Surface waves generated by a travelling pressure point

The case in which the water has infinite depth is considered, the effect of surface tension as well.

Problem Prescription

Disturbance: pressure point, and finite depth: \(h\)

velocity potential \[ \begin{equation} \Phi=U x+\phi \end{equation} \] Laplace Equation \[ \begin{equation} \phi_{x x}+\phi_{y y}+\phi_{z z}=0, \end{equation} \] boundary condition \[ \begin{equation} \begin{aligned} &\phi_z=0 \quad \text { on } \quad z=-h\\ &P \delta(x) \delta(y)+\frac{1}{2} \rho\left(U^2+2 U \phi_x\right)+\rho g \eta-T\left(\eta_{x x}+\eta_{y y}\right)=\text { constant }=\frac{1}{2} \rho U^2 \quad \text { on } \quad z=0 \end{aligned} \end{equation} \] Kinematic condition \[ \begin{equation} \frac{\mathrm{d} \eta}{\mathrm{~d} t}=U \eta_x=\phi_z \end{equation} \] Using Fourier Transform, we have \[ \begin{equation} \eta(x, y, z)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathrm{e}^{\mathrm{i}(\alpha x+\beta y)} \frac{F(\alpha, \beta)}{G(\alpha, \beta)} \mathrm{d} \alpha \mathrm{~d} \beta, \end{equation} \] where \[ \begin{equation} \begin{gathered} F(\alpha, \beta)=\frac{P k \sinh \{k(h+z)\}}{4 \pi^2 \rho \cosh (k h)} ; \\ G(\alpha, \beta)=\alpha^2-k\left(1+\sigma k^2\right) \tanh (k h) ; \\ k=\sqrt{\alpha^2+\beta^2} \end{gathered} \end{equation} \] It is evident form inspection that \(\alpha =k_x,\;\beta=k_y\), and \(\sigma=T/\rho\).

Implementation steps

Rotating \(\alpha\) and \(\beta\) axes through an angle \(\theta\), where \(x=r\cos \theta,\;y=r\sin \theta\), for the particular direction \(\mathbf{r}=(x,y)\) along which we evaluate \(\eta\) \[ \begin{equation} \eta=\int_{-\infty}^{\infty} \mathrm{d} \bar{\beta} \int_{-\infty}^{\infty} \mathrm{e}^{\mathrm{i} \bar{\alpha} r} \frac{F(\bar{\alpha}, \bar{\beta})}{G(\bar{\alpha}, \bar{\beta})} \mathrm{d} \bar{\alpha}, \end{equation} \] where \(\alpha=\bar{\alpha} \cos \theta-\bar{\beta} \sin \theta ; \quad \beta=\bar{\alpha} \sin \theta+\bar{\beta} \cos \theta\).

--- note -------------------------------------------------

? why is there only \(\bar{\alpha}\)? how do we vanish \(\bar\beta\) in exponential term \(\mathrm{e}^{\mathrm{i} \bar{\alpha} r}\)

✏️A: \[ \alpha x+\beta y=(\bar{\alpha} \cos \theta-\bar{\beta} \sin \theta) r \cos \theta+(\bar{\alpha} \sin \theta+\bar{\beta} \cos \theta) r \sin \theta . \] collect the \(\bar{\alpha}\) and \(\bar{\beta}\) terms \[ \begin{aligned} \alpha x+\beta y & =\bar{\alpha} r\left(\cos ^2 \theta+\sin ^2 \theta\right)+\bar{\beta} r(-\sin \theta \cos \theta+\cos \theta \sin \theta) \\ & =\bar{\alpha} r+\bar{\beta} r(0) \\ & =\bar{\alpha} r . \end{aligned} \] Also, regarding the differential term, it follows that \[ \begin{aligned} & d \alpha=\cos \theta d \bar{\alpha}-\sin \theta d \bar{\beta} \\ & d \beta=\sin \theta d \bar{\alpha}+\cos \theta d \bar{\beta} \end{aligned} \] subsequently \[ \begin{aligned} d \alpha \wedge d \beta & =(\cos \theta d \bar{\alpha}-\sin \theta d \bar{\beta}) \wedge(\sin \theta d \bar{\alpha}+\cos \theta d \bar{\beta}) \\ & =-\sin ^2 \theta d \bar{\beta} \wedge d \bar{\alpha}+\cos ^2 \theta d \bar{\alpha} \wedge d \bar{\beta} \\ & =d \bar{\alpha} \wedge d \bar{\beta} \end{aligned} \]

The integrals depends on contributions from the singularities. Here it is \[ \alpha^2=k \tanh (k h) \] when surface tension is neglected, and \(h\) is infinite, \[ \begin{aligned} & \alpha^2=\sqrt{\alpha^2+\beta^2} \\ & \alpha^4-\alpha^2-\beta^2=0 \\ & \alpha^2=\frac{1 \pm \sqrt{1+4 \beta^2}}{2} \\ & \alpha^2 \geq 0 \Rightarrow \\ & \alpha^2=\frac{1+\sqrt{1+4 \beta^2}}{2} \\ & \alpha= \pm \frac{1}{\sqrt{2}}\left[1+\left(1+4 \beta^2\right)^{\frac{1}{2}}\right]^{\frac{1}{2}} \end{aligned} \]

Because of using Cauchy principal value of the integral, it leads to gravity waves appearing ahead of the pressure point, and some means must be found to give the correct physical solution. Therefore to solve this, we apply a radiation condition which allow the pressure increase exponentially with time like \(e^{\epsilon t}(\epsilon>0)\). Then \(G\) is replaced by \[ \begin{equation} G-\mathrm{i} 2 \alpha \epsilon-\epsilon^2 \end{equation} \] the poles are now at complex values of \(\bar{\alpha}\). We next shift the path of integration to the line \(\Im \bar{\alpha}=H>0\). This introduces a factor \(\mathrm{e}^{-H r}\)

