Math 6410 • ODE

Consider the problem of a thin layer of paint of thickness \(h(x,t)\) and speed
\(u(x,y,t)\) flowing down a wall, see Fig. 1. The paint is assumed to be uniform in
the \(z\)-direction. The balance between gravity and viscosity (fluid friction) means
that the velocity satisfies the equation
\[
  \frac{\partial^2u}{\partial y^2}=-c,
\]
where \(c\) is a positive constant. This is supplemented by the boundary conditions
\[
  u(x,0,t)=0,\qquad
  \left.\frac{\partial u(x,y,t)}{\partial y}\right|_{y=h}=0.
\]
The density of paint per unit length in the \(x\)-direction is \(\rho_0h(x,t)\) where
\(\rho_0\) is a constant, and the corresponding flux is
\[
  q(x,t)=\rho_0\int_0^h u(x,y,t)\dd y.
\]

\begin{center}
\begin{tikzpicture}[scale=0.9,>=Latex,line cap=round,line join=round]
  \draw[->] (0,0) -- (0,-3.2) node[below] {\(x\)};
  \draw[->] (0,0) -- (2.25,0) node[right] {\(y\)};
  \draw[thick] (0,0) -- (0,-3.0);
  \node[left] at (-0.2,-1.65) {wall};
  \draw[thick] (1.45,0) .. controls (1.45,-0.7) and (1.8,-1.25) .. (1.68,-1.65)
    .. controls (1.55,-2.05) and (1.35,-2.45) .. (1.35,-3.0);
  \node[right,align=left] at (1.75,-0.95) {surface of\\paint layer};
  \draw[->] (0,-1.15) -- (1.55,-1.15);
  \node[below] at (0.8,-1.15) {\(h(x,t)\)};
  \draw (0,-2.05) -- (1.58,-2.05);
  \draw (1.58,-2.05) -- (1.35,-2.45);
  \draw[->] (0.40,-2.05) -- (0.40,-2.65);
  \draw[->] (0.85,-2.05) -- (0.85,-2.75);
  \draw[->] (1.25,-2.05) -- (1.25,-2.78);
  \node at (0.8,-2.95) {\(u(x,y,t)\)};
  \node at (1.05,-3.65) {Fig. 1.};
\end{tikzpicture}
\end{center}

\begin{enumerate}
\item[(a)] Using conservation of paint, and solving for \(u(x,y,t)\) in terms of
\(h(x,t)\) and \(y\), derive the following PDE for the thickness \(h\):
\[
  \frac{\partial h}{\partial t}+ch^2\frac{\partial h}{\partial x}=0.
\]
\item[(b)] Set \(c=1\). Show that the characteristics are straight lines and that the
Rankine-Hugoniot condition on a shock \(x=S(t)\) is
\[
  \frac{dS}{dt}=\frac{[h^3/3]_-^+}{[h]_-^+}.
\]
\item[(c)] A stripe of paint is applied at \(t=0\) so that
\[
  h(x,0)=
  \begin{cases}
    0, & x<0\text{ or }x>1,\\
    1, & 0<x<1.
  \end{cases}
\]
Show that, for small enough \(t\),
\[
  h=
  \begin{cases}
    0, & x<0,\\
    (x/t)^{1/2}, & 0<x<t,\\
    1, & t<x<S(t),\\
    0, & S(t)<x,
  \end{cases}
\]
where the shock is \(x=S(t)=1+t/3\).
\item[(d)] Explain why this solution changes at \(t=3/2\), and show that thereafter
\[
  \frac{dS}{dt}=\frac{S}{3t}.
\]
\end{enumerate}