Tsunami Generation

Abstract

This chapter theoretically investigates tsunami generation. When an earthquake occurs in an offshore region, seismic waves, ocean acoustic waves, and tsunami are excited. Although the compressibility and elasticity of the sea layer are important for the propagation of ocean acoustic waves and high-frequency seismic waves, we may assume that the sea layer is incompressible for tsunami. This chapter is based on incompressible fluid dynamics. The theory gives the analytical solutions for tsunami generation and propagation, by which we would be able to understand the mechanism behind these phenomena in addition to describing the motion. Section 5.1 explains the difference between ocean acoustic waves and tsunami. In Sect. 5.2, a linear potential theory is formulated for the tsunami generation process in a water with uniform depth. Analytical solutions for the sea-surface displacement, velocity, and pressure field in the seawater are derived. In Sect. 5.3, we examine the analytical solutions for tsunami generation and propagation. The mathematical equations can directly provide us with a clear perspective on the tsunami mechanism. In Sect. 5.4, we bridge the gap between the analytical solutions derived under a constant sea-depth assumption and tsunami simulations with realistic bathymetry. The theoretical background of the initial conditions in the numerical simulations is explained.

Appendices

Appendices

5.1.1 Appendix 5A: Equation (5.28)

We derive Eq. (5.28) from Eq. (5.27). In Eq. (5.27), the residue theorem is used with respect to the integration over the angular frequency. Since the poles are located on the path of integration, we introduce artificial damping parameters to shift the poles from the path and to integrate with the residue theorem. These artificial damping parameters will be taken to be zero after the integration using the residue theorem. An artificial damping parameter (ε ₁ > 0) is introduced in Eq. (5.10) as

$$ {\left.\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial t}\right|}_{z=0}+{g}_0\eta \left(x,y,t\right)=-{\left.2{\varepsilon}_1\phi \left(\mathbf{x},t\right)\right|}_{z=0}. $$

(5.5A.1)

Using another surface boundary condition (Eq. (5.9)):

$$ \frac{\partial \eta \left(x,y,t\right)}{\partial t}={\left.\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial z}\right|}_{z=0}, $$

(5.5A.2)

and (5.5A.1), we obtain

$$ {\left.\frac{\partial^2\phi \left(\mathbf{x},t\right)}{\partial {t}^2}\right|}_{z=0}+{g}_0{\left.\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial z}\right|}_{z=0}=-{\left.2{\varepsilon}_1\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial t}\right|}_{z=0}, $$

(5.5A.3)

instead of Eq. (5.11) and

$$ {\left.\left(\frac{d}{dz}-\frac{\omega^2+2i{\upepsilon}_1\omega }{g_0}\right)\widehat{\phi}\left({k}_x,{k}_y,z,\omega \right)\right|}_{z=0}=0, $$

(5.5A.4)

instead of Eq. (5.20).

Additionally, another artificial damping parameter (ε ₂ > 0) is introduced in Eq. (5.24) as

$$ \chi (t)=\frac{1}{T}\left[H(t)-H\left(t-T\right)\right]{e}^{-{\varepsilon}_2t}. $$

(5.5A.5)

Then, we obtain

$$ \widehat{\chi}\left(\omega \right)={\int}_{-\infty}^{\infty}\chi (t){e}^{i\omega t} dt=\frac{1}{i\left(\omega +i{\upepsilon}_2\right)T}\left({e}^{i\left(\omega +i{\epsilon}_2\right)T}-1\right). $$

(5.5A.6)

