2.1.14 \(u_t+ c u_x = 0\) and \(u(x,0)=e^{-x^2}\)

problem number 14

Taken from Mathematica help pages

Solve for \(u(x,t)\)

\[ u_t+ c u_x = 0 \]

With initial conditions \(u(x,0)=e^{-x^2}\)

Mathematica ✓

ClearAll["Global`*"]; 
ic  = u[x, 0] == Exp[-x^2]; 
pde =  D[u[x, t], {t}] + c*D[u[x, t], {x}] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[x, t], {x, t}], 60*10]];
 

\[\left \{\left \{u(x,t)\to e^{-(x-c t)^2}\right \}\right \}\]

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x, t), t) + c* diff(u(x, t),x) =0; 
ic  := u(x,0)=exp(-x^2); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t))),output='realtime'));
 

\[u \left (x , t\right ) = {\mathrm e}^{-\left (t c -x \right )^{2}}\]

Hand solution

Solve

\begin{equation} u_{t}+cu_{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} & =c\tag {4}\end{align}

Solving (3) gives

\begin{align} u & =u\left ( x\left ( 0\right ) \right ) \nonumber \\ & =e^{-x\left ( 0\right ) ^{2}}\tag {5}\end{align}

We need to find \(x\left ( 0\right ) \). From (4)

\begin{align*} x & =ct+x\left ( 0\right ) \\ x\left ( 0\right ) & =x-ct \end{align*}

Then (5) becomes

\[ u\left ( x\left ( t\right ) ,t\right ) =e^{-\left ( x-ct\right ) ^{2}}\]

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