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I'm currently struggling to proof the saddle-path stability in the Ramsey-Cass-Koopmans model. Based on the books Introduction to Modern Economic Growth by D. Acemoglu (chapter 7 and 8) and Mathematics for Economists by Blume and Simon (chapter 25.4) , the following was given:

Consider the boundary value problem, corresponding to the equilibrium conditions, of the baseline neoclassical (Ramsey) growth model

$ \dot k(t) =f(k(t))-(\delta+n)k(t)-c(t) \\ \dot c(t) =\frac{c(t)}{\epsilon_u(c(t))}(f'(k(t))-\delta-\rho) $

subject to $ k(t) , ~ c(t) ≥ 0 $, the initial value condition $ k(0) = k_0 > 0 $ and the transversality condition

$ \lim_{t \rightarrow \infty} \big[k(t)e^{-\int_0^t(f'(k(s))-\delta-n)ds}\big]=0 $

where

$ \epsilon_u(c(t)) \equiv - \frac{u''(c(t))~c(t)}{u'(c(t))} $

is the elasticity of marginal utility of consumption, and $ ρ > 0 $ and $ 0 < δ < 1.$ We assume that the production function $ f (·) $ satisfies $ f (0) = 0,~ f ′(k) > 0,~ f ′′(k) < 0 $ for all $ k > 0 $ (so it features diminishing marginal returns to capital), and the two Inada conditions, $ \lim_{k→0} f ′(k) = ∞ $ and $\lim_{k→∞} f ′(k) = 0 $.

Here's my attempt:

First, I linearized the two differential equations $ \dot c(t) $ and $ \dot k(t)$. Then, I did a first-order Taylor expansion around the steady-state $(k^∗, c^∗) $ which gives

$ \begin{bmatrix} \dot k(t)\\ ~ \\ \dot c(t) \end{bmatrix} \approx \begin{bmatrix} \frac{\partial \dot k(t)}{\partial k(t)} \big\rvert_{k(t) = k^*,c(t)=c^*} & \frac{\partial \dot k(t)}{\partial c(t)} \big\rvert_{k(t) = k^*,c(t)=c^*} \\ ~ \\ \frac{\partial \dot c(t)}{\partial k(t)}\big\rvert_{k(t) = k^*,c(t)=c^*} & \frac{\partial \dot c(t)}{\partial c(t)} \big\rvert_{k(t) = k^*,c(t)=c^*} \end{bmatrix} \begin{bmatrix} k(t)- k^* \\ ~ \\ c(t)- c^* \end{bmatrix} \\ ~ \\ = \begin{bmatrix} f'(k^*) - (\delta + n) & - 1 \\ ~ \\ \frac{c^*}{\epsilon_u(c^*)} f''(k^*) & \frac{1}{\epsilon_u(c^*)} \big(f'(k^*)-\delta - \rho \big) \end{bmatrix} \begin{bmatrix} k(t)- k^* \\ ~ \\ c(t)- c^* \end{bmatrix} \\ ~ \\ $

Since we know

$ \dot k(t)=0 \implies c^*=f(k^*)-(\delta+n)k^* \\ \dot c(t)=0 \implies f'(k^*)=\delta + \rho $

the linearized model simplifies to

$ \begin{bmatrix} \dot k(t)\\ ~ \\ \dot c(t) \end{bmatrix} \approx \begin{bmatrix} \rho - n & - 1 \\ ~ \\ \frac{c^*}{\epsilon_u(c^*)} f''(k^*) & 0 \end{bmatrix} \begin{bmatrix} k(t)- k^* \\ ~ \\ c(t)- c^* \end{bmatrix} \\ ~ \\ $

where $ c^*f''(k^*)/\epsilon_u(c^*) < 0$ and $\rho - n > 0$.

Then I defined

$ A \equiv \begin{bmatrix} \rho - n & - 1 \\ ~ \\ \frac{c^*}{\epsilon_u(c^*)} f''(k^*) & 0 \end{bmatrix} $

Thus, we get

$ \begin{equation} det(A) =\frac{c^*}{\epsilon_u(c^*)}f''(k^*) < 0 \\ tr(A) = \rho - n > 0 \end{equation}$

Then, the characteristic polynomial is

$ \begin{equation} det(A-\lambda I_2) = 0 \iff \lambda^2 - tr(A)\lambda + det(A) \overset{!}{=} 0 \iff \lambda^2 - (\rho-n)\lambda + \frac{c^*f''(k^*)}{\epsilon_u(c^*)} \overset{!}{=} 0 \end{equation} $

This is how far I've come. The next step would be to compute the eigenvalues $λ_{1,2}$ and the eigenvectors $V_1,~V_2$. We were told that if $ tr(A) = λ_1 + λ_2 > 0 $, then the system is unstable. The general solution of the linearized system should look like this

