Analytical solution of wave-induced pore pressure under solitary waves

Abstract

Previous studies have found that waves are similar to solitary waves when introduced in a flat shallow water area. Therefore, solitary waves are often used to analyze waves near the shore. In this study, according to the relevant assumptions of solitary waves, the derivation of the wave free surface equation of the solitary wave is discussed in detail. Then, using the boundary conditions, a new analytical expression of solitary wave pressure is obtained. The seabed is treated as a porous medium and characterized by Biot’s theory. By using the method of separation of variables, an analytical solution of wave-induced pore pressure is derived. Then, the analytical results of the present solution are compared to experimental data. Finally, the effects of wave and soil characteristics, such as relative wave depth, relative wave height, wave period, soil permeability, soil elastic modulus and Poisson’s ratio are investigated.

Appendices

Appendix A

1.1 Analytic expression of first-order approximate solitary wave theory

Substituting expression (A.1) into Eq. (1) yields.

$$\phi = \phi_{0} - \frac{{z^{2} }}{2!}\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }} + \frac{{z^{4} }}{4!}\frac{{\partial^{4} \phi_{0} }}{{\partial x^{4} }} - \frac{{z^{6} }}{6!}\frac{{\partial^{6} \phi_{0} }}{{\partial x^{6} }} + \cdot \cdot \cdot$$

(A.1)

$$\frac{\partial \phi }{{\partial z}} = - z\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }} + \frac{{z^{3} }}{3!}\frac{{\partial^{4} \phi_{0} }}{{\partial x^{4} }} - \frac{{z^{5} }}{5!}\frac{{\partial^{6} \phi_{0} }}{{\partial x^{6} }} + \cdot \cdot \cdot$$

(A.2)

Assuming that the wave height is much smaller than the water depth, z is replaced with d. Equations (A.1) and (A.2) can be expressed as,

$$\phi = \phi_{0} - \frac{{d^{2} }}{2!}\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }} + \frac{{d^{4} }}{4!}\frac{{\partial^{4} \phi_{0} }}{{\partial x^{4} }} - \frac{{d^{6} }}{6!}\frac{{\partial^{6} \phi_{0} }}{{\partial x^{6} }} + \cdot \cdot \cdot$$

(A.3)

$$\frac{\partial \phi }{{\partial z}} = - d\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }} + \frac{{d^{3} }}{3!}\frac{{\partial^{4} \phi_{0} }}{{\partial x^{4} }} - \frac{{d^{5} }}{5!}\frac{{\partial^{6} \phi_{0} }}{{\partial x^{6} }} + \cdot \cdot \cdot$$

(A.4)

Substituting the first item on the right side of equation (A.3) into Eq. (4) yields.

$$\frac{{\partial \phi_{0} }}{\partial t} + g\eta = 0$$

(A.5)

Substituting the first item on the right side of equation (A.4) into Eq. (6) yields.

$$d\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }} + \frac{\partial \eta }{{\partial t}} = 0$$

(A.6)

The second derivative for x at both ends of equation (A.5) is

$$\frac{\partial }{\partial t}\left( {\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }}} \right) + g\frac{{\partial^{2} \eta }}{{\partial x^{2} }} = 0$$

(A.7)

The first partial derivative for t at both ends of equation (A.6) is

$$d\frac{\partial }{\partial t}\left( {\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }}} \right) + \frac{{\partial^{2} \eta }}{{\partial t^{2} }} = 0$$

(A.8)

According to equation (A.7) and equation (A.8),

$$\frac{{\partial^{2} \eta }}{{\partial t^{2} }} = gd\frac{{\partial^{2} \eta }}{{\partial x^{2} }}$$

(A.9)

Similarly, the first partial derivatives for t at both ends of equation (A.5) is

$$\frac{{\partial^{2} \phi_{0} }}{{\partial t^{2} }} + g\frac{\partial \eta }{{\partial t}} = 0$$

(A.10)

According to equations (A.10) and (A.6),

$$\frac{{\partial^{2} \phi_{0} }}{{\partial t^{2} }} = gd\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }}$$

(A.11)

Using the assumption of linearization, the velocity along the cross section is constant and equal to the bottom velocity. The horizontal velocity of the water quality point under the linearization condition is $u_{0} = \frac{{\partial \phi_{0} }}{\partial x}$,then,

$$\frac{{\partial^{2} u_{0} }}{{\partial t^{2} }} = gd\frac{{\partial^{2} u_{0} }}{{\partial x^{2} }}$$

(A.12)

According to Eqs. (17) and (20), $u_{0}$ is

$$u_{0} = \sqrt{\frac{g}{d}} \eta$$

(A.13)

Assuming that the waveforms remain unchanged during wave propagation, the horizontal velocity of the water quality point u can be replaced with $u_{0}$; then $u = \frac{\partial \phi }{{\partial x}} = \frac{{\partial \phi_{0} }}{\partial x} = u_{0}$.