--- note ---------------------------------------------

By shifting the integration contour to the line where \(\operatorname{Im} \bar{\alpha}=H\) \[ \bar{\alpha}=x+i H, \quad x \in(-\infty, \infty), \]

the differential becomes \(d \bar{\alpha}=d x\). Substituting this transformation into the integral yields:

\[ \int_{-\infty}^{\infty} e^{i \bar{\alpha} r} \frac{F(\bar{\alpha}, \bar{\beta})}{G(\bar{\alpha}, \bar{\beta})} d \bar{\alpha}=\int_{-\infty}^{\infty} e^{i(x+i H) r} \frac{F(x+i H, \bar{\beta})}{G(x+i H, \bar{\beta})} d x \]

The exponential term can be decomposed as: \[ e^{i(x+i H) r}=e^{i x r} e^{-H r} \]

This factorization reveals that \(e^{Hr}\) emerges as a constant multiplier independent of \(x\). Thus, the translated contour integral reduces to: \[ e^{-H r} \int_{-\infty}^{\infty} e^{i x r} \frac{F(x+i H, \bar{\beta})}{G(x+i H, \bar{\beta})} d x \]

This derivation clarifies the origin of the \(e^{Hr}\) term.

into the integral, making it negligible, but we must add on \(2 \pi i\) times the sum of the residues at poles with \(0<\Im \bar{\alpha}<H\).

Singularities

Three canonical categories of singularities: Removable, Pole, and Branch Points.

A residue exists if and only if a function’s Laurent expansion about \(z_0\) contains a finite \((z−z_0)^{−1}\)term. This definition fails when:

• Removable singularities (all \(\left(z-z_0\right)^{-n}\) coefficients vanish),
• Essential singularities (non-terminating negative-power terms),
• No single-valued Laurent expansion exists (branch points).

Residue theorem:

Suppose \(f\) is holomorphic within the region enclosed by a simple closed curve \(C\), with the exception of isolated singularities at points \(\left\{a_k\right\}\). Under these conditions:

\[ \oint_C f(z) d z=2 \pi i \sum_k \operatorname{Res}\left(f, a_k\right) \]

explanation on singularities in complex analysis

Isolated Singularity

Definition: A point \(z_0\) is called an isolated singularity of \(f(z)\) if \(f(z)\) is analytic in a punctured neighborhood \(0 < |z - z_0| < R\) but not at \(z_0\).

Subclasses:

• Removable Singularity
  
• Pole
  
• Essential Singularity

Discriminant: Classified by the structure of negative-power terms in the Laurent series expansion:

• Removable Singularity: No negative-power terms (all coefficients \(a_{-n} = 0\)).
  
• Pole: Finite negative-power terms (highest negative power: \((z - z_0)^{-n}\), \(n \in \mathbb{N}^+\)).
  
• Essential Singularity: Infinitely many negative-power terms.

Removable Singularity

Definition: A singularity \(z_0\) is removable if there exists an analytic function \(g(z)\) such that \(f(z) = g(z)\) for all \(z \neq z_0\) in a neighborhood of \(z_0\).

Key Properties:

• \(\displaystyle \lim_{z \to z_0} f(z)\) exists and is finite.
  
• The singularity is eliminated by redefining \(f(z_0) = \lim_{z \to z_0} f(z)\). Example: \(f(z) = \frac{\sin z}{z}\) has a removable singularity at \(z = 0\).

Pole

Definition: A singularity \(z_0\) is a pole of order \(m\) if there exists \(m \in \mathbb{N}^+\) such that \((z - z_0)^m f(z)\) is analytic and non-zero at \(z_0\). Key Properties:
- \(\displaystyle \lim_{z \to z_0} |f(z)| = +\infty\).
- The Laurent expansion terminates at \((z - z_0)^{-m}\). Example: \(f(z) = \frac{1}{(z - 1)^3}\) has a third-order pole at \(z = 1\).

Essential Singularity

Definition: A singularity \(z_0\) is essential if it is isolated but neither removable nor a pole.

Key Properties (Weierstrass-Casorati Theorem):

• The image of any punctured neighborhood of \(z_0\) is dense in \(\mathbb{C} \cup \{\infty\}\).
  
• The Laurent series contains infinitely many negative-power terms. Example: \(f(z) = e^{1/z}\) has an essential singularity at \(z = 0\).

Branch Singularity

Definition: A point \(z_0\) is a branch singularity (or branch point) if analytic continuation of \(f(z)\) around a closed path encircling \(z_0\) results in a multivalued function. Key Properties:

• Non-isolated: Multivaluedness persists in any neighborhood of \(z_0\).
  
• Requires branch cut(s) to define single-valued branches.
  Example: \(f(z) = \sqrt{z}\) has a branch point at \(z = 0\).

Distinctions and Connections

Core Differences

Connections

• Isolated Singularity Family: Removable singularities, poles, and essential singularities are subclasses of isolated singularities, distinguished by Laurent series structures.
  
• Uniqueness of Branch Singularities:
  • Non-isolated nature precludes Laurent series analysis; requires Riemann surfaces for multivaluedness.
    
  • Topological distinction: Branch singularities alter the global structure of the complex plane, while isolated singularities are local phenomena.

Comparative Examples

1. Removable vs. Pole:
   • \(f(z) = \frac{\sin z}{z}\) (removable): \(\lim_{z \to 0} f(z) = 1\).
     
   • \(f(z) = \frac{1}{z^2}\) (pole): \(\lim_{z \to 0} |f(z)| = +\infty\).
2. Pole vs. Essential Singularity:
   • \(f(z) = \frac{1}{z^3}\) (pole): Laurent series contains only \(z^{-3}\).
     
   • \(f(z) = e^{1/z}\) (essential): Laurent series \(\sum_{n=0}^\infty \frac{z^{-n}}{n!}\).
3. Essential vs. Branch Singularity:
   • \(e^{1/z}\) (essential): Locally dense but single-valued.
     
   • \(\sqrt{z}\) (branch): Monodromy \(\sqrt{z} \to -\sqrt{z}\) upon encircling \(z = 0\).

Applications

1. Residue Theorem: Applies only to isolated singularities (poles and essential singularities); poles yield straightforward residue calculations.
   
