Deriving the Speed of Solow Convergence

Why Derive a Convergence Speed?

The Solow model’s steady state identifies the capital stock toward which an economy adjusts, but it does not by itself say how quickly adjustment occurs. The local convergence rate answers a more specific question: when capital per effective worker is close to its steady-state level, how quickly does a small gap shrink? The answer depends on the slope of net investment at the steady state, not just on the steady-state capital stock.

The analysis here uses the continuous-time Solow model with constant saving rate, population growth, labor-augmenting technological progress, and depreciation. Capital and output are measured per effective worker. This framework makes the distinction between a local exponential rate and stability in general explicit.

Definition: Let $k$ be capital per effective worker, $f(k)$ output per effective worker, $s$ the constant saving rate, and $b=n+g+\delta$ the break-even investment rate, where $n$ is population growth, $g$ is labor-augmenting technological progress, and $\delta$ is depreciation. In continuous time, capital evolves according to $$ \dot{k}=h(k)=s f(k)-b k. $$ A steady state $k^*$ satisfies $h(k^*)=0$, or equivalently $s f(k^*)=b k^*$. The local convergence rate is $\lambda=-h'(k^*)$ when $h'(k^*)<0$; in that case, small deviations from $k^*$ shrink approximately at the exponential rate $\lambda$.

Linearizing Around the Steady State

Write the deviation from steady state as $x(t)=k(t)-k^*$. Since $h(k^*)=0$, a first-order Taylor expansion of net investment around $k^*$ gives

\dot{x}=h(k^*+x)\approx h(k^*)+h'(k^*)x=h'(k^*)x.

The resulting linear differential equation has the solution $x(t)\approx x(0)e^{h'(k^*)t}$. Thus, if $h'(k^*)<0$, define $\lambda=-h'(k^*)>0$; then the size of the local gap follows $ |x(t)|\approx |x(0)|e^{-\lambda t}$. A larger positive $\lambda$ means faster local adjustment. Because $h'(k)=s f'(k)-b$, the rate is

\lambda=b-s f'(k^*)=n+g+\delta-s f'(k^*).

This is a local result. It describes behavior sufficiently close to the steady state; it does not claim that a large initial gap follows the exponential formula exactly. The steady state and its stability are established in the earlier material on stability of the Solow steady state. Here, the derivative gives the additional information about the first-order speed.

Write net investment.
Use $h(k)=s f(k)-b k$, with the model’s break-even investment rate $b=n+g+\delta$.

Use the steady-state condition.
Confirm $s f(k^*)=b k^*$. This identifies the point around which the equation is being linearized.

Differentiate and evaluate.
Calculate $h'(k^*)=s f'(k^*)-b$, then check its sign.

Interpret the result.
If $h'(k^*)<0$, the local exponential convergence rate is $\lambda=-h'(k^*)$. If $h'(k^*)=0$, the first-order calculation gives no positive exponential rate; higher-order terms must be examined.

Why the Formula Holds

The linearization follows from the local shape of net investment. A deviation $x$ changes net investment, to first order, by $h'(k^*)x$. If the slope is negative, a positive deviation produces negative net investment, while a negative deviation produces positive net investment. In either case the change points back toward the steady state. The linear equation keeps this first-order response and omits terms that become relatively small as $x$ approaches zero.

Solving the linear equation also shows why the derivative’s sign matters. Separating variables gives $dx/x=h'(k^*)dt$. Integrating from time 0 to $t$ yields $\ln|x(t)|-\ln|x(0)|=h'(k^*)t$, and therefore $x(t)=x(0)e^{h'(k^*)t}$. A negative derivative makes this gap decay exponentially; a positive derivative makes small deviations grow. A zero derivative makes the first-order equation uninformative about whether the original nonlinear system is stable.

When the production function is twice differentiable and $f''(k)<0$, diminishing returns mean that the marginal product $f'(k)$ falls as capital rises. This curvature affects the slope of net investment at the steady state through $f'(k^*)$. It does not mean that curvature alone determines the speed: the rate also depends on break-even investment, and the marginal product must be evaluated at the particular steady state.

Connection to Capital’s Output Elasticity

For a positive steady-state capital stock and output, define capital’s output elasticity at the steady state as $\alpha^*=k^* f'(k^*)/f(k^*)$. It measures the percentage change in output associated locally with a one-percent change in capital, holding the production technology fixed. Combining this definition with the steady-state condition gives a useful equivalent expression for the local rate.

s f'(k^*)=s\frac{\alpha^* f(k^*)}{k^*} =\alpha^*\frac{s f(k^*)}{k^*} =\alpha^* b, \qquad \lambda=b(1-\alpha^*).

