#### 2.1.2 Transport equation $$u_t-3 u_x = 0$$ IC $$u(0,x)=e^{-x^2}$$. Peter Olver textbook, 2.2.2 (a)

problem number 2

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

Solve for $$u(t,x)$$ in $$u_t-3 u_x = 0$$ with IC $$u(0,x)=e^{-x^2}$$

Mathematica

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

Maple

$u \left (t , x\right ) = {\mathrm e}^{-\left (3 t +x \right )^{2}}$

Hand solution

Solve \begin {equation} u_{t}-3u_{x}=0 \tag {1} \end {equation} With initial conditions $$u\left ( x,0\right ) =e^{-x^{2}}$$

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} & =-3 \tag {4} \end {align}

Eq (3) gives $$u=u\left ( x\left ( 0\right ) \right )$$. Using the given initial conditions, this becomes \begin {equation} u=e^{-x\left ( 0\right ) ^{2}} \tag {5} \end {equation} Eq (4) is now used to ﬁnd $$x\left ( 0\right )$$. Soving (4) gives $$x=x\left ( 0\right ) -3t$$. Hence $$x\left ( 0\right ) =x-3t$$. Therefore (5) becomes$u\left ( x,t\right ) =e^{-\left ( x-3t\right ) ^{2}}$ The following is an animation of the solution

 3D 2D

Source code used for the above  ________________________________________________________________________________________