Variable Depth Shallow Water Wave Equation

Introduction

We consider here the problem of waves reflected by a region of variable depth in an otherwise uniform depth region assuming the equations of Shallow Depth.

Equations

We begin with the shallow depth equation

[math]\displaystyle{ \rho(x)\partial_t^2 \zeta = \partial_x \left(h(x) \partial_x \zeta \right). }[/math]

subject to the initial conditions

[math]\displaystyle{ \zeta_{t=0} = \zeta_0(x)\,\,\,{\rm and}\,\,\, \partial_t\zeta_{t=0} = \partial_t\zeta_0(x) }[/math]

where [math]\displaystyle{ \zeta }[/math] is the displacement, [math]\displaystyle{ \rho }[/math] is the string density and [math]\displaystyle{ h(x) }[/math] is the variable depth (note that we are unifying the variable density string and the wave equation in variable depth because the mathematical treatment is identical).

We assume that the depth is constant and equal to one outside the region [math]\displaystyle{ 0\lt x\lt 1 }[/math]. We can therefore write the wave as

Solution using Separation of Variables

Taking a separable solution [math]\displaystyle{ \ w (x,t) = \Tau (t) \hat{w} (x) }[/math] gives the eigenvalue problem

[math]\displaystyle{ \partial_x \left( c(x)^2 \partial_x\hat{w} \right) = \lambda\hat{w} \quad (2) }[/math]

Given boundary conditions [math]\displaystyle{ \hat{w} (0) = a }[/math] and [math]\displaystyle{ \hat{w} (1) = b }[/math] we can take

[math]\displaystyle{ \hat{w} = (b-a)x + a + u \quad (3) }[/math]

With [math]\displaystyle{ u = \sum_{n=1}^{N} a_n \sin (n \pi x) }[/math] a series solution satisfying [math]\displaystyle{ u (0) = u (1) = 0 }[/math]

We wish to solve for [math]\displaystyle{ a_n }[/math] given [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math]. Equation (2) can be transformed into the Sturm-Liouville problem of minimizing the functional

[math]\displaystyle{ J [u] = \int_{0}^{1} c^2 \left(\frac{d \hat{w}}{d x}\right)^2 + \lambda \hat{w}^2 \,dx }[/math]

which is minimal exactly when [math]\displaystyle{ \partial_{a_n} J = 0 \quad \forall a_n }[/math]

Substituting in [math]\displaystyle{ \hat{w} = (b-a)x + a + \sum_{n=1}^{N} a_n \sin (n \pi x) }[/math] and [math]\displaystyle{ \frac {d \hat{w}}{d x} = (b-a) + \sum_{n=1}^{N} a_n n \pi \cos (n \pi x) }[/math]

gives

[math]\displaystyle{ \partial_{a_n} J = \int_{0}^{1} c^2 \left( n \pi \cos (n \pi x) \left((b-a) + \sum_{m=1}^{N} a_m m \pi \cos (m \pi x) \right) \right) \, dx + \int_{0}^{1} \lambda \left( \sin (n \pi x) \left((b-a)x + a + \sum_{m=1}^{N} a_m \sin (m \pi x) \right) \right) \, dx = 0 \quad (4) }[/math]

Since [math]\displaystyle{ \sin(n \pi x) }[/math] and [math]\displaystyle{ \sin(m \pi x) }[/math] are orthogonal the second integral in (4) can be calculated.