2. Integral Evaluation:
   • Branch singularities require contour deformation around branch cuts (e.g., keyhole contours).
     
   • Essential singularities may involve infinite series for residue computation.
     
3. Geometric Function Theory:
   • Branch singularities correspond to Riemann surface branching.
     
   • Essential singularities reflect chaotic behavior in complex dynamics.

Summary

• Isolated Singularities (removable, poles, essential) are classified by local Laurent series properties.
  
• Branch Singularities are global, multivalued phenomena requiring topological tools (branch cuts, Riemann surfaces).
  
• Essential singularities represent the "most pathological" isolated singularities, while branch singularities exist outside the isolated singularity framework.

--- note --------------------------------

Q: why is \(\sqrt{z}\) single-valued function?

A:

1. Commencing with the definition of the complex square root
   • For a complex number \(z = re^{i\theta}\) (expressed in exponential form, where \(r\) is the modulus and \(\theta\) the argument), its square root is defined as \(\sqrt{z} = \sqrt{r}e^{i\theta/2}\).
   • The argument \(\theta\) of a complex number exhibits periodicity, i.e., \(\theta\) and \(\theta + 2k\pi\) (\(k \in \mathbb{Z}\)) represent equivalent angular orientations. For instance, when \(k = 1\), \(\theta + 2\pi\) corresponds to the same direction as \(\theta\).
2. Analysis of the multivalued nature of the square root function
   • Substituting into the square root expression yields \(\sqrt{z}=\sqrt{r} e^{i(\theta+2 k \pi) / 2}=\sqrt{r} e^{i \theta / 2} e^{i k \pi}\).
   • For distinct integer values of \(k, e^{i k \pi}\) assumes two distinct cases:
     • When \(k\) is even (e.g., \(k=0,2,4, \ldots\) ), \(e^{i k \pi}=1\), resulting in \(\sqrt{z}=\sqrt{r} e^{i \theta / 2}\).
     • When \(k\) is odd (e.g., \(k=1,3,5, \ldots\) ), \(e^{i k \pi}=-1\), yielding \(\sqrt{z}=-\sqrt{r} e^{i \theta / 2}\).
   • Thus, for any nonzero \(z\), the square root function bifurcates into two distinct values, demonstrating its multivalued character.
3. Multivaluedness through the lens of analytic continuation
   • Consider a circular path encircling the origin, starting from the positive real axis. As \(z\) traverses this path with \(\theta\) increasing from \(0\) to \(2\pi\), the argument of the square root evolves from \(0\) to \(\pi\).
   • Upon completing a full rotation (\(\theta \rightarrow \theta + 2\pi\)), the square root function does not revert to its initial value but instead assumes the negative of its original value. This monodromy phenomenon underscores the multivaluedness inherent to the square root function.
4. Branch cuts as a resolution to multivaluedness
   • To enforce single-valuedness, a branch cut is introduced. For the square root function, a conventional choice is the ray extending from the origin along the negative real axis to infinity.
   • This cut partitions the complex plane into two regions. By restricting the argument to a principal branch (e.g., \(-\pi < \theta \leq \pi\)), the square root becomes single-valued within each region. Crossing the branch cut induces a discontinuous transition between branches, further illustrating the multivalued nature.

surface displacement without surface tension

Evaluating the inner integral, we get \[ \begin{equation} \eta=\int_{-\infty}^{\infty}\left\{2 \pi \mathrm{i} \Sigma \mathrm{e}^{\mathrm{i} \bar{\alpha} r} \frac{F}{\partial G / \partial \bar{\alpha}}+o\left(r^{-N}\right)\right\} \mathrm{d} \bar{\beta}, \end{equation} \]

The inner integral in approached by contour integration along the lines (this refers to the sectorial integration domain encompassing singularities), but the only other contribution is from a branch point at \(\bar{\alpha} =i\abs{\beta}\), and this term decays like \(e^{-\abs{\beta}r}\), which can be neglected.

--- note --------------------------------------------------

Q: branch point \(\bar{\alpha}=i\abs{\beta}\)

A: The square root introduces multivaluedness in \(k\), rendering it a two-valued function. The condition \(\bar{\alpha}^2+\) \(\bar{\beta}^2=0\), which yields \[ \bar{\alpha}= \pm i|\bar{\beta}| \]

identifies the branch points of the equation \(k=0\). Conventionally, these branch points are joined by a branch cut along the imaginary axis, inducing a sign change in \(k\) when crossing the cut.

we put \(s=\bar{\alpha}-i|\bar{\beta}|\), we may rewrite \(s=s e^{ \pm i \pi}\), and, \(s=0\) is the branch points, then \[ \begin{equation} \begin{aligned} \bar{\alpha}^2+\bar{\beta}^2&=(i \abs{\beta})^2+2 i b s e^{ \pm i \pi}+O\left(s^2\right)+\abs{\beta}^2\\ &=2 i b s+O\left(s^2\right) \end{aligned} \end{equation} \] Neglecting higher-order infinitesimal terms, it follows that \[ \begin{equation} k \equiv \sqrt{\bar{\alpha}^2+\bar{\beta}^2} \simeq \sqrt{2 b} s^{1 / 2} e^{\pm i \pi / 4} \end{equation} \] Increasing \(\arg s\) by \(2\pi\) flips the sign of \(e^{\pm i \pi / 4}\) (e.g. from \(-\pi\) to \(\pi\)), accounting for the sign difference across the branch point.

Next, we consider \(e^{i\bar{\alpha}r}\), traversing the branch cut, yields: \[ \begin{equation} e^{i \bar{\alpha} r}=e^{i\left(i \abs{\beta}+s e^{ \pm i \pi}\right) r}=e^{-\abs{\beta} r} e^{-s r} e^{ \pm 0^{+} r} \end{equation} \]

• shared prefactor \(e^{-\abs{\beta}r}\) which is the dominant exponential decay
• the phase difference between upper/lower sides is infinitesimal

The main contribution for large \(r\) comes from the points according to stationary phase method where \[ \begin{equation} \mathrm{d} \bar{\alpha} / \mathrm{d} \bar{\beta}=0 \quad(\mathrm{~d} \alpha / \mathrm{d} \beta=-\tan \theta=-y / x) \end{equation} \] At these points normal to the line \(G=0\) parallel to the \(\bar\alpha\) axis.