This form is valid when $k^*>0$, $f(k^*)>0$, and the steady-state condition holds. It makes the role of the output elasticity transparent: for a fixed break-even rate, a smaller $\alpha^*$ implies a larger local convergence rate. In the Cobb-Douglas case $f(k)=A k^\alpha$, the elasticity is constant and equals $\alpha$, so $\lambda=(1-\alpha)b$. Do not assume this simplified expression for every production function without checking its elasticity at the steady state.

Worked Example: Calculating the Local Rate and a Gap Estimate

Consider a continuous-time economy with $f(k)=\sqrt{k}$, saving rate $s=0.20$, and break-even investment rate $b=0.10$ per year. Find the steady state and local convergence rate. Then use the local approximation to estimate capital per effective worker after 10 years if it begins at $k(0)=4.4$.

Method. First solve the steady-state condition, then differentiate net investment at that steady state. The model assumes the stated rates and production function remain constant.

At a steady state, $0.20\sqrt{k^*}=0.10k^*$. Since $k^*>0$, divide by $0.10\sqrt{k^*}$ to obtain $2=\sqrt{k^*}$, hence $k^*=4$. Check: output is $f(4)=2$, saving is $0.20(2)=0.40$, and break-even investment is $0.10(4)=0.40$, so the steady-state condition holds.

The marginal product is $f'(k)=1/(2\sqrt{k})$. Thus $f'(4)=1/4=0.25$, and $s f'(4)=0.20(0.25)=0.05$. Therefore $h'(4)=0.05-0.10=-0.05$ per year, so $\lambda=0.05$ per year. The negative derivative indicates local exponential convergence.

The initial deviation is $x(0)=4.4-4=0.4$. At 10 years, the local estimate is $x(10)\approx0.4e^{-0.05(10)}=0.4e^{-0.5}\approx0.2426$. Therefore $k(10)\approx4+0.2426=4.2426$. The gap has shrunk, as expected. This is a local approximation, not an exact solution for the nonlinear transition path.

Check. The predicted capital stock remains above $k^*=4$, and the gap $0.2426$ is smaller than the initial gap $0.4$, consistent with the negative slope of net investment at the steady state. The method avoids treating the local estimate as an exact path for a sizeable deviation.

Worked Example: Comparing Elasticities and Curvature

Compare two economies, each with break-even investment $b=0.08$ per year, saving rate $s=0.08$, and steady state $k^*=1$. Their production functions are $f_1(k)=k^{0.25}$ and $f_2(k)=k^{0.50}$. Find each local convergence rate and compare their curvature at the steady state.

Method. Verify the steady state, calculate each marginal product, and apply $\lambda=b-sf'(k^*)$. At $k=1$, both production functions equal 1. Hence saving is $0.08$, equal to break-even investment $0.08(1)=0.08$; $k^*=1$ is the steady state in both cases.

For $f_1$, $f_1'(k)=0.25k^{-0.75}$, so $f_1'(1)=0.25$. Thus $s f_1'(1)=0.08(0.25)=0.02$ and $\lambda_1=0.08-0.02=0.06$ per year. Its second derivative is $f_1''(k)=0.25(-0.75)k^{-1.75}$, giving $f_1''(1)=-0.1875$.

For $f_2$, $f_2'(k)=0.50k^{-0.50}$, so $f_2'(1)=0.50$. Thus $s f_2'(1)=0.08(0.50)=0.04$ and $\lambda_2=0.08-0.04=0.04$ per year. Its second derivative is $f_2''(k)=0.50(-0.50)k^{-1.50}$, giving $f_2''(1)=-0.25$.

Both functions have diminishing returns at the steady state, and the second has the more negative second derivative there. Yet its local convergence rate is smaller: $0.04<0.06$. The direct rate calculation depends on the marginal product $f'(k^*)$, not on comparing second derivatives alone. Curvature describes how marginal product changes with capital; the marginal product at the steady state enters the first-order speed.

Check. Each rate is positive, so both steady states have negative net-investment slopes. The calculation avoids assuming that a more negative second derivative automatically means faster convergence.

Zero Derivative: Stability Is Not the Same as Exponential Speed

A negative derivative at the steady state is sufficient for local exponential convergence, but it is not necessary for stability. If $h'(k^*)=0$, linearization gives a zero first-order rate and cannot settle the stability question. Higher-order terms may still point toward the steady state, or they may push the system away. The sign and form of those terms matter.