[math]\displaystyle{ \int_{0}^{1} \lambda \left( \sin (n \pi x) \left((b-a)x + a + \sum_{m=1}^{N} a_m \sin (m \pi x) \right) \right) \, dx =\lambda \int_{0}^{1} \left( (b-a)x+a \right) \sin (n \pi x) \, dx + \frac{\lambda a_n}{2} }[/math]

and

[math]\displaystyle{ \int_{0}^{1} \lambda \left( \sin (n \pi x) \left((b-a)x + a + \sum_{m=1}^{N} a_m \sin (m \pi x) \right) \right) \, dx }[/math] [math]\displaystyle{ = \lambda \left((b-a) \int_{0}^{1} x \sin(n \pi x) \, dx + a \int_{0}^{1} \sin(n \pi x) \, dx \right) }[/math] [math]\displaystyle{ = \frac{\lambda (b-a)}{(n \pi)^2} \Big[ \sin(n \pi x) -n \pi x \cos( n \pi x) \Big]_{0}^{1} - \frac{a \lambda}{n \pi} \Big[\cos(n \pi x) \Big]_{0}^{1} }[/math] [math]\displaystyle{ = \frac{\lambda (b-a)}{(n \pi)^2} \left(n \pi (-1)^{n+1} \right) + \frac{a \lambda}{n \pi} \left(1 - (-1)^n \right) }[/math] [math]\displaystyle{ = \frac{\lambda}{n \pi} \left( b (-1)^{n+1}+a \right) }[/math]

So equation (4) can be written as [math]\displaystyle{ \int_{0}^{1} c^2 \left( n \pi \cos (n \pi x) \left((b-a) + \sum_{m=1}^{N} a_m m \pi \cos (m \pi x) \right) \right) \, dx + \frac{\lambda}{n \pi} \left( b (-1)^{n+1}+a \right) + \frac{\lambda a_n}{2}= 0 }[/math]

The remaining integral can be split into two parts and with the sum taken outside the integral we obtain [math]\displaystyle{ \int_{0}^{1} c^2 (b-a) n \pi \cos (n \pi x) + \sum_{m=1}^{N} \left( \int_{0}^{1} c^2 n m \pi^2 \cos(n \pi x) \cos (m \pi x) \, dx \right) a_m + \frac{\lambda}{n \pi} \left( b (-1)^{n+1}+a \right) + \frac{\lambda a_n}{2}= 0 }[/math]

or, on rearranging [math]\displaystyle{ \sum_{m=1}^{N} \left( \int_{0}^{1} c^2 n m \pi^2 \cos(n \pi x) \cos (m \pi x) \, dx \right) a_m+ \frac{\lambda a_n}{2}= -\int_{0}^{1} c^2 (b-a) n \pi \cos (n \pi x) - \frac{\lambda}{n \pi} \left( b (-1)^{n+1}+a \right) }[/math]

Now if we form the vector [math]\displaystyle{ \mathbf{a} = \begin{pmatrix} a_1 \\ \vdots \\ a_N \end{pmatrix} }[/math] and recalling that we have the above expression for all n, we can write the above as a matrix multiplication of [math]\displaystyle{ \mathbf{a} }[/math]

[math]\displaystyle{ \mathrm{M} \mathbf{a} = \mathbf{f} }[/math]

with [math]\displaystyle{ \mathrm{M}_{(n,m)} = \int_{0}^{1} c^2 n m \pi^2 \cos(n \pi x) \cos (m \pi x) \, dx + \delta_{nm} \frac{\lambda}{2} \quad \delta_{nm} = \left\{ \begin{matrix} 1 & \mathrm{if} \quad n = m \\ 0 & \mathrm{if} \quad n \neq m \end{matrix} \right. }[/math]

and [math]\displaystyle{ \mathbf{f} = \begin{pmatrix} f_1 \\ \vdots \\ f_N \end{pmatrix} }[/math] with [math]\displaystyle{ f_n = -\int_{0}^{1} c^2 (b-a) n \pi \cos (n \pi x) + \frac{\lambda}{n \pi} \left( b (-1)^n-a \right) }[/math]

So, given a suitable [math]\displaystyle{ c(x) }[/math] and boundary conditions [math]\displaystyle{ \hat{w} (0) = a }[/math] and [math]\displaystyle{ \hat{w} (1) = b }[/math] we have a system of linear equations that can be solved to give the coefficients [math]\displaystyle{ a_n }[/math] which in turn define the function [math]\displaystyle{ \hat{w} }[/math].