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

Added Sept 12, 2019.

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

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Figure 2.1:Source code 3D

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Figure 2.2:Source code 2D

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