Resonant and non-resonate solution of a flow with a rigid lid at z=H

A) FORCED SOLUTION:

Since the flow is hydrostatic the vertical velocity equations in steady state is:

Integrating in x the solution is:

To find W0, we should apply the boundary condition at z=0

The full solution is then:

This is a general solution for the steady state response. Note that this solution is out of phase with h, as you well know from your numerical results. Since ux=-wz, it is easy to see that u is proportional to h.The solution is valid for NH/U different than nπ. If NH/U=nπ , τhe solution for W would be unbounded in the interior of the flow, in such case the solution is resonant.

B) RESONANT SOLUTION

Since the resonant solution is transient (it grows in time) we should inspect the transient equation for W:

The method to solve this equation is by Laplace transform the time dependence and applying initial conditions for W(t=0). We first find the solution in x and z for the transform, we anti-transform this solution to the time space and obtain the full expression for W as a function of x, z, and t that satisfy the boundary conditions and the initial condition. This is a very elaborated proseadure. However it is not difficult to find out that the Laplace transform has a double pole for W. This implies that the resonant solution will grow linear in time. Assuming that we can try a solution of the form:

Notice the solution has terms proportional to t and z. We need to find now the expressions for Ws and Wc.

It is easy to show that if we substitute in R.1 W for R.2 and after a lot of cancellations, we find a simple condition for Ws=f(Wc). (careful with signs)

The boundary condition at z=0 the same as in the non-resonant case, provides the value of Wc(x)

The full Resonant solution will be:

Note that the forced amplitude (second term RHS) is proportional to dh/dx and it is out of phase with h, whereas the resonant term that grows with t, is proportional to hxx and it is in phase with the orography. This solution satisfy a particular initial value, for a more general initial value solution it should be use the Laplace transform method. The initial value in the inviscid case is not very important because the linear resonant solution will keep growing and will be the prevalent solution after a long time. In numerical experiment however, where dissipation is important, the solution will became stationary with the amplitude for W to have both contributions, resonant and forced. Depending how much the resonant mode grows, W display a phase between 90 to 180 degrees respect to h.