Example
We can now completely solve an example.
Utility and production functions
We assume that utility is logarithmic and we take a Cobb-Douglas production function.
$$\begin{eqnarray} u(c_{t}) = \log (c_{t}), \\\ F(K_{t},L_{t}) = K_{t}^{\alpha}L_{t}^{1-\alpha}, \\\ \alpha \in (0,1). \end{eqnarray} $$ We can check that both functions satisfy the Inada conditions:
$$\begin{eqnarray}\lim_{c_{t} \rightarrow 0}u^{\prime}(c_{t}) = \lim_{c_{t} \rightarrow 0} \frac{1}{c_{t}} = +\infty, \\\ \lim_{c_{t} \rightarrow +\infty}u^{\prime}(c_{t}) = \lim_{c_{t} \rightarrow +\infty} \frac{1}{c_{t}} = 0, \\\ \lim_{K_{t} \rightarrow 0} F^{\prime}_{K_{t}}(K_{t}, L_{t}) = \lim_{K_{t} \rightarrow 0} \alpha K_{t}^{\alpha - 1}L_{t}^{1-\alpha} = +\infty, \\\ \lim_{K_{t} \rightarrow +\infty} F^{\prime}_{K_{t}}(K_{t}, L_{t}) = \lim_{K_{t} \rightarrow +\infty} \alpha K_{t}^{\alpha - 1}L_{t}^{1-\alpha} = 0, \\\ \lim_{L_{t} \rightarrow 0} F^{\prime}_{L_{t}}(K_{t}, L_{t}) = \lim_{L_{t} \rightarrow 0} (1-\alpha) K_{t}^{\alpha}L_{t}^{\alpha} = +\infty, \\\ \lim_{L_{t} \rightarrow +\infty} F^{\prime}_{L_{t}}(K_{t}, L_{t}) = \lim_{L_{t} \rightarrow +\infty} (1- \alpha) K_{t}^{\alpha}L_{t}^{\alpha} = 0. \end{eqnarray} $$
The production function, expressed in intensive terms, becomes:
$$F(K_{t}, L_{t}) = L_{T} F\left(\frac{K_{t}}{L_{t}},1\right) = L_{t} f(k_{t}) = L_{t} k_{t}^{\alpha}, k_{t} \equiv \frac{K_{t}}{L_{t}}.$$ with total production $Y_{t} = F(K_{t},L_{t})$ and production per capita is equal to $y_{t} \equiv \frac{Y_{t}}{L_{t}} = f(k_{t}).$
Household’s optimisation
Instead of using the results from previous sections, we develop again the utility maximisation. First, we write down the household’s budget constraint:
Household’s budget constraint
At each period, the total received income is composed of wages and interest. It can be spent on consumption or saving. Remember that capital depreciates at a rate $\delta \in [0,1].$ Hence, we receive back from firms $1-\delta$ times the capital we lend. Therefore:
$$ w_{t} + r_{t}k_{t} + (1-\delta)k_{t} = c_{t} + k _{t+1}.$$
Intertemporal utility maximisation: Lagrangian
The intertemporal utility maximisation problem is:
$$\begin{eqnarray} \max_{c_{t}, k_{t+1}} \sum_{t=0}^{\infty} \beta^{t} u(c_{t}) & \\\ \mathrm{s.t.} \quad w_{t} + r_{t} k_{t} + (1-\delta)k_{t} = c_{t} + k_{t+1}, \\\ c_{t}, k_{t+1} >0, k_{t=0} = k_{0} > 0. \end{eqnarray} $$
The Lagrangian becomes:
$$ \mathcal{L} = \sum_{t=0}^{\infty} \beta^{t} u(c_{t}) + \sum_{t=0}^{\infty} \lambda_{t}(w_{t} + (r_{t} + 1 - \delta)k_{t} - c_{t} - k_{t+1}).$$
We use the first-order conditions with respect to $c_{t}$ and $k_{t+1}$ to find the optimal consumption path: the Euler equation, which we combine later with the transversality condition.
$$\frac{\partial \mathcal{L}}{\partial c_{t}} = \beta^{t} u^{\prime}(c_{t}) - \lambda_{t} = \beta^{t}\frac{1}{c_{t}} - \lambda_{t} = 0$$ $$\frac{\partial \mathcal{L}}{\partial k_{t+1}} = -\lambda_{t} + \lambda_{t+1} (r_{t+1} + 1 - \delta) = 0.$$
Combining both, we obtain the Euler equation:
$$\beta^{t}\frac{1}{c_{t}} = r_{t+1} \beta^{t+1} \frac{1}{c_{t+1}} \implies c_{t+1} = \beta (r_{t} + 1 - \delta)c_{t}.$$
Finally, we should remember to impose the transversality condition:
$$\lim_{t \rightarrow +\infty} \beta^{t}u^{\prime}(c_{t})k_{t+1} = \lim_{t \rightarrow +\infty}\beta^{t}\frac{1}{c_{t}}k_{t+1} = 0.$$
Firm’s optimisation
In this model, firms operate under perfect competition, making zero profits. Moreover, factors are paid their marginal productivity.