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• Q:❓ At these points normal to the line \(G=0\) parallel to the \(\bar\alpha\) axis.

A: Recall that the dispersion curve \[ \mathcal{C}: G(\bar{\alpha}, \bar{\beta})=0 \]

is a line in the two-dimensional \((\bar{\alpha}, \bar{\beta})\)-plane. Along that line we can regard \(\bar{\alpha}\) as a function of \(\bar{\beta}\) :

\[ (\bar{\alpha}, \bar{\beta})=(\bar{\alpha}(\bar{\beta}), \bar{\beta}), \quad \text { with } \quad \frac{d \bar{\alpha}}{d \bar{\beta}}=-\frac{G_{\bar{\beta}}}{G_{\bar{\alpha}}} . \] then normal to the line \(G=0\), then we consider tangent vector to the curve \[ \mathbf{t}=\left(\frac{d \bar{\alpha}}{d \bar{\beta}}, 1\right) \]

When \(d \bar{\alpha} / d \bar{\beta}=0\) this becomes \(\mathbf{t}=(0,1)\), i.e. it points straight up the \(\bar{\beta}\)-axis. A line whose tangent is vertical has its normal horizontal. In other words \[ \mathbf{n} \| \bar{\alpha} \text {-axis } \quad(\mathbf{n} \cdot \mathbf{t}=0) \]

• tangent vector

For a plane curve given implicitly by \(F(x,y)=0\), pick any regular point \((x_0,y_0)\)

where the gradient \(\nabla F(x_0,y_0)=(F_x,F_y)\neq(0,0)\). A tangent vector \(\mathbf{t}\) at that point must be orthogonal to the gradient, because the gradient points in the normal direction:

\[ \nabla F\cdot\mathbf{t}=0 \;\Longrightarrow\; F_x\,t_x+F_y\,t_y=0. \]

A convenient choice (unique up to a non‑zero scalar factor) is

\[ \mathbf{t}=(-F_y,\;F_x)\qquad\text{evaluated at the point }(x_0,y_0). \]

Equivalently, if \(F_y\neq 0\) we can write the slope form

\[ \frac{dy}{dx}=-\frac{F_x}{F_y}, \quad\text{so}\quad \mathbf{t}=(1,\,-F_x/F_y), \]

and if \(F_x\neq 0\) swap the roles. Any non‑zero scalar multiple of these vectors points along the tangent line to the curve at the chosen point.

Then rewrite \(\bar \alpha\) as Taylor series \[ \begin{equation} \bar{\alpha}=\bar{\alpha}_m+\frac{1}{2} \kappa\left(\bar{\beta}-\bar{\beta}_m\right)^2, \end{equation} \] where \(\kappa=\left(\mathrm{d}^2 \bar{\alpha} / \mathrm{d} \bar{\beta}^2\right)_m\) is the curvature of the line \(G=0\) (assumed non-zero here) and we get \[ \begin{equation} \eta \sim 2 \pi \mathrm{i} \sum_m \frac{F}{\partial G / \partial \bar{\alpha}} \exp \left(\mathrm{i} \bar{\alpha}_m r+\frac{1}{4} \pi \mathrm{i} \operatorname{sgn} \kappa\right) \sqrt{ \frac{2 \pi}{|\kappa| r}}+O\left(r^{-1}\right) \end{equation} \]

--- note -------------------------------------------------

Q: Where does the \(O(r^{-1})\) error term originate?

A: Outside the small neighbourhoods of the stationary points \(\varphi^{\prime}(\bar{\beta})\) is bounded away from zero. On that set we integrate by parts once: (path: non-stationary) \[ \int_{\text {non-stationary }} a e^{i r \varphi} d \bar{\beta}=\frac{1}{i r}\left[\frac{a}{\varphi^{\prime}} e^{i r \varphi}\right]_{\text {ends }}-\frac{1}{i r} \int e^{i r \varphi} \frac{d}{d \bar{\beta}}\left(\frac{a}{\varphi^{\prime}}\right) d \bar{\beta} . \]

• The bracketed end terms vanish because \(a(\bar{\beta})\) falls off quicker than any power as \(|\bar{\beta}| \rightarrow \infty\).
• The remaining integral is uniformly bounded, so the whole piece is \(O\left(r^{-1}\right)\).

at points \(m\), \(\partial G / \partial \bar{\alpha}= \pm|\nabla G|\).

proof:

we start from the derivatives \[ \frac{\partial(\bar{\alpha}, \bar{\beta})}{\partial(\alpha, \beta)}=\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)=R \] then \[ \begin{aligned} \nabla G & =\partial_{(\alpha, \beta)} G=\partial_{(\bar{\alpha}, \beta)} G \partial_{(\alpha, \beta)}(\bar{\alpha}, \bar{\beta}) \\ & =R \partial_{(\bar{\alpha}, \bar{\beta})} G \\ & =\binom{\partial_\bar{\alpha} G \cos \theta-\partial_{\bar{\beta}} G \sin \theta}{\partial_\bar{\alpha} G \sin \theta+\partial_{\bar{\beta}} G \cos \theta} \end{aligned} \] since \(\frac{d \bar{\alpha}}{d \bar{\beta}}=-\frac{\partial_{\bar\beta} G}{\partial_{\bar\alpha} G}=0\) then

\[ \partial_\bar\beta G=0 \]

therefore

\[ \begin{aligned} \nabla G & =\left(\partial_{\overline{\alpha}} G \cos \theta, \partial_{\overline{\alpha}} G \sin \theta\right) \\ & |\nabla G|=\left|\partial_{\overline{\alpha}} G\right| \\ \Rightarrow & \partial_\bar{\alpha} G= \pm|\nabla G| \end{aligned} \] Now we get the radiation condition \[ \frac{\partial G}{\partial \bar{\alpha}}=\frac{1}{\cos \theta} \frac{\partial G}{\partial \alpha}=\frac{1}{\sin \theta} \frac{\partial G}{\partial \beta} \] Subsequently, at the stationary points \(m\) \[ \kappa=-\frac{\partial^2 G}{\partial \bar{\beta}^2}/\frac{\partial G}{\partial \bar{\alpha}} \] at points \(m\). When the depth is infinite this is

\[ \alpha^2 \cos \theta /\left(2 \alpha^2-1\right)>0 \]

