rigid-free boundary

The solution for the case when the top surface is free and the bottom surface is rigid can be deduced from the odd solution of the rigid-rigid case. The problem is defined by

$$(D^2-a^2)^3W=-a^2 R_a W$$ (67)

subject to boundary condition

$$W=DW = (D^2-a^2)^2 W = 0 \quad z=0$$ (68)

$$W=D^2 W = D^4 W = 0 \quad z=1$$ (69)

The boundary conditions at the mid height for the odd solution is the (69). Accordingly an odd solution for the rigid-rigid boundary at depth $d$ provides solution for the rigid-free boundary for a depth $d/2$. Thus using the stability results from the rigid-rigid case, we have

$$a_c=5.365/2\approx 2.682 \quad R_c=17610.39/2^4\approx 1100.65$$ (70)

[Note: From $\Delta_1^2 f + k^2 f$ in dimensional form, we have $a=kL$ ($L$ is the scaling length) as the nodimensional wave number.]