Boundary conditions (3) and (4) can be expressed as,

$$\frac{\partial \phi }{{\partial t}} + g\eta + \frac{1}{2}\eta^{2} \frac{g}{d} = 0$$

(A.14)

$$\left. {\frac{\partial \phi }{{\partial z}}} \right|_{z = \eta + d} = \frac{\partial \eta }{{\partial t}} + \frac{\partial \eta }{{\partial x}} \cdot \left. {\frac{\partial \phi }{{\partial x}}} \right|_{z = \eta + d} = \frac{\partial \eta }{{\partial t}} + u_{0} \frac{\partial \eta }{{\partial x}} = \frac{\partial \eta }{{\partial t}} + \sqrt{\frac{g}{d}} \eta \frac{\partial \eta }{{\partial x}}$$

(A.15)

Choosing the first two items on the right side of equation (A.3), and the first partial derivatives for t at both ends of it, we obtain,

$$\frac{\partial \phi }{{\partial t}} = \frac{{\partial \phi_{0} }}{\partial t} - \frac{{d^{2} }}{2}\frac{\partial }{\partial t}\left( {\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }}} \right)$$

(A.16)

According to equation (A.7), equation (A.16) can be expressed as

$$\frac{\partial \phi }{{\partial t}} = \frac{{\partial \phi_{0} }}{\partial t} + \frac{{gd^{2} }}{2}\frac{{\partial^{2} \eta }}{{\partial x^{2} }}$$

(A.17)

Substitute equation (A.17) into equation (A.14); then,

$$\frac{{\partial \phi_{0} }}{\partial t} + g\eta + \frac{g}{2}\left( {\frac{{\eta^{2} }}{d} + d^{2} \frac{{\partial^{2} \eta }}{{\partial x^{2} }}} \right) = 0$$

(A.18)

Substitute the first two items on the right side of equation (A.4) into equation (A.15).

Then,

$$\frac{\partial \eta }{{\partial t}} + d\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }} + \sqrt{\frac{g}{d}} \eta \frac{\partial \eta }{{\partial x}} - \frac{{d^{3} }}{6}\sqrt{\frac{g}{d}} \frac{{\partial^{3} \eta }}{{\partial x^{3} }} = 0$$

(A.19)

Then,

$$\frac{\partial \eta }{{\partial t}} + d\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }} + \sqrt{\frac{g}{d}} \frac{\partial }{\partial x}\left( {\frac{1}{2}\eta^{2} - \frac{{d^{3} }}{6}\frac{{\partial^{2} \eta }}{{\partial x^{2} }}} \right) = 0$$

(A.20)

In order to obtain the wave equation, the assumption that the waveform remains unchanged in the process of propagation is assumed. That is,

$$\left\{ \begin{gathered} \frac{\partial }{\partial t} = - c\frac{\partial }{\partial x} \hfill \\ \frac{{\partial^{2} }}{{\partial t^{2} }} = c^{2} \frac{{\partial^{2} }}{{\partial x^{2} }} \hfill \\ \end{gathered} \right.$$

(A.21)

where $c = \sqrt {g\left( {d + H} \right)}$ is the propagation speed of waves.

The second derivative for x at both ends of equation (A.18) is

$$\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\frac{{\partial \phi_{0} }}{\partial t}} \right) + g\frac{{\partial^{2} \eta }}{{\partial x^{2} }} + \frac{g}{2}\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\frac{{\eta^{2} }}{d} + d^{2} \frac{{\partial^{2} \eta }}{{\partial x^{2} }}} \right) = 0$$

(A.22)

The first partial derivative for t at both ends of equation (A.20) is

$$\frac{{\partial^{2} \eta }}{{\partial t^{2} }} + d\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\frac{{\partial \phi_{0} }}{\partial t}} \right) + \sqrt{\frac{g}{d}} \frac{{\partial^{2} }}{\partial x\partial t}\left( {\frac{1}{2}\eta^{2} - \frac{{d^{3} }}{6}\frac{{\partial^{2} \eta }}{{\partial x^{2} }}} \right) = 0$$

