Ocean wave solution

Ocean wave solution is analytical expression of the idealized ocean wave in the Lagrangian coordinates.

This article is just rewrite of the article Exact waves at surface of liquid with slightly mofidied notations; the origin of coordinates is displaced in such a way, that point $$x=0$$, $$t=0$$ corresponds to the top of the wave.

WARNING

This is very draft; I begin to change notations but ..

Equation

The Navier–Stokes equation, the term with $$\nu \nabla^2 \vec V$$ where $$\vec V=\vec V(\vec X, t)$$ describes viscosity of the fluid; in some cases it can be neglected, converting the Navier–Stokes equation to the Euler equation

$$\!\!\!\!\!\!\!\!\!\!\!\! (1) ~ ~ ~ ~ \displaystyle \dot{\vec V} + \vec V_{,a} V_a= - \frac{1}{\rho} \nabla p + f$$

$$f$$ is interpreted as gravitational force, that, in principle, may depend on all the coordinates $$\vec X$$ and on time $$t$$.

The additional subscript denotes the derivation with respect to one of arguments; the subscript indicates the number of this argument. In order to avoid confusion, in this formalism, the partial derivative of any physical quantity with respect to any physical quantity is forbidden, prohibited. The only functions can be differentiated, and the name of this function should be explicitly declared. However, for simplification, in some formulas, the arguments are omitted, but every formula should allow the unambiguous recovery of these arguments. In particular, $$\nabla$$ differentiates with respect to coordinates $$\vec X$$, giving an additional valence to the dimension of array.

For the non–compressive liquid, $$\nabla \vec V =0$$, and the non–trivial solution can be parametrized in terms of the elementary functions, assuming the translational symmetry with respect to one of coordinates. In this case, the only two spacial coordinates should be taken in to account, and one time as third coordinate.

Parametrization

Map of the ArcSerega function

The only two spatial coordinates $$Z$$ and $$Y$$ are used; and the letter $$Z$$ is not necessary for the third spatial coordinate. There fore it is used below for another meaning.

The solution of equation (1) can be parametrized with the Serega function

$$\!\!\!\!\!\!\!\!\!\!\!(2) ~ ~ ~ ~ Z=\mathrm{Serega}(z)=z+\mathrm i ~ \exp(\mathrm i z^*)$$

the real and imaginaginary part of $$Z=X+\mathrm i Y$$ are interpreted as coordinates of particles, while the real and imaginary parts of $$z=x+\mathrm i y$$ are interpreted as two parameters specifying the particle of the fluid.

The Serega function is not holomorphic; to, the separate expressions for the real and imaginary parts are used. The coordinates of the particles of the liquid can be written in such a way:

$$\!\!\!\!\!\!\!\!\!\!\!(3) ~ ~ ~ ~ X=X(x,y,t)= \Re( \mathrm{Serega}( k x - \omega t + \mathrm i y))$$
$$\!\!\!\!\!\!\!\!\!\!\!(4) ~ ~ ~ ~ Y=Y(x,y,t)= \Im( \mathrm{Serega}( k x - \omega t + \mathrm i y))$$

Here $$x$$ and $$y$$ determine the particle, while $$X$$ and $$Y$$ are Cartesian coordinates of the particle at time $$t$$.

Parameter $$k$$ has sense of wavenymber.

Parameter $$\omega$$ has sense of frequency.

Particles with $$y=y_s=\mathrm {constant}$$ can be interpreted as surface of the liquid, at least while $$y<0$$.

The components of velocity

$$\!\!\!\!\!\!\!\!\!\!\!(5) ~ ~ ~ ~ u=\dot X=u(x,y,t)$$
$$\!\!\!\!\!\!\!\!\!\!\!(6) ~ ~ ~ ~ v=\dot Y=v(x,y,t)$$

are derivatives of functions $$X$$ and $$Y$$ with respect to the last argument. $$u$$ has sense of the horizontal component of velocity while $$v$$ has sense of the vertical component.

