Stability of the Solow Steady State

Why Stability Matters

Finding a steady state identifies a capital stock at which investment just covers depreciation and the dilution of capital across a growing workforce. That calculation alone does not tell us what happens when the economy starts elsewhere. Stability answers the next question: if capital per worker is above or below its steady-state level, does the model move it back toward that level?

In the basic Solow model without technological progress, the answer follows from the shape of the production function and the sign of net investment. With diminishing marginal product, investment per worker rises less than proportionately with capital per worker, while break-even investment rises in direct proportion. Under the conditions stated below, their intersection is unique, and capital moves toward it from either side.

The Equation of Motion and the Steady State

Let \(k\) denote capital per worker, \(f(k)\) output per worker, \(s\) the constant fraction of output saved and invested, \(n\) the constant population growth rate, and \(\delta\) the constant depreciation rate. Assume \(0<s<1\), \(n+\delta>0\), and a differentiable production function with \(f(0)=0\), \(f'(k)>0\), and \(f''(k)<0\) for positive \(k\). There is no technological progress in this version of the model.

Investment per worker is \(s f(k)\). Break-even investment per worker is \((n+\delta)k\): it replaces depreciated capital and provides the capital needed to equip new workers. The resulting continuous-time equation of motion is:

Write \(b=n+\delta\) for the break-even investment rate and define \(g(k)=s f(k)-bk\). Thus \(\dot{k}=g(k)\): a positive value of \(g(k)\) means capital per worker is increasing, while a negative value means it is decreasing. A steady state is a value \(k^*\) for which \(\dot{k}=0\).

For a positive steady state, the equation \(s f(k^*)=bk^*\) can be divided by \(k^*\). The investment condition is therefore equivalent to \(f(k^*)/k^*=b/s\). Under strict concavity and \(f(0)=0\), the average product \(f(k)/k\) decreases as \(k\) increases. If it starts above \(b/s\) and eventually falls below \(b/s\), it crosses that value exactly once, giving a unique positive steady state. These conditions hold, for example, when the average product tends to infinity as \(k\) approaches zero and to zero as \(k\) grows without bound. They are stated to make clear that diminishing returns alone does not guarantee a positive steady state for every imaginable production function and parameter combination.

Why Capital Moves Toward the Steady State

The central sign result comes directly from comparing average product with the break-even requirement. For \(k>0\), factor \(k\) out of net investment:

At the steady state, \(s f(k^*)/k^*=b\). Since \(f(k)/k\) decreases with \(k\), it exceeds its steady-state value below \(k^*\) and falls below it above \(k^*\). Because \(k\) is positive, this gives a positive sign for \(g(k)\) below the steady state and a negative sign above it.

This sign pattern establishes more than a tendency to move in the right direction. If the initial capital stock is below \(k^*\), it rises but cannot pass through the steady state: at \(k^*\), the rate of change is zero, and a smooth, uniquely determined equation of motion does not allow a trajectory to cross the steady-state solution. It is increasing and bounded above by \(k^*\), so it has a limit. That limit must itself satisfy \(g(k)=0\); otherwise, continuity would keep the rate of change bounded away from zero near the limit, contradicting convergence to it. By uniqueness of the positive steady state, the limit is \(k^*\). The same reasoning applies from above: capital falls, remains bounded below by \(k^*\), and converges to it.

There is also a local check. For a strictly concave production function with \(f(0)=0\), the marginal product at a positive \(k^*\) is less than the average product there. Since \(s f(k^*)/k^*=b\), it follows that:

A small increase above \(k^*\) therefore makes net investment negative, and a small decrease below it makes net investment positive. This local result agrees with the global sign argument. It confirms stability, but the sign argument is more informative: it establishes the direction of movement for every positive capital stock on either side, not only for small departures.

What the Stability Result Does—and Does Not—Say

In this model, stability is a directional result: if capital per worker is below the unique positive steady state, net investment is positive; if it is above, net investment is negative. The argument does not calculate how long convergence takes or the exact capital stock at each future date. Those questions require further analysis of the equation of motion.

The assumptions matter. If production were linear rather than strictly concave, average product might not decline, and there need not be a unique attracting positive steady state. If saving were too low relative to break-even investment and the production function did not generate a positive intersection, an interior steady state might not exist. Thus, do not infer stability merely from the word “steady state”; establish existence and check the direction of net investment on both sides.

Common Sign and Interpretation Traps

• Reversing the direction. When investment exceeds break-even investment, \(\dot{k}>0\), so capital per worker rises—not falls.
• Leaving out population growth. Break-even investment is \((n+\delta)k\), not just \(\delta k\), when the workforce is growing.
• Checking only the equality. Solving \(s f(k^*)=(n+\delta)k^*\) finds a steady state, but stability requires checking signs away from it or using an appropriate derivative argument.
• Claiming that diminishing marginal product guarantees everything. Strict concavity gives a decreasing average product under the stated assumptions, but existence of a positive intersection still depends on the production function and parameters.
• Confusing stable with already steady. An economy away from \(k^*\) is not at equilibrium; it moves toward the steady state according to the sign of \(\dot{k}\).