[next] [prev] [prev-tail] [tail] [up]

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 ✓ 

ClearAll["Global`*"]; 
pde =  D[u[t, x], {t}] -3* D[u[t, x], {x}] == 0; 
ic = u[0,x]==Exp[-x^2]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[t, x], {t, x}], 60*10]];
\[\left \{\left \{u(t,x)\to e^{-(3 t+x)^2}\right \}\right \}\]

Maple ✓ 

restart; 
pde := diff(u(t, x), t) - 3*diff(u(t, x),x) =0; 
ic:=u(0,x)=exp(-x^2); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(t,x))),output='realtime'));
\[u \left (t , x\right ) = {\mathrm e}^{-\left (3 t +x \right )^{2}}\]

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

___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

[next] [prev] [prev-tail] [front] [up]