$ \begin{equation} \begin{bmatrix} k(t)\\c(t) \end{bmatrix} = \begin{bmatrix} k^*\\ c^* \end{bmatrix} + \gamma_1 ~ e^{\lambda_1 t} ~V_1 + \gamma_2 ~ e^{\lambda_2 t} ~ V_2 \end{equation}$

(cf. Simon, Blume), where $\lambda_1,~ \lambda_2$ are the coefficients of the roots $λ_1,~ λ_2$. We were also told the following:

"Recall that the goal was to solve the system of the two differential equations in the neighborhood of the steady-state. If the eigenvalues of $A$ are smaller than one in absolute values, the system is (globally) asymptotically stable. If, on the other hand, all eigenvalues of $A$ are greater than one in absolute values, the system can be thought of as “completely unstable,” since all trajectories explode when they start from an initial position other than the steady-state itself. The interesting case occurs when, as here, some of the eigenvalues of $A$ are greater and some are smaller than one in absolute values. In order for the system to be saddle-path stable, we require $|λ_1| > 1$ and $|λ_2| < 1$. That means there exist solution trajectories that, starting arbitrarily close to the steady-state, get arbitrarily far from it as time goes to infinity. However, there exists a region such that any trajectory that starts in the region (or is found in this region at some point in time) converges to the steady-state. In that case the trajectory remains inside this region while converging to the steady-state.

In case of $|λ_2| < 1$, the saddle-path will correspond to the eigenvector $V_2$ associated with $λ_2$, allowing us to ignore the unstable path corresponding to the eigenvector associated with $λ_1$. This means that the coefficient $γ_1$ of the explosive (unstable) root, namely $|λ_1| > 1$, must be set to $γ_1 = 0$."

In order to finish the proof, I need to find the solution to the (simplified) linearized system.

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  • $\begingroup$ What is your question exactly? You have a quadratic equation in $\lambda$ which gives you two roots. These will correspond to the eigenvalues you are looking for. Also, for a continuous system, you want to look for eigenvalues with positive or negative sign (not for ones where the absolute values are greater or less than one). $\endgroup$
    – tdm
    Jan 11 at 14:42
  • $\begingroup$ I have problems with the calculation of the eigenvalues and vectors. How do I calculate them in this particular case? $\endgroup$ Jan 11 at 14:51
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You got to the quadratic equation $$ \lambda^2 - (\rho - n)\lambda + \frac{c^\ast f''(k^\ast)}{\varepsilon} $$ The discriminant is given by: $$ \Delta = (\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{\varepsilon} $$ So the two roots are: $$ \lambda_1, \lambda_2 = \frac{(\rho - n) \pm \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}}{2} $$ As $f''(k) < 0$ and $\rho > n$ we see that both roots are real, that one of them is larger than $0$ and one is smaller than zero, so we have a saddle path.

The eigenvectors $V_1$ and $V_2$ satisfy $A V_1 = \lambda_1 V_1$ and $A V_2 = \lambda_2 V_2$, so for the first eigenvector: $$ \begin{bmatrix} \rho - n & -1 \\ \frac{c^\ast f''(k^\ast)}{\varepsilon} & 0\end{bmatrix}\begin{bmatrix}v_1\\ 1\end{bmatrix} = \lambda_1 \begin{bmatrix} v_1\\ 1\end{bmatrix} $$ Here we normalized the second component to 1 which we can do as eigenvectors are only given up to scale. So: $$ \begin{align*} &(\rho - n) v_1 - 1 = \frac{\left[(\rho - n) + \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}v_1\\ \leftrightarrow &v_1 = \frac{2}{(\rho - n) - \sqrt{(\rho - n)^2) - 4 \frac{c^\ast f''k)}{\varepsilon}}} \end{align*} $$ Equivalently, we can also use the second condition: $$ v_1 =\frac{\varepsilon}{c^\ast f''(k^\ast)}\frac{\left[(\rho - n) + \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2} $$ It is easy to see that these two are the same.

For the second eigenvector we have: $$ \begin{bmatrix} \rho - n & -1 \\ \frac{c^\ast f''(k^\ast)}{\varepsilon} & 0\end{bmatrix}\begin{bmatrix}v_2\\ 1\end{bmatrix} = \lambda_2 \begin{bmatrix} v_2\\ 1\end{bmatrix} $$ So: $$ \begin{align*} &(\rho - n) v_2 - 1 = \frac{\left[(\rho - n) - \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}v_2\\ \leftrightarrow &v_2 = \frac{2}{(\rho - n) + \sqrt{(\rho - n)^2) - 4 \frac{c^\ast f''k)}{\varepsilon}}} \end{align*} $$

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  • $\begingroup$ Amazing, thank you :) $\endgroup$ Jan 11 at 15:56

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