4.7.1 Problem 2

Solve

\begin{equation} u_{t}+\left ( xt\right ) u_{x}=0 \tag {1}\end{equation}

\(u\left ( x,0\right ) =2x\)

Solution

Seek solution where \(u\left ( s\right ) =u\left ( t\left ( s\right ) ,x\left ( s\right ) \right ) =\) constant,hence

\[ \frac {du}{ds}=\frac {\partial u}{\partial t}\frac {dt}{ds}+\frac {\partial u}{\partial x}\frac {dx}{ds}=0 \]

Compare to (1) we see that \(\frac {dt}{ds}=1\) or \(t=s\) and \(\frac {dx}{ds}=xt\), but since \(t=s\) then \(\frac {dx}{ds}=xs\), and this has solution \(x=x_{0}\exp \left ( \frac {s^{2}}{2}\right ) \) but \(s=t\) \(,\) hence

\begin{equation} x=x_{0}\exp \left ( \frac {t^{2}}{2}\right ) \tag {2}\end{equation}

Now at \(t=0\), the solution is \(2x_{0},\) but this solution is valid any where on this characteristic line and not just when \(t=0\). hence

\[ u\left ( x,t\right ) =2x_{0}\]

But \(x_{0}=x\exp \left ( \frac {-t^{2}}{2}\right ) \) from (2), hence

\[ u\left ( x,t\right ) =2x\exp \left ( \frac {-t^{2}}{2}\right ) \]