WIth the inverse function $$\mathrm{ArcSerega}$$, velocities can be expressed in terms of the Cartesian coordinates:

$$\!\!\!\!\!\!\!\!\!\!\!(7) ~ ~ ~ ~U=U(x(X,Y,t), y(X,Y,t), t)=u(x,y,t)$$
$$\!\!\!\!\!\!\!\!\!\!\!(8) ~ ~ ~ ~V=V(x(X,Y,t), y(X,Y,t), t)=v(x,y,t)$$

where

$$\!\!\!\!\!\!\!\!\!\!\!(9) ~ ~ ~ ~x(X,Y,t)= \frac{1}{k}\left( -\omega t + \Re(\mathrm{ArcSerega} ( kX - \omega t +\mathrm i k Y )) \right)$$
$$\!\!\!\!\!\!\!\!\!\!\!\!(10) ~ ~ ~ ~y(X,Y,t)= \frac{1}{k} \Im(\mathrm{ArcSerega} ( kX - \omega t +\mathrm i k Y ))$$

These velocities satisfy the Navier-Stokes equation for the Eulerian liquid (without viscosity). The surface tension of the liquid is neglected. The liquid is treated as non–compressible. The gravitational force with acceleration $$g=\omega/k$$ should be applied as the "external force".

The 15digit implementation of functions $$\mathrm{Serega}$$ and $$\mathrm{ArcSerega}$$ in C++ are suggested in the description of the complex map at right.

At the matching of coordinates $$x$$, $$y$$ with coordinates $$X$$, $$Y$$ by equations (3,4), the relations $$u=U$$ and $$v=V$$ holds. However, the derivatives, in general, are not the same; $$u_1\! \ne\! U_1$$, $$u_2 \!\ne\! U_2$$, $$\dot u \!\ne\! \dot U$$, $$v_1 \!\ne\! V_1$$, $$V_2\! \ne\! V_2$$, $$\dot v\! \ne\! \dot V$$. These derivatives are analyzed in the next section

Distribution of velocities

The parameterization of movement with Serega function above refers to the particles of the fluid. Velocities can be evaluated as the time derivatives of $$X(x,y,t)$$ and $$Y(x,y,t)$$. Properties of these velocities are described in this section.

Notations

In order to make the deduction compact, the following notations are used::

$$c=\cos(kx\!-\!\omega t)$$ ; $$~c'=\!-k \sin(kx\!-\!\omega t)=\!-ks$$; $$\dot c=\omega \sin(kx\!-\!\omega t)= \omega s$$;
$$s=\sin(kx\!-\!\omega t)$$ ; $$~s'= k \cos(kx\!-\!\omega t)=-kc~$$; $$~\dot s=\omega \cos(kx\!-\!\omega t)= \omega c$$;
$$r=\frac{1}{k}\exp(ky)$$ ; $$~r'=\exp(y)=kr$$

Then

$$\!\!\!\!\!\!\!\!\!\!\!\! (4) ~ ~ ~ ~ \displaystyle X=X(x,y,t) = x - r s$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (5) ~ ~ ~ ~ \displaystyle Y=Y(x,y,t) = y + r c$$

Velocities:

$$\!\!\!\!\!\!\!\!\!\!\!\! (6) ~ ~ ~ ~ \displaystyle \dot X=X(x,y,t) = - r c = U(x,y,t)$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (7) ~ ~ ~ ~ \displaystyle \dot Y=Y(x,y,t) = - r s = V(x,y,t)$$

The goal is to express $$u(X,Y,t)=U(x,y,t)$$ and $$v(X,Y,t)=V(x,y,t)$$ in such a way that $$X=X(x,y,t)$$ and $$Y=Y(x,y,t)$$.

Derivatives of $$x$$ and $$y$$ at constant $$t$$

Consider variation of the spatial coordinates at $$t=$$const. Then

$$\!\!\!\!\!\!\!\!\!\!\!\! (8) ~ ~ ~ ~ \mathrm d X= (1-rck) ~\mathrm d x - k r s ~ \mathrm d y$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (9) ~ ~ ~ ~ \mathrm d Y= -rsk ~ \mathrm d x + (1+krc) ~ \mathrm d y$$

Then

$$\!\!\!\!\!\!\!\!\!\!\!\! (10) ~ ~ ~ ~ \mathrm d x= \frac{1+krc}{1-(krc)^2} ~\mathrm d X + \frac{krs}{1-(krc)^2} ~ \mathrm d Y$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (11) ~ ~ ~ ~ \mathrm d y= \frac{krs}{1-(krc)^2} ~ \mathrm d X + \frac{1-krc}{1-(krc)^2} ~ \mathrm d Y$$

In such a way,

$$\!\!\!\!\!\!\!\!\!\!\!\! (11) ~ ~ ~ ~ x_1= \frac{1+krc}{1-(kr)^2}~$$; $$~x_2=\frac{krs}{1-(kr)^2}$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (12) ~ ~ ~ ~ y_1= \frac{krs}{1-(kr)^2}~ ~$$; $$~ y_2=\frac{1-krc}{1-(kr)^2}$$

Ia it is declared above, $$x=x(X,Y,t)$$, $$y=y(X,Y,t)$$; and the subscript indicates the number of argument to evaluate the derivative.

