delta )k]^{-\theta } e^{-(\rho -(1-\tau _k)(r-\delta ))t} . \end{aligned}$$
(51)

The partial Lagrangian approach yielded three first integrals. Now we utilize these first integrals to find closed-form solution for k(t). We can utilize either any one of these first integrals or any two first integrals to construct closed-form solution for k(t). Setting \(I_3=B\), we have

$$\begin{aligned} \left( -\dot{k}+(1-\tau _k)(r-\delta )k\right) ^{-\theta } e^{-(\rho -(1-\tau _k)(r-\delta ))t}=B, \end{aligned}$$
(52)

where B is an arbitrary constant. Equation (52) yields first-order ODE in k(t)

$$\begin{aligned} -\dot{k}+(1-\tau _k)(r-\delta )k =B^{-\frac{1}{\theta }}e^{-\frac{(\rho -(1-\tau _k)(r-\delta ))t}{\theta }} \end{aligned}$$
(53)

and this can be solved to obtain closed-form solution for k(t). We can make use of \(I_1\) to directly obtain k(t) without solving ODE (53). We set \(I_1=A\) to arrive at

$$\begin{aligned}&\left[ -\dot{k}+(1-\tau _k)(r-\delta )k\right] ^{-\theta } k \left[ \rho +(\theta -1)(1-\tau _k)(r-\delta )\right] \nonumber \\&\quad -\,{\theta }\left[ -\dot{k}+(1-\tau _k)(r-\delta )k\right] ^{1-\theta } =A(1-\theta )e^{\rho t},\nonumber \\ \end{aligned}$$
(54)

where A is an arbitrary constant. After some simplifications, equation (54) with the help of (53) gives following solution for capital stock k(t)

$$\begin{aligned} k(t)=\frac{\theta B^{-\frac{1}{\theta }}e^{-\frac{\rho -(1-\tau _k)(r-\delta )}{\theta }t} +\frac{A}{B}(1-\theta )e^{(1-\tau _k)(r-\delta )t}}{\rho -(1-\theta )(1-\tau _k)(r-\delta )}. \end{aligned}$$
(55)

The consumption is \( c(t)=-\dot{k}+(1-\tau _k)(r-\delta )k\) and thus

$$\begin{aligned} c(t)=B^{-\frac{1}{\theta }}e^{-\frac{(\rho -(1-\tau _k)(r-\delta ))t}{\theta }}. \end{aligned}$$
(56)

Equation (32), with the aid of Eqs. (55) and (56), finally gives closed-form solution for investment i(t). Now we make use of initial conditions in order to specify values of arbitrary constant A and B. The initial conditions \( c(0)=c_0, \;k(0)=k_0\) give

$$\begin{aligned} A= & {} \frac{k_0 c_0^{-\theta }}{1-\theta }\left[ \rho - (1-\theta )(1-\tau _k)(r-\delta )\right] \nonumber \\&-\frac{\theta }{1-\theta }c_0^{1-\theta },\; B=c_0^{-\theta }. \end{aligned}$$
(57)

Next we check whether the derived closed-form solutions for c(t) and k(t) satisfy the transversality conditions or not. The transversality condition (36), with the aid of Eqs. (55) and (56), becomes

$$\begin{aligned} \lim _{t\rightarrow \infty } \frac{\theta B^{1-\frac{1}{\theta }}e^{-\frac{\rho -(1-\theta )(1-\tau _k)(r-\delta )}{\theta }t} +A(1-\theta )}{\rho -(1-\theta )(1-\tau _k)(r-\delta )}=0 , \end{aligned}$$
(58)

and it goes to zero provided \(\frac{\rho -(1-\theta )(1-\tau _k)(r-\delta )}{\theta }>0\) and \(A=0\). Equation (57) for \(A=0\) yields

$$\begin{aligned} \frac{\rho -(1-\theta )(1-\tau _k)(r-\delta )}{\theta } =\frac{c_0}{k_0}. \end{aligned}$$
(59)