From the foregoing analysis, \[ G(\alpha, \beta)=\alpha^2-k=0, \quad k=\sqrt{\alpha^2+\beta^2} \]

Set \(k=\alpha^2\) in the definition of \(k\) :

\[ k^2=\alpha^4=\alpha^2+\beta^2 \Longrightarrow \beta^2=\alpha^2\left(\alpha^2-1\right) . \] therefore \(\alpha^2-1 \geqslant 0\) and \(\cos \theta >0\).

Because of the symmetry of the curves \(G=0\) about the axes, it is clear that the points \(\left(\alpha_m, \beta_m\right)\) and \(\left(-\alpha_m,-\beta_m\right)\) have normals in the same direction.

Therefore, we can write \[ \begin{equation} \eta \sim-\frac{2^{\frac{5}{2}} \pi^{\frac{3}{2}}}{r^{\frac{1}{2}}} \sum_{m+} \frac{F \cos \theta}{\partial G / \partial \alpha(\sqrt{|\kappa|)}} \sin \left(\alpha x+\beta y+\frac{1}{4} \pi \operatorname{sgn} \kappa\right)+O\left(\frac{1}{r}\right), \end{equation} \]

where \(\kappa=-\frac{\partial^2 G}{\partial \bar{\beta}^2} / \frac{\partial G}{\partial \bar{\alpha}}\)

we shall begin with the case of water of infinite depth, for this case \(G=0\), then we get

\[ \begin{equation} \begin{gathered} \alpha_m=\frac{1}{2 \sqrt{2}|\tan \theta|}\left\{4 \tan ^2 \theta+1 \pm \sqrt{\left(1-8 \tan ^2 \theta\right) }\right\}^{\frac{1}{2}} \\ \beta_m=-\alpha_m\left(2 \alpha_m^2-1\right) \tan \theta \end{gathered} \end{equation} \]

Consequently, we derive,

\[ \begin{equation} \eta \sim-\frac{P}{\rho} \sqrt{\frac{2}{\pi r}}\sum_{m+} \mathrm{e}^{\alpha^2 z}\left\{\frac{\alpha^7 \cos \theta}{\left(2 \alpha^2-1\right)\left|1-6 \alpha^2 \sin ^2 \theta\right|}\right\}^{\frac{1}{2}} \sin \left(\alpha x+\beta y+\frac{1}{4} \pi \operatorname{sgn} \kappa\right)+O\left(\frac{1}{r}\right). \end{equation} \]

From equation \((18)\) we can see that outside this wedge our approximation gives a disturbance of at most \(O\left(r^{-1}\right)\). However, from previous theories (Ursell, F. 1960), we know that the disturbance there is actually \(O\left(r^{-3}\right)\). The two values of \(\alpha_m\) in (17) correspond to the diverging and transverse wave systems identified in earlier theories, depending on whether the + or - sign is chosen.

For precise wave-crest plotting, we express \(\alpha_m=k \cos \phi\) and \(\beta_m=k \sin \phi\), transforming the condition \(G=0\) into:

\[ k=\sec ^2 \phi \] at constant phase, we have \[ \begin{equation} \alpha_m x+\beta_m y=k r \cos (\theta-\phi)=k p=N=\mathrm{constant} \end{equation} \] Note that \(N\) represent the phase number.

In this definition, \(p\) denotes the length of the perpendicular dropped from the origin to the tangent line of constant phase. This perpendicular aligns with \(\mathbf{k}\), forming angle \(\phi\) with the \(x\)-axis (see Figure).

in terms of \(p\) and \(\phi\) the equation of a line is

\[ \begin{equation} x=p \cos \phi-\frac{\mathrm{d} p}{\mathrm{~d} \phi} \sin \phi ; \quad y=p \sin \phi+\frac{\mathrm{d} p}{\mathrm{~d} \phi} \cos \phi . \end{equation} \]

which gives

\[ p=N / k=N \cos ^2 \phi \]

The line of constant phase is equation \((20)\). Solution (16) fails at \(\theta= \pm \theta_c\) because \(\kappa=0\), despite the presence of a single stationary point \(\left(\sqrt{\frac{3}{2}}, \mp \frac{\sqrt{3}}{2}\right)\). Here, the sign change in \(\kappa\) produces a \(\frac{\pi}{2}\) phase offset between the diverging and transverse wave systems."

To find a solution when \(\abs{\theta}=\theta_c\) we must replace \((14)\) by \[ \begin{equation} \bar{\alpha}=\bar{\alpha}_m+\frac{1}{6}\left(\frac{\mathrm{~d}^3 \bar{\alpha}}{\mathrm{~d} \bar{\beta}^3}\right)_m\left(\bar{\beta}-\bar{\beta}_m\right)^3 \end{equation} \]

When \(|\theta|\) is near to \(\theta_c\) the solution (28) is not adequate because \(\kappa\) is small. Here both the \(\kappa\) and \(\mathrm{d}^3 \bar{\alpha} / \mathrm{d} \bar{\beta}^3\) terms of (22) and (36) must be retained when approximating for \(\bar{\alpha}\), and there are additional difficulties as the two points of stationary phase are close together.

