<p>You outline the assumption that <span class="math-container">$f$</span> is strictly concave in <span class="math-container">$k$</span>, so
the average slope
<span class="math-container">$$
\frac{f(k) - f(0)}{k - 0}
$$</span>
is larger than <span class="math-container">$f'(k)$</span>.</p>
<p>I will also use the frequently made assumption that <span class="math-container">$f(0) = 0$</span>, though <span class="math-container">$f(0) \geq 0$</span> would be sufficient.</p>
<p>Taking these two together, we get
<span class="math-container">$$
f'(k) &lt; \frac{f(k) - f(0)}{k - 0} = \frac{f(k)}{k}
$$</span>
or
<span class="math-container">$$
f'(k)k &lt; f(k).
$$</span>
In the steady state
<span class="math-container">$$
sf( \bar k ) - (n+g+\delta) \bar k = 0.
$$</span>
Substituting in our inequality
<span class="math-container">$$
sf'( \bar k ) \bar k  - (n+g+\delta) \bar k &lt; 0.
$$</span>
<strong>if</strong> <span class="math-container">$\bar k \neq 0$</span>, so we are not in the 'trivial' steady state, we can divide by <span class="math-container">$\bar k$</span> and we get
<span class="math-container">$$
sf'( \bar k ) - (n+g+\delta) &lt; 0.
$$</span></p>
<hr />
<p>In the case of  <span class="math-container">$f(0) &lt; 0$</span> you would have a steady state where <span class="math-container">$f(k)$</span> intersects the line <span class="math-container">$(n+g+\delta)k$</span> from below, with a higher slope. This steady state would indeed be unstable. The same is true for <span class="math-container">$\bar k = 0$</span> when <span class="math-container">$f(0) = 0$</span>.</p>

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2024-01-07T10:18:04.957Z
2024-01-07T11:12:43.453Z
2024-01-07T11:12:43.453Z
Giskard
<p>I'm currently reading Romer's advanced macroeconomics, and encounter a question of linear stability of the model at the steady state.</p>
<p>We have the key equation <span class="math-container">$k'(t)=sf(k(t))-(n+g+\delta)k(t)$</span>, where <span class="math-container">$k$</span> is capital per effective labour, <span class="math-container">$n,g,\delta$</span> each represents constant growth rate of labour, knowledge and depreciation rate of capital, respectively.</p>
<p>Use linear stability analysis, steady state is where <span class="math-container">$k'(t)$</span> is zero, denoted by <span class="math-container">$\bar k$</span>. We can check the nature of the steady state by using <span class="math-container">$sf'(\bar k)-(n+g+\delta)$</span>. We want to prove that the sign of it is negative.</p>
<p>We know <span class="math-container">$f'(k) &gt;0$</span> and <span class="math-container">$f''(k)&lt;0$</span>, also other assumptions of <span class="math-container">$f$</span> given in the Solow model. Is it possible to prove it mathematically?</p>

<mathematical-economics><solow>
linear stability analysis of basic Solow model
Nonenicht
# linear stability analysis of basic Solow model

(Asked by: username_0 on 2024-01-07)

I'm currently reading Romer's advanced macroeconomics, and encounter a question of linear stability of the model at the steady state.

We have the key equation $k'(t)=sf(k(t))-(n+g+\delta)k(t)$, where $k$ is capital per effective labour, $n,g,\delta$ each represents constant growth rate of labour, knowledge and depreciation rate of capital, respectively.

Use linear stability analysis, steady state is where $k'(t)$ is zero, denoted by $\bar k$. We can check the nature of the steady state by using $sf'(\bar k)-(n+g+\delta)$. We want to prove that the sign of it is negative.

We know $f'(k) >0$ and $f''(k)<0$, also other assumptions of $f$ given in the Solow model. Is it possible to prove it mathematically?

**Tags:** mathematical-economics, solow


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## 1 answers

### Answer by username_1 on 2024-01-07. Score: 0 | Accepted answer

You outline the assumption that $f$ is strictly concave in $k$, so
the average slope
$$
\frac{f(k) - f(0)}{k - 0}
$$
is larger than $f'(k)$.

I will also use the frequently made assumption that $f(0) = 0$, though $f(0) \geq 0$ would be sufficient.

Taking these two together, we get
$$
f'(k) < \frac{f(k) - f(0)}{k - 0} = \frac{f(k)}{k}
$$
or
$$
f'(k)k < f(k).
$$
In the steady state
$$
sf( \bar k ) - (n+g+\delta) \bar k = 0.
$$
Substituting in our inequality
$$
sf'( \bar k ) \bar k - (n+g+\delta) \bar k < 0.
$$
**if** $\bar k \neq 0$, so we are not in the 'trivial' steady state, we can divide by $\bar k$ and we get
$$
sf'( \bar k ) - (n+g+\delta) < 0.
$$

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In the case of $f(0) < 0$ you would have a steady state where $f(k)$ intersects the line $(n+g+\delta)k$ from below, with a higher slope. This steady state would indeed be unstable. The same is true for $\bar k = 0$ when $f(0) = 0$.


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