Now substituting \(A=0,\; B= c_0^{-\theta }\) into (55) and (56), we obtain following final form of solution for k(t) and c(t)

$$\begin{aligned} k(t)=k_0e^{-\frac{\rho -(1-\tau _k)(r-\delta )}{\theta }t},\quad c(t)=c_0e^{-\frac{\rho -(1-\tau _k)(r-\delta )}{\theta }t}, \end{aligned}$$
(60)

provided parameter restriction (59) holds. The growth rates for capital and consumption are given by

$$\begin{aligned} \frac{\dot{c}}{c}=\frac{ \dot{k}}{k}=\frac{(1-\tau _k)(r-\delta )-\rho }{\theta }. \end{aligned}$$
(61)

4.2 Ramsey neoclassical growth model with CRRA utility function

We consider the following Ramsey neoclassical growth model [6], where the representative consumer’s utility maximization problem is defined as

$$\begin{aligned} \mathrm{Max}_c \quad \int _0^{\infty } e^{-r t}c^{1-\sigma } \hbox {d}t ,\quad \sigma \not =0,1 \end{aligned}$$
(62)

subject to the capital accumulation equation

$$\begin{aligned} \dot{k}(t) =k^\beta -\delta k-c, \; k(0)=k_0,\quad 0<\beta <1, \end{aligned}$$
(63)

where c(t) is the consumption per person, k(t) is the capital labor ratio, and \(\beta , \delta , r\) are the capital share, depreciation rate, rate of time preferences, respectively. The intertemporal elasticity of substitution is given by \(1/{\sigma }\) and \(k_0\) is the initial capital stock. The optimal control problem (62) and (63) in the calculus of variations takes the following form:

$$\begin{aligned} \mathrm{Max_k} \quad \int _0^{\infty } e^{-r t}\left[ k^\beta -\delta k-\dot{k}\right] ^{1-\sigma }\hbox {d}t, \quad k(0)=k_0. \end{aligned}$$
(64)

The partial Lagrangian for this model is

$$\begin{aligned} L=-\left[ k^\beta -\delta k-\dot{k}\right] ^{1-\sigma } \end{aligned}$$
(65)

which is free of the discount factor and the partial Euler–Lagrange equation is

$$\begin{aligned} \frac{\delta L}{\delta k}=-r(1-\sigma )\left[ k^\beta -\delta k-\dot{k}\right] ^{-\sigma }. \end{aligned}$$
(66)

The partial Noether operators determining Eq. (22) for the partial Lagrangian (65) yields

$$\begin{aligned}&\eta (\sigma -1)(\beta k^{\beta -1}-\delta )+\left[ \eta _t+\dot{k}(\eta _k-\xi _t)-\dot{k}^2 \xi _k\right] (1-\sigma )\nonumber \\&\qquad -\,(\xi _t+\dot{k} \xi _k)\left[ k^\beta -\delta k-\dot{k}\right] \nonumber \\&\quad =(B_t+\dot{k} B_k)[k^\beta -\delta k-\dot{k}]^{\sigma }+(\eta -\xi \dot{k})r(\sigma -1),\nonumber \\ \end{aligned}$$
(67)

where B(t, k) is the gauge term. By separating equation (67) with respect to powers of \(\dot{k}\), we easily have

$$\begin{aligned}&\left[ k^\beta -\delta k-\dot{k} \right] ^{\sigma }: B_t=0,\; B_k=0,\end{aligned}$$
(68)
$$\begin{aligned}&\quad \dot{k}^2:\xi _k=0, \end{aligned}$$
(69)
$$\begin{aligned}&\quad \dot{k}: (1-\sigma )\eta _k+\sigma \dot{\xi }-r(1-\sigma ) \xi =0, \end{aligned}$$
(70)
$$\begin{aligned}&\quad \dot{k}^0: \eta (\sigma -1)(\beta k^{\beta -1}-\delta )+\eta _t(1-\sigma ) \nonumber \\&\qquad -\,\dot{\xi }(k^{\beta }-\delta k)+r(1-\sigma ) \eta =0. \end{aligned}$$
(71)