When Eqs. (5.5A.4) and (5.5A.6) are used, the equation corresponding to Eq. (5.27) is written as

$$ {\displaystyle \begin{array}{c}\phi \left(x,y,z,t\right)\kern0.75em \\ {}=\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}\frac{1}{k}\frac{\widehat{d}\left({k}_x,{k}_y\right)}{\cosh \left({kh}_0\right)}\\ {}\kern1.25em \frac{1}{2\pi}\frac{i}{T}\ {\int}_{-\infty}^{\infty } d\omega {e}^{- i\omega t}\frac{1-{e}^{i\left(\omega +i{\epsilon}_2\right)T}}{\omega +i{\epsilon}_2}\frac{\left({\omega}^2+i2{\upepsilon}_1\omega \right)\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\left({\omega}^2+i2{\upepsilon}_1\omega \right)-{g}_0k\kern0.15em \tanh \left({kh}_0\right)}\\ {}=\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}\frac{1}{k}\frac{\widehat{d}\left({k}_x,{k}_y\right)}{\cosh \left({kh}_0\right)}\\ {}\kern1.25em \frac{1}{2\pi}\frac{i}{T}\left[{\int}_{-\infty}^{\infty } d\omega \frac{e^{- i\omega t}}{\omega +i{\epsilon}_2}\frac{\left({\omega}^2+i2{\upepsilon}_1\omega \right)\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\left({\omega}^2+i2{\upepsilon}_1\omega \right)-{g}_0k\kern0.15em \tanh \left({kh}_0\right)}\right.\\ {}\left.\kern3.75em -\underset{-\infty }{\overset{\infty }{\int }} d\omega \frac{e^{- i\omega \left(t-T\right)}{e}^{-{\epsilon}_2T}}{\omega +i{\upepsilon}_2}\frac{\left({\omega}^2+i2{\epsilon}_1\omega \right)\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\left({\omega}^2+i2{\upepsilon}_1\omega \right)-{g}_0k\kern0.15em \tanh \left({kh}_0\right)}\right].\end{array}} $$

(5.5A.7)

We first consider the integration in the second term:

$$ {\int}_{-\infty}^{\infty } d\omega \frac{e^{- i\omega \left(t-T\right)}{e}^{-{\varepsilon}_2T}}{\omega +i{\varepsilon}_2}\frac{\left({\omega}^2+i2{\varepsilon}_1\omega \right)\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\omega^2+i2{\varepsilon}_1\omega -{g}_0k\kern0.15em \tanh \left({kh}_0\right)}. $$

(5.5A.8)

Considering that the artificial damping is very small or the artificial damping parameters are much smaller than the angular frequency ω for tsunamis, we employ the residue theorem. The poles of Eq. (5.5A.8) are located at ω = − iε ₁ ± ω ₀ and − iε ₂ in the lower half of the ω-plane (Fig. 5.A.1), where

Fig. 5.A.1

Paths of integral in the complex ω-plane for Eq. (5.5A.8). The poles (crosses) are located at ω = − iε ₁ ± ω ₀ and − iε ₂ in the lower half plane

$$ {\omega}_0\equiv \sqrt{g_0k\kern0.15em \tanh \left({kh}_0\right)}. $$

(5.5A.9)

When t − T < 0, we take the path of the integral in the upper half plane (path I in Fig. 5.A.1) including no poles. In this case, then, the integration is zero as

$$ {\int}_{-\infty}^{\infty } d\omega \frac{e^{- i\omega \left(t-T\right)}{e}^{-{\varepsilon}_2T}}{\omega +i{\varepsilon}_2}\frac{\left({\omega}^2+i2{\varepsilon}_1\omega \right)\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\omega^2+i2{\varepsilon}_1\omega -{g}_0k\kern0.15em \tanh \left({kh}_0\right)}=0\kern0.5em \mathrm{for}\ t-T<0. $$

(5.5A.10)

