Consider the continuous-time Solow model without technological progress and with constant population growth rate equal to $n$. If capitalists don't save and the workers save a fraction $s \in (0,1)$ of their income, show that there can be multiple steady state equilibria possible. Assume that the production function satisfies all the neo-classical assumptions.

The usual start:

\begin{align} \dot k = \frac{\dot{K}}{K} - n = \frac{swL - \delta K}{K} - n = \frac{s(f(k) - kf'(k))}{k} - (n + \delta) = 0 \\ \implies g(k) \stackrel{def}{:=} f(k) - kf'(k) - \left(\frac{n + \delta}{s}\right)k = 0 \end{align}

I have to show that there are more than one $k$ that satify $g(k) = 0$.

How do I proceed from here?

I constructed an example with $f(k) = k^{0.6}$ and I obtain three steady states. But one is negative, the other is $0$ and the third one is positive, so effectively, I only have two steady states, out of which $k=0$ is an example I am trying to avoid.

Here's a graph of $g(k)$ from the above construction: enter image description here

  • $\begingroup$ Not quite sure what you are asking here, there are details missing. Proving that $k = 0$ is a steady state and that a positive steady state also exists would work? What are your so called "necessary conditions"? Is $f$ continuous? $\endgroup$
    – Giskard
    Commented Apr 1, 2023 at 18:42
  • $\begingroup$ Also, do you know any fixed point theorems? $\endgroup$
    – Giskard
    Commented Apr 1, 2023 at 18:43
  • $\begingroup$ Some parts are missing in your question but $k=0$ is probably an unstable equilibrium. Normally, the basic Solow model has an analytical solution and there is a unique equilibrium $\endgroup$ Commented Apr 1, 2023 at 20:58
  • $\begingroup$ @Giskard Necessary assumptions to mean the neoclassical assumptions like $F$ is differentiable, $F_i > 0, F_{ii} < 0$, Inada conditions, CRS, etc. I don't know fixed point theorems as such, but if you use any, I'll look that up. If you show that $k=0$ is a steady state and there's another positive steady state, it would work, yes (but I was trying to find out at least two positive steady state equilibria). $\endgroup$
    – user43302
    Commented Apr 2, 2023 at 2:58
  • $\begingroup$ @optimalcontrol I have modified it so there's less confusion now. The question asks us to show that multiple steady states are possible, not that it's always the case. But from reading the comments, it seems that $k=0$ is always a steady state for this model; can you please explain why? $\endgroup$
    – user43302
    Commented Apr 2, 2023 at 3:00

1 Answer 1


I think that it is very difficult to find an example of production function which in this case violates the unicity of equilibrium.

Instead, I thought of a different approach, that I sketch below, and that I hope can help.

I take, to begin, the last equation you wrote:

$$\dot k= s[f(k) - kf'(k) ]- \left({n + \delta}\right)k = 0 \quad (1)$$

In our problem the usual neoclassical assumptions about the production function hold, in particular

$$f'(k)>0\qquad (2)$$ and $$f''(k)<0\quad (3).$$

In the standard Solow model the motion equation, set equal to $0$ to derive the steady state value of $k$, is:

$$\dot k= sf(k) - \left({n + \delta}\right)k = 0 \quad (4)$$

The assumptions $(2)$ and $(3)$, together with the other neoclassical assumptions, ensure that the steady state in the standard Solow model exists and it is unique$^1$.

And we can draw the usual nice graph of the usual Solow model, with our nice functions:

enter image description here

$$Fig. 1$$

In our non standard problem, instead, according to $(1)$, the equation that describes the steady state is

$$s[f(k) - kf'(k) ]= \left({n + \delta}\right)k \quad (5)$$

so that, even if the production function is always the same, we are now equating to $\left({n + \delta}\right)k$ not $sf(k)$ but the left side of $(5)$.

Nothing ensures that this latter function gives rise to a unique steady state, because it is possible that it doesn't maintain the nice properties of $f(k)$.

Consider the first derivative of the function in the brackets of the left side of $(5)$, that I call $g(t)$:

$$g'(t)= (f(k) - kf'(k))'= f'(k)- [f'(k)+f''(k) k]=-f''(k)k>0\qquad (6) $$

which is positive, according to $(3)$.

Now consider the second derivative of $g(t)$:

$$g''(t)= [-f''(k)k]'= -f''' (k)k-f''(k)\qquad (7) $$

What's the sign of this second derivative of $(g(t)$? We don't know, we don't know anything about the third derivative $f'''(k)$.

Therefore, we can have a function not so nice as in the standard model, the second derivative of which changes its sign.

And we can have a situation as depicted in the picture below, where a 'bad' function leads to multiple equilibria:

enter image description here


$^1$ Apart the trivial solution $k(t) \equiv 0$.

  • $\begingroup$ Thanks, can you also tell if $k=0$ is always an equilibrium? Possibly it is, because when there's zero capital, the production function will be $0$ (neoclassical assumption) and so will be $\bar{A}k$ variable (that determines the number of machines/etc. required). $\endgroup$
    – user43302
    Commented Apr 3, 2023 at 5:09
  • $\begingroup$ I do not see the economic intuition in your equation regarding the part $s(f(k)-kf'(k))$. Another way to generate multiple equilibria in this setting would be to consider a depreciation rate that depends on the capital accumulation $k$. This is like in environmental economics literature where the absorption of pollution in the atmosphere depends on the pollution stock itself. Franz Wirl has some nice papers in Journal of Economic Dynamics and Control in 90s using this approach. $\endgroup$ Commented Apr 3, 2023 at 8:05
  • $\begingroup$ @ Solow supremacy If you consider the differential equation of the standard Solow model $\dot k= sf(k) - \left({n + \delta}\right)k = 0 $ you see that, if $k=0$, $f(k)=0$ (by assumption on the production function) and the second term $(n+\delta) k$ is obviously equal to 0. $\endgroup$ Commented Apr 3, 2023 at 10:49
  • $\begingroup$ @ optimal control I answered the specific question by Solow supremacy, how can be shown that multipla equilibria can appear when there is the assumption that only workers save, and @Solow supremacy asked how to proceed after the equation they wrote. I didn't address the general problem 'how there can be multiple equilibria'. Of course, if we look at textbooks, there are many example of assumptions that lead to multipla equilibria. – $\endgroup$ Commented Apr 3, 2023 at 11:13

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