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, $r=r_0 \exp(kr)$
$V(X,Y,t)=v(x,y,t)$
$p=\frac{g}{\rho}(h-y)$
$P=\rho \mu$
$\mu=-gy+r_0 \omega^2(\exp(ky)-1)$
$\omega^2=gk$
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