When t − T > 0, we take the integral path in the lower half plane including the poles (path II in Fig. 5.A.1). By setting ε ₁ → 0, ε ₂ → 0 and using the residue theorem, we calculate Eq. (5.5A.8) as

$$ {\displaystyle \begin{array}{l}{\int}_{-\infty}^{\infty } d\omega \frac{e^{- i\omega \left(t-T\right)}{e}^{-{\varepsilon}_2T}}{\omega +i{\varepsilon}_2}\frac{\left({\omega}^2+i2{\varepsilon}_1\omega \right)\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\left({\omega}^2+i2{\varepsilon}_1\omega \right)-{g}_0k\kern0.15em \tanh \left({kh}_0\right)}\\ {}\kern1em \approx \underset{-\infty }{\overset{\infty }{\int }} d\omega \frac{e^{- i\omega \left(t-T\right)}}{\omega +i{\varepsilon}_2}\frac{\omega^2\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\left[\omega -\left({\omega}_0-i{\varepsilon}_1\right)\right]\left[\omega -\left(-{\omega}_0-i{\varepsilon}_1\right)\right]}\\ {}\kern1em =-2\pi i\left[\mathrm{Res}\left(\omega =-i{\varepsilon}_2\right)+\mathrm{Res}\left(\omega ={\omega}_0-i{\varepsilon}_1\right)+\mathrm{Res}\left(\omega =-{\omega}_0-i{\varepsilon}_1\right)\right]\\ {}\kern1em =-2\pi i\left[\frac{-{g}_0k\kern0.15em \cosh \kern0.15em kz}{\omega_0^2}+\frac{\omega_0^2\kern0.15em \sinh \kern0.15em kz+{g}_0k\kern0.15em \cosh \kern0.15em kz}{\omega_0^2}\cos \left[{\omega}_0\left(t-T\right)\right]\right]\end{array}} $$

(5.5A.11)

Summarizing Eqs. (5.5A.10) and (5.5A.11), we obtain

$$ {\displaystyle \begin{array}{l}{\int}_{-\infty}^{\infty } d\omega \frac{e^{- i\omega \left(t-T\right)}}{\omega}\frac{\omega^2\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\omega^2-{g}_0k\kern0.15em \tanh \left({kh}_0\right)}\\ {}\kern1em =2\pi i\left[\frac{g_0k\kern0.15em \cosh \kern0.15em kz}{\omega_0^2}-\frac{\omega_0^2\kern0.15em \sinh \kern0.15em kz+{g}_0k\kern0.15em \cosh \kern0.15em kz}{\omega_0^2}\cos \left[{\omega}_0\left(t-T\right)\right]\right]H\left(t-T\right).\end{array}} $$

Using (5.5A.9), we have

$$ {\displaystyle \begin{array}{l}{\int}_{-\infty}^{\infty } d\omega \frac{e^{- i\omega \left(t-T\right)}}{\omega}\frac{\omega^2\sinh (kz)+{g}_0k\kern0.15em \cosh (kz)}{\omega^2-{g}_0k\kern0.15em \tanh \left({kh}_0\right)}\\ {}\kern0.75em =2\pi i\left[\frac{\cosh \kern0.15em kz}{\tanh \kern0.15em {kh}_0\ }\hbox{--} \left(\frac{\cosh \kern0.15em kz}{\tanh \kern0.15em {kh}_0}+\sinh \kern0.15em kz\right)\cos \left[{\omega}_0\left(t-T\right)\right]\right]H\left(t-T\right).\end{array}} $$

(5.5A.12)

The first term in the bracket in Eq. (5.5A.7) can also be calculated in a similar way. The velocity potential (5.5A.7) is then given by

$$ {\displaystyle \begin{array}{l}\phi \left(x,y,z,t\right)\\ {}=-\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}\frac{\widehat{d}\left({k}_x,{k}_y\right)}{k\kern0.15em \cosh \kern0.15em {kh}_0}\\ {}\kern1em \left[\frac{\cosh \kern0.15em kz}{\tanh \kern0.15em {kh}_0}\frac{H(t)-H\left(t-T\right)}{T}\right.\\ {}\kern3em \left.-\left(\frac{\cosh \kern0.15em kz}{\tanh \kern0.15em {kh}_0}+\sinh \kern0.15em kz\right)\frac{H(t)\cos \left({\omega}_0t\right)-H\left(t-T\right)\cos \left({\omega}_0\left(t-T\right)\right)}{T}\right]\\ {}=-\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}\frac{\widehat{d}\left({k}_x,{k}_y\right)}{k}\\ {}\kern1em \left[\frac{\cosh \kern0.15em kz}{\sinh \kern0.15em {kh}_0}\frac{H(t)-H\left(t-T\right)}{T}\right.\\ {}\kern3em \left.-\left(\frac{\cosh \kern0.15em kz}{\sinh \kern0.15em {kh}_0}+\frac{\sinh \kern0.15em kz}{\cosh \kern0.15em {kh}_0}\right)\frac{H(t)\cos \left({\omega}_0t\right)-H\left(t-T\right)\cos \left({\omega}_0\left(t-T\right)\right)}{T}\right]\end{array}} $$

