\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 79, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/79\hfil Degeneracy in the Blasius problem] {Degeneracy in the Blasius problem} \author[F. Ahmad\hfil EJDE-2007/79\hfilneg] {Faiz Ahmad} \address{Faiz Ahmad \newline Department of Mathematics, Faculty of Science, King Abdulaziz University \\ P. O. Box 80203, Jeddah 21589, Saudi Arabia} \email{faizmath@hotmail.com \quad Phone:+966 2 2523205} \thanks{Submitted June 27, 2006. Published May 25, 2007.} \subjclass[2000]{34A12, 34A34, 47J06} \keywords{Navier-Stokes equations; Blasius problem; degeneracy; \hfill\break\indent Wang equation; well-posed problem; ill-posed problem} \begin{abstract} The Navier-Stokes equations for the boundary layer are transformed, by a similarity transformation, into the ordinary Blasius differential equation which, together with appropriate boundary conditions constitutes the Blasius problem, \begin{equation*} f'''(\eta )+\frac{1}{2}f(\eta )f''(\eta)=0,\quad f(0)=0,\; f'(0)=0,\;f'(\infty )=1. \end{equation*} The well-posedness of the Navier-Stokes equations is an open problem. We solve this problem, in the case of constant flow in a boundary layer, by showing that the Blasius problem is ill-posed. If the second condition is replaced by $f'(0)=-\lambda$, then degeneracy occurs for $0<\lambda <\lambda _{c}\simeq 0.354$. We investigate the problem analytically to explain this phenomenon. We derive a simple equation $g(\alpha ,\lambda )=0$, whose roots, for a fixed $\lambda$, determine the solutions of the problem. It is found that the equation has exactly two roots for $0<\lambda <\lambda _{c}$ and no root beyond this point. Since an arbitrarily small perturbation of the boundary condition gives rise to an additional solution, which can be markedly different from the unperturbed solution, the Blasius problem is ill-posed. \end{abstract} \maketitle \numberwithin{equation}{section} \section{Introduction} A boundary value problem is said to be \textit{well-posed} if a solution to the problem exists, this solution is unique and depends continuously on the boundary conditions. Otherwise the problem is called \textit{ill-posed}. A boundary value problem, describing a physical situation, must be well-posed because the conditions at a boundary are only satisfied approximately also various approximations are involved in the derivation of the equations governing a problem. Well-posedness of a problem governed by the Poisson equation with Dirichlet boundary conditions in a bounded domain follows easily by an application of the maximum principle for harmonic functions. However the above problem may be ill-posed on an unbounded domain. A famous open problem of applied mathematics is the well-posedness or otherwise of the Navier-Stokes equations, the fundamental equations of fluid mechanics . In this paper we shall investigate the well-posedness of the Blasius problem which deals with the Navier-Stokes equations specialized to the fluid flow in a boundary layer \cite{b2,s1}. The two dimensional constant viscous flow over a semi-infinite flat plate is modeled by the Blasius problem $$f'''(\eta )+\beta _{0}f(\eta )f''(\eta)=0,\\ f(0)=0,\quad f'(0)=0,\quad f'(\infty )=1,\quad \beta _{0}>0,\quad \eta \in [0,\infty ) \label{e1.1}$$ Here $\eta$ represents a \textit{similarity variable} introduced by Blasius \cite{b2} to transform a pair of partial differential Navier-Stokes equations into a single ordinary differential equation contained in \eqref{e1.1}. For a thorough discussion of this problem see Schlichting and Gersten \cite{s1}. If the third condition in \eqref{e1.1} is replaced by $f''(0)=\alpha$, where $\alpha >0$, then the initial-value problem $$f'''(\eta )+\beta _{0}f(\eta )f''(\eta)=0,\\ f(0)=0,\quad f'(0)=0,\quad f^{''}(0)=\alpha , \quad \beta_{0}>0,\quad \eta \in [0,\infty ) \label{e1.2}$$ will be called the Blasius initial value problem. It is well-known that any solution of \eqref{e1.2} corresponding to a fixed $\alpha$ has the property that as $\eta \rightarrow \infty$, the first derivative of $f(\eta )$ approaches a constant limit. There is a unique value of $\alpha$, say $\alpha _{0}$, for which the solution of the initial-value problem \eqref{e1.2} is also the solution of the Blasius problem \eqref{e1.1}. Several authors have devised numerical algorithms to find a good approximate value of this number, see Asaithambi \cite{a4} and references therein. For $\beta _{0}=\frac{1}{2}$ it is found that $$\alpha _{0}=0.33206 \label{e1.3}$$ Recently Wang \cite{w1} developed an analytical method for finding $\alpha_{0}$. Fang et al. \cite{f1} showed that for arbitrary $\beta _{0}$, $$\alpha _{0}=0.469600\sqrt{\beta _{0}}. \label{e1.4}$$ In view of the above result, it is sufficient to consider the Blasius problem for a fixed $\beta _{0}$, say $\beta _{0}=\frac{1}{2}$. We shall do this in this paper. Let $\lambda \geq 0$, then the boundary-value problem $$f'''(\eta )+\frac{1}{2}f(\eta )f''(\eta)=0,\\ f(0)=0,\quad f'(0)=-\lambda ,\quad f'(\infty )=1,\quad \eta \in [0,\infty ) \label{e1.5}$$ will be called the modified Blasius problem and the related problem $$f'''(\eta )+\frac{1}{2}f(\eta )f''(\eta)=0,\\ f(0)=0,\quad f'(0)=-\lambda ,\quad f^{''}(0)=\alpha, \quad \eta \in [0,\infty ) \label{e1.6}$$ will be named the modified Blasius initial value problem. Physically problem \eqref{e1.5} now models a boundary layer flow over a moving plate with constant velocity $\lambda$. Every solution of the problem \eqref{e1.2} increases from $0$\ to $\infty$ remaining convex throughout until $f'(\eta )$ approaches a constant limit. On the other hand a solution of the problem \eqref{e1.6} first decreases, attains a minimum and then increases to infinity. A striking difference between the two problems is that whereas the Blasius problem % \eqref{e1.1} possesses a unique solution, the modified Blasius problem % \eqref{e1.5} possesses two solutions when $$0<\lambda <\lambda _{c}\simeq 0.354, \label{e1.7}$$ these solutions coalesce into a unique solution for $\lambda =\lambda _{c}$ and there is no solution beyond this critical value \cite{a1,a2}. Thus the modified Blasius problem becomes degenerate as soon as $\lambda >0$. The evidence in favor of this remarkable phenomenon is based on numerical results and it is not clear why it should appear at $\lambda =0$ and then disappear at $\lambda =\lambda _{c}$. In this paper we shall provide a theory for this. In Section 2 we shall present a qualitative theory of problem \eqref{e1.6} which will be used in Section 4 to derive a simple equation of the form $$g(\lambda ,\alpha )=0. \label{e1.8}$$ For any fixed $\lambda \in (0,\lambda _{c})$, this equation has exactly two solutions $\alpha _{1}$\ and $\alpha _{2}$ both of them lying in a finite interval $(0,\alpha _{0})$. When $\lambda >\lambda _{c}$, Equation \eqref{e1.8} fails to have any real solution. The occurrence of a second solution as soon as $\lambda >0$, makes the Blasius problem ill-posed. \section{Qualitative theory of the modified Blasius problem} We consider the initial value problem \eqref{e1.6} where $\alpha >0$ and $% \lambda>0$. Anticipating the result $f''(\eta )>0$\ for $% 0\leq \eta <\infty$, we write the differential equation