Exact waves at surface of liquid

Exact waves at surface of liquid refers to the solution of the equations

$$\!\!\!\!\!\!\!\!\!\!\!\! (1) ~ ~ ~ ~ \displaystyle \frac {\mathrm d \vec v}{\mathrm d t}= - \frac{1}{\rho} \nabla p + f$$

where

$$\displaystyle \frac {\mathrm d \vec v}{\mathrm d t} = \dot{\vec v} + \vec v_{,a} v_a$$

This equation neglects viscosity $$\nu \nabla^2 \vec v$$. $$f$$ is interpreted as gravitational force.

Assume the non–compressive liquid, $$\nabla \vec v =0$$.

The Blue tetrad (1993) suggests the solution

$$\!\!\!\!\!\!\!\!\!\!\!\! (2) ~ ~ ~ ~ \displaystyle X(x,y,t) = x + r(y) \cos(kx-\omega t)$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (3) ~ ~ ~ ~ \displaystyle Y(x,y,t) = y + r(y) \sin(kx-\omega t)$$

In this notations, at $$t=0$$, $$x=0$$, the vertical speed is maximal. Perhaps, this is not best choice; it may have sense to have the crest at $$x=0$$.

Notations:

$$c=\cos(kx-\omega t)$$
$$s=\sin(kx-\omega t)$$
$$r=r(y)$$ ; $$~ r'=r'(y)$$

Let $$t=$$const. Then

$$\!\!\!\!\!\!\!\!\!\!\!\! (4) ~ ~ ~ ~ \displaystyle X=X(x,y,t) = x + r c$$
$$\!\!\!\!\!\!\!\!\!\!\!\! (5) ~ ~ ~ ~ \displaystyle Y=Y(x,y,t) = y + r s$$
$$\mathrm d X= (1-rsk) ~\mathrm d x + r' c ~ \mathrm d y$$
$$\mathrm d Y= rck ~ \mathrm d x + (1+r's) ~ \mathrm d y$$
$$(1+r's) ~\mathrm d X - r' c ~ \mathrm d Y = \Big( (1-rsk)(1+r's) - rr' c^2 k \Big)~ \mathrm d x$$
$$-rck ~ \mathrm d X + (1-rsk)~ \mathrm d Y = \Big( - r'rc^2 k (1-rsk)(1+r's) \Big)~ \mathrm d y$$

Let $$D=(1-rsk)(1+r'c)-r r- k c^2$$

Then $$D=1+ (r'-kr) s -k r r'$$

$$\frac{\partial x}{\partial X}=\frac{1}{D}(1+r's) ~ ~$$, $$~ ~ \frac{\partial x}{\partial Y}= \frac{-1}{D} r'c$$
$$\frac{\partial y}{\partial X}= \frac{-1}{D}r'ck ~ ~$$, $$~ ~\frac{\partial x}{\partial Y}= \frac{1}{D}(1-rck)$$

Then

$$\nabla \vec v= \frac{\partial v_x}{\partial x} \frac{ \partial x}{\partial X}+ \frac{\partial v_x}{\partial y} \frac{ \partial y}{\partial X}+ \frac{\partial v_y}{\partial x} \frac{ \partial x}{\partial Y}+ \frac{\partial v_y}{\partial y} \frac{ \partial y}{\partial Y}=$$ $$(kr'-r)\omega c$$

Hence, $$r=r_0 \exp(kr)$$

$$V(X,Y,t)=v(x,y,t)$$

$$p=\frac{g}{\rho}(h-y)$$

$$P=\rho \mu$$

$$\mu=-gy+r_0 \omega^2(\exp(ky)-1)$$

$$\omega^2=gk$$

References

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http://www.sciencedirect.com/science/article/pii/S0021999105002196 Dmitry Chalikova, Dmitry Sheinin. Modeling extreme waves based on equations of potential flow with a free surface. Journal of Computational Physics, Volume 210, Issue 1, 20 November 2005, Pages 247–273.

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