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3-5Karman.tex
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\section{Karman's momentum integral approach}
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Ref: Schlichting: {\bf Boundary layer theory}\\
 
With a general pressure gradient the boundary layer equations can be solved   by a variety of modern numerical means. An alternative which can still be  employed to  simplify   calculations is the momentum integral method of Karman. We explain this method for a transient boundary layer along the $x$-axis forced by  an unsteady pressure gradient outside. This pressure gradient can be due to some unsteady and nonuniform flow such as waves or gust. 

\subsection{General ideas}
Let us limit  our discussoin to a two dimentional  flow.  Recall the continuity    \be u_x+w_z=0 \label{cont}\ee
and boundary layer equation:
\be 
\rho(  u_t + uu_x + wu_z  ) =   - p_x + \mu u_{zz} \label{x-momentum}\ee \be  0=   - p_z\label{z-momentum}\ee
The boundary conditions are 
\be u = w = 0, \qquad z = 0; ~~u \rightarrow U,~ w \rightarrow 0,  \qquad z  \rightarrow   \infty .
\ee
For general $U(x,t)$ the longitudinal pressure gradient is 
\be -p_x = \rho(U_t+UU_x)\ee
so that \be 
\rho(  u_t + uu_x + wu_z  ) =   \rho(U_t+UU_x) + \mu u_{zz} \label{x-momentuma}\ee 
Instead of solving the initial-boundary-value problem accurately we introduce a moment method by integrating the momentum equation across the boundary layer, then make a reasonable assumption on the velocity profile to get the longitudinal variation of the boundary layer  thickness. This is called the Karman's momentum integral  approximation in boundary  layer theory. 

 
Let us add the following two equations 
\[ 
-\mu u_{zz} = \rho\lcb U_t+UU_x -(u_t+uu_x+wu_z)
\rcb\]
and \[ 0=(U-u)u_x+(U-u)w_z\]
to get, \[ 
-\mu u_{zz} = \rho\lcb \f{\p}{\p t}(U-u) +\f{\p}{\p x}[u(U-u)]+(U-u)\f{\p U}{\p x}+\f{\p}{\p z}[(w(U-u)] 
\rcb\]
Now let us integrate across the boundary  layer, 
\[ \mu u_z|_0=\rho\lcb \f{\p }{\p t} \int^\infty_0 (U-u)dz + \f{\p  }{\p x} \int_0^\infty u(U-u)dz +\f{\p U}{\p x}\int_0^\infty (U-u)dz\rcb + \rho[w(U-u)]_0^\infty\]
 The boundary terms   again vanish. 
Finally, we have, 
\be \fbox{\parbox{13.5cm}{
\[ \f{\p}{\p t} \int^\infty_0 \rho (U-u)dz + \f{\p}{\p x} \int^\infty_0 \rho \lp Uu-u^2\rp  dz +\f{\p U}{\p x}\int^\infty_0(U-u)dz= \left. \mu \f{\p u}{\p z} \right|_0 \] 
} \label{Karman-mom}}\ee
This is  the K\'arm\'an momentum integral equation, representing the momentum balance across the thickness of the  boundary layer.  

\begin{figure} [h]
\begin{center}
\includegraphics[scale=1]{3-displBL.eps}\end{center}
\caption{Displacement thickness}. 
\label{fig:displBL}
\end{figure}
Karman also introduces the displacement thickness as a measure of  the lost volume  
\be U\delta_1 = \int^\infty_0 \lp U - u \rp dz, ~~\mbox{or}~~ \delta_1 = \int^\infty_0 \lp 1 - \f{u}{U} \rp dz, \ee 
and the momentum  thickness as a measure of the lost momentum 
 \be U^2\delta_2 = \int^\infty_0  u(U-u)dz, ~~\mbox{or}~~ \delta_2 = \int^\infty_0 \f{u}{U} \lp 1 - \f{u}{U} \rp dz, \ee
 both due to the slowing down of the fluid in the boundary layer. See Figure \ref{fig:displBL}.
 
