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2-4-apile.tex\\
{\bf Reference:} \\

 Landau \& Lifshitz, {\em Fluid Mechanics}, 1959.

 
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\section*{2.4.a. Release of a fluid pile from a horizontal plane}
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On a horizontal plane $\theta=0$. Eq (2.2.22) reduces to 
\be \f{\p h}{\p t} =\alpha\f{\p}{\p x}\lp h^3\f{\p h}{\p x} \rp \label{NLheat}\ee
where  $\alpha=\rho g/3\mu$. 
This is a nonlinear  diffusion equation where the diffusivity   $ah^3$ increases with the unknown $h$. 

Consider the  release of a fluid  pile with  the initial volume $Q$ confined in a small   region. Note that
\[ \f{\p}{\p t}\int_{-\infty}^\infty h(x,t)\, dx = \alpha\lb h^3\f{\p h}{\p x}\rb_{-\infty}^\infty = 0\]
Thus \be \int_{-\infty}^\infty h(x,t)\, dx =\int_{-\infty}^\infty h(x,0)\, dx   =Q\label{initialQ}\ee
 i.e., the total mass is  constant. The initial-value problem governed by (\ref{NLheat}) 
 and (\ref{initialQ}) can be solved by the method of similarity, which reduces the PDE to an ODE.
 
 \subsection*{2.4.a.1. The similarity method}
 Consider the one-parameter transformation:
 \be h=\lambda^ah',~~x=\lambda^b x',~~~t=\lambda^c t'\label{affine}\ee where $a,b,c)$ will be chosen to leave the initial value problem ( PDE and initial condition)  unchanged  (invariant). Substituting (\ref{affine}) into (\ref{NLheat}), 
  \be \lambda^{a-c}\f{\p h'}{\p t'} =\lambda^{4a-2b}\alpha\f{\p}{\p x'}\lp {h'}^3\f{\p h'}{\p x'} \rp \label{NLheat'}\ee 
  and then in (\ref{initialQ}),
 \be  \lambda^{a+b}\int_{-\infty}^\infty h'(x,0)\, dx'   =Q\label{intitalQ}\ee
  we find the conditions for invariance, 
  \be a-c=4a-2b; ~~~a+b=0; ~~~i.e.,~~~ b=-a, ~~c=-5a; \ee
 implying also
  \be h=\lambda^ah',~~x=\lambda^{-a} x',~~~t=\lambda^{-5a} t'\label{affine2}\ee
 This suggests the followng combination of variables: 
  \be \xi=\f{x}{(At)^{1/5}}, ~~~f(\xi) = h(Bt)^{1/5} , \label{similarity variables} \ee
  $\xi$ is  called a {\it similarity } variable. In the space-time plane,   at all points  on the curve of constant $\xi=x/(Bt)^{1/5}$, $f  =h(Bt)^{1/5}$ is the same. Thus for a fixed $\xi$, $h(Bt)^{1/5}$ is fixed, implying that $h(Bt)^{1/5}$ is a fun$h(Bt)^{1/5}$ is ction of $\xi$, or 
  \be h=\f{f(\xi)}{(Bt)^{1/5}}\label{sim-soln}\ee 
  Physically the maximum of $h$ decays in time as $t^{-1/5}$, while the pile spreads in time as  $t^{1/5}$.
  
