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).

Waves in a finite basin

We consider the problem of waves in a finite basin [math]\displaystyle{ 0\lt x\lt 1 }[/math]. At the edge of the basin the boundary conditions are

[math]\displaystyle{ \left.\partial_x \zeta\right|_{x=0} = \left.\partial_x \zeta\right|_{x=1} =0 }[/math]

.

We solve the equations by expanding in the modes for the basin which satisfy

[math]\displaystyle{ \partial_x \left(h(x) \partial_x \zeta_n \right) = -\lambda_n, }[/math]

normalised so that

[math]\displaystyle{ \int_0^1 \zeta_n \zeta_m = \delta_{mn}. }[/math]

The solution is then given by

[math]\displaystyle{ \zeta(x,t) = \sum_{n=0}^{\infty} \left(\int_0^1 \zeta_n(x^\prime) \zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \cos(\sqrt{\lambda_n} t ) }[/math] [math]\displaystyle{ + \sum_{n=1} ^{\infty} \left(\int_0^1 \zeta_n(x^\prime) \partial_t\zeta_0 (x^\prime) dx^\prime \right) \zeta_n(x) \frac{\sin(\sqrt{\lambda_n} t )}{\sqrt{\lambda_n}} }[/math]

where we have assumed that [math]\displaystyle{ \lambda_0 = 0 }[/math].

Calculation of [math]\displaystyle{ \zeta_n }[/math]

We can calculate the eigenfunctions [math]\displaystyle{ \zeta_n }[/math] by an expansion in the modes for the case of uniform depth. We use the Rayleigh-Ritz method. The eigenfunctions are local minimums of

[math]\displaystyle{ J[\zeta] = \int_0^1 \frac{1}{2}\left\{ \left(h(x) \partial_x \zeta\right)^2 - \lambda \zeta^2 \right\} }[/math]

subject to the boundary conditions that the normal derivative vanishes (where [math]\displaystyle{ \lambda }[/math] is the eigenvalue).

We expand the displacement in the eigenfunctions for constant depth [math]\displaystyle{ h=1 }[/math]

[math]\displaystyle{ \zeta = \sum_{n=1}^{N} a_n \psi_n(x) }[/math]

where

[math]\displaystyle{ \psi_n = \sqrt{2} \cos( (n-1) \pi x),\,\,n\ne 1 }[/math]

[math]\displaystyle{ \psi_0 = 1,\, }[/math]

and substitute this expansion into the variational equation we obtain

[math]\displaystyle{ \mathbf{M} \vec{a} = \lambda \vec{a} }[/math]

where the elements of the matrix M are

[math]\displaystyle{ m_{mn} = \int_0^1 \left\{ \left(\partial_x \psi_m h(x) \partial_x \psi_n\right) \right\} }[/math]

Waves in an infinite basin

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 \,\mathrm{d}x }[/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 {\mathrm{d} \hat{w}}{\mathrm{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) \, \mathrm{d}x + \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) \, \mathrm{d}x = 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) \, \mathrm{d}x =\lambda \int_{0}^{1} \left( (b-a)x+a \right) \sin (n \pi x) \, \mathrm{d}x + \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) \, \mathrm{d}x }[/math] [math]\displaystyle{ = \lambda \left((b-a) \int_{0}^{1} x \sin(n \pi x) \, dx + a \int_{0}^{1} \sin(n \pi x) \, \mathrm{d}x \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) \, \mathrm{d}x + \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) \, \mathrm{d}x \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) \, \mathrm{d}x \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) \, \mathrm{d}x + \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].