The method of stationary phase then gives \[ \begin{equation} \eta \sim\left\{\frac{-4 \pi \sqrt{3}\Gamma\left(\frac{4}{3}\right)F \cos \theta}{\partial G / \partial \alpha}\left(\frac{6}{r\left|\mathrm{~d}^3 \bar{\alpha} / \mathrm{d} \bar{\beta}^3\right|}\right)^{\frac{1}{3}} \sin (\alpha x+\beta y)\right\}_{\substack{\alpha=\alpha_m \\ \beta=\beta_m}} . \end{equation} \] When \(|\theta|\) is near to \(\theta_c\) the solution (16) is not adequate because \(\kappa\) is small. Here both the \(\kappa\) and \(d^3 \bar{\alpha} / d \bar{\beta}^3\) terms must be retained when approximating for \(\bar{\alpha}\), and there are additional difficulties as the two points of stationary phase are close together.

The waves along the line \(\theta=0\) have also given difficulty in previous methods. Here the transverse system gives the contribution

\[ -\frac{P}{\rho} \sqrt{\frac{2}{\pi x}} \mathrm{e}^z \sin \left(x+\frac{1}{4} \pi\right) \]

to \(\eta\), and the diverging system gives a term in which \(\alpha_m, \beta_m \rightarrow \infty\), since \(\mathrm{d} \alpha / \mathrm{d} \beta \rightarrow 0\) on \(G=0\) as \(\alpha, \beta \rightarrow \infty\). Thus as \(\theta \rightarrow 0\) the diverging wave amplitude \(\rightarrow \infty\) and its wave length \(\left(2 \pi / \sqrt{\alpha_m^2+\beta_m^2}\right) \rightarrow 0\);

the effect of surface tension

for the case infinite depth, the poles of the inner integral are now given by \[ \begin{equation} \alpha^2-k-\sigma k^3=0 \end{equation} \] If we again write \(\alpha=k \cos \phi, \beta=k \sin \phi\), (23) can be solved for \(k\) :

\[ \begin{equation} k=\frac{\cos ^2 \phi \pm \sqrt{\cos ^4 \phi-4 \sigma}}{2 \sigma} \end{equation} \]

These two solutions will be referred to as capillary waves (upper sign) and gravity waves. The two curves given by (24) meet at points \(M\), say, where \(k_M=\sigma^{-\frac{1}{2}}\), \(\cos \phi_M=(4 \sigma)^{\frac{1}{4}}\).

When \(\cos ^4 \phi<4 \sigma\) there are no real solutions of the equation \(G=0\), and the poles will be at complex values of \(\bar{\alpha}\); any waves in this range will therefore have an exponential decay factor when the first integration has been carried out, and they will be neglected.

When the outer Fourier integral for the free-surface elevation is closed in the lower half of the complex \(\bar{\alpha}-\) plane, the far-field contribution is obtained from the poles of the integrand,

\[ \eta(r)=\int_{-\infty}^{\infty} \frac{F(\bar{\alpha}, \bar{\beta})}{G(\bar{\alpha}, \bar{\beta})} e^{i \bar{\alpha} r} d \bar{\alpha} \quad \longrightarrow \quad \sum_p \operatorname{Res}_{\bar{\alpha}=\bar{\alpha}_p}[F / G] e^{i \bar{\alpha}_p r}, \quad r=\sqrt{x^2+y^2} \]

where each pole satisfies \(G\left(\bar{\alpha}_p, \bar{\beta}\right)=0\) and lies below the real axis:

\[ \bar{\alpha}_p=a-i b, \quad b>0 \]

The exponential factor for such a pole is

\[ e^{i \bar{\alpha}_p r}=e^{i a r} e^{-b r} \]

so the spatial attenuation is governed solely by the imaginary part \(b\) : the amplitude decays as \(e^{-b r}\).

For certain values of the transverse wavenumber \(\bar{\beta}\) the root of \(G=0\) approaches the real axis, i.e. \(b \ll 1\). In this limit the decay length \(1 / b\) becomes very large and the associated mode behaves almost like a freely propagating wave, producing only weak attenuation in the physical field. Conversely, for other \(\bar{\beta}\) the pole sits deeper in the lower half-plane, \(b=O(1)\); the factor \(e^{-b r}\) then suppresses the contribution rapidly and the corresponding mode is essentially evanescent.

Hence, when the poles generated by \(G=0\) lie close to the \(\bar{\alpha}\) axis, the imaginary part \(b\) is small and the exponential decay of the residue term is correspondingly weak, giving a slowly attenuating wave component in the far field.

The tangent at \(M\) passes through the origin, making an angle \(\phi_M\), say, with the \(x(\alpha)\) axis; the wave crests therefore make an angle \(\frac{1}{2} \pi-\phi_M\) with the \(x\) axis since they are perpendicular to the tangent (k). Thus the velocity normal to the wave crest at points \(M\) is \[ U \sin \left(\frac{1}{2} \pi-\phi_M\right)=U \cos \phi_M=(4 T g / \rho)^{\frac{1}{4}}=U_{\min .} \]

Parallelism of \(k_x\) to \(x\)-axis?

Typically, we don't explicitly state that \(k_x\) is parallel to the physical \(x\)-axis; the equivalence is implicit in the Fourier representation. Immediately after the transform is written as

\[ \eta(x, y, z)=\iint e^{i(\alpha x+\beta y)} \frac{F(\alpha, \beta)}{G(\alpha, \beta)} d \alpha d \beta \]

It remarks "It is evident from inspection that \(\alpha=k_x, \beta=k_{y .}\)." Once \(\alpha \equiv k_x\) and \(\beta \equiv k_y\) are identified, the axes of the \((\alpha, \beta)_{-}\)(i.e. \(k_{-}\)) plane are, by definition, parallel to the spatial \(\mathscr{x}\) - and \(\mathscr{y}\)-directions. Hence, when the discussion later refers to the tangent at the intersection point \(M\) making an angle \(\phi_M\) with the " \(x(\alpha)\)-axis," it is simply invoking this original identification; no additional statement about the parallelism of \(k_x\) and the physical \(x\)-axis is required.

Wave-Crest Orthogonality to the Wave Vector （no matter the Presence of Uniform Currents or not)

The wave number vector \(\mathbf{k}\) is, by definition, normal to isophase surfaces; hence wave crests remain perpendicular to \(\mathbf{k}\) even in a uniform background flow \(\mathbf{U}\). The current modifies phase and group velocities through a Doppler shift but leaves the geometric relationship between \(\mathbf{k}\) and the crest intact.

