2.1.72 \(u_t + x u_x = 0\) with \(u(x,0)=x^2\). Math 5587

problem number 72

Added May 23, 2019.

From Math 5587 midterm I, Fall 2016, practice exam, problem 3.

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

\begin{align*} u_t + x u_x = 0 \end{align*}

With with \(u(x,0)=x^2\)

Mathematica ✓

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

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

Maple ✓

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

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

Hand solution

Solve \(u_{t}+xu_{x}=0\) with \(u\left ( x,0\right ) =x^{2}\). Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)

\begin{align} \frac {dt}{ds} & =1\tag {1}\\ \frac {dx}{ds} & =x\tag {2}\\ \frac {du}{ds} & =0\tag {3}\end{align}

With initial conditions at \(s=0\)

\[ t\left ( 0\right ) =0,x\left ( 0\right ) =\xi ,u\left ( 0\right ) =\xi ^{2}\]

Equation (1) gives

\begin{align} t & =s+t\left ( 0\right ) \nonumber \\ & =s\tag {5}\end{align}

Equation (2) gives

\begin{align} \ln x & =s+x\left ( 0\right ) \nonumber \