in the form $$\frac{f'''(\eta )}{f''(\eta )}=-\frac{1}{% 2}f(\eta ). \label{e2.1}$$ An integration from $0$ to $\eta$ gives $$f''(\eta )=\alpha \exp \big(-\frac{1}{2}\int_{0}^{\eta }f(u)du% \big). \label{e2.2}$$ It is clear that $$f''(\eta )>0,\quad 0\leq \eta <\infty , \label{e2.3}$$ and we have justified the step leading to \eqref{e2.1}. Since $f(0)=0$ and $f'(0)<0$, a solution of \eqref{e2.1} is negative on some interval. Suppose $f(\eta )\leq 0$ on $(0,\infty )$. Then \eqref{e2.2} gives \begin{equation*} f''(\eta )>\alpha ,\quad 0\leq \eta <\infty . \end{equation*} Integration of the above result twice on $[0,\eta ]$ gives \begin{equation*} f(\eta )>\alpha \frac{\eta ^{2}}{2}-\lambda \eta . \end{equation*} Since the right hand side of the above inequality is positive for sufficiently large $\eta$\ we have a contradiction of the assumption that $f(\eta )\leq 0$\ on $[0,\infty )$. Hence there is a point $\eta _{1}$ such that $f$ changes sign at this point. Thus \begin{gather} f(\eta )<0,\;0<\eta <\eta _{1}, \label{e2.4} \\ f(\eta _{1})=0,\;f'(\eta _{1})>0. \label{e2.5} \end{gather} Due to \eqref{e2.3}, $f'(\eta )$ increases on $(\eta _{1},\infty )$ and remains positive, therefore the function $f(\eta )$ also remains positive on this interval. On $[0,\eta _{1}]$, the continuous function $% f'(\eta )$ has changed sign hence, due to the well-known intermediate value theorem \cite{b1}, there is a point $\eta_{0},\;0<\eta_{0}<\eta _{1}$ where $$f'(\eta _{0})=0 \label{e2.6}$$ Using \eqref{e2.3}, \eqref{e2.4} and \eqref{e2.5} in the differential equation \eqref{e2.1}, we obtain \begin{gather} f'''(\eta )>0,\quad 0<\eta <\eta _{1}, \label{e2.7} \\ f'''(\eta _{1})=0. \label{e2.8} \end{gather} Thus we have proved the following result: \begin{quote} Every solution of problem \eqref{e1.6} has a unique minimum at a point $\eta _{0}$ and a unique zero $\eta _{1}$. On $[0,\eta_{1}]$ the second derivative of the solution increases and has a maximum at $\eta _{1}$. \end{quote} For a typical solution, the graphs of $f(\eta )$ and $f''(\eta )$ are shown in Fig.1. This Figure depicts a solution of the problem \eqref{e1.6} for $\lambda =0.2$ and $\alpha =0.016$. \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(109,75)(-5,-35) \put(0,-30){\line(1,0){102}} \put(0,0){\line(1,0){100}} \put(0,-30){\line(0,1){70}} \qbezier(0,0)(65,-70)(100,40) \qbezier(0,0)(50,2)(65,6) \qbezier(65,6)(82,12)(100,5) \multiput(0,-30.5)(17,0){7}{\line(0,1){1}} \multiput(-.5,-17)(0,17){4}{\line(1,0){1}} {\scriptsize \put(-1,-33.5){0} \put(16,-33.5){2} \put(33,-33.5){4} \put(50,-33.5){6} \put(67,-33.5){8} \put(83,-33.5){10} \put(100,-33.5){12} \put(-8,-17.5){$-0.5$} \put(-4,-.8){0} \put(-6,16.2){0.5} \put(-4,33){1}} \put(20,-15){$f(\eta)$} \put(65,10){$f''(\eta)$} \end{picture} \end{center} \caption{Graphs of $f(\protect\eta)$ and $f''(\eta )$, the solution of \eqref{e1.6} with $\lambda =0.2$} \end{figure} \subsection*{Asymptotic behavior} There exist positive constants $K$ and $c$ such that $$f(\eta )>c>0,\quad \eta \geq K \label{e2.9}$$ Using this in \eqref{e2.2}, we find \begin{equation*} f''(\eta )<\alpha \exp \Big( -\frac{1}{2}\Big[ \int_{0}^{K}f(\eta )d\eta +\int_{K}^{\eta }c\,d\eta \Big] \Big) ,\quad \eta \geq K \end{equation*} or $$f''(\eta )<\alpha F\exp \big( -\frac{1}{2}c(\eta -K)\big),\quad \eta \geq K, \label{e2.10}$$ where we have set $F=\exp [-\frac{1}{2}\int_{0}^{K}f(\eta )d\eta ]$. From % \eqref{e2.3} and \eqref{e2.10}, we get $$\lim_{\eta \to \infty }f''(\eta )=0 \label{e2.11}$$ Integrating \eqref{e2.10} on $[K,\eta )$. We get f'(\eta )0. \label{e2.13} There exists a constant $L$\ such