(5.5A.13)

This equation is identical to Eq. (5.28).

5.1.2 Appendix 5B: Tsunami Propagation from an Initial Height Distribution and Zero Velocity Distribution

Setting the initial conditions as

$$ \eta \left(x,y,t=0\right)={\eta}_0\left(x,y\right), $$

(5.5B.1)

and

$$ {\mathbf{v}}_{\mathbf{H}}\left(\mathbf{x},t=0\right)=0,\mathrm{and}\ {v}_z\left(\mathbf{x},t=0\right)=0, $$

(5.5B.2)

the tsunami propagation is solved in the Cartesian coordinates (Fig. 5.B.1).

Fig. 5.B.1

Coordinates used in the formulation

The velocity potential (v = ∇ϕ) satisfies the Laplace equation (Eq. (5.3)):

$$ \Delta \phi \left(\mathbf{x},t\right)=0. $$

(5.5B.3)

The velocity potential satisfies the boundary condition at the sea surface (Eq. (5.11)):

$$ {\left.\frac{\partial^2\phi \left(\mathbf{x},t\right)}{\partial {t}^2}\right|}_{z=0}+{g}_0{\left.\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial z}\right|}_{z=0}=0, $$

(5.5B.4)

and the boundary condition at the sea bottom:

$$ {\left.\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial z}\right|}_{z=-{h}_0}=0. $$

(5.5B.5)

Note that the initial conditions (Eqs. (5.5B.1) and (5.5B.2)) and the bottom boundary condition (Eq. (5.5B.5)) are different when compared with the tsunami generation problem treated in 5.2.1.

In order to solve this problem, we introduce the Fourier transform as

$$ \widehat{\upphi}\left({k}_x,{k}_y,z,t\right)=\int {\int}_{-\infty}^{\infty } dxdy\kern0.15em {e}^{-i\left({k}_xx+{k}_yy\right)}\phi \left(x,y,z,t\right). $$

(5.5B.6)

The corresponding inverse Fourier transform is

$$ \phi \left(x,y,z,t\right)=\frac{1}{{\left(2\pi \right)}^2}\int {\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}\ \widehat{\phi}\left({k}_x,{k}_y,z,t\right). $$

(5.5B.7)

Then, the Laplace equation (Eq. (5.5B.3)) and the boundary conditions (Eqs. (5.5B.4) and (5.5B.5)) are rewritten as

$$ \frac{d^2}{dz^2}\widehat{\phi}\left({k}_x,{k}_y,z,t\right)={k}^2\widehat{\phi}\left({k}_x,{k}_y,z,t\right), $$

(5.5B.8)

where k ² = k _x ² + k _y ²,

$$ {\left.\frac{\partial^2\widehat{\phi}\left({k}_x,{k}_y,z,t\right)}{\partial {t}^2}\right|}_{z=0}+{g}_0{\left.\frac{\partial \widehat{\phi}\left({k}_x,{k}_y,z,t\right)}{\partial z}\right|}_{z=0}=0, $$

(5.5B.9)

and

$$ {\left.\frac{\partial \widehat{\phi}\left({k}_x,{k}_y,z,t\right)}{\partial z}\right|}_{z=-{h}_0}=0. $$

(5.5B.10)

The general solution of Eq. (5.5B.8) is given by

$$ \widehat{\phi}\left({k}_x,{k}_y,z,t\right)=A\left({k}_x,{k}_y,t\right)\cosh (kz)+B\left({k}_x,{k}_y,t\right)\sinh (kz). $$