$$r_{t} = F^{\prime}_{K_{t}}(K_{t}, L_{t}) = \alpha K_{t}^{\alpha - 1}L_{t}^{1-\alpha} = \alpha \left(\frac{K_{t}}{L_{t}}\right)^{\alpha - 1} = \alpha k_{t}^{\alpha -1},$$ $$w_{t} = F^{\prime}_{L_{t}}(K_{t}, L_{t}) = (1-\alpha) K_{t}^{\alpha} L_{t}^{-\alpha} = (1-\alpha)\left(\frac{K_{t}}{L_{t}}\right)^{\alpha} = (1-\alpha) k_{t}^{\alpha}.$$
Alternatively, using the information in the Appendix and working directly with the intensive-form production function:
$$r_{t} = f^{\prime}(k_{t}) = \frac{\partial f(k_{t})}{\partial k_{t}} = \alpha k_{t}^{\alpha -1},$$ $$w_{t} = f(k_{t}) - f^{\prime}(k_{t})k_{t} = k_{t}^{\alpha} - \alpha k_{t}^{\alpha - 1} k_{t} = (1-\alpha)k_{t}^{\alpha}.$$
The dynamic system
We are now in a position to solve the model. We have the following equations:
$$ c_{t+1} = \beta (r_{t+1} + 1 - \delta) c_{t}, \\\ w_{t} + r_{t}k_{t} + (1-\delta) k_{t} = c_{t} + k_{t+1}, \\\ w_{t} = (1-\alpha) k^{\alpha}_{t}, \\\ r_{t} = \alpha k^{\alpha-1}, \\\ \lim_{t \rightarrow +\infty} \beta^{t}\frac{1}{c_{t}} k_{t+1} = 0.$$
Since markets clear, we substitute the wage and interest rate into the budget constraint.
$$c_{t+1} = \beta (\alpha k^{\alpha-1} + 1 - \delta) c_{t},$$ $$k_{t}^{\alpha} + (1-\delta) k_{t} = c_{t} + k_{t+1},$$ $$\lim_{t \rightarrow +\infty} \beta^{t}\frac{1}{c_{t}} k_{t+1} = 0.$$
Steady state
Before trying to solve for the optimal trajectory, we analyse the steady states of this economy. A steady-state $(\bar{k}, \bar{c})$ is such that capital and consumption remain constant over time: $c_{t+1} = c_{t}$ and $k_{t+1} = k_{t}.$ Applying this idea to the equations above we get —here we can temporarily forget about the transversality condition.
$$\bar{c} = \beta(\alpha \bar{k}^{\alpha-1} + 1 - \delta) \bar{c},$$ $$\bar{k}^{\alpha} + (1-\delta)\bar{k} = \bar{c} + \bar{k}.$$
Using the first equation we can easily get the steady level of capital:
$$ \bar{k} = \left(\frac{\alpha \beta}{1-\beta(1-\delta)}\right)^{\frac{1}{1-\alpha}}.$$
Applying $\bar{k}$ in the second equation yields the steady-state level of consumption:
$$ \bar{c} = \left(\frac{\alpha \beta}{1-\beta (1-\delta)}\right)^{\frac{\alpha}{1-\alpha}} - \delta \left(\frac{\alpha \beta}{1-\beta (1- \delta)}\right)^{\frac{1}{1-\alpha}}.$$
Stability
First, we analyse the stability of the steady state, as we did in the general case Since we are using log-utility, the coefficient of relative risk aversion $\epsilon_{c} = 1.$
Instead of plugging the correct values in the Jacobian matrix $\bar{A}$, for the sake of clarity, we derive everything again:
$$\bar{A} = \begin{pmatrix} \frac{\partial c_{t+1}}{\partial c_{t}} & \frac{\partial c_{t+1}}{\partial k_{t}} \\\ \frac{\partial k_{t+1}}{\partial c_{t}} & \frac{\partial k_{t+1}}{\partial k_{t}} \end{pmatrix} $$ $$ = $$ $$\begin{pmatrix} \beta \left( \alpha k_{t+1}^{\alpha -1} + 1 - \delta \right) - \frac{\beta c_{t} \alpha (\alpha-1) }{k_{t+1}^{2-\alpha}} & \frac{\beta c_{t} \alpha (\alpha -1)}{k_{t+1}^{2-\alpha}} \left( \alpha k_{t}^{\alpha-1} + 1 - \delta \right) \\\ -1 & \alpha k_{t}^{\alpha -1} + 1 - \delta \end{pmatrix}$$
Evaluating it at the steady state: $$\bar{A}\bigr\rvert_{\substack{ k_{t} = k_{t+1} = \bar{k} \\\ c_{t} = c_{t+1} = \bar{c} } } = \begin{pmatrix} 1 - \beta \bar{c} \alpha (\alpha -1) \bar{k}^{\alpha -2} & \alpha (\alpha -1 ) \bar{c} \bar{k}^{\alpha -2} \\\ -1 & \frac{1}{\beta} \end{pmatrix},$$
where we have used $1 = \beta \left( \alpha \bar{k}^{\alpha -1} + 1 - \delta \right)$ at the steady state.