(A.23)

Assuming that the wave height is much smaller than the water depth, then $c \approx \sqrt {gd}$. According to equation (A.21), equation (A.2) can be expressed as,

$$\frac{{\partial^{2} \eta }}{{\partial t^{2} }} + d\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\frac{{\partial \phi_{0} }}{\partial t}} \right) - \frac{g}{2}\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\eta^{2} - \frac{{d^{3} }}{3}\frac{{\partial^{2} \eta }}{{\partial x^{2} }}} \right) = 0$$

(A.24)

Multiplying d at both ends of Eq. (30) yields.

$$d\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\frac{{\partial \phi_{0} }}{\partial t}} \right) + gd\frac{{\partial^{2} \eta }}{{\partial x^{2} }} + \frac{gd}{2}\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\frac{{\eta^{2} }}{d} + d^{2} \frac{{\partial^{2} \eta }}{{\partial x^{2} }}} \right) = 0$$

(A.25)

Expression (A.24) minus equation (A.25) results in the wave surface equation of solitary wave as,

$$\frac{{\partial^{2} \eta }}{{\partial t^{2} }} = gd\frac{{\partial^{2} \eta }}{{\partial x^{2} }} + g\frac{{\partial^{2} }}{{\partial x^{2} }}\left( {\eta^{2} + \frac{{d^{3} }}{3}\frac{{\partial^{2} \eta }}{{\partial x^{2} }}} \right)$$

(A.26)

In order to solve equation (A.26), three non-dimensional parameters need to be introduced.

$$\left\{ \begin{gathered} \xi d = x - ct \hfill \\ \overline{\eta } = \frac{\eta }{d} \hfill \\ b = \frac{{c^{2} }}{gd} - 1 \hfill \\ \end{gathered} \right.$$

(A.27)

Substituting the non-dimensional parameters into equation (A.26) yields.

$$\frac{{\partial^{2} \overline{\eta } }}{{\partial t^{2} }} = g\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }} + gd\frac{{\partial^{2} }}{{\partial \xi^{2} }}\left( {\overline{\eta }^{2} + \frac{1}{3}\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }}} \right)$$

(A.28)

Substituting the non-dimensional parameters into equation (A.21) yields.

$$\frac{{\partial^{2} }}{{\partial t^{2} }} = g\left( {1 + b} \right)\frac{{\partial^{2} }}{{\partial \xi^{2} }}$$

(A.29)

Then, equation (A.28) can be expressed as,

$$g\left( {1 + b} \right)\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }} - g\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }} - gd\frac{{\partial^{2} }}{{\partial \xi^{2} }}\left( {\overline{\eta }^{2} + \frac{1}{3}\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }}} \right) = 0$$

(A.30)

Simplifying equation (A.30), we obtain,

$$\frac{{\partial^{2} }}{{\partial \xi^{2} }}\left( {gb\overline{\eta } - gd\overline{\eta }^{2} - \frac{gd}{3}\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }}} \right) = 0$$

(A.31)

Integrating equation (A.31) twice, we obtain,

$$gb\overline{\eta } - gd\overline{\eta }^{2} - \frac{gd}{3}\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }} + A_{1} \xi + A_{2} = 0$$

(A.32)

Assume that the waves disappear in the infinite distance of the x direction, when $x = \pm \infty$,

$$\eta = \frac{{\partial^{2} \eta }}{{\partial x^{2} }} = 0$$

(A.33)

According to formula (A.33)A₁ = A₂ = 0. Then, equation (A.32) can be expressed as,

$$b\overline{\eta } - d\overline{\eta }^{2} - \frac{d}{3}\frac{{\partial^{2} \overline{\eta } }}{{\partial \xi^{2} }} = 0$$

(A.34)

At the peak of the wave $\xi = 0$, $\overline{\eta } = \frac{H}{d}$, and $\frac{{d\overline{\eta } }}{d\xi } = 0$.Then,

$$b = \frac{H}{d}$$

(A.35)

where H is the wave height.

Solving equation (A.34), the analytic expression of $\overline{\eta }$ can be obtained as

$$\overline{\eta } = bdsech^{2} \left[ {\sqrt{\frac{3bd}{2}} \xi } \right]$$

(A.36)

Substitute equation (A.27) and (A.35) into equation (A.36). The wave surface equation of isolated wave can be obtained as Eq. (8).

