# A simple wave for the linear shallow water equations

I'm looking for a simple, right-traveling wave for the linear shallow water equations (1D). My question: what are the initial conditions (velocity $$U_0(x)$$ and/or average water height, average velocity) for a simple (single!) right-traveling wave?

I want to create a simple wave solution for the 1D shallow water equations by choosing appropriate initial conditions, just like in this other post. I'm comparing a staggered grid with a non-staggered grid. On the staggered grid, if I have a simple wave traveling to the right, there should not be a left-traveling wave, whereas on the non-staggered grid, there can still be a component with negative group velocity. Just like in the other question, I'm also working with a Gaussian initial wave profile. The problem is that when I use their equation for $$m_0(x)$$, I still get a (small) left-traveling component on both grids. I don't have access to the book by Leveque, so I don't understand where the equation for $$m_0(x)$$ comes from. Could someone please tell me what the reasoning behind $$m_0(x)$$ is or share with me the relevant page(s) from Leveque?

I'm working with linearized shallow water equations, which, I think, is different from their example. It's pretty much the simplest case of shallow water equations: linear equations and no bottom topography. On the staggered grid, the dispersion relation with my discretization scheme (centered-space, leapfrog-time) is

$$\sin(\omega\Delta t) = \frac{(U \pm 2c)\Delta t}{\Delta x}\sin\Big(\frac{k\Delta x}{2}\Big)$$

Where $$c = \sqrt{gH}$$ with $$H$$ the average water height and $$g$$ the gravitational constant. I thought I could just set my initial velocity $$U_0(x)$$ equal to $$2\sqrt{g H}$$ or to $$2\sqrt{g H_0(x)}$$ to eliminate the "-" solution, and have only the right-traveling wave left, but when I do there is a left-traveling trough and I don't know what to make of it. The pictures below are for $$U_0(x) = 2\sqrt{g H_0(x)}$$. The other thing I tried gives a forward and a backward wave.

What are the initial conditions for a simple (single!) right-traveling wave?

Your insights will be very much appreciated.

• Could you please rephrase your question so that it includes at least one question mark "?" making it clear what it is you are asking? Dec 14, 2021 at 21:11

To get a purely right-going solution of the 1D wave equation, your initial condition $$(\eta, u)^T$$ at each value of $$x$$ should be a multiple of a certain vector. For the linearized shallow water equations with gravitational constant $$g$$, that vector is
$$\begin{pmatrix} 1 \\ \sqrt{g/H(x)} \end{pmatrix}$$
Thus if $$\eta(x)$$ and $$u(x)$$ are your initial surface height and velocity, you should have $$u(x) = \eta(x)\sqrt{g/H(x)}$$. For this initial condition, the exact solution is purely right-going. Numerically, if you are using a multistep method (it sounds like you are) then you may see a very small part going to the left. The magnitude of that part will decrease as you refine your grid.