Example. Assistance can maintain or erode expertise [ftip-00OO]
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    Fix maintenance \(m\geq 0\), assistance intensity \(a\geq 0\), and parameters \(\eta ,\delta _0>0\), \(\xi ,\delta _1\geq 0\). Consider

    \[\dot H=\eta m+\xi a-(\delta _0+\delta _1a)H,\qquad H_* =\frac {\eta m+\xi a}{\delta _0+\delta _1a}.\]

    Here \(\xi a\) models learning produced by assistance and \(\delta _1aH\) models lost practice. The solution is \(H(t)=H_*+(H_0-H_*)e^{-(\delta _0+\delta _1a)t}\). If \(H_0\geq H_{\min }\), the stock stays above that floor for every \(t\geq 0\) exactly when \(H_*\geq H_{\min }\); if the stationary stock is smaller, it eventually crosses below. For a finite deadline \(T\), monotonicity instead makes viability equivalent to \(H(T)\geq H_{\min }\), which may hold even when the stationary stock is smaller. Moreover,

    \[\frac {dH_*}{da}= \frac {\xi \delta _0-\delta _1\eta m}{(\delta _0+\delta _1a)^2}.\]

    Differentiation shows that the sign depends on the stated parameters. Neither automatic deskilling nor automatic skill improvement follows from assistance alone. Bastani et al. study learning outcomes under different assistance designs; their particular experiment does not identify universal coefficients for this stock law. Assistance and maintenance costs must also fit the resource account.