that when $\eta \geq L$, $f'(\eta )>\frac{l}{2}$. An integration from $L$ to $\eta$ gives \begin{equation*} f(\eta )>f(L)+\frac{l}{2}(\eta -L),\quad \eta \geq L. \end{equation*} It follows that $$\lim_{\eta \to \infty }f(\eta )=\infty \label{e2.14}$$ Thus solutions of \eqref{e1.2} and \eqref{e1.6} asymptotically behave in a similar manner. \section{Wang's transformation} Wang \cite{w1} used the change of variables $$x=f'(\eta ),\quad y=f''(\eta ) \label{e3.1}$$ to transform the Blasius equation $f'''(\eta )+\frac{1}{2}f(\eta )f''(\eta )=0$ into the second order Wang's equation $$yy''+\frac{1}{2}x=0. \label{e3.2}$$ \newline If we make another change of variable $z=x+\lambda$ then \eqref{e3.2} will become $$yy''+\frac{1}{2}(z-\lambda )=0, \label{e3.3}$$ where differentiation now is with respect to $z$. The initial conditions are $$y(0)=\alpha ,\quad y'(0)=0. \label{e3.4}$$ The above problem can be solved by the Adomian decomposition method \cite{w1} or by a technique used by Ahmad \cite{a3}. The solution up to the term $z^{8}$ is found to be \label{e3.5} \begin{aligned} y(z) &= \alpha +\frac{\lambda }{4\alpha }z^{2}-\frac{1}{12\alpha }z^{3}- \frac{\lambda ^{2}}{96\alpha ^{3}}z^{4}+\frac{\lambda }{120\alpha ^{3}}z^{5}+ \frac{7\lambda ^{3}-8\alpha ^{2}}{5760\alpha ^{5}}z^{6} \\ &\quad -\frac{59\lambda ^{2}}{40320\alpha ^{5}}z^{7}-\frac{127\lambda ^{4}-336\lambda \alpha ^{2}}{645120\alpha ^{7}}z^{8}+\dots \end{aligned} Substitute $z=2\lambda$ in \eqref{e3.5}. This corresponds to $x=\lambda$. Keeping terms only up to $\lambda ^{6}$ we find $$y(2\lambda )=\alpha +\frac{\lambda ^{3}}{3\alpha }+\frac{\lambda ^{6}}{ 90\alpha ^{3}} \label{e3.6}$$ Concerning a solution of \eqref{e1.6}, Equation \eqref{e3.6} indicates that there is some point $\eta =\eta _{2}$ such that $$f'(\eta _{2})=\lambda ,\quad f''(\eta _{2})=\alpha + \frac{\lambda ^{3}}{3\alpha }+\frac{\lambda ^{6}}{90\alpha ^{3}} \label{e3.7}$$ We shall utilize \eqref{e3.7} in the next section to prove the existence of degeneracy in the Blasius problem. \section{Degeneracy} Consider a solution of the problem \eqref{e1.6} which also happens to be a solution of the modified Blasius problem \eqref{e1.5}. Also let $\eta \geq \eta _{2}$. Keeping in view \eqref{e2.13}, we approximate the function $f(\eta )$ between the points $\eta _{2}$ and $\eta$ by the equation of the line segment joining them and we take the slope of this segment as the average of the slope at the point $\eta _{2}$, which is $\lambda$, and the slope at $\eta$ which we take as unity. The larger $\eta _{2}$ is, the better this approximation becomes. The area under the curve representing $f(\eta )$ from the point $\eta _{2}$ to $\eta$ is approximately $\frac{1}{4}(1+\lambda )(\eta -\eta _{2})^{2}$. Using this and \eqref{e3.7} in \eqref{e2.2}, we find $$f''(\eta )=(\alpha +\frac{\lambda ^{3}}{3\alpha }+\frac{ \lambda ^{6}}{90\alpha ^{3}})\exp \big( -\frac{1}{8}(1+\lambda )(\eta -\eta _{2})^{2}\big) ,\quad \eta \geq \eta _{2} \label{e4.1}$$ Integrating \eqref{e4.1} from $\eta _{2}$ to $\infty$, we get \begin{equation*} f'(\infty )-f'(\eta _{2})=\sqrt{\frac{2\pi }{1+\lambda }} \Big( \alpha +\frac{\lambda ^{3}}{3\alpha }+\frac{\lambda ^{6}}{90\alpha ^{3} }\Big) \end{equation*} Putting values of $f'(\infty )$ and $f'(\eta _{2})$ and dividing with the term under the square-root sign, we obtain $$\frac{(1-\lambda )\sqrt{1+\lambda }}{\sqrt{2\pi }}=\alpha +\frac{\lambda ^{3} }{3\alpha }+\frac{\lambda ^{6}}{90\alpha ^{3}}\; \label{e4.2}$$ For a fixed $\lambda$, the number of real positive roots determines the number of solutions possessed by the modified Blasius problem under