Unsteady waves created by a disturbance on the surface of a running stream

Schedule 5 Nov 2024

IELTS writing -1
IELTS speaking -1
IELTS wordlist -3
water wave chapter 7 and part of chapter 8

Addressing the Disturbance Potential through the Application of Fourier Transforms and Complex Analysis

boundary conditions

free surface, the disturbance potential $\varphi(x, y ; t)$

\[ \begin{gathered} \frac{p}{\rho}+g \eta+\varphi_t+U \varphi_x+\frac{U^2}{2}=0 \quad \text{(7.4.1)}\\ \eta_t+U \eta_x-\varphi_y=0\quad \text{(7.4.2)} \end{gathered} \]

where $y=0, t>0$

bottom $y=-h$

\[ \varphi_y=0 \quad t \geq 0 \quad \text{(7.4.3)} \]

initial instant $t=0$

\[ \begin{aligned} & \varphi(x, y ; 0)=\eta(x ; 0)=p(x ; 0)=0 \quad \text{(7.4.4)}\\ & \varphi_t(x, y ; 0)=0\quad \text{(7.4.5)} \end{aligned} \]

set surface pressure $p$ for $t>0$

\[ p=p(x), t>0 \quad \text{(7.4.6)} \]

at $\infty$

boundedness assumptions

At $\infty$ we make no assumptions other than boundedness assumptions. We shall not formulate these boundedness conditions explicitly: instead, they are used implicitly in what follows because of the fact that Fourier transforms in $x$ for $-\infty<x<\infty$ are applied to $\varphi$ and $p$ and their derivatives. Of course, this means that these quantities must not only be bounded but also must tend to zero at $\infty$, and this seems reasonable since the initial conditions leave the water undisturbed at $\infty$.

notes

Continuous functions that are Fourier transformable possess the following properties. 1. $\int_{-\infty}^{\infty}|f(t)| d t<\infty$ 2. Behavior at Infinity. if $f(t)$ does not vanish at infinity, there exists a positive constants $\epsilon>0$ and an infinite interval $(\beta, \infty)$ where $|f(t)|>t$, Consequently,

\[\int_\beta^{\infty}|f(t)| d t \rightarrow \infty\]

which can result in a divergent integral.

Overall
A continuous function that is Fourier transformable is guaranteed to

\[f(t)\left\{\begin{array}{l}\text { boundedness } \\\sim O(0) \text { as } t \rightarrow \infty\end{array}\right.\]

find a solution of $\varphi$

Using Equations $(7.4.1)$ to $(7.4.6)$ together, we eliminate $\eta$ \[ \quad \varphi_{t t}+U^2 \varphi_{x x}+2 U \varphi_{x t}+g \varphi_y=-\frac{U}{\rho} p_x \quad\text{at } y=0 \tag{7.4.7} \] important: On surface $(y=0$ or $y=\eta)$

2-D \[ u_y=\eta_t+u_x \eta_x \quad \eta=\eta(x) \]
3-D \[ u_z=\eta_t+u_\perp \cdot \nabla_{\perp} \eta \quad \eta=\eta(x, y) \]

notes

take derivative $(7.4 .1)$ with respect to $x$ and $t$ respectively. we arrive at

\[\begin{array}{ll}\frac{p_x}{\rho}+g \eta_x+\varphi_{t x}+U\varphi_{x x}=0\quad \text{(N.7.4.1)}\\g \eta_t+\varphi_{t t}+U\varphi_{x t}=0 \quad \text{(N.7.4.2)}\end{array}\]

where $p_t=0$, substituting (N.7.4.1) and (N.7.4.2) into (7.4.2), we derive

\[\begin{aligned}& -\frac{\varphi_{t t}+U \varphi_{x t}}{g}+U \frac{\varphi_{t x}+U \varphi_{x x}+\frac{p_x}{\rho}}{-g}-\varphi_y=0 \\& -\left(\varphi_{t t}+U \varphi_{x t}\right)-U \varphi_{t x}-U^2 \varphi_{x x}-\frac{p_x}{\rho} U-\varphi_y=0 \\& \Rightarrow \varphi_{t t}+2 U \varphi_{x t}+U^2 \varphi_{x x}+\varphi_y=-\frac{U}{\rho} p_x\end{aligned}\]

Applying the Fourier transform to $\nabla^2 \varphi=0$ with respect to $x$ to yield

\[ \bar{\varphi}_{y y}-s^2 \bar{\varphi}=0 \quad\text{(7.4.8)} \]

where $\bar{\varphi}=F[\varphi] \quad \bar{\varphi}_{y y}=F\left[\varphi_{y y}\right]$, In the same way we have

\[ \bar{\varphi}_y=0, \text { at } y=-h\quad\text{(7.4.8-1)} \]

Hence, $\bar{\varphi}$ must be of the form

\[ \bar{\varphi}(s, y ; t)=A(s ; t) \cosh s(y+h)\quad\text{(7.4.9)} \] for equation (7.4.8), we posit a solution of the form

\[ \bar{\varphi}=A(s ; t) \cosh s y \]

Referring to equation (7.4.8-1), the solution may be written as

\[ \bar{\varphi}=A(s ; t) \cosh s(y+h) \] perform a Fourier tranform on (7.4.7) yields (in terms of $x$)

\[ \bar{\varphi}_{t t}+2 i s U \bar{\varphi}_t+g \bar{\varphi}_y-U^2 s^2 \bar{\varphi}=-\frac{i s U}{\rho} \bar{p} \quad \text { at } y=0 \]

notes

The general solution to a non-homogeneous ordinary differential equation (ODE) is obtained by adding the general solution of the corresponding homogeneous ODE to a particular solution of the non-homogeneous ODE. Specifically, consider a non-homogeneous linear ODE of the form: \[L[y]=g(x)\]

where $L$ is a linear differential operator, $y$ is the unknown function, and $g(x)$ is the non-homogeneous term. The general solution $y$ to this equation can be expressed as:

\[y=y_h+y_p\]

where:

$y_h$ is the general solution to the corresponding homogeneous ODE $L[y]=0$, which encompasses all possible solutions to the homogeneous equation.
$y_p$ is a particular solution to the non-homogeneous ODE $L[y]=g(x)$, satisfying the non-homogeneous equation.