The solution of equations (68)- (71) yield the following partial Noether operator and gauge terms

$$\begin{aligned} X=e^{\delta \beta (1-\sigma )t}\frac{\partial }{\partial t}-\delta e^{\delta \beta (1-\sigma )t}k\frac{\partial }{\partial k},\; B=0, \end{aligned}$$
(72)

with \(\sigma \) satisfying

$$\begin{aligned} \sigma =\frac{r+\delta }{\beta \delta } . \end{aligned}$$
(73)

The first integral corresponding to the partial Noether operator and gauge terms given in (72) is

$$\begin{aligned} I= & {} e^{\delta \beta (1-\sigma )t}\left[ -\sigma \left[ k^\beta -\delta k-\dot{k}\right] ^{1-\sigma }+(\sigma -1)\right. \nonumber \\&\left. \left[ k^\beta -\,\delta k-\dot{k}\right] ^{-\sigma }k^{\beta }\right] . \end{aligned}$$
(74)

Naz et al. [28] derived the following first integral by utilizing the partial Hamiltonian approach

$$\begin{aligned} I= e^{\delta \beta (1-\sigma )t}[-\sigma c^{1-\sigma }+(\sigma -1)c^{-\sigma }k^{\beta }]. \end{aligned}$$
(75)

A closer look at the first integrals given in (74) and (75) shows that both first integrals are identical by replacing

$$\begin{aligned} c=k^\beta -\delta k-\dot{k} \end{aligned}$$
(76)

in it. These lead to the solution of the underlying model as shown in [28] and are given by

$$\begin{aligned} c(t)= & {} \left( 1-\frac{\beta \delta }{r+\delta }\right) k^{\beta }, \end{aligned}$$
(77)
$$\begin{aligned} k(t)= & {} \left[ \frac{\beta }{r+\delta }+\left( k_0^{1-\beta } -\frac{\beta }{r+\delta }\right) e^{-(1-\beta )\delta t}\right] ^{\frac{1}{1-\beta }}.\nonumber \\ \end{aligned}$$
(78)

The steady state for this model is

$$\begin{aligned} \bar{k}=\left( \frac{r+\delta }{\beta }\right) ^{\frac{1}{\beta -1}} \end{aligned}$$
(79)

and

$$\begin{aligned} \bar{c}=\left( 1-\frac{\beta \delta }{r+\delta }\right) \bar{k}^{\beta }. \end{aligned}$$
(80)

In order to determine the dynamics of the system, we now differentiate (77) with respect to t

$$\begin{aligned} \frac{\dot{c}}{c}= \beta \frac{\dot{k}}{k}. \end{aligned}$$
(81)

From (81), we have the growth rate of capital and consumption is the same as for \(\beta >0\). The growth rate of capital is given by

$$\begin{aligned} \frac{\dot{k}}{k}= \frac{-\delta \left( k_0^{1-\beta } -\frac{\beta }{r+\delta }\right) e^{-(1-\beta )\delta t}}{\frac{\beta }{r+\delta } +\left( k_0^{1-\beta }-\frac{\beta }{r+\delta }\right) e^{-(1-\beta )\delta t}} \end{aligned}$$
(82)

or

$$\begin{aligned} \frac{\dot{k}}{k} =\delta \bigg [\bigg (\frac{k}{\bar{k}}\bigg )^{\beta -1}-1\bigg ], \end{aligned}$$
(83)

where \(\bar{k}=\left( \frac{r+\delta }{\beta }\right) ^{\frac{1}{\beta -1}}\). From (83), clearly the growth rate of capital decreases over time and goes to zero as the steady state is reached for \(t\mapsto \infty \). The growth rate of consumption also decreases over time as the steady state is approached. For the case when the capital’s share is equal to the reciprocal of the intertemporal elasticity of substitution studied by [33], the growth rate of capital decreases over time, but the growth rate of consumption grows linearly with capital stock.