Worked Example: Stable Adjustment With Zero First-Order Rate

As a local mathematical illustration, take $s=B=1$ and $f(k)=k-(k-1)^3$ near $k^*=1$, so net investment is $h(k)=s f(k)-Bk$. Determine whether $k^*=1$ is locally stable and find the exact path for an initial deviation $x(0)=0.20$.

At $k=1$, $f(1)=1$, so $h(1)=1-1=0$. Writing $x=k-1$ gives $h(k)=-x^3$. The derivative is $h'(k)=-3(k-1)^2$, so $h'(1)=0$. The first-order formula therefore gives no positive exponential convergence rate.

The nonlinear equation is $\dot{x}=-x^3$. For a nonzero $x$, differentiate $x^{-2}$: $d(x^{-2})/dt=-2x^{-3}\dot{x}=2$. Integrating gives $x(t)^{-2}=x(0)^{-2}+2t$, and the solution with the same sign as its initial value is $x(t)=x(0)/\sqrt{1+2x(0)^2t}$. With $x(0)=0.20$ and $t=12.5$ years, $x(12.5)=0.20/\sqrt{1+2(0.20)^2(12.5)}=0.20/\sqrt{2}\approx0.1414$. The deviation is smaller and tends toward zero, but it does not follow a positive first-order exponential rate.

Check. For positive $x$, $\dot{x}<0$; for negative $x$, $\dot{x}>0$. Both signs move toward zero, confirming local asymptotic stability. This is an illustration of the zero-derivative caveat, not an assertion that this production function satisfies the standard diminishing-returns assumption throughout a neighborhood.

Common Mistakes and Why the Rate Matters

A common sign error is to calculate $s f'(k^*)-b=-0.05$ in the first example and report “convergence speed $-0.05$.” That quantity is $h'(k^*)$, the slope of net investment. The positive convergence rate is its negative: $\lambda=b-s f'(k^*)=0.05$ per year. Another error is to use $b$ itself as the convergence rate. Break-even investment is not the speed; the marginal-product term $s f'(k^*)$ must be subtracted.

For a positive exponential rate, the half-life is the time required for the local gap to fall to half its initial size. Solving $e^{-\lambda t}=1/2$ gives $t=\ln(2)/\lambda$. The units must match: if $\lambda$ is per year, this half-life is in years. This calculation is a local estimate and should not be interpreted as an exact time-to-adjustment formula for a large deviation.

The local speed helps distinguish the location of a steady state from the strength of adjustment around it. Two economies can each have a stable steady state but have different marginal products there and therefore different local rates. Conversely, a zero derivative means the first-order rate is zero or unavailable, not that the economy is necessarily unstable. Always calculate the derivative, state its sign, and use higher-order analysis when the derivative vanishes.

Key takeaway: In the continuous-time Solow model, $\lambda=b-s f'(k^*)$. A negative net-investment derivative yields local exponential convergence at positive rate $\lambda$. If that derivative is zero, linearization alone cannot determine stability or a positive speed.

Check Your Understanding

For each problem, use the continuous-time Solow model and report rates per year. Treat any exponential path as a local approximation.

At a steady state, $b=0.14$ and $s f'(k^*)=0.09$. Calculate $h'(k^*)$, determine whether the first-order dynamics converge, and state the local rate. (Answer: $h'(k^*)=-0.05$ per year; local convergence at $0.05$ per year.)
A Cobb-Douglas economy has capital output elasticity $\alpha=0.30$ and break-even investment rate $b=0.10$. Calculate its local convergence rate and the half-life of a small capital gap. (Answer: $\lambda=0.07$ per year; half-life $=\ln(2)/0.07\approx9.90$ years.)
At a steady state, $b=0.12$, $k^*=2$, and $s f(k^*)=0.24$. The capital output elasticity there is $0.40$. Use the steady-state condition to calculate $s f'(k^*)$ and the local rate. (Answer: $s f'(k^*)=0.40(0.12)=0.048$; $\lambda=0.12-0.048=0.072$ per year.)
A locally stable steady state has $\lambda=0.04$ per year, and the initial capital gap is $0.30$. Estimate the gap after 5 years using the linearized path, then state the estimated capital deviation’s sign if the initial capital stock was below steady state. (Answer: $0.30e^{-0.20}\approx0.2456$; the deviation remains negative, approximately $-0.2456$.)
Near $k^*=2$, net investment is $h(k)=-(k-2)^3$. Find $h'(k^*)$, determine the direction of motion for a small deviation on each side, and state whether a positive exponential rate follows from linearization. (Answer: $h'(2)=0$; deviations on both sides move toward 2, so the steady state is locally stable in this model, but no positive first-order exponential rate follows from linearization.)

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