= Solution
Integrate once and use $h=e^{-\alpha x}$ to obtain $p'=6e^{2\alpha x}+C e^{3\alpha x}$. The two zero endpoint <pressures> determine
$$
C=-9\frac{e^{2\alpha}-1}{e^{3\alpha}-1}=-6M,\qquad M=\frac{3(e^{2\alpha}-1)}{2(e^{3\alpha}-1)}.
$$
Consequently
$$
p'=6e^{3\alpha x}(e^{-\alpha x}-M),\qquad
p=\frac3\alpha(e^{2\alpha x}-1)-\frac{2M}\alpha(e^{3\alpha x}-1).
$$
The constant $M=\int_0^1h^{-2}\,dx/\int_0^1h^{-3}\,dx$ is a <weighted mean>[weighted average] of $h$ with positive weight $h^{-3}$, so $e^{-\alpha}<M<1$. Since $e^{-\alpha x}$ decreases strictly, $p'$ changes sign exactly once, from positive to negative. Thus the unique maximum is at
$$
\boxed{x_m=-\frac1\alpha\log M=-\frac1\alpha\log\frac{3(e^{2\alpha}-1)}{2(e^{3\alpha}-1)}.}
$$
For small $\alpha$, $\log M=-\alpha/2-5\alpha^2/24+O(\alpha^4)$, so $\boxed{x_m=\tfrac12+\tfrac5{24}\alpha+O(\alpha^3)\sim\tfrac12}$. The <exponential-gap slider bearing> provides a particularly simple example where a converging gap produces positive load-bearing <pressure>.
Back to article page
