“If $\lambda$ is greater than than 1, the system explodes.” Why does the system explode?

Question

David Romer in his textbook Advanced Macroeconomics (Third Edition) writes regarding the speed of convergence of the Diamond model the following:

(Pg. 83)

Equation (2.60) [$k_{t+1}={1\over{(1+n)(1+g)}}{1\over{2+\rho}}(1-\alpha)k_t^\alpha$] gives $k_{t+1}$ as a function of $k_t$. The economy is on its balanced growth path when these two are equal, that is $k^*$ is defined by

$k^*={1\over{(1+n)(1+g)}}{1\over{2+\rho}}(1-\alpha)k^{* \alpha}\space\space\space\space\space\space\space\space\space\space\space\space$ (2.61)

Solving this expression for $k^*$ yields

$k^*=\left[{(1-\alpha)\over{(1+n)(1+g)(2+\rho)}}\right]^{1 \over{1-\alpha}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ (2.62)

[...] We can also find out how quickly the economy converges to the balanced growth path. To do this, we again linearize around the balanced growth path. [...] Thus:

$k_{t+1} \simeq k^*+\left( {{dk_{t+1}}\over{dk_t}}\Big{|}_{k_t=k^*}\right) \left(k_t-k^*\right)\space\space\space\space$ (2.64)

Let $\lambda$ denote $dk_{t+1}/dk_t$ evaluated at $k_t=k^*$. With this definition, we can rewrite (2.64) [...]

$k_t-k^* \simeq \lambda^t (k_0-k^*)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ (2.65)

Where $k_0$ is the initial value of $k$

My Question

(Pg 84)

[...] if $\lambda$ is greater than 1, the system explodes

What does it mean the system explodes when $\lambda$ is greater than 1? Why does it explode when $\lambda$ is greater than 1?

Answer 1

Eq. (2.64) can be written as (at first order)

$$ k_{t+1} - k^* = \lambda (k_t - k^*) \tag{1a} $$

Define the quantity $\kappa_t$ as

$$ \kappa_t \stackrel{\rm def}{=} k_t - k^* \tag{2} $$

So that

$$ \kappa_{t+1} = \lambda \kappa_t \tag{1b} $$

This can be solved very easy by realizing that

\begin{eqnarray} \kappa_1 &=&\lambda \kappa_0 \\ \kappa_2 &=&\lambda \kappa_1 = \lambda(\lambda\kappa_0) = \lambda^2\kappa_0 \\ &\vdots& \\ \kappa_t &=& \lambda^t \kappa_0 \tag{3} \end{eqnarray}

So the dynamics is $\kappa$ is driven by the value $\lambda^t$. You can see that if $\lambda = 1$ then $\lambda^t=\lambda$ for each $t$ and $\kappa$ remains constant. However, if $\lambda > 1$ the values $k^t$ will become increasingly large (explodes) as $t$ increases, that is

$$ \lim_{t\to \infty} \kappa_t \stackrel{\lambda > 1}{=} \infty $$

Similarly

$$ \lim_{t\to \infty} \kappa_t \stackrel{0< \lambda < 1}{=} 0 $$

Answer 2

A system is explosive if its coefficients are non-stationary.

Stationary is an important property to have in dynamic models as it tells us that an equilibrium value is obtainable (which is important in finding the BGP). If your system is explosive no equilibrium (and BGP) exists.

In your example where you have the equality: $$k_t-k^* \simeq \lambda^t (k_0-k^*)$$

If $\lambda\le1$, then as time progresses ($t=1,2,...,n$) the impact of a policy change approaches a new equilibrium and BGP.

However if $\lambda>1$ then your system is explosive and equilibrium will not be reached, thus making the system unsolvable as $t$ tends to $\infty$.