Consider a water tank shown in the figure. It has one wall at \(x = L\) and can be taken to be very wide in the z direction. When filled with a liquid of surface tension \(S\) and density \( \rho \), the liquid surface makes angle \( \theta_0 \) (\( \theta_0 < < 1 \)) with the x-axis at \(x = L\). If \(y(x)\) is the height of the surface then the equation for \(y(x)\) is: (take \(g\) as the acceleration due to gravity)
To derive the differential equation for the height \(y(x)\) of the liquid surface, we need to consider the forces involved due to surface tension and gravity. The surface tension causes a force that depends on the curvature of the surface, while gravity causes a downward force proportional to the height of the liquid column.
The Young-Laplace equation relates the curvature of a surface to the pressure difference across the surface due to surface tension. For a surface described by a function \(y(x)\), the curvature \(\kappa\) can be approximated by the second derivative \(\kappa \approx \frac{d^2y}{dx^2}\) when \(\theta_0 \ll 1\).
The pressure difference \(\Delta P\) across the liquid surface due to surface tension is given by:
\(\Delta P = S \kappa = S \frac{d^2y}{dx^2}\)
The hydrostatic pressure due to gravity at height \(y(x)\) is:
\(P = \rho g y(x)\)
For equilibrium, set the pressure due to surface tension equal to the hydrostatic pressure:
\[ S \frac{d^2y}{dx^2} = \rho g y(x) \]
Dividing both sides by \(S\), we obtain the differential equation:
\[ \frac{d^2y}{dx^2} = \frac{\rho g}{S} y(x) \]
This is a linear second-order differential equation describing the shape of the surface of the liquid.