Eigenvalues
We can use the fact $\mathrm{tr}(\bar{A}) = \lambda_{1} + \lambda_{2}, \mathrm{det}(\bar{A}) = \lambda_{1} \lambda_{2}.$ Therefore:
$$\lambda_{1} \lambda_{2} = \frac{1}{\beta} $$ $$\lambda_{1} + \lambda_{2} = \frac{1}{\beta} + 1 - \beta \bar{c} \alpha (\alpha-1) \bar{k}^{\alpha -2}.$$ The first equation implies that either a) $\lambda_{1} > 0, \lambda_{2} > 0$ or b) $\lambda_{1} < 0, \lambda_{2} < 0.$ However, since $\lambda_{1} + \lambda_{2} > 1$ b) is impossible. Then, $\lambda_{1} > 0, \lambda_{2} > 0.$ Substitute $\lambda_{1} \lambda_{2} = \frac{1}{\beta} $ in the second equation:
$$ \lambda_{1} + \lambda_{2} = 1+ \lambda_{1} \lambda_{2} - \beta ( \alpha (\alpha - 1) \bar{k}^{\alpha -2}\frac{\bar{c}}{\epsilon_{\bar{c}}}.$$
Rearranging: $$ \lambda_{1} + \lambda_{2} - \lambda_{1} \lambda_{2} = 1+ - \beta \alpha (\alpha -1) \bar{k}^{\alpha -2}\frac{\bar{c}}{\epsilon_{\bar{c}}} > 1 \implies \lambda_{1} + \lambda_{2} - \lambda_{1}\lambda_{2} - 1 >0.$$ Factorisation leads to: $$(1 - \lambda_{2})(\lambda_{1} - 1) > 0.$$ Therefore, since $\lambda_1,, \lambda_2 >0: $$ \lambda_{2} < 1 \implies \lambda_{1} \in (0,1)$$ $$ \lambda_{1} > 1 \implies \lambda_{2} \in (1,+\infty)$$ and the steady state is saddle.
Trajectory
Describing the trajectory around the steady state analytically would be cumbersome. One possibility is to introduce an alternative production function of the $Ak$ family which alleviates this problem, see the notes by Maurice Obsfeld
Instead, we solve this section numerically. In particular, assume that $\alpha = 0.3, \, \beta = 0.9, \, \delta = 0.1.$ The initial level of capital $k_{t=0} = 1.$
With these values, we have the following:
$$\bar{k} = 1.65202, \, \bar{c}=0.997329, \, \bar{A} = \begin{pmatrix} 1.08029 & -0.089214 \\\ -1 & 1.11111 \end{pmatrix}.$$
The eigenvalues of the matrix $\bar{A}$ are $$\lambda_{1} = 0.796618 \in (-1,1)$$ $$\lambda_{2} =1.39479 > 1.$$
One possible set of eigenvectors is given by:
$$v_{1} = (0.314494 , 1), v_{2} = (-0.283675, 1)$$ which are associated with $\lambda_{1}$ and $\lambda_{2}$, respectively. Hence, the matrix $X$ becomes:
$$X = \begin{pmatrix} 0.314494 & -0.283675 \\\ 1 & 1 \end{pmatrix}.$$
Solving the system
Applying the results from before the system is:
$$\begin{pmatrix} c_{t} - \bar{c} \\\ k_{t} - \bar{k} \end{pmatrix} = z_{0}^{1}, 0.796618^{t} \begin{pmatrix} 0.314494 \\\ 1 \end{pmatrix} + z_{0}^{2}, 1.39479 \begin{pmatrix} -0.283675 \\\ 1 \end{pmatrix}.$$
Since $\lambda_{2} > 1$ we must set the value $z_{0}^{2} = 0$ to avoid explosive behaviour. Consequently:
$$\begin{pmatrix} c_{t} - \bar{c} \\\ k_{t} - \bar{k} \end{pmatrix} = z_{0}^{1}, 0.796618^{t} \begin{pmatrix} 0.314494 \\\ 1 \end{pmatrix}.$$
To solve the system of equations, first set $t=0$ and focus on $k_t - \bar{k}.$
$$\underbrace{k_{0}}_{=1} - \underbrace{\bar{k}}_{=1.65202} = z_{0}^{1} \times 1= -0.652017.$$
Hence, capital dynamics are governed by
$k_{t} - \bar{k} = -0.652017 \times 0.796618^{t}$
The law of motion of consumption becomes
$c_{t} - \bar{c} = -0.652017 \times 0.314494 \times 0.796618^{t} = -0.205055 \times 0.786618^{t}.$
Finally, we solve for the initial level of consumption compatible with the saddle path. For $t=0$ we have:
$c_{0} - \underbrace{\bar{c}}_{=0.997329} = \underbrace{z_{0}^{1}}_{= -0.652017} \times 0.314494 \implies c_{0} = 0.792273.$