Derivatives of $$x$$ and $$y$$ at constant $$X$$ and $$Y$$

Consider variation of time at $$x=\mathrm{const}$$ and $$y=\mathrm{const}$$

$$\!\!\!\!\!\!\!\!\!\!\!\! (13) ~ ~ ~ ~ x_1 \dot X + x_2 \dot Y + \dot x = 0$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (14) ~ ~ ~ ~ y_1 \dot X + y_2 \dot Y + \dot y = 0$$

then

$$\!\!\!\!\!\!\!\!\!\!\!\! (15) ~ ~ ~ ~ \dot x= -x_1 u -x_2 v = - r \omega \frac{c+kr}{q-(k r)^2}$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (16) ~ ~ ~ ~ \dot y= -y_1 u - y_1 v= - r \omega \frac{s}{q-(k r)^2}$$

Euler coordinates

Historically, the coordinates $$x$$, $$y$$, related with the elements of the fluid, are called Lagrange's coordinates, and the cartesian coordinates $$X$$, $$Y$$, in which the Laws of Newton are valid, are called Euler's coordinates. In particular the Navier-Stokes equation (1) is written in the Euler's coordinates. The results of the excercises above relate these two systems of coordinates and allow to verify the solution (4),(5).

Consider $$U=U(X,Y,t)$$ and $$V=V(X,Y,t)$$ such that

$$\!\!\!\!\!\!\!\!\!(17)~ ~ ~ U(X(x,y,t),Y(x,y,t),t)=u(x,y,t)$$
$$\!\!\!\!\!\!\!\!\!(18)~ ~ ~ V(X(x,y,t),Y(x,y,t),t)=v(x,y,t)$$

and

$$\!\!\!\!\!\!\!\!\!(19)~ ~ ~ U(X,Y,t)=u(x(X,Y,t),y(X,Y,x),t)$$
$$\!\!\!\!\!\!\!\!\!(20)~ ~ ~ V(X,Y,t)=v(x(X,Y,t),y(X,Y,x),t)$$

Then

$$\!\!\!\!\!\!\!\!\!(21)\displaystyle~ ~ ~ U_1=u_1 x_1 + u_2 y_1 = - \omega krs \frac{1+krc}{1-(kr)^2}+\omega krc \frac{krs}{1-(kr)^2}=$$ $$\displaystyle \frac{\omega kr}{1-(kr)^2} \left( -s-krsc+krcs\right)=\frac{-\omega krs}{1-(kr)^2}$$
$$\!\!\!\!\!\!\!\!\!(22)\displaystyle~ ~ ~ U_2=u_1 x_2 + u_2 y_2 = - \omega krs \frac{krs}{1-(kr)^2}+\omega krc \frac{1-krc}{1-(kr)^2}=$$ $$\displaystyle \frac{\omega kr}{1-(kr)^2} \left(-krs^2+c-krc^2\right)=\omega kr\frac{c+kr}{1-(kr)^2}$$
$$\!\!\!\!\!\!\!\!\!(23)\displaystyle~ ~ ~ V_1=v_1 x_1 + v_2 y_1 = - \omega krc \frac{1+krc}{1-(kr)^2}+\omega krs \frac{krs}{1-(kr)^2}=$$ $$\displaystyle \frac{\omega kr}{1-(kr)^2} \left( c+krc^2+krs^2 \right)=\omega kr\frac{c+kr}{1-(kr)^2}$$
$$\!\!\!\!\!\!\!\!\!(24)\displaystyle~ ~ ~ V_2=v_1 x_2 + v_2 y_2 = - \omega krc \frac{krs}{1-(kr)^2}+\omega krs \frac{1-krc}{1-(kr)^2}=$$ $$\displaystyle \frac{\omega kr}{1-(kr)^2} \left(-krcs+s-krsc\right)=\omega kr\frac{s}{1-(kr)^2}$$

The spacial derivatives of velocities satisfy the condition of continuity

$$\!\!\!\!\!\!\!\!\!(25)\displaystyle~ ~ ~ U_1+V_2=0$$

for incompressible fluid.

