#### 2.1.8 Transport equation $$u_t-x u_x= 0$$ IC $$u(x,0)=\frac {1}{1+x^2}$$. Peter Olver textbook, problem 2.2.17

problem number 8

Taken from Peter Olver textbook, Introduction to Partial diﬀerential equations. problem 2.2.17

Solve $$u_t-x u_x= 0$$ with IC $$u(x,0)=\frac {1}{1+x^2}$$

Mathematica

$\left \{\left \{u(x,t)\to \frac {1}{e^{2 t} x^2+1}\right \}\right \}$

Maple

$u \left (x , t\right ) = \frac {1}{x^{2} {\mathrm e}^{2 t}+1}$

Hand solution

Solve the initial value problem $u_{t}-xu_{x}=0$ With initial conditions $$u\left ( 0,x\right ) =\frac {1}{1+x^{2}}$$

Solution

Let $$u=u\left ( x\left ( t\right ) ,t\right )$$. Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =-x\tag {4} \end {align}

Solving (3) gives\begin {align} u & =u\left ( x\left ( 0\right ) \right ) \nonumber \\ & =\frac {1}{1+x\left ( 0\right ) ^{2}}\tag {5} \end {align}

We just need to ﬁnd $$x\left ( 0\right )$$ to ﬁnish the solution. From (4)\begin {align} \ln \left \vert x\right \vert & =-t+C\nonumber \\ x & =x\left ( 0\right ) e^{-t}\nonumber \\ x\left ( 0\right ) & =xe^{t}\tag {6} \end {align}

Substituting (6) in (5) gives$u\left ( x\left ( t\right ) ,t\right ) =\frac {1}{1+x^{2}e^{2t}}$ The following is an animation of the solution

Source code used for the above

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