| T ( | d²z | + | d²z | ) + (ρ − σ)gz. |
| dx² | dy² |
If this quantity is of the same sign as z, the displacement will be increased, and the equilibrium will be unstable. If it is of the opposite sign from z, the equilibrium will be stable. The limiting condition may be found by putting it equal to zero. One form of the solution of the equation, and that which is applicable to the case of a rectangular orifice, is
z = C sin px sin qy.
Substituting in the equation we find the condition
| (p² + q²)T − (ρ − σ)g ={ | +ve stable. |
| 0 neutral. | |
| −ve unstable. |
That the surface may coincide with the edge of the orifice, which is a rectangle, whose sides are a and b, we must have
pa = mπ, qb = nπ,
when m and n are integral numbers. Also, if m and n are both unity, the displacement will be entirely positive, and the volume of the liquid will not be constant. That the volume may be constant, either n or m must be an even number. We have, therefore, to consider the conditions under which
| π² ( | m² | + | n² | ) T − (ρ − σ)g |
| a² | b² |
cannot be made negative. Under these conditions the equilibrium is stable for all small displacements of the surface. The smallest admissible value of m²/a² + n²/b² is 4/a² + 1/b², where a is the longer side of the rectangle. Hence the condition of stability is that