In certain cases when the velocity profile in the boundary layer  can be reasonably  guessed in advance, the Karman momentum intergal equation can be the basis of obtaining an approximate solution. The procedure is to assume a reasonable  profile :
\be u=Uf\lp\f{y}{\delta}\rp\label{vel-profile} \ee
and substitute (\ref{vel-profile} ) into (\ref{Karman-mom}) equation. After evaluating the integrals a differential equation is obtained for the boundary layer thickness $\delta(x,t) $,  which can be more easily solved for certain initial and boundary conditions.   
%%%%%%%%%%%%%%%%%%%%%%%
 \subsection{Application to a transient boundary layer  along a flat  plate}
 
Consider the two dimensional boundary layer near edge of half infinite plane along the $x$ axis due to the  impulsive start along its own plane.   There is no motion anywhere for $t<0$. At   $t=0$ the plane suddenly advances from right to left perpendicular to its leading edge. What is the boundary layer flow?
 \begin{figure} [h]
\begin{center}\includegraphics[scale=1]{3-7-1.eps}
\end{center} 
\caption{Front of boundary layer  }. 
\label{fig:Blasius}
\end{figure}
 Referring to Figure \ref{fig:Blasius}
where  the coordinate system is fixed on the plane. For all $t>0$, 
     \[ \f{\p U}{\p t}= \f{\p U}{\p x} = 0 \]
The boundary  layer equation reads:
 \be 
\rho(  u_t + uu_x + wu_z  ) =    \mu u_{zz} \label{x-momentumb}\ee 
and the Karman momentum inetgral equation reads
\be   \f{\p}{\p t} \int^\infty_0 \rho (U-u)dz + \f{\p}{\p x} \int^\infty_0 \rho \lp Uu-u^2\rp  dz  = \left. \mu \f{\p u}{\p z} \right|_0   \label{Karman-mom0}\ee

Let us  assume \be \f{u}{U} = f (\eta),  \qquad \eta\equiv \f{z}{\delta (x,t)} \ee
then \be  \delta_1=\delta \int^\infty_0(1 - f (\eta)) d\eta = \alpha_1 \delta \ee
\be \delta_2 = \delta \int^\infty_0 f(\eta)(1-f(\eta)) d\eta = \alpha_2 \delta \ee
and \be u_z = \f{U}{\delta} f'(0). 
\ee  Substituting into Eq. (\ref{Karman-mom0}) 
\[ \alpha_1 U \f{\p \delta}{\p t} + \alpha_2 U^2 \f{\p \delta}{\p x} = \nu \f{U}{\delta} f'(0) \]
Therefore,
\be \lp \alpha_1 \rp \f{\p \lp \delta^2/2 \rp}{\p t} + \lp \alpha_2 U \rp \f{\p \lp \delta^2/2 \rp}{\p x} =  {\nu f'(0)}   \label{Eqn.(9)} \ee
which is a first order wave (hyperbolic) equation for $\delta^2/2$.  We must add the boundary and initial conditions: 
\be \delta( 0, t) = 0 , ~~~\forall t > 0 \label{boundary} \ee
\be \delta(x, 0) = 0,~~~\forall x > 0 \label{initial} \ee

\subsubsection{Heuristic solution}

The fact that the boundary value   in (\ref{boundary})  is independent of $t$ for all $t$ suggests that the solution is independent $t$ for sufficiently large $x$, i.e.,
\be \f{\p}{\p x}\lp\f{\delta^2}{2}\rp= \f{2\nu f'(0)}{\alpha_2 U}\ee 
subject to the boundary condition (\ref{boundary}). The solution is simply 
\be \delta=\sqrt{\f{2\nu f'(0)x}{\alpha_2U}}\label{Blasius-solution}\ee 
This is the approximate version of  the solution by Blasius who  solved the  steady boundary layer equation
\be uu_x+vu_y=\nu u_{yy} ~~~\mbox{for} ~~x>0, ~~y>0\label{Blasius problem}\ee
by the method of similarity.


On the other hand, the fact that the initial  value 
in (\ref{boundary}) is independent of $x$ for all $x$ suggests that the solution is independent of $x$ for sufficiently large $t$,  i.e., 
\be \f{\p}{\p t} \lp\f{\delta^2}{2}\rp= \f{2\nu f'(0)}{\alpha_1}\ee 
subject to the initial condition (\ref{initial}). The solution is simply 
\be \delta=\sqrt{\f{2\nu f'(0)t}{\alpha_1}}\label{Rayleigh-solution}\ee 
This is the approximate version of  the solution by Rayleigh's problem which is known to be governed by 
\be u_t=\nu u_{yy} ~~~\mbox{for} ~~t>0, ~~y>0\label{Rayleigh problem}\ee

To find the range of each solution in the $x\sim t$ plane, we equate the two $\delta$'s and get 
\be x=\f{\alpha_2Ut}{\alpha_1}\ee
Thus (\ref{Blasius-solution})   holds in the wedge $x> \f{\alpha_2Ut}{\alpha_1}>0$ and (\ref{Rayleigh-solution}) holds in the wedge   $0<x<\f{\alpha_2Ut}{\alpha_1}$. See Figure \ref{fig:BL-quart1}.

