#### 2.1.5 Transport equation $$u_t-4u_x+u = 0$$ IC $$u(0,x)=\frac {1}{1+x^2}$$. Peter Olver textbook, 2.2.2 (d)

problem number 5

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

Solve $$u_t-4u_x+u = 0$$ with IC $$u(0,x)=\frac {1}{1+x^2}$$

Mathematica

$\left \{\left \{u(t,x)\to \frac {e^{-t}}{16 t^2+8 t x+x^2+1}\right \}\right \}$

Maple

$u \left (t , x\right ) = \frac {{\mathrm e}^{-t}}{\left (4 t +x \right )^{2}+1}$

Hand solution

\begin {align} u_{t}-4u_{x}+u & =0\tag {1}\\ u\left ( x,0\right ) & =\frac {1}{1+x^{2}}\nonumber \end {align}

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} & =-u\tag {3}\\ \frac {dx}{dt} & =-4\tag {4} \end {align}

Solving (3) gives\begin {align*} \frac {du}{u} & =-dt\\ \ln \left \vert u\right \vert & =-t+c\\ u & =u\left ( x\left ( 0\right ) \right ) e^{-t} \end {align*}

Using the given initial conditions, this becomes \begin {equation} u=\frac {1}{1+x\left ( 0\right ) ^{2}}e^{-t}\tag {5} \end {equation} Now we just need to ﬁnd $$x\left ( 0\right )$$. From (4)\begin {align*} x & =-4t+x\left ( 0\right ) \\ x\left ( 0\right ) & =x+4t \end {align*}

Substituting the above into (5) gives$u=\frac {e^{-t}}{1+\left ( x+4t\right ) ^{2}}$ The above is the method I prefer to solve these problems. Here are some alternative ways

Alternative approach

Let $$\xi$$ be the characteristic variable deﬁned such that $$\xi =x-ct$$. Where characteristic lines are given by $$x=x_{0}+ct$$. But $$c=-4$$ in this problem. Hence characteristic lines are $x=x_{0}-4t$ And$\xi =x+4t$ Then $$u_{t}-4u_{x}$$ are transformed to $$v\left ( t,\xi \right )$$ as was done in part (a) (will not be repeated) which results in$u_{t}-4u_{x}=\frac {\partial v}{\partial t}$ Substituting the above into (1) gives (where now $$v$$ is used in place of $$u$$).$\frac {\partial v}{\partial t}+v=0$ This is now ﬁrst order ODE since it only depends on $$t$$. Therefore $$v^{\prime }+v=0$$. This is linear in $$v$$. Hence the solution is $$\frac {d}{dt}\left ( ve^{\int dt}\right ) =0$$ or $$ve^{t}=F\left ( \xi \right )$$ where $$F$$ is arbitrary function of $$\xi$$. Hence$v\left ( t,\xi \right ) =e^{-t}F\left ( \xi \right )$ Converting to $$u\left ( t,x\right )$$ gives\begin {equation} u\left ( t,x\right ) =e^{-t}F\left ( x+4t\right ) \tag {2} \end {equation} At $$u\left ( 0,x\right ) =\frac {1}{1+x^{2}}$$ the above becomes$\frac {1}{1+x_{0}^{2}}=F\left ( x_{0}\right )$ From the above then (2) can be written as$u\left ( t,x\right ) =\frac {e^{-t}}{1+\left ( x+4t\right ) ^{2}}$ An alternative approach to solve transport PDE is by using Lagrange-charpit method$\frac {dt}{1}=-\frac {dx}{4}=\frac {-du}{u}$ Integrating $$\frac {dt}{1}=\frac {-dx}{4}$$ gives $$t=-\frac {1}{4}x+C_{2}$$ or $C_{2}=t+\frac {1}{4}x$ Now either $$\frac {dt}{1}=\frac {-du}{u}$$ or $$\frac {-dx}{4}=\frac {-du}{u}$$ can be integrated. The choice is not important. Integrating $$\frac {dt}{1}=\frac {-du}{u}$$ gives $$t=-\ln u+C_{1}$$ or $C_{1}=t+\ln u$ But $$C_{1}=F\left ( C_{2}\right )$$ always, where $$F$$ is arbitrary function therefore

\begin {align} t+\ln u & =F\left ( C_{2}\right ) \nonumber \\ t+\ln u & =F\left ( t+\frac {1}{4}x\right ) \nonumber \\ \ln u & =F\left ( t+\frac {1}{4}x\right ) -t\nonumber \\ u & =e^{-t}e^{F\left ( t+\frac {1}{4}x\right ) } \tag {1} \end {align}

At $$u\left ( 0,x\right ) =\frac {1}{1+x^{2}}$$ the above becomes\begin {align*} \frac {1}{1+x^{2}} & =e^{F\left ( \frac {1}{4}x\right ) }\\ F\left ( \frac {x}{4}\right ) & =\ln \left ( \frac {1}{1+x^{2}}\right ) \end {align*}

Let $$z=\frac {x}{4}$$, then $$x=4z$$. The above becomes$F\left ( z\right ) =\ln \left ( \frac {1}{1+\left ( 4z\right ) ^{2}}\right )$ From the above then (1) can be written as\begin {align*} u\left ( t,x\right ) & =e^{-t}e^{\ln \left ( \frac {1}{1+\left ( 4\left ( t+\frac {1}{4}x\right ) \right ) ^{2}}\right ) }\\ & =\frac {e^{-t}}{1+\left ( 4\left ( t+\frac {1}{4}x\right ) \right ) ^{2}}\\ & =\frac {e^{-t}}{1+\left ( 4t+x\right ) ^{2}} \end {align*}

The following is an animation of the solution

 3D 2D

Source code used for the above

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