  Subsituting (\ref{sim-soln}) into (\ref{NLheat}), and using the facts,
  \[ \f{\p h}{\p t} =- \f{1}{5}\f{1}{B^{1/5}}\f{f}{t^{6/5}}-\f{1}{5}\f{\xi f'}{B^{1/5} t^{6/5}}~ ~~\f{\p h}{\p x}=\f{1}{B^{1/5} t^{1/5}}\f{f'}{A^{1/5} t^{1/5}}\]
  we find
    \[ -\f{1}{5}\f{1}{B^{1/5}t^{6/5}}
  \lp f+\xi \f{df}{d\xi}\rp
  =\f{\alpha}{B^{4/5}t^{4/5} A^{2/5}t^{2/5}}\f{d}{d \xi}\lp f^3\f{df}{ d \xi} \rp 
  \]
  From the initial condition, we get 
  \[\lp \f{At}{Bt}\rp^{1/5} \int_{-\infty}^\infty f(\xi)d\xi = Q\]For simpliicty let us set
  \be A^{2/5}B^{3/5} = \alpha  \label{condition1}\ee
  and 
   \be  \lp \f{A}{B}\rp^{1/5} = Q,  \label{condition2}\ee
   We then get an  ODE:
   \be  5\f{d}{d \xi}\lp f^3\f{df}{ d \xi} \rp +\f{d}{d\xi}\lp \xi f  \rp=0 \label{sim-ODE}
  \ee
  subject to  
  \be \int_{-\infty}^\infty f(\xi)d\xi = 1.\label{Sim_IC}\ee
  at $t\to 0$. Solving (\ref{condition1}) and (\ref{condition2}) we get 
  \be A=\alpha Q^3,~~~B=\f{\alpha}{Q^2}\ee
  
  \subsection*{2.4.a.2. Solution}
  Integrating (\ref{sim-ODE}), 
  \[ 5f^3\f{d f}{d\xi} +\xi f = 0\]
  The integration constant  is zero because of symemetry  at $x=0$ so that   $f'(0)=0$. 
  Now
  \[ 5f^2\f{df}{d\xi}+\xi=0, ~~\mbox{implying}~~\f{5}{3}df^3+\xi d\xi =0\]
  Integrating,
  \[ f^3=\f{3}{5}\f{\xi_0^2-\xi^2}{2}, \]
  where $f(\xi_0)=0$, we get
  \be f=\lb \f{3}{10}(\xi_0^2-\xi^2) \rb^{1/3}, ~~-\xi_0<\xi<\xi_0; ~~~=0, ~~ \mbox{othewise}.\ee 
  
  To find $\xi_0$ we use (\ref{Sim_IC}), 
  \[ \int_{-\xi_0}^{\xi_0} f(\xi)d\xi = 1. \]
  or \[ \int_{-\xi}^{\xi_0} \lb \xi_0^2-\xi^2) \rb^{1/3}d\xi =
  \lp \f{10}{3}\rp^{1/3}  \]
  This determines   $\xi_0$. The rest is algebra.
  
Let $\zeta=(\xi/\xi_0)^2$, then 
\[ d\zeta=2\xi d\xi/{\xi_0}^2, ~~~d\xi=\f{\xi_0}{2}\f{ d\zeta }{\zeta^{1/2}}\]
\[ \int_{-\xi}^{\xi_0} \lb \xi_0^2-\xi^2) \rb^{1/3}d\xi =\lp \xi_0 \rp^{5/3}\int_0^1(1-\zeta)^{1/3} \zeta^{-1/2}d\zeta=\lp \xi_0 \rp^{5/3}\f{2\sqrt{\pi}}{5}\f{\Gamma\lp\f{1}{3}\rp}{\Gamma\lp\f{5}{6}\rp} \]
where $\Gamma(z)$ is the  Gamma function, hence 
\be   \xi_0= \f{x_0}{(\alpha Q^3t)^{1/5}}=\lp \f{10}{3}\rp^{1/5}\lp   \f{5}{2\sqrt{\pi}}  \f{\Gamma\lp\f{5}{6}\rp}{\Gamma\lp\f{1}{3}\rp}\rp^{3/5} 
\ee 
The maximum width is 
\be x_0 = \lp \f{10}{3}\rp^{1/5}\lp   \f{5}{2\sqrt{\pi}}  \f{\Gamma\lp\f{5}{6}\rp}{\Gamma\lp\f{1}{3}\rp}\rp^{3/5}(\alpha Q^3t)^{1/5}\ee
The maximum depth is 
 \be h(0,t)=\f{f(0)}{(Bt)^{1/5}}=\lb\f{3\xi_0^2}{10}\rb^{1/3}\lp\f{Q^2}{\alpha t}\rp^{1/5}\ee
  
  \end{document}