#### 2.1.9 Transport equation $$u_t+(1-2 t) u_x= 0$$ IC $$u(x,0)=\frac {1}{1+x^2}$$. Peter Olver textbook, problem 2.2.29

problem number 9

Taken from Peter Olver textbook, Introduction to Partial diﬀerential equations. problem 2.2.29

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

Mathematica

$\left \{\left \{u(x,t)\to \frac {1}{2 t^2 x+t^4-2 t^3+t^2-2 t x+x^2+1}\right \}\right \}$

Maple

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

Hand solution

Solve $u_{t}+\left ( 1-2t\right ) u_{x}=0$ 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} & =\left ( 1-2t\right ) \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)$x=t-t^{2}+C$ At $$t=0$$$x\left ( 0\right ) =C$ Hence \begin {align*} x & =t-t^{2}+x\left ( 0\right ) \\ x\left ( 0\right ) & =x-t+t^{2} \end {align*}

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

Source code used for the above

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