1. Geometric Necessity

For a monochromatic plane wave

\[ \phi(\mathbf{x}, t)=\mathbf{k} \cdot \mathbf{x}-\omega t \]

wave crests satisfy \(\phi=\) const. The spatial gradient is

\[ \nabla \phi=\mathbf{k} \]

which is orthogonal to every isophase (crest) surface. This definition is independent of reference frame, isotropy, or the presence of a uniform current.

2. Uniform Background Flow

With a steady current \(\mathbf{U}\), the intrinsic frequency \(\sigma\) in the dispersion relation is Doppler-shifted:

\[ \omega=\sigma+\mathbf{k} \cdot \mathbf{U} \]

The orientation of \(\mathbf{k}\) and thus the crest geometry are unchanged. Only the following vectors acquire an advective component:

a. Phase velocity: \(\mathbf{c}_p=\mathbf{U}+\frac{\sigma}{|\mathbf{k}|} \hat{\mathbf{k}}, \quad\)
b. Group velocity: \(\mathbf{c}_g=\mathbf{U}+\nabla_{\mathbf{k}} \sigma\).

3. Directional Distinctions

Thus, while energy packets may advance obliquely, the instantaneous crest orientation remains normal to \(\mathbf{k}\).

4. Extension to Shear Flows

For spatially varying currents, \(\mathbf{U}=\mathbf{U}(\mathbf{x})\), the ray equations give

\[ \frac{d \mathbf{k}}{d t}=-(\nabla \mathbf{U})^{\top} \mathbf{k} . \]

The wave vector (and hence the local crest normal) evolves in time, but its orthogonality to the local crest is preserved at every instant.

5. Conclusions

Wave crests are perpetually perpendicular to the wave number vector by geometric definition. Uniform currents alter phase and group velocities through advection yet do not tilt the crests relative to \(\mathbf{k}\). Misinterpretations generally arise from conflating wave-vector direction with energy or phase propagation directions.

why \(U_{\min}\)

the dispersion relation is

\[ \omega^2=g k+\frac{T}{\rho} k^3 \quad \text { (deep-water gravity-capillary) } \]

so that the group speed (velocity normal to the wave-crest)

\[ U_g(k)=\frac{d \omega}{d k}=\frac{g+3 T k^2 / \rho}{2 \sqrt{g k+T k^3 / \rho}} \]

Locate the extremum, Set \(d U_g / d k=0\) :

\[ \frac{d U_g}{d k}=0 \Longrightarrow k^2=\frac{\rho g}{T} \frac{1}{2} . \]

Hence the critical wavenumber is \[ k_M=\sqrt{\frac{\rho g}{2 T}}=\sigma^{-1 / 2}, \quad \sigma \equiv \frac{T}{\rho U^2} . \]

Evaluate group velocity \(U_g\) at \(k=k_M\)

\[ U_g\left(k_M\right)=\frac{g+3 T k_M^2 / \rho}{2 \sqrt{g k_M+T k_M^3 / \rho}}=\frac{g+3 g / 2}{2 \sqrt{g k_M+g k_M / 2}}=(4 T g / \rho)^{1 / 4} \equiv U_{\min } . \]

For \(k \ll k_M\) (pure-gravity range) \(U_g \sim \sqrt{g / 4 k} \rightarrow \infty\); for \(k \gg k_M\) (pure-capillary range) \(U_g \sim \sqrt{3 T / 4 \rho} k^{1 / 2} \rightarrow \infty\). Thus the stationary value at \(k_M\) is the global minimum.

Geometric link with \(\phi_M\)

At the intersection point \(M\) of the two dispersion branches

\[ \cos ^4 \phi_M=4 \sigma, \text { so } \cos \phi_M=(4 T g / \rho)^{1 / 4} / U \]

Taking the component of the ship speed \(U\) normal to the wave-crest

\[ U \sin \left(\frac{\pi}{2}-\phi_M\right)=U \cos \phi_M=(4 T g / \rho)^{1 / 4}=U_{\min } \]

which reproduces the analytic minimum derived above.

So, here, because a Doppler shift adds only a linear term \(-\mathbf{U} \cdot \mathbf{k}\) to the dispersion relation, it translates the \(\omega(k)\) surface without altering its slope. The group speed, defined by \(\mathbf{c}_g=\nabla_k \omega\), therefore remains the derivative of the intrinsic frequency

\[ \omega_0^2=g k+T k^3 / \rho \]

even in the ship-fixed frame. Hence \(c_g(k)=\partial \omega_0 / \partial k\) can be compared directly with the projected ship speed \(U \cos \theta\) to determine which wavenumbers satisfy the steady-wake condition.

The radiation condition is

\[ \frac{\bar{\alpha} \cos \theta-\bar{\beta} \sin \theta}{\partial G / \partial \bar{\alpha}}>0 . \]

For the points of stationary phase, we have

\[ \begin{aligned} \partial G / \partial \alpha & =(\alpha / k)\left\{2 k-\left(1+3 \sigma k^2\right)\right\} \\ & =(3 \sigma \alpha / k)\left(k_{+}-k\right)\left(k-k_{-}\right) \end{aligned} \]

where \(k_{+}, k_{-} =(1 / 3 \sigma)(1 \pm \sqrt{1-3 \sigma})\)

Note that \(k_{-}<1\left(\sigma<\frac{1}{4}\right)\), while at \(k=k_{+}, \mathrm{d} \alpha / \mathrm{d} \beta=\partial G / \partial \beta / \partial G / \partial \alpha=\infty\). Hence \(k=k_{+}\) gives the points at which the curve \(G=0\) is parallel to the \(\alpha\) axis. Since \(k \geqslant 1\) for any \(\sigma, k=k_{-}\)is not a point on the curve. Now the radiation condition becomes

\[ \frac{k \cos \theta}{3 \sigma\left(k_{+}-k\right)\left(k-k_{-}\right)}>0 \]

which is equivalent to

\[ \left.\begin{array}{rlrl} \cos \theta & >0 & & k_{-}<1 \leqslant k<k_{+} \\ & <0 & & k>k_{+} . \end{array}\right\} \]

This decomposition plays a critical role in analyzing the radiation condition:
1. Sign analysis:

The expression \(\partial G / \partial \alpha = (3 \sigma \alpha / k)(k_{+}-k)(k-k_{-})\) explicitly reveals sign changes in the derivative, corresponding to the tangent directions along the \(G=0\) curve (Figures 4–5).