Thus, the general solution to a non-homogeneous ODE is the sum of the general solution to the homogeneous ODE and a particular solution to the non-homogeneous ODE. This method is commonly employed in solving non-homogeneous linear ODEs and is considered a standard approach.

Substituting $(7,4,9)$ into it results in \[ A_{t t}+2 i s U A_t+\left[g s \tanh s h-s^2 U^2\right] A=-\frac{i s U \bar{p}}{\varrho \cosh s h}\text { (7.4.11) } \] Here $\bar{p}(s)$ is the transform of $p(x)$, the conditions: \[ A(s ; 0)=A_t(s ; 0)=0 \] put it all together and yield \[ A(s ; t)=\frac{i s U \bar{p}}{\varrho \cosh s h} \cdot\left\{\begin{array}{l} \frac{1}{s^2 U^2-g s \tanh s h} \\ +\frac{1}{2 \sqrt{g s \tanh s h}} \cdot \frac{e^{-i t(s U+\sqrt{g s \tanh s h})}}{s U+\sqrt{g s \tanh s h}} \\ -\frac{1}{2 \sqrt{g s \tanh s h}} \cdot \frac{e^{-i t(s U-\sqrt{g s \tanh s h})}}{s U-\sqrt{g s \tanh s h}} \end{array}\right\} \]

notes

A solution for the homogeneous form of (7.4.11) is presented:

\[\begin{aligned}& \Rightarrow \quad \lambda^2+2 i s U \lambda+g s \tanh s h-s^2 U^2=0 \\& \Rightarrow \quad \lambda=\frac{-2 i s U \pm \sqrt{-4 s^2 U^2-4\left(g s \tanh s h-s^2 U^2\right)}}{2 \times 1} \\& =-i s U \pm \sqrt{g s \tanh s h} \\& =-i(s U \pm \sqrt{gs\tanh sh }) \\& A^h=c_1 e^{-i t(s U+\sqrt{g s \tanh s h})}+c_2 e^{-i t(s U-\sqrt{g s \tanh s h})}\end{aligned}\]

particular solution

Sine the RHS of ODE is a constant with respect to $t$, we can assume a constant particular solution $A^p$, substitute $A^p$ into (7.4.11)

\[\begin{aligned}& \left(g s \tanh sh-s^2 U^2\right) A^p=-\frac{i s U\bar{p}}{\rho \cosh sh} \\&A^p= -\frac{i s U \bar{p}}{\rho \cosh sh\left(gs\tanh{sh}-s^2 U^2\right)}\end{aligned}\]

Genevan solution

\[A=A^h+A^p\]

Applying initial condition

\[\begin{aligned}& A(s ; 0)=0 \\\Rightarrow & c_1+c_2+A^p=0 \\& A_t(s ; 0)=0 \\\Rightarrow & -i(s U+\sqrt{g s \tanh s h}) c_1-i(s U-\sqrt{gs \tanh {sh}}) c_2=0\end{aligned}\]

solving for $c_1$ and $c_2$

\[\begin{aligned}& c_1=-\frac{i \bar{p} U s}{\rho \cosh s h}\left(\frac{1}{2 \sqrt{g s \tanh s h}} \frac{(-s U+\sqrt{g s \tanh s h})}{-s^2 U^2+g s \tanh s h}\right) \\& c_2=-\frac{i \bar{p} U \quad(s U+\sqrt{g s \tanh s h})}{2 \rho \cosh s h} \sqrt{g s \tanh s h\left(s U^2-g \tanh s h\right)}\end{aligned}\]

final solution \[ \begin{aligned} A&= -\frac{i \bar{p} U s}{\rho \cosh s h}\left(\frac{-s U+\sqrt{g s \tanh s h}}{2 \sqrt{g s \tanh s h}\left(-s^2 U^2+g s \tanh s h\right)}\right) e^{-i t(s U+\sqrt{g s \tanh h})} \\ &-\frac{i \bar{p} U s}{\rho \cosh s h}\left(\frac{s U+\sqrt{g s \tanh s h}}{2 \sqrt{g s \tanh s h}\left(s^2 U^2-g \tanh s h\right)}\right) e^{-i t(s U-\sqrt{g s \tanh s h})} \\ &+\frac{i s U \bar{p}}{\rho \cosh {sh}\left(gs \tanh{sh}-s^2 U^2\right)}\\ &\begin{aligned} = & -\frac{i \bar{p} U s}{\rho\cosh{sh}}\left[\frac{1}{s^2 U^2-g s \tanh s h}+\frac{1}{2\sqrt{g s \tanh s h}} \frac{e^{-i t(s U+\sqrt{g s \tanh s h})}}{s U+g s \tanh s h}\right. \\ & \left.-\frac{1}{2 \sqrt{g s \tanh s h}} \frac{e^{-i t(s U-\sqrt{g s \tanh s h})}}{s U-g s \tanh s h}\right] \end{aligned} \end{aligned} \]