consideration. The graph of the function $$g(\alpha )=\alpha +\frac{\lambda ^{3}}{3\alpha }+\frac{\lambda ^{6}}{ 90\alpha ^{3}} \label{e4.3}$$ for a fixed $\lambda$\ is shown in Figure 2. The minimum value of $g(\alpha)$ is $1.2032\lambda ^{3/2}$ and this occurs when $\alpha=0.6433\lambda ^{3/2}$. It is clear that exactly two roots of \eqref{e4.2} will exist as long as $$\frac{(1-\lambda )\sqrt{1+\lambda }}{\sqrt{2\pi }}>0.6433\;\lambda ^{3/2} \label{e4.4}$$ This happens for $$0<\lambda <\lambda _{c}=0.386 \label{e4.5}$$ For $\lambda =\lambda _{c}$ the two solutions overlap and if $\lambda>\lambda _{c}$, Equation \eqref{e4.2} has no real root implying that no solution of the modified Blasius problem exists. Thus the problem becomes degenerate between $\lambda =0$ and $\lambda =\lambda _{c}$. \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(88,45)(-5,-5) \put(-5,0){\line(1,0){85}} \put(0,-5){\line(0,1){45}} \qbezier(1,40)(1,0)(5,2) \qbezier(5,2)(9,2)(80,40) \multiput(25,-.5)(25,0){3}{\line(0,1){1}} \multiput(-.5,12)(0,12){3}{\line(1,0){1}} {\scriptsize \put(23.2,-3){0.1} \put(48,-3){0.2} \put(73,-3){0.3} \put(-5,11){0.1} \put(-5,23){0.2} \put(-5,35){0.3}} \put(50,30){$g(\alpha)$} \put(81,-1){$\alpha$} \end{picture} \end{center} \caption{Graph of $g(\protect\alpha)$ defined by \eqref{e4.3} when $\lambda =0.1$} \end{figure} \section{Numerical results} In Table 1 we present two pairs of values, $\alpha _{1}$, $\alpha _{2}$ and $\alpha _{1}^{e}$, $\alpha _{2}^{e}$ of the parameter $\alpha$ each of one produces a solution of the problem \eqref{e1.6} which is also a solution of the modified Blasius problem \eqref{e1.5}. The numbers $\alpha _{1}$ and $\alpha_{2}$ have been found by solving \eqref{e4.2} while $\alpha _{1}^{e}$ and $\alpha _{2}^{e}$ refer to their exact values found by a numerical solution of \eqref{e1.6} for various $\alpha$ and searching for those values for which $f'(R)\simeq 1$ for sufficiently large $R$. \begin{table}[tbp] \begin{center} \begin{tabular}{|l|l|l|l|l|} \hline $\lambda$ & $\alpha _{1}$ & $\alpha _{2}$ & $\alpha _{1}^{e}$ & $\alpha _{2}^{e}$ \\ \hline 0.1 & 0.0034 & 0.376 & 0.00137 & 0.326 \\ \hline 0.15 & 0.0083 & 0.361 & 0.0061 & 0.317 \\ \hline 0.20 & 0.016 & 0.342 & 0.0157 & 0.304 \\ \hline 0.25 & 0.028 & 0.318 & 0.0321 & 0.284 \\ \hline 0.30 & 0.0468 & 0.287 & 0.0600 & 0.252 \\ \hline 0.34 & 0.0706 & 0.252 & 0.1034 & 0.206 \\ \hline 0.354 & 0.083 & 0.235 & 0.149 & 0.150 \\ \hline \end{tabular}% \end{center} \caption{Approximate and exact values of $\protect\alpha$ that produce solutions of the modified Blasius problem} \end{table} The approximate critical value $\lambda _{c}$ was found above as 0.386. Its exact value is 0.3546. Considering the nature of approximations involved in obtaining \eqref{e4.2}, agreement between approximate and exact results is remarkable. \subsection*{Discussion} A study of the qualitative behavior of the solutions of \eqref{e1.6} paved the way for a proper understanding of degeneracy in the Blasius problem. All pairs of values of $\alpha _{1}^{e}$ and $\alpha _{2}^{e}$ lie in the interval $(0,0.33206)$. In \cite{a1} an analytical solution of the problem was obtained by employing the \textit{homotopy analysis method} \cite{l1} and it was erroneously claimed that the above interval is $(0.33206, 0.52)$. Existence of multiple solutions of \eqref{e1.5} for $\lambda \in (0,\lambda _{c})$ demonstrates that the Blasius problem \eqref{e1.1} is ill-posed. A solution of a well-posed problem must depend \textit{continuously} on the initial or boundary conditions. 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