The following problem is taken from Question 1, Chapter 5 of:
D.R. Durran, Numerical Methods for Fluid Dynamics: With Applications to Geophysics,. 1. Texts in Applied Mathematics 32, DOI 10.1007/978-1-4419-6412-0 1
It states the following:
Consider two sets of equations that might be supposed to govern one-dimensional shallow-water flow:
system 1:
$ \frac{\partial u}{\partial t} + \frac{\partial }{\partial x}(\frac{u^{2}}{2} + gh) = 0 $ with $ \frac{\partial h}{\partial t} + \frac{\partial (hu)}{\partial x} = 0 $
system 2:
$ \frac{\partial hu}{\partial t} + \frac{\partial }{\partial x}(hu^{2} + \frac{gh^{2}}{2}) = 0 $ with $ \frac{\partial h}{\partial t} + \frac{\partial (hu)}{\partial x} = 0 $
a) Under what conditions do the systems have identical solutions?
b) Give an example, including initial conditions and expressions for the time dependant solutions, for which these systems have different solutions.
c) In those situations where these systems have different solutions, which one serves as the correct mathematical model for shallow-water flow?
My solutions:
a)
system 2) is equivalent to system 1) under the condition that:
$ \frac{\partial h}{\partial t}u + \frac{\partial h}{\partial x}u= 0 $
which leads to equality between the systems if $ \frac{\partial u}{\partial x} = 0 $, i.e. $ u = f(t) $.
b) For different solutions we thus need that $ u = f(x, t) $.
If we try a solution of the form $ u = f(x - ct) $ and $ h = \sin(x) $ then system 1 results in:
$ -cf^{'} + 2ff^{'} + g \cos (x) = 0 $ and $ \sin (x) f^{'} + \cos(x) f = 0 $
whilst system 2 gives:
$ -\sin(x)cf^{'} + 2\sin(x)ff^{'} + \cos (x)f^{2} + g\sin(x)\cos(x) = 0 $ and $ \sin (x) f^{'} + \cos(x) f = 0 $
Now the flux equations both give an ODE for f which can be solved to give $ f = A \operatorname{cosec} (x) $
but the 2 momentum equations are clearly different and thus will give another distinct solution for f.
c) System 1) is the conventionally used form fort shallow water flow. Rearranging the momentum equation for system 2 gives:
$ \frac{\mathrm{d} hu}{\mathrm{d} t} = - hu\frac{\partial u}{\partial x} - gh\frac{\partial h}{\partial x} $
The last term here can be recognised as the force associated with hydrostatic pressure whereas the first term is an unphysical contribution from somewhere else. Thus I would reject system 2) on physical grounds.
I would appreciate people's views on my approach here and any modification required. In particular, for part b) I am unconfident that my approach is correct as we seem to have two completely different equations, for both systems, that specify f. In part c) I am also unsure. What does this 'unphysical' extra term represent?