Subsections


24.2 A parabolic example

This example illustrates Laplace transform solution for a parabolic partial differential equation.


24.2.1 The physical problem

Find the flow velocity in a viscous fluid being dragged along by an accelerating plate.

Figure 24.1: Viscous flow next to a moving plate
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24.2.2 The mathematical problem

Figure 24.2: Viscous flow next to a moving plate
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Try a Laplace transform in $t$.


24.2.3 Transform the problem

Transform the partial differential equation:

\begin{displaymath}
\strut u_t = \kappa u_{xx}
\quad \stackrel{\hbox{Table 6...
... - \overline{\smash{u(x,0)}\vphantom{.}} = \kappa \hat u_{xx}
\end{displaymath}

Transform the boundary condition:

\begin{displaymath}
u_x = g(t)
\quad
\Relbar\joinrel\Relbar\joinrel\Relbar...
...\Relbar\joinrel\Longrightarrow
\quad
\hat u_x = \hat g(s)
\end{displaymath}


24.2.4 Solve the transformed problem

Solve the partial differential equation:

\begin{displaymath}
s \hat u = \kappa \hat u_{xx}
\end{displaymath}

This is a constant coefficient ordinary differential equation in $x$, with $s$ simply a parameter. Solve from the characteristic equation:

\begin{displaymath}
s = \kappa k^2 \quad\quad\Rightarrow\quad\quad k = \pm \sqrt{s/\kappa}
\end{displaymath}


\begin{displaymath}
\hat u = A e^{\sqrt{s/\kappa}  x} + B e^{-\sqrt{s/\kappa}  x}
\end{displaymath}

Apply the boundary condition at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\infty$ that $u$ must be regular there:

\begin{displaymath}
A=0
\end{displaymath}

Apply the given boundary condition at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0:

\begin{displaymath}
\hat u_x = \hat g(s) \quad\quad\Rightarrow\quad\quad - B \sqrt{\frac{s}{\kappa}} = \hat g
\end{displaymath}

Solving for $B$ and plugging it into the solution of the ordinary differential equation, $\hat u$ has been found:

\begin{displaymath}
\hat u = - \sqrt{\frac{\kappa}{s}} e^{-\sqrt{s/\kappa}  x} \hat g
\end{displaymath}


24.2.5 Transform back

We need to find the original function $u$ corresponding to the transformed

\begin{displaymath}
\hat u = - \sqrt{\frac{\kappa}{s}} e^{-\sqrt{s/\kappa}  x} \hat g
\end{displaymath}

We do not really know what $\hat g$ is, just that it transforms back to $g$. However, we can find the other part of $\hat u$ in the tables.

\begin{displaymath}
- \sqrt{\frac{\kappa}{s}} e^{-\sqrt{s/\kappa}  x}
\quad...
...w}
\quad
- \sqrt{\frac{\kappa}{\pi t}} e^{-x^2/4\kappa t}
\end{displaymath}

How does $\hat g$ times this function transform back? The product of two functions, say $\hat f(s)\hat g(s)$, does not transform back to $f(t)g(t)$. The convolution theorem Table 6.3 # 7 is needed:

\begin{displaymath}
u(x,t) = - \int_0^t
\sqrt{\frac{\kappa}{\pi (t-\tau)}}
e^{-x^2/4\kappa(t-\tau)} g(\tau) { \rm d}\tau
\end{displaymath}