Parametric Lamb Waves

Define $k$ to be a single wavelength of zonally propagating lamb wave.

Relevant quantities: $$p' = p_0 \exp\left(- \int_{z=0}^z \frac{g}{R_d\overline{T}}\, \mathrm{d} z\right) \cos(k(x-ct))$$

$$c = \sqrt{\gamma R_d \overline{T}}$$

$$u'(z) = - \frac{p_0}{(-c)\overline{\rho}} \exp\left(\frac{g}{R_d \overline{T}} z\right) \cos(k(x-ct))$$

$$u'(z) = \frac{p_0}{c\rho_0} \cos(k(x-ct))$$

From one of Christiane's presentations we get

$$\omega = \mathrm{D}_t p = \partial_t p + \mathbf{v} \cdot \nabla p + w \partial_z p$$

$$\omega = \left(\frac{R_dT}{pc_p} - \partial_p T\right)^{-1} \left(\partial_t T + u \partial_x T + v \partial_y T \right)$$

Allow variation in horizontal structure of temperature but not vertical. Assume that density is constant in horizontal, vary in vertical.

$$(\overline{p}(z) + p'(x, z)) = \overline{\rho}(z) R_d (\overline{T} + T'(x))$$ $$p'(x, z) = \overline{\rho}(z) R_d T'(x)$$

And canceling gives $$T'(x) = \frac{p_0}{\rho_0R_d} \cos(k(x-c t))$$

Let's linearize the nonhydrostatic equation in height coordinates to try to figure this out: $$\mathrm{D}_t w = -\rho^{-1}\partial_z p - g$$ $$\overline{\rho}\partial_t w = -\partial_z\overline{p} - \partial_z p' - \overline{\rho} g$$ $$\overline{\rho}\partial_t w = - \partial_z p'$$

So I think this gives us full linearized closure of our equations. This can now be implemented.