(5.5B.11)

The coefficient B(k _x, k _y, t) is represented by A(k _x, k _y, t) by considering the bottom boundary condition (Eq. (5.5B.10)). Then, Eq. (5.5B.11) is calculated as

$$ \widehat{\phi}\left({k}_x,{k}_y,z,t\right)=A\left({k}_x,{k}_y,t\right)\frac{\cosh \left[k\left(z+{h}_0\right)\right]}{\cosh \left({kh}_0\right)}. $$

(5.5B.12)

Substituting (5.5B.12) into (5.5B.9), we obtain

$$ \frac{d^2A\left({k}_x,{k}_y,t\right)}{dt^2}+{\omega}_0^2\ A\left({k}_x,{k}_y,t\right)=0, $$

(5.5B.13)

where

$$ {\omega}_0^2={g}_0k\kern0.15em \tanh \left({kh}_0\right). $$

(5.5B.14)

Equation (5.5B.14) represents the well-known dispersion relation (Eq. (3.47)). The general solution of Eq. (5.5B.13) is

$$ A\left({k}_x,{k}_y,t\right)=C\left({k}_x,{k}_y\right)\cos \left({\omega}_0t\right)+D\left({k}_x,{k}_y\right)\sin \left({\omega}_0t\right). $$

(5.5B.15)

Then, the velocity potential of Eq. (5.5B.11) is given by

$$ \widehat{\phi}\left({k}_x,{k}_y,z,t\right)=\left[C\cos \left({\omega}_0t\right)+D\sin \left({\omega}_0t\right)\right]\frac{\cosh \left[k\left(z+{h}_0\right)\right]}{\cosh \left({kh}_0\right)}. $$

Considering the initial velocity condition of Eq. (5.5B.2), the coefficient of C needs to be zero:

$$ \widehat{\phi}\left({k}_x,{k}_y,z,t\right)=D\left({k}_x,{k}_y\right)\sin \left({\omega}_0t\right)\frac{\cosh \left[k\left(z+{h}_0\right)\right]}{\cosh \left({kh}_0\right)}. $$

(5.5B.16)

Then, the velocity potential is represented by the inverse Fourier transform as

$$ \phi \left(x,y,z,t\right)=\frac{1}{{\left(2\pi \right)}^2}\int {\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}D\left({k}_x,{k}_y\right)\sin \left({\omega}_0t\right)\frac{\cosh \left[k\left(z+{h}_0\right)\right]}{\cosh \left({kh}_0\right)}. $$

(5.5B.17)

The tsunami height is represented by the velocity potential as

$$ \eta \left(x,y,t\right)=-\frac{1}{g_0}{\left.\frac{\partial \phi }{\partial t}\right|}_{z=0}. $$

(5.5B.18)

By substituting (5.5B.17) into (5.5B.18), we get

$$ \eta \left(x,y,t\right)=-\frac{1}{{\left(2\pi \right)}^2}\frac{1}{g_0}\int {\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}{\omega}_0D\left({k}_x,{k}_y\right)\cos \left({\omega}_0t\right). $$

(5.5B.19)

Considering the initial tsunami height distribution (5.5B.1), we obtain

$$ {\widehat{\eta}}_0\left({k}_x,{k}_y\right)=-\frac{\omega_0D\left({k}_x,{k}_y\right)}{g_0}, $$

or

$$ D\left({k}_x,{k}_y\right)=-\frac{g_0}{\omega_0}{\widehat{\eta}}_0\left({k}_x,{k}_y\right), $$