 
Physically:   if $t >  {\alpha_1x}/{\alpha_2 U} $, (\ref{Blasius-solution})  applies and one has the steady Blasius flow   past a semi-infinite plate; the boundary layer is already in the steady state. See Figure \ref{fig:BL-quart2},   If $t <  {\alpha_1 x}/{\alpha_2 U} $, (\ref{Rayleigh-solution}) applies and one has Rayleigh's problem of impulsively started plane    infinite in both $x<0$ and $x>0$. The boundary layer is still in the initial  stage and the effect of the leading edge is  not felt elsewhere.  The result  is summarized in Figure   \ref{fig:BL-quart3}.

 
To calculate $\alpha_1 $ and $\alpha_2$ let us make a special choice of the velocity profile so that $f(0) = 0, f'(1) = f''(1) = 0$  (due to Karman and Polhausen)
\be f(\eta) = 2\eta - 2\eta^3 + \eta^4, ~~ 0 < \eta < 1 \ee
Note that \be f(0)= 0 , ~~
f'(\eta) = 2 - 6 \eta^2 + 4 \eta^3,~~f''(\eta) = - 12 \eta + 12 \eta
\ee
then
\begin{eqnarray*}
\alpha_1 & = & \int^1_0 \lp 1 - 2\eta + 2\eta^3 - \eta^4 \rp d\eta = 3/10=0.3 \\
\alpha_2 & = & \int^1_0 \lp 2\eta - 2\eta^3 + \eta^4 \rp \lp 1 - 2\eta + 2\eta^3 - \eta^4 \rp d\eta = 37/315\cong 0.117 \end{eqnarray*}
and \[ f'(0)=2, ~~~\mbox{or}~~ \left. \f{\p u}{\p z} \right|_{z = 0}=\f{2U}{\delta} 
 \] 
Hence the displacement thickness of the Blasius boundary layer is 
\be \delta_1=\alpha_1\delta = \alpha_1\sqrt{\f{2\nu f'(0)x}{\alpha_2 U}}=1.754 \sqrt{\f{\nu x}{U}}\ee
 
 
\begin{figure} [h]
\begin{center}\includegraphics[scale=1]{3-7-2.eps}
\end{center} 
\caption{  Evolution of a transient boundary layer.  (a)   Blasius region and Rayleigh region} 
\label{fig:BL-quart1}
\end{figure}
\begin{figure} [h]
\begin{center}\includegraphics[scale=1]{3-7-21.eps}
\end{center} 
 \caption{ Evolution of a transient boundary layer. (b) Blasius region} 
\label{fig:BL-quart2}
\end{figure}
\begin{figure} [h]
\begin{center}\includegraphics[scale=1]{3-7-22.eps}
\end{center}
 \caption{ Evolution of a transient boundary layer. (c) Rayleigh  region} 
\label{fig:BL-quart3}
\end{figure}
The momentum integral method is the special case of the moment method, since the Karman  equation is the  zeroth moment of the boundary layer equation. For the classical steady boundary layer problem solved exactly by Blasius using the  similarity method, the momentum integral approximation gives fairly good results, even  with various crude profiles, see Table \ref{table:Karman}.
\begin{table}[h]\hskip 1in
\begin{tabular}{|c|c|}\hline\hline\
{Velocty profile}& {Displacment boundary layer thickness}\\
\hline 
$\f{u}{U}=f(\eta)$, \,\, $\eta=\f{y}{\delta(x)}$ & $\delta_1\sqrt{\f{U}{\nu x}}$ \\
\hline
$\eta$ & 1.732\\
\hline
$\sin \f{\pi \eta}{2}$ & 1.741\\
\hline
$
2\eta-2\
\eta^3+\eta^4$ & 1.754\\ \hline
Blausius & 1.721\\
\hline\hline
\end{tabular}
\caption{Comparson of approximate solution by momentum integral method with the exact solution of Blasius}
\label{table:Karman}
\end{table}
  
