Example 3.5.1.
Consider the first order PDE
\begin{equation*}
y_t = - \alpha y_x, \qquad \text{for } x > 0, \enspace t > 0,
\end{equation*}
with side conditions
\begin{equation*}
y(0,t) = C, \qquad y(x,0) = 0 .
\end{equation*}
This equation is called the convection equation or sometimes the transport equation, and it already made an appearance in Section 1.9, with different conditions. See Figure 3.5 for a diagram of the setup.
A physical setup of this equation is a river of solid goo, as we do not want anything to diffuse. The function \(y\) is the concentration of some toxic substance 2 . The variable \(x\) denotes position where \(x=0\) is the location of a factory spewing the toxic substance into the river. The toxic substance flows into the river so that at \(x=0\) the concentration is always \(C\text{.}\) We wish to see what happens past the factory, that is at \(x > 0\text{.}\) Let \(t\) be the time, and assume the factory started operations at \(t=0\text{,}\) so that at \(t=0\) the river is just pure goo.
Consider a function of two variables \(y(x,t)\text{.}\) Let us fix \(x\) and transform the \(t\) variable. For convenience, we treat the transformed \(s\) variable as a parameter, since there are no derivatives in \(s\text{.}\) That is, we write \(Y(x)\) for the transformed function, and treat it as a function of \(x\text{,}\) leaving \(s\) as a parameter.
\begin{equation*}
Y(x)
= {\mathcal L} \bigl\{ y(x,t) \bigr\}
= \int_0^\infty y(x,t) e^{-st} \,ds .
\end{equation*}
The transform of a derivative with respect to \(x\) is just differentiating the transformed function:
\begin{equation*}
{\mathcal L} \bigl\{ y_x(x,t) \bigr\} =
\int_0^\infty y_x(x,t) e^{-st} \,ds
=
\frac{d}{dx} \left[\int_0^\infty y(x,t) e^{-st} \,ds \right]
=
Y'(x) .
\end{equation*}
To transform the derivative in \(t\) (the variable being transformed), we use the rules from Section 3.2:
\begin{equation*}
{\mathcal L} \bigl\{ y_t(x,t) \bigr\}
=
sY(x) - y(x,0) .
\end{equation*}
In our specific case, \(y(x,0)=0\text{,}\) and so \({\mathcal L} \bigl\{ y_t(x,t) \bigr\} = sY(x)\text{.}\) We transform the equation to find
\begin{equation*}
sY(x) = -\alpha Y'(x) .
\end{equation*}
This ODE needs an initial condition. The initial condition is the other side condition of the PDE, the one that depends on \(x\text{.}\) Everything is transformed, so we must also transform this condition
\begin{equation*}
Y(0) =
{\mathcal L} \bigl\{ y(0,t) \bigr\}
=
{\mathcal L} \bigl\{ C \bigr\}
=
\frac{C}{s} .
\end{equation*}
We solve the ODE problem \(sY(x) = -\alpha Y'(x)\text{,}\) \(Y(0) = \frac{C}{s}\text{,}\) to find
\begin{equation*}
Y(x) = \frac{C}{s} e^{-\frac{s}{\alpha} x} .
\end{equation*}
We are not done, we have \(Y(x)\text{,}\) but we really want \(y(x,t)\text{.}\) We transform the \(s\) variable back to \(t\text{.}\) Let
\begin{equation*}
u(t) = \begin{cases} 0 & \text{if $t < 0$}, \\ 1 & \text{otherwise}
\end{cases}
\end{equation*}
be the Heaviside function. As
\begin{equation*}
{\mathcal L} \bigl\{ u(t-a) \bigr\} =
\int_0^\infty u(t-a) \, e^{-st} \,dt
=
\int_a^\infty e^{-st} \,dt
=
\frac{e^{-as}}{s} ,
\end{equation*}
then
\begin{equation*}
y(x,t) = {\mathcal L}^{-1} \left\{
\frac{C}{s} e^{-\frac{s}{\alpha} x}
\right\}
=
C u\bigl(t-\nicefrac{x}{\alpha}\bigr) .
\end{equation*}
In other words,
\begin{equation*}
y(x,t) =
\begin{cases}
0 & \text{if $t < \nicefrac{x}{\alpha}$}, \\
C & \text{otherwise.}
\end{cases}
\end{equation*}
See Figure 3.6 for a diagram of this solution. The line of slope \(\nicefrac{1}{\alpha}\) indicates the wavefront of the toxic substance in the picture as it is leaving the factory. What the equation does is simply move the initial condition to the right at speed \(\alpha\text{.}\)
Shhh... \(y\) is not differentiable, it is not even continuous (nobody ever seems to notice). How could we plug something that’s not differentiable into the equation? Well, just think of a differentiable function very very close to \(y\text{.}\) Or, if you recognize the derivative of the Heaviside function as the delta function, then all is well too:
\begin{equation*}
y_t (x,t) = \frac{\partial}{\partial t} \left[
C u\bigl(t-\nicefrac{x}{\alpha}\bigr)
\right]
=
C u'\bigl(t-\nicefrac{x}{\alpha}\bigr)
=
C \delta\bigl(t-\nicefrac{x}{\alpha}\bigr)
\end{equation*}
and
\begin{equation*}
y_x (x,t) = \frac{\partial}{\partial x} \left[
C u\bigl(t-\nicefrac{x}{\alpha}\bigr)
\right]
=
- \frac{C}{\alpha} u'\bigl(t-\nicefrac{x}{\alpha}\bigr)
=
- \frac{C}{\alpha} \delta\bigl(t-\nicefrac{x}{\alpha}\bigr) .
\end{equation*}
So \(y_t = - \alpha y_x\text{.}\)