(5.5B.20)

where ${\widehat{\eta}}_0\left({k}_x,{k}_y\right)$ is the Fourier transform of the initial tsunami height distribution η ₀(x, y). Substituting Eq. (5.5B.20), we obtain the solution of the velocity potential as

$$ {\displaystyle \begin{array}{c}\phi \left(x,y,z,t\right)=\frac{-1}{{\left(2\pi \right)}^2}\int {\int}_{-\infty}^{\infty }{dk}_x{dk}_y\ {e}^{i\left({k}_xx+{k}_yy\right)}\\ {}\frac{g_0}{\omega_0}{\widehat{\eta}}_0\left({k}_x,{k}_y\right)\sin \left({\omega}_0t\right)\frac{\cosh \left[k\left(z+{h}_0\right)\right]}{\cosh \left({kh}_0\right)}.\end{array}} $$

Using the dispersion relation (Eq. (5.5B.14)), we calculate

$$ \phi \left(x,y,z,t\right)=\frac{-1}{{\left(2\pi \right)}^2}\int {\int}_{-\infty}^{\infty }{dk}_x{dk}_y\ {e}^{i\left({k}_xx+{k}_yy\right)}\frac{\omega_0}{k}{\widehat{\eta}}_0\left({k}_x,{k}_y\right)\frac{\cosh \left[k\left(z+{h}_0\right)\right]}{\sinh \left({kh}_0\right)}\sin \left({\omega}_0t\right). $$

(5.5B.21)

Using the velocity potential of Eq. (5.5B.21), we can represent the velocity, the tsunami height, and the pressure change in the following.

$$ {\displaystyle \begin{array}{c}{\mathbf{v}}_H\left(\mathbf{x},t\right)={\nabla}_H\phi \left(\mathbf{x},t\right)=\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}{\widehat{\eta}}_0\left({k}_x,{k}_y\right)\\ {}\ \left[-i{\omega}_0\frac{{\mathbf{k}}_H}{k}\frac{\cosh \left[k\left(z+{h}_0\right)\right]}{\sinh \kern0.15em {kh}_0}\sin \left({\omega}_0t\right)\right],\end{array}} $$

(5.5B.22)

where ∇ _H = (∂/∂x)e _x + (∂/∂y)e _y, and k _H = k _xe _x + k _ye _y.

$$ {\displaystyle \begin{array}{c}{\mathrm{v}}_z\left(\mathbf{x},t\right)=\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial z}=\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}{\widehat{\eta}}_0\left({k}_x,{k}_y\right)\\ {}\left[-{\omega}_0\frac{\sinh \left[k\left(z+{h}_0\right)\right]}{\sinh \kern0.15em {kh}_0}\sin \left({\omega}_0t\right)\right],\end{array}} $$

(5.5B.23)

$$ \eta \left(\mathbf{x},t\right)=-\frac{1}{g_0}{\left.\frac{\partial \phi }{\partial t}\right|}_{z=0}=\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}{\widehat{\eta}}_0\left({k}_x,{k}_y\right)\cos \left({\omega}_0t\right), $$

(5.5B.24)

and

$$ {\displaystyle \begin{array}{c}{p}_e\left(x,y,z,t\right)=-{\rho}_0\frac{\partial \phi \left(\mathbf{x},t\right)}{\partial t}\\ {}=\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}{\rho}_0{\widehat{\eta}}_0\left({k}_x,{k}_y\right)\\ {}\frac{\omega_0^2}{k}\frac{\cosh \left(z+{h}_0\right)}{\sinh \left({kh}_0\right)}\cos \left({\omega}_0t\right).\end{array}} $$

(5.5B.25)

Using the dispersion relation (Eq. (5.5B.14)), we calculate

$$ {\displaystyle \begin{array}{c}{p}_e\left(x,y,z,t\right)=\frac{1}{{\left(2\pi \right)}^2}{\int}_{-\infty}^{\infty }{\int}_{-\infty}^{\infty }{dk}_x{dk}_y\kern0.15em {e}^{i\left({k}_xx+{k}_yy\right)}\\ {}{\rho}_0{g}_0{\widehat{\eta}}_0\left({k}_x,{k}_y\right)\frac{\cosh \left(z+{h}_0\right)}{\cosh \left({kh}_0\right)}\cos \left({\omega}_0t\right).\end{array}} $$

(5.5B.26)