  
 

\subsubsection{Formal solution}
 A more formal solution can be obtained as follows:
Let 
\[ a = \alpha_1 \qquad b = \alpha_2 U \qquad c = \nu f'(0) \]
then 
\be  a \f{\p}{\p t}\lp \f{\delta^2}{2} \rp+ b  \f{\p}{\p x}\lp \f{\delta^2}{2} \rp= c \ee 
The solution can be facilitated by a change of coordinates from $(x,t)$ to $(\xi, \zeta)$  where
\be \xi = ax + bt \qquad \zeta = ax - bt \ee 
then 
\[ \f{\p}{\p t}= \f{\p}{\p \xi}\f{\p \xi}{\p t} + \f{\p}{\p \zeta}\f{\p \zeta}{\p t}; ~~
 \f{\p}{\p x}= \f{\p}{\p \xi}\f{\p \xi}{\p x} + \f{\p}{\p \zeta}\f{\p \zeta}{\p x}
 \] 
\begin{eqnarray*}
\lp \delta^2 \rp_t & = & \lp \delta^2 \rp_\xi b + \lp \delta^2 \rp_\zeta (-b) \\
\lp \delta^2 \rp_x & = & \lp \delta^2 \rp_\xi a + \lp \delta^2 \rp_\zeta a\end{eqnarray*}
Therefore, from (\ref{Eqn.(9)}), 
\[\f{1}{2}\lb  ab \lp \delta^2 \rp_\xi - ab \lp \delta^2 \rp_\zeta + ab \lp \delta^2 \rp_\xi+ ab \lp \delta^2 \rp_\zeta \rb = c\]
\[ \f{1}{2} \lp \delta^2 \rp_\xi= \f{c}{2ab} =   \f{\nu f'(0)}{2\alpha_1 \alpha_2 U }   \]
Integrate once
\be \f{\delta^2}{2} = \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} \xi+ G (\zeta) \ee
where $G$ is an arbitrary function of $ \zeta $. 

To determine $G(\zeta)$ we first use the initial condition that $\delta = 0$ at $t = 0$ \underline{for all $x > 0$}.
\[ 0 = \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} (\xi)_{t = 0} + G(\zeta)_{t = 0} \]
or
\[  0 = \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} (ax) + G(ax) ~~ \mbox{for} ~~ x>0. \]
Therefore
\be G(\zeta) = - \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} \zeta, \qquad \zeta > 0 \label{Eqn. (11)} \ee


What is $G(\zeta)$  for $\zeta < 0$? Let us use the boundary condition at $x=0$ that  $\delta = 0 $ for  all  $t>0$. 
 \[ 0 = \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} (\xi)_{x = 0} + G(\zeta)_{x = 0} \]
or
\[  0 =- \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} (-bt) + G(-bt),  \qquad t > 0 \]
Note that for $t > 0$, $-bt <0$.  Therefore,
\be G(\zeta) =  \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} \zeta \qquad \mbox{for all} \quad \zeta < 0 . \label{Eqn. (12)}\ee
Eqs. (\ref{Eqn. (11)}) and (\ref{Eqn. (12)}) complete the solution for all $\zeta > 0$ and $\zeta < 0$.

Let us return to $x$ and $t$
\begin{eqnarray*}
\f{\delta^2}{2} & = &  \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} \xi - 
 \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} \zeta = \f{\nu f'(0)}{2\alpha_1  \alpha_2 U} 2 bt \qquad \mbox{for} \quad ax - bt > 0 \\
& = &  \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} \xi + 
 \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} \zeta = \f{\nu f'(0)}{2\alpha_1 \alpha_2 U} 2ax \qquad \mbox{for} \quad ax - bt < 0
\end{eqnarray*}
Therefore,
\be \delta = \sqrt{\f{2\nu f'(0)}{\alpha_1} t} \qquad \mbox{for} \quad x > \f{\alpha_2 U}{\alpha_1} t \label{Eqn. (13.a)} \ee
and
\be \delta = \sqrt{\f{2\nu f'(0)}{\alpha_2 U} x} \qquad \mbox{for} \quad x < \f{\alpha_2 U}{\alpha_1} t \label{Eqn. (13.b)} \ee
These are the same as (\ref{Rayleigh-solution}) and (\ref{Blasius-solution}). 

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