4.8.1 Problem 4

Solve

\(u_{t}-\left ( tx^{2}\right ) u_{x}=0\)

\(u\left ( x,0\right ) =1+x\)

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 \(\frac {dx}{ds}=-tx^{2}\), but since \(t=s\) then \(\frac {dx}{ds}=-sx^{2}\) hence we need to solve

\begin{align*} \frac {dx}{x^{2}} & =-sds\\ -\frac {1}{x} & =-\frac {s^{2}}{2}+x_{0}\end{align*}

but \(s=t\) hence

\begin{equation} -\frac {1}{x}=-\frac {t^{2}}{2}+x_{0} \tag {1}\end{equation}

Hence

\begin{equation} x_{0}=-\left ( \frac {1}{x}-\frac {t^{2}}{2}\right ) \tag {2}\end{equation}

At \(t=0\,\),

\[ x_{0}=-\frac {1}{x}\]

Now we are told the solution at \(t=0\) is \(1+x\), or \(1-\frac {1}{x_{0}}\)but this solution is valid any where on this characteristic line and not just when \(t=0\). hence

\[ u\left ( x,t\right ) =1-\frac {1}{x_{0}}\]

Replace the value of \(x_{0}\) obtained in (2) we obtain

\begin{align*} u\left ( x,t\right ) & =1-\frac {1}{-\left ( \frac {1}{x}-\frac {t^{2}}{2}\right ) }\\ & =1+\frac {2x}{2-xt^{2}}\end{align*}

Hence

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

To avoid a solution \(u\) which blow up, we need \(2-xt^{2}\neq 0\), hence \(xt^{2}\neq 2\) \(,\,\) for example, \(x=2\) and \(t=1\) will not give a valid solution. so all region in \(x-t\) plane in which \(xt^{2}=2\) is not a valid region to apply this solution at.

The solution breaks down along this line in the \(x-t\) plane

To see it in 3D, here is the \(u\left ( x,t\right ) \) solution that includes the above line, and we see that the solution below the line and the above the line are not continuous across it. ( I think there is a name to this phenomena that I remember reading about sometime, may be related to shockwaves but do not now know how this would happen in reality)