Substituting the first two items on the right of equation (A.1) into Eq. (2), we obtain,

$$\frac{{p_{w} }}{{\rho_{w} }} = \frac{{z^{2} }}{2}\frac{\partial }{\partial t}\left( {\frac{{\partial^{2} \phi_{0} }}{{\partial x^{2} }}} \right) - \frac{\partial \phi }{{\partial t}} - g\left( {z - d} \right) - \frac{1}{2}\left( {u^{2} + v^{2} } \right)$$

(A.37)

According to equation (A.21), equation (A.37) can be written as,

$$\frac{{p_{w} }}{{\rho_{w} }} = cu_{0} - \frac{{z^{2} }}{2}c\left( {\frac{{\partial^{2} u_{0} }}{{\partial x^{2} }}} \right) - g\left( {z - d} \right) - \frac{1}{2}\left( {u^{2} + v^{2} } \right)$$

(A.38)

Assuming that the wave height is much smaller than the water depth, according to equation (A.13).

$$cu_{0} = g\eta$$

(A.39)

Substitute equation (A.39) into equation (A.38),then,

$$\frac{{p_{w} }}{{\rho_{w} }} = g\eta - \frac{{z^{2} }}{2}c\left( {\frac{{\partial^{2} u_{0} }}{{\partial x^{2} }}} \right) - g\left( {z - d} \right) - \frac{1}{2}\left( {u^{2} + v^{2} } \right)$$

(A.40)

Assume that the velocity potential is slow, and the amplitude of fluctuation is very small. Compared to other items, the terms $\frac{{z^{2} }}{2}c\left( {\frac{{\partial^{2} u_{0} }}{{\partial x^{2} }}} \right)$ and $\frac{1}{2}\left( {u^{2} + v^{2} } \right)$ can be ignored. Then,

$$\frac{{p_{w} }}{{\rho_{w} g}} = \left( {\eta + d} \right) - z$$

(A.41)

Substitute Eq. (8) into equation (A.41), the analytical expression for the first-order approximate solitary wave pressure is obtained as Eq. (11).

Appendix B

2.1 Analytic expression of wave-induced pore pressure under solitary waves

Substituting Eq. (24) into Eq. (18), we obtain,

$$C_{s} \frac{{T^{\prime}(t)}}{T(t)} = \frac{{X^{\prime\prime}(x)}}{X(x)} + \frac{{Z^{\prime\prime}(z)}}{Z(z)}$$

(B.1)

Because the left side of equation (B.1) depends only on T and the right side depends on x and z, the two sides of equation (B.1) must be constant, which is $- \lambda$. Then,

$$C_{s} \frac{{T^{\prime}(t)}}{T(t)} = \frac{{X^{\prime\prime}(x)}}{X(x)} + \frac{{Z^{\prime\prime}(z)}}{Z(z)} = - \lambda$$

(B.2)

where $\lambda$ is a constant.

According to equation (B.2),

$$\frac{{Z^{\prime\prime}(z)}}{Z(z)} + \lambda = - \frac{{X^{\prime\prime}(x)}}{X(x)}$$

(B.3)

Because the right of equation (B.3) depends only on x and the left side depends on z, the two sides of equation (B.3) must be constant, which is $- \beta$. Then,

$$\frac{{Z^{\prime\prime}(z)}}{Z(z)} + \lambda = - \frac{{X^{\prime\prime}(x)}}{X(x)} = \beta$$

(B.4)

where $\beta$ is a constant.

Three ordinary differential equations are obtained:

$$X^{\prime\prime}(x) + \beta X(x) = 0$$

(B.5)

$$Z^{\prime\prime}(z) + \left( {\lambda - \beta } \right)Z(z) = 0$$

(B.6)

$$C_{s} T^{\prime}(t) + \lambda T(t) = 0$$

(B.7)

According to Eqs. (22) and (23),

$$X^{\prime}(0) = X^{\prime}(l) = 0$$

(B.8)

According to equation (B.8), the ordinary differential equation (B.5) is solved as Eq. (25).

According to expression (21),

$$Z^{\prime}( - h) = 0$$

(B.9)

When $\lambda = \beta$,

$$\lambda = \left( {\frac{n\pi }{l}} \right)^{2} ,Z(z) = C_{2}$$

(B.10)

This does not conform to objective reality.

When $\lambda < \beta$,

$$Z(z) = 0$$

(B.11)

The above problem only has a zero solution, and it does not conform to the conditions.

When $\lambda > \beta$, the expression of Z(z) is Eq. (26) or Eq. (27).

To solve equation (B.7). The expression of T(t) is Eq. (28) or Eq. (29).