Section 11 - Out of steady state

Phase Diagram & Saddle-Path Dynamics

Out of steady state, the joint behaviour of $k$ and $c$ is best read off a **phase diagram** in $(k, c)$ space. Two zero-loci divide the plane into four regions, each with a characteristic direction of motion.

The two zero-loci

Step 1
$\dot k = 0 \;\Longleftrightarrow\; c = f(k) - (n + \delta) k$
An inverted-U-shaped curve in $(k, c)$ space, peaked at the golden-rule $k_{GR}$ .
Step 2
$\dot c = 0 \;\Longleftrightarrow\; k = k^* \quad (\text{a vertical line})$
From the Euler equation: $\dot c = 0$ iff $f'(k) = \rho + \delta$ .

The two loci intersect at $(k^*, c^*)$ . The vertical $\dot c = 0$ locus sits *to the left* of the peak of the $\dot k = 0$ curve - because $k^* < k_{GR}$ . Four regions emerge:

Region	$k$ relative to $k^*$	$c$ relative to $\dot k = 0$ curve	Direction of motion
I - NW	$k < k^*$	Above curve	$\dot k < 0,\; \dot c > 0$
II - NE	$k > k^*$	Above curve	$\dot k < 0,\; \dot c < 0$
III - SE	$k > k^*$	Below curve	$\dot k > 0,\; \dot c < 0$
IV - SW	$k < k^*$	Below curve	$\dot k > 0,\; \dot c > 0$

Two regions (I and III) carry trajectories *away* from the steady state; two regions (II and IV) toward it. The saddle path is the boundary between these basins.

Live phase diagram

The plot below shows the $\dot k = 0$ curve, the saddle path through $(k^*, c^*)$ , and a vertical reference line at $k^*$ . Drag the sliders to see how the picture deforms.

Phase diagram in $(k, c)$ space

Linearization around the steady state

To extract convergence speeds, linearize the system around $(k^*, c^*)$ . Let $\tilde k = k - k^*$ , $\tilde c = c - c^*$ :

Step 1
$\begin{pmatrix} \dot{\tilde k} \\ \dot{\tilde c} \end{pmatrix} = J \begin{pmatrix} \tilde k \\ \tilde c \end{pmatrix}$
Linearized system, with $J$ the Jacobian of the RHS evaluated at the steady state.
Step 2
$J = \begin{pmatrix} \partial \dot k / \partial k & \partial \dot k / \partial c \\ \partial \dot c / \partial k & \partial \dot c / \partial c \end{pmatrix}_{(k^*, c^*)}= \begin{pmatrix} f'(k^*) - (n + \delta) & -1 \\ \tfrac{c^*}{\theta} f''(k^*) & 0 \end{pmatrix}$
Compute the four partials.
Step 3
$= \begin{pmatrix} \rho - n & -1 \\ \tfrac{c^*}{\theta} f''(k^*) & 0 \end{pmatrix}$
Use $f'(k^*) = \rho + \delta$ , so $f'(k^*) - (n + \delta) = \rho - n > 0$ .

The Jacobian's trace and determinant determine its eigenvalues:

Step 1
$\operatorname{tr} J = \rho - n > 0$
Effective discount rate.
Step 2
$\det J = \frac{c^*}{\theta}\, f''(k^*) < 0$
Since $f''(k^*) < 0$ by diminishing returns.
Step 3
$\lambda^{\pm} = \frac{\rho - n \pm \sqrt{(\rho - n)^2 - 4 \det J}}{2}$
Discriminant exceeds $(\rho - n)^2$ , so $\lambda^- < 0 < \lambda^+$ .
Step 4
$\boxed{\;\lambda^- < 0 < \lambda^+\;}$
One stable and one unstable eigenvalue - saddle structure.

Saddle path slope and convergence speed

The stable manifold is spanned by the eigenvector associated with $\lambda^-$ . Solving $(J - \lambda^- I) v = 0$ :

Step 1
$(\rho - n - \lambda^-) v_k - v_c = 0$
First row of $(J - \lambda^- I) v = 0$ .
Step 2
$\left.\frac{dc}{dk}\right|_{\text{SP}} = \frac{v_c}{v_k} = \rho - n - \lambda^- > 0$
Saddle path slopes *upward* in $(k, c)$ space.
Step 3
$k(t) - k^* = (k_0 - k^*)\, e^{\lambda^- t}$
Capital approaches $k^*$ exponentially at rate $|\lambda^-|$ .
Step 4
$\text{Half-life} \;=\; \frac{\ln 2}{|\lambda^-|}$
Periods for the gap to halve.