However, the condition of potential flow cannot be satisvied, because the rotor

$$\!\!\!\!\!\!\!\!\!(26)\displaystyle~ ~ ~ V_1-U_2=\frac{\omega kr}{1-(kr)^2} 2 kr=\frac{2\omega k^2r^2}{1-(kr)^2}$$

is not zero.

Pressure

In order to keep the Navier-Stokes equation the pressure $$p=p(x,y,t)=P(X(x,y,t),Y(x,y,t),t)$$ should satisfy equations

$$\!\!\!\!\!\!\!\!\!(27)\displaystyle~ ~ ~ P_1=-\dot U- U_1U-U_2V=-\omega^2 rs$$
$$\!\!\!\!\!\!\!\!\!(28)\displaystyle~ ~ ~ P_2=-\dot V- V_1U-V_2V= ~\omega^2 rc$$

As usually, $$P_1$$ means derivative of function $$P$$ with respect to its first argument, and $$P_2$$ means its derivative with respect to the seconr argument. Dot differentiates with respect to the last argument (in this case, it is third argument); numeration of arguments begin with unity. (The zero'th argument is reserved for the name of the called function).

The explicit expressions (21(-(24) are used to get the last equalities in (27),(28). The equations for $$p$$ can be written as follows:

$$\!\!\!\!\!\!\!\!\!(29)\displaystyle~ ~ ~ \frac{1}{\rho}p_1=P_1 X_1+P_2 Y_1= - \omega^2 r s$$
$$\!\!\!\!\!\!\!\!\!(30)\displaystyle~ ~ ~ \frac{1}{\rho}p_1=P_1 X_1+P_2 Y_1=\omega^2 r \cdot(1+c)$$

The solution is

$$\!\!\!\!\!\!\!\!\!(31)\displaystyle~ ~ ~ p=p(x,y,z)=\frac{\omega^2}{k} r~(1+c)$$

Now, apply the gravitational force. Let the new pressure

$$\!\!\!\!\!\!\!\!\!(32)\displaystyle~ ~ ~ Q=Q(X,Y,t)=P-\rho g Y$$
$$\!\!\!\!\!\!\!\!\!(33)\displaystyle~ ~ ~ q=q(x,y,t)=Q(X(x,y,t),Y(x,y,t),t)= \rho \frac{\omega^2}{k} r (1+c) - \rho g (u+rc)$$

Let at some level $$y=\frac{\alpha}{k}$$, the condition $$q=0$$ is realized; $$\alpha$$ is supposed to be a real negative constant. Then

$$\!\!\!\!\!\!\!\!\!(34)\displaystyle~ ~ ~ \frac{\omega^2}{k} r (1+c) = \rho g \left(\alpha/k+rc \right)$$

giving

$$\!\!\!\!\!\!\!\!\!(35)\displaystyle~ ~ ~\frac{\omega^2}{k} r = g \alpha/k$$
$$\!\!\!\!\!\!\!\!\!(36)\displaystyle~ ~ ~\frac{\omega^2}{k} = g$$

so, $$r=\alpha/k$$; $$\alpha=kr=\exp(ky)$$ corresponds to the level of the surface of the liquid, id est, $$y=\ln(alpha)/k$$. In such a way, any level $$y=\mathrm {const}<0$$ can be chosen as surface of the fluid.

The frequency $$\omega$$ and wavelength $$k$$ are related with

$$\!\!\!\!\!\!\!\!\!(37)\displaystyle~ ~ ~ \frac{\omega^2}{k}= g$$

Note that in this approximation (no dissipation, no surface tension) the speed of waves does not depend on the amplitude $$\exp(ky)/k$$.

References

http://www.jetpletters.ac.ru/ps/1940/article_29409.pdf A. I. Dyachenko, V. E. Zakharov. Compact equation for gravity waves on deep water. JETP Letters, VOLUME 93 | ISSUE 12 | PAGE 782