# Solow model, time and steady state

Suppose we have a Solow model:

$$Y(t)=C(t)+I(t)$$

$$I(t)=sY(t)$$

$$\dot K=I(t)-δK(t)$$

With a given Cobb-Douglas:

$$Y(t)=Z(t)K^aL^{1-a}$$

$$y(t)=Y(t)/L(t)$$

$$k(t)=K(t)/L(t)$$

$$y=Zk^a$$

We also know these:

$$L=L(t), \dot L/L=n$$

$$Z=Z(t), \dot Z/Z=g$$

Therefore we have:

$$L(t)=L(0)e^{nt}$$

$$Z(t)=Z(0)e^{gt}$$

We get to this point:

$$\dot k=sZk^a-(n+δ)k$$

In the steady state:

$$k=k^*$$

$$k^*=\left(\frac{sZ}{n+δ}\right)^{1/(1-a)}$$

The question has to do with the time dimension in Z. Time enters the final expression if we substitute Z with what we found above, but I'm not sure what to make of this.

$$k^*=\left[\frac{sZ(0)e^{gt}}{n+δ}\right]^{1/(1-a)}$$

Surely there's something this neophyte is mixing up. I thought that in a steady state capital per worker remains the same.

• You can write $Z$ only in case where it grows "exogenously". This means that the differential equation $\dot{Z}$ should be an independent equation from the dynamic system. Equivalently, $Z$ evolves in an independant way from the other key variables of the model. Feb 3, 2017 at 1:09

Kshitiz is right. If you define effective capital per worker as $\hat k(t) \equiv \dfrac{K(t)}{L(t)Z(t)}$, the equation for the dynamic of $\hat k(t)$ is:
$$\dot {\hat k} (t) = s {\hat k}^a - (n+g+\delta){\hat k}$$
$${\hat k} = \left(\frac{s}{n+g+\delta}\right)^{1/(1-a)}$$
Intuitively, every period the marginal product of capital is increasing because of the exogenous expansion in $Z(t)$. In the standard Solow diagram with capital per worker in the horizontal axis, this exogenous technical change means an outward expansion of the $y$ and $sy$ curves. This higher return to capital induces more investment and an expansion of capital per worker. As technical change occurs every period, this process continues forever.