Secondly, the gross saving rate \(s=1-c/k^{\beta }\) for this model is constant and is given by

$$\begin{aligned} s=\frac{\beta \delta }{r+\delta }=\sigma . \end{aligned}$$
(84)

4.3 Economic growth model with logarithmic utility function

We consider the following neoclassical growth model [6], where the representative consumer’s utility maximization problem is defined as

$$\begin{aligned} \mathrm{Max}_c \quad \int _0^{\infty } e^{-r t}\ln (c) \hbox {d}t , \end{aligned}$$
(85)

subject to the capital accumulation equation

$$\begin{aligned} \dot{k}(t) =k^\beta -(\delta -A) k-c, \; k(0)=k_0,\; 0<\beta <1, \end{aligned}$$
(86)

where c(t) is the consumption per person, k(t) the capital labor ratio, \(A>0\) the marginal product of capital, \(\beta \) the capital share, \(\delta \) the depreciation rate, and r is the rate of time preferences. In the capital accumulation equation (86), we consider a Cobb–Douglas version of the A-K technology: \(y=k^\beta +Ak\). The current value Hamiltonian for this model is

$$\begin{aligned} H=\ln (c)+\lambda (k^\beta -(\delta -A) k-c) \end{aligned}$$
(87)

where \(\lambda \) is the costate variable. The necessary first-order conditions for optimal control are

$$\begin{aligned} \lambda= & {} \frac{1}{c}, \end{aligned}$$
(88)
$$\begin{aligned} \dot{k}(t)= & {} k^\beta -(\delta -A) k-c, \end{aligned}$$
(89)
$$\begin{aligned} \dot{\lambda }= & {} - (\beta k^{\beta -1}-\delta +A-r)\lambda , \end{aligned}$$
(90)

and the growth rate of consumption is

$$\begin{aligned} \dot{c}= \left( \beta k^{\beta -1}-\delta +A-r\right) c. \end{aligned}$$
(91)

We do not apply the partial Hamiltonian approach to solving this model. We solve this model by the partial Lagrangian approach, and first we transform the problem to an equivalent calculus of variations problem.

The optimal control problems (85) and (86) in the calculus of variations can be expressed as

$$\begin{aligned} \mathrm{Max}_k \quad \int _0^{\infty } e^{-r t}\ln \bigg (k^\beta -\delta k+A k-\dot{k}\bigg )\hbox {d}t, \;k(0)=k_0. \end{aligned}$$
(92)

The partial Lagrangian for this model is

$$\begin{aligned} L=- \ln \bigg (k^\beta -\delta k+A k -\dot{k}\bigg )=\ln \bigg (\frac{1}{k^\beta -\delta k+Ak-\dot{k}}\bigg ) \end{aligned}$$
(93)

and the partial Euler–Lagrange equation is

$$\begin{aligned} \frac{\delta L}{\delta k}=-r\bigg (k^\beta -\delta k+A k-\dot{k}\bigg )^{-1}. \end{aligned}$$
(94)

For the partial Lagrangian (93), the partial Noether operators determining Eq. (22) results in

$$\begin{aligned}&-\eta (\beta k^{\beta -1}-\delta +A)+\eta _t +\dot{k}(\eta _k-\xi _t)-\dot{k}^2 \xi _k \nonumber \\&\qquad +\,(\xi _t+\dot{k} \xi _k)(k^\beta -\delta k+A k-\dot{k})\ln \nonumber \\&\qquad \times \,\bigg (\frac{1}{k^\beta -\delta k+A k-\dot{k}}\bigg ) \nonumber \\&\quad =(B_t+\dot{k} B_k)(k^\beta -\delta k+A k-\dot{k})-r(\eta -\xi \dot{k}),\nonumber \\ \end{aligned}$$
(95)

where B(t, k) is the gauge term. The term \((\xi _t+\dot{k} \xi _k)(k^\beta -\delta k+A k-\dot{k})\ln ({k^\beta -\delta k+A k-\dot{k}})^{-1}\) in Eq. (95) after power series expansion in \(\dot{k}\) contains higher powers of \(\dot{k}\) and thus yields \(\xi _k=0,\;\xi _t=0\). Now separating the rest of the terms in Eq. (95) with respect to powers of \(\dot{k}\), we have

$$\begin{aligned}&\dot{k}^2:B_k=0, \end{aligned}$$
(96)
$$\begin{aligned}&\dot{k}: \eta _k+B_t-r \xi =0, \end{aligned}$$
(97)
$$\begin{aligned}&\dot{k}^0:-\eta (\beta k^{\beta -1}-\delta +A)+\eta _t -B_t\left( k^{\beta }-\delta k+A k\right) \nonumber \\&\quad +\,r \eta =0. \end{aligned}$$
(98)

