4.8 Problem 3

Solve

\(u_{t}+\left ( x\sin t\right ) u_{x}=0\)

\(u\left ( x,0\right ) =\frac {1}{1+x^{2}}\)

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 \(\frac {dx}{ds}=x\sin t\), but since \(t=s\) then \(\frac {dx}{ds}=x\sin s\), and this has solution

\begin{align*} \ln x & =\int \sin \left ( s\right ) ds\\ x & =x_{0}\exp (-\cos \left ( s\right ) ) \end{align*}

but \(s=t\) hence

\begin{equation} x=x_{0}\exp (-\cos \left ( t\right ) ) \tag {1}\end{equation}

Hence

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

At \(t=0\,\),

\[ x=x_{0}\exp \left ( -1\right ) \]

Now we are told the solution at \(t=0\) is \(\frac {1}{1+x^{2}}\), or \(\frac {1}{1+\left [ x_{0}\exp \left ( -1\right ) \right ] ^{2}}\)but this solution is valid any where on this characteristic line and not just when \(t=0\). hence

\[ u\left ( x,t\right ) =\frac {1}{1+\left [ x_{0}\exp \left ( -1\right ) \right ] ^{2}}\]

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

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

Hence

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