2. Radiation condition:

   Through Equations (48)–(49):
   \[ \frac{\cos \theta}{\partial G / \partial \alpha} > 0 \implies \operatorname{sign}\left( \partial G / \partial \alpha \right) = \operatorname{sign}(\cos \theta), \]
   the factorization enables direct sign determination via comparison of \(k\) with \(k_{+}\) and \(k_{-}\) (e.g., \(\partial G / \partial \alpha > 0\) when \(k<k_{+}\)).

3. Wave system separation:

   • \(k<k_{+}\) corresponds to gravity waves (\(\cos \theta > 0\), propagating behind the pressure point).
     
   • \(k>k_{+}\) corresponds to capillary waves (\(\cos \theta < 0\), which may appear ahead of the pressure point).

Why \(\nabla G\) is Parallel to \(r\)

#parallel

Let the dispersion curve be defined by \(G(\alpha, \beta)=0\). For an observation point \(\mathbf{r}=(x, y)=r(\cos \theta, \sin \theta)\) the Fourier kernel reads

\[ e^{i(\alpha x+\beta y)} \]

Rotate \((\alpha, \beta)\) through the angle \(\theta\) :

\[ \bar{\alpha}=\alpha \cos \theta+\beta \sin \theta, \quad \bar{\beta}=-\alpha \sin \theta+\beta \cos \theta, \]

so that

\[ \alpha x+\beta y=\bar{\alpha} r . \]

1. Stationary point with respect to \(\bar{\beta}\)

Applying the method of steepest descents to the outer \(\bar{\beta}\)-integration gives

\[ \frac{\partial \bar{\alpha}}{\partial \bar{\beta}}=0 .\tag{J1} \]

2. Relating \(\partial \bar{\alpha} / \partial \bar{\beta}\) to \(\nabla_k G\)

Along the curve \(G=0\) we have \(G_{\bar{\alpha}} d \bar{\alpha}+G_{\bar{\beta}} d \bar{\beta}=0\); hence

\[ \frac{d \bar{\alpha}}{d \bar{\beta}}=-\frac{G_{\bar{\beta}}}{G_{\bar{\alpha}}}\tag{J2} \]

Combining (J1) and (J2) yields the stationary-phase condition

\[ G_{\bar{\beta}}=0 .\tag{J3} \]

3. Gradient parallelism

The Jacobian of the rotation is unity, so

\[ \binom{G_{\bar{\alpha}}}{G_{\bar{\beta}}}=\left(\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)\binom{G_\alpha}{G_\beta} . \]

Condition (J3) gives

\[ -\sin \theta G_\alpha+\cos \theta G_\beta=0 \Longrightarrow \nabla_{(\alpha, \beta)} G \|(\cos \theta, \sin \theta)=\hat{\mathbf{r}} .\tag{J4} \]

4. Result

Equation (J4) shows that a wave-number point contributes to the far field in direction \(\theta\) if and only if the normal to the dispersion curve \(G=0\) is parallel to the observation vector. In other words,

\[ \nabla G \| \mathbf{r} \quad \Longleftrightarrow \quad \text { normal direction angle }=\theta . \]

near point \(C\)

Around point \(C\), "Which side is 'ahead of the pressure point' ?"

The pressure source (or a ship) is placed at the origin and assumed to move with speed \(U\) in the \(+x\) direction. In the body-fixed frame the external flow appears to come from \(-x\); therefore

(the convention is standard in Lighthill-type ship-wake studies, even if not reiterated).

How the radiation condition selects \(\theta= \pm \pi\) The radiation test reads

\[ \frac{\bar{\alpha} \cos \theta-\bar{\beta} \sin \theta}{\partial G / \partial \bar{\alpha}}>0 \Longrightarrow \alpha \cos \theta\left(\frac{\partial G}{\partial \alpha}\right)^{-1}>0 \tag{J5} \]

For the capillary branch in the limit \(\phi \rightarrow 0^{+}\)

\[ \left(\beta \rightarrow 0^{+}, k \gg 1\right): \]

• \(\alpha>0\) and very large;
• a direct evaluation gives \(\partial G / \partial \alpha<0\). Insert the signs into (J5):

\[ (+) \cos \theta(-)>0 \Longrightarrow \cos \theta<0 \Rightarrow \theta=\pi(\text { or }-\pi) . \]

Hence the only observation angles that survive the radiation test are \(\theta= \pm \pi\), the two directions straight upstream.

Physical picture

• As \(\phi \rightarrow 0^{+}\)the wave vector is nearly parallel to the \(\alpha\)-axis, and the group velocity \(\mathbf{c}_g\) points toward \(-x\) because \(\partial G / \partial \alpha<0, \partial G / \partial \beta<0\).
• Requiring outward radiation forces one to keep the same-direction phase-speed branch; the radiation condition effects precisely this choice.
• Therefore the first observable capillary waves appear at \(\theta= \pm \pi\).

As \(\phi\) increases the normal rotates, the crest lines sweep around the hull, and finally become asymptotic to the Kelvin angles

\[ \theta= \pm\left(\frac{\pi}{2}-\phi_M\right) \text { downstream. } \]

Summary

"The radiation condition shows that the capillary wave starts ahead of the pressure point" means: in the large- \(k\) section of the dispersion loop only the normals pointing to \(-x(\theta= \pm \pi)\) satisfy (J5), hence they form the first segment of the radiating capillary wave.

capillary waves likes something a cross between a parabola and a hyperbola; the gravity wave-crests are more like true hyperbolae