The solution of Eqs. (96)–(98) yields the following partial Noether operator and gauge terms

$$\begin{aligned} \xi =c_1,\;\; \eta =\frac{r c_1 k}{1-\beta },\;\;B=\frac{c_1\beta t r}{\beta -1}+c_2, \end{aligned}$$
(99)

with \(\beta \) satisfying

$$\begin{aligned} \beta =\frac{r+\delta -A}{ \delta -A}. \end{aligned}$$
(100)

The first integral corresponding to the partial Noether operator and gauge terms given in (96) under the parameter restriction (100) is

$$\begin{aligned} I= & {} \frac{\beta r t}{\beta -1}+\ln (k^\beta -\delta k+A k-\dot{k})\nonumber \\&+\,k^{\beta }(k^\beta -\delta k+A k-\dot{k})^{-1}-1. \end{aligned}$$
(101)

We set this integral to be a constant as

$$\begin{aligned} I= & {} \frac{\beta r t}{\beta -1}+\ln \left( k^\beta -\delta k+A k-\dot{k}\right) \nonumber \\&+\,k^{\beta }\left( k^\beta -\delta k+A k-\dot{k}\right) ^{-1}-1=e_1 \end{aligned}$$
(102)

in which \(e_1\) is an arbitrary constant. From this equation (102), we deduce k which is written in terms of the control variable c as

$$\begin{aligned} k=\left[ (e_1+1-\ln c + \delta \beta t-A\beta t)c\right] ^{1\over \beta }. \end{aligned}$$
(103)

Now Eq. (91) after substitution of the value of k in (103) becomes

$$\begin{aligned}&\dot{c}+(\delta -A)\beta c\nonumber \\&\quad =\beta c \left[ \left( e_1+1-\ln c+ \delta \beta t-A\beta t\right) c\right] ^{\beta -1\over \beta }. \end{aligned}$$
(104)

We set \(S=c\exp (\delta -A)\beta t\). Then, (104) becomes

$$\begin{aligned} \dot{S}=\beta S(e_1+1-\ln S)^{\beta -1\over \beta }S^{\beta -1\over \beta }\exp (-rt) \end{aligned}$$
(105)

which is integrable by quadratures

$$\begin{aligned} \int {\hbox {d}S\over \beta S^{2\beta -1\over \beta }(e_1+1-\ln S)^{\beta -1\over \beta }}=-{1\over r}\exp (-rt)+e_2, \end{aligned}$$
(106)

where \(e_2\) is another arbitrary constant. This is a new solution.

5 Concluding remarks

We have developed a new approach termed as a partial or discount free Lagrangian approach for dynamical systems. We have shown how a partial Lagrangian approach provides solutions for a range of equations of economics. This naturally arises from the allied notion of a current value Hamiltonian via the partial Legendre transformation in a generalized setting when the costate and control variables can be expressed in terms of the state variables and their derivatives. This approach, when it works, is simpler than the more general partial Hamiltonian approach as the control variables do not enter into the calculations which are therefore much more simplified. Notwithstanding, the partial or discount free Lagrangian that arises is independent of the discount factor which makes it desirable in the same way as the current value Hamiltonian is desirable for not having the discount factor.

We worked out three model examples, one linear and two nonlinear, using this approach. We have demonstrated the simplicity of this method as compared to the worked out example of [28] which involves more variables in the form of the control variables. In the third example, we derived a first integral and the new exact solutions for the agents maximization problem with logarithmic time preference.

The utility of the method advocated here for economic models is thus advantageous when a discount free Lagrangian exists. This is indeed the case for a number of economic models which need investigation in future research. It is worthy to mention here that for the case of more control variables it is sometimes convenient to apply the more generalized approach of partial Hamiltonians.