Consider a shallow-water system with mean depth $H$, where the base state consists of the flow $(u,v)=(u_{0},0)$, with a sloped water surface $\eta_{0}(x,y) = - \gamma y$, where $u_{0}$ and $\gamma$ are constants. Consider now small amplitude deviations $u'$ $v'$ $\eta'$ from this base state and assume for simplicity that none of the deviations vary with $y$ (ie the partials of the primes wrt to $y$ are zero). With this assumption linearise the shallow water system about the mean state.
For this question do we have to linearise the equations
$u=u_{0}+u'$
$v= 0+v' $
$\eta = -\gamma y + \eta'$
if i am right here, i am unsure how to proceed.
any pointers please?, A previous example I have
used the momentum equation $\frac{DU'}{DT}+g\frac{\partial\eta'}{\partial x}-fv'=0$
and expanded it to $\frac{\partial u'}{\partial t}+u'\frac{\partial u'}{\partial x} +v' \frac{\partial u'}{\partial y} +g \frac{\partial\eta'}{\partial x}-fv'=0$
then neglected the middle two terms and dropped the primes
Im a little unsure of this process and help understanding would be
appreciated.
This can only be answered denpending on what formulation of the shallow water equations you consider. Based on the formula you presented, I suspect you use the shallow water equations for constant density $\rho$ with viscous effects but respected coriolis force $f$ in non-conservative form. The notation for the different heights is not really consistent, I assume you apply the one used in this resource, which also deals with linearization around a certain state. \begin{align} \partial_t \eta+ \partial_x \Big((\eta +h) u \Big) + \partial_y \Big( (\eta +h) v \Big) & = 0 \\ \frac{D u}{Dt} - fv + g \partial_x \eta & = 0 \\ \frac{D v}{Dt} - fu + g \partial_y \eta & = 0 \end{align}
Since you have three unknowns, you must have three equations at hand - that's why only the momentum equations do not suffice. For $u_0 \neq u_0(x, y, t), \gamma \neq \gamma(x, y, t)$ you get for the presented linearizations: \begin{align} \partial_t \eta'+ \partial_x \Big((-\gamma y + \eta' +h) (u_0 + u') \Big) + \partial_y \Big( (-\gamma y + \eta' +h) v' \Big) & = 0 \\ \partial_tu' + (u_0 + u') \partial_x u' + v' \partial_y u' - fv' + g \partial_x \eta' & = 0 \\ \partial_t v' + (u_0 + u') \partial_x v' + v' \partial_y v' - f(u_0 + u') + g (- \gamma \partial_y \eta') &= 0 \end{align} To obtain truly linear equations, no products of $(\cdot)'$ quantities (and derivatives) may remain. They are typically neglected under the reasoning that for varaitions of order $\mathcal{O}(\Delta)$, variations of order $\mathcal{O}(\Delta^2)$ are negligible for $\Delta \ll 1$. This gives \begin{align} \partial_t \eta'+ \partial_x \Big(-\gamma y u_0 + \eta' u_0 + h u_0 - \gamma y u' + hu' \Big) + \partial_y \Big( -\gamma yv' +h v' \Big) & = 0 \\ \partial_tu' + u_0 \partial_x u' - fv' + g \partial_x \eta' & = 0 \\ \partial_t v' + u_0 \partial_x v' - f(u_0 + u') + g (- \gamma \partial_y \eta') &= 0 \end{align} and after further simplification of the first equation you obtain a linear system of unknowns $\eta', u', v'$ with parameters $f, g, h, u_0, \gamma$. \begin{align} \partial_t \eta'+ u_0 \partial_x\eta' - \gamma y \partial_x u' + h \partial_x u' -\gamma ( v' + y \partial_y v') +h \partial_y v'& = 0 \\ \partial_tu' + u_0 \partial_x u' - fv' + g \partial_x \eta' & = 0 \\ \partial_t v' + u_0 \partial_x v' - f(u_0 + u') + g (- \gamma \partial_y \eta') &= 0 \end{align}