Economic growth · Derivation

The Solow equation, derived and solved

Lessons 5 and 6 use one equation again and again. This page shows where it comes from, one small step at a time with every symbol defined, adds technical progress, and then solves it exactly for Cobb–Douglas, which gives the speed of convergence its proof.

The short answer

How do you derive and solve the Solow equation?

The Solow model is one differential equation. Deriving it shows that nothing is hidden: it is a statement about how capital per worker changes when saving adds to it, depreciation and new workers subtract from it. Solving it shows the whole path of the economy, not just where it ends up.

K ˙ = s Y - δ K ⟹ k ˙ = s f (k) - (n + δ) k; u = k^{1 - α} ⟹ u ˙ = (1 - α) (s - (n + δ) u)

Derivation: differentiate k = K/L, substitute K̇ = sY − δK, and use Y/L = f(k).
Technical progress only replaces n + δ with n + g + δ, when capital is measured per effective worker.
Exact solution for Cobb–Douglas: u = k^1−α follows a linear equation, so u(t) = u* + (u(0) − u*)e^−λt.
Speed: λ = (1 − α)(n + g + δ), and the half-life of the gap is ln 2 / λ.

New to this? The few things you need first

Derivatives with respect to time: A dot above a variable means its derivative with respect to time: K̇ = dK/dt. The growth rate of a variable is its dot over itself, so K̇/K = d ln K/dt.
Growth rate of a ratio: d ln(K/L)/dt = K̇/K − L̇/L. This is the one rule of calculus the derivation needs.
The Solow model: Output Y = F(K, L) with constant returns, a constant saving rate s, depreciation rate δ, and labour growing at n.

Start here

The setup, with every symbol

Y is output, K is the capital stock and L is the labour force, all at time t. Production has constant returns to scale, $Y = F (K, L)$ .
s is the saving rate, a constant between 0 and 1. All saving is invested, so investment is sY.
δ is the depreciation rate: a fraction δ of the capital stock wears out each moment.
n is the growth rate of the labour force: L̇/L = n.
k = K/L is capital per worker, y = Y/L is output per worker.

Capital grows by investment and shrinks by depreciation, which is the one piece of accounting in the model:

K ˙ = s Y - δ K

The derivation

Deriving the equation, one step at a time

Derivation

From K̇ = sY − δK and L̇/L = n, show that k̇ = s f(k) − (n + δ)k.

Take the log of k = K/L and differentiate with respect to time.
$\frac{d l n k}{d t} = \frac{K ˙}{K} - \frac{L ˙}{L} ⟹ \frac{k ˙}{k} = \frac{K ˙}{K} - n$
Multiply both sides by k = K/L.
$k ˙ = \frac{K ˙}{K} \cdot k - nk = \frac{K ˙}{L} - nk$
Substitute the accounting identity for K̇.
$\frac{K ˙}{L} = \frac{s Y - δ K}{L} = s \frac{Y}{L} - δ \frac{K}{L} = s y - δ k$
Use constant returns: Y/L = F(K/L, 1) = f(k).
$y = f (k)$
Collect terms.
$k ˙ = s f (k) - δ k - nk = s f (k) - (n + δ) k$

Answer

That is the fundamental equation of the Solow model. Constant returns are used once, in step 4: without them output per worker would depend on both K and L, not on k alone.

Read the result as a budget for each worker. Saving per worker, s f(k), is the new capital. The term nk is the capital that must be set aside to give each of the new workers the same amount as everyone else, and δk is the capital that wore out. Their sum is break-even investment. k is constant when the budget balances.

One change

Adding technical progress

Let production be $Y = F (K, A L)$ with technology growing at a constant rate, Ȧ/A = g. Define capital and output per effective worker, $k^= K / (A L)$ and $y^= Y / (A L) = f (k^)$ . Repeat the derivation with one change: the denominator, AL, grows at n + g.

\frac{k ^ ˙}{k ^} = \frac{K ˙}{K} - (n + g) ⟹ k^˙ = \frac{s Y - δ K}{A L} - (n + g) k^= s f (k^) - (n + g + δ) k^

Every result of the lesson without technical progress carries over with n replaced by n + g and k, y by k̂, ŷ. That is why the page works with n + δ below and the answer for the general case is obtained by writing n + g + δ.

The exact path

Solving it exactly for Cobb–Douglas

Let f(k) = k^α. The equation k̇ = s k^α − (n + δ)k is not linear in k, but a change of variable makes it so. Define $u = k^{1 - α}$ . Then

Solution

Solve k̇ = s k^α − (n + δ)k exactly.

Differentiate u = k^(1−α).
$u ˙ = (1 - α) k^{- α} k ˙$
Substitute k̇.
$u ˙ = (1 - α) k^{- α} (s k^{α} - (n + δ) k) = (1 - α) (s - (n + δ) k^{1 - α})$
Recognise u.
$u ˙ = (1 - α) (s - (n + δ) u)$
This is linear: its steady state is u* = s/(n + δ), and the gap shrinks exponentially.
$u (t) = u^{*} + (u (0) - u^{*}) e^{- λ t}, λ = (1 - α) (n + δ)$
Undo the substitution.
$k (t) = [u^{*} + (k (0)^{1 - α} - u^{*}) e^{- λ t}]^{\frac{1}{1 - α}}$

Answer

As t → ∞, u(t) → u* = s/(n + δ) and k(t) → (s/(n + δ))^1/(1−α), the steady state of Lesson 5. The path is exact: no approximation was made.

The exact path of capital per worker (Cobb–Douglas)

k (t)^{1 - α} = \frac{s}{n + δ} + (k (0)^{1 - α} - \frac{s}{n + δ}) e^{- (1 - α) (n + δ) t}

The whole transition in one line. With technical progress, replace k by k̂ and n by n + g.

Numerical check

α = 1/3, s = 0.25, n + g + δ = 0.08, starting at k̂(0) = 1. Find k̂ after 20 years.

Steady state of u.
u* = 0.25 / 0.08 = 3.125, so k̂* = 3.125^1.5 = 5.52.
Start.
u(0) = 1^2/3 = 1.
Speed.
λ = (2/3)(0.08) = 0.0533, so e^−λ·20 = e^−1.067 = 0.344.
Path.
u(20) = 3.125 + (1 − 3.125)(0.344) = 3.125 − 0.731 = 2.394.
Back to k̂.
k̂(20) = 2.394^1.5 = 3.70.

Answer

About 3.70, which is about 60% of the way from 1 to the steady state of 5.52 in 20 years. The exact formula and a step-by-step simulation of k̇ = sk̂^α − 0.08k̂ agree.

What it means

Reading off the speed of convergence

The gap in u shrinks by a fixed fraction every instant, at rate λ. So the gap halves every ln 2 / λ years, whatever the starting point. With λ = 5.3% the half-life is 13 years; with the human-capital version of Lesson 8, λ = 2.7% and it is 26 years. The saving rate s appears in u* (where the economy is heading) but not in λ (how fast it gets there), which is why raising saving changes the destination and not the speed.

In terms of the log of capital, which is what a regression sees, the same result comes from expanding the growth rate of k̂ around the steady state: the slope with respect to ln k̂ is −(1 − α)(n + g + δ), the same λ. The expansion is exact in u and approximate in ln k̂, which is why the two agree only near the steady state.

Where this is used

Lesson 6 states λ = (1 − α)(n + g + δ); Lesson 8 turns it into the regression coefficient β = (1 − e^−λT)/T and compares it with the data. Both depend on this one substitution.

Practice

Check yourself

0 of 6 correct so far

Numbers

Problem 1

α = 0.5, s = 0.3, n + δ = 0.1, k(0) = 1. Find u* = s/(n + δ) and k* (steady-state capital per worker).

u*k*

Problem 2

Same economy. Find the speed λ (per cent a year) and k(10), capital per worker after 10 years (one decimal).

λ (%)k(10)

Concepts

1In deriving k̇ = s f(k) − (n + δ)k, the term nk appears because

2The substitution u = k^(1−α) makes the Cobb–Douglas Solow equation

Quick answers

Frequently asked questions

How is the Solow capital accumulation equation derived?

Start from K̇ = sY − δK, capital grows by saving less depreciation. Define capital per worker k = K/L and take its time derivative: k̇ = K̇/L − nk, because the labour force grows at rate n. Substitute K̇ and use Y/L = f(k), which holds with constant returns to scale. The result is k̇ = s f(k) − (n + δ)k.

Can the Solow equation be solved exactly?

Yes, for Cobb–Douglas production. With f(k) = k^α, the substitution u = k^(1−α) turns the equation into a linear one, u̇ = (1 − α)(s − (n + δ)u), whose solution is u(t) = u* + (u(0) − u*)e^(−λt) with λ = (1 − α)(n + δ). Capital per worker is then k(t) = u(t)^(1/(1 − α)).

Where does the speed of convergence (1 − α)(n + g + δ) come from?

From the exact solution above: in the variable k̂^(1−α) the gap to the steady state shrinks by the factor e^(−λt), with λ = (1 − α)(n + g + δ). Equivalently, a first-order expansion of the growth rate of k̂ around the steady state gives a slope of −(1 − α)(n + g + δ) on ln k̂.

Back to the lessons: the Solow model and Solow in motion.

The Solow equation, derived and solved

The setup, with every symbol

Deriving the equation, one step at a time

Adding technical progress

Solving it exactly for Cobb–Douglas

Reading off the speed of convergence

Check yourself

Numbers

Concepts

Frequently asked questions

Read next

The Solow equation, derived and solved

The setup, with every symbol

Deriving the equation, one step at a time

Adding technical progress

Solving it exactly for Cobb–Douglas

Reading off the speed of convergence

Check yourself

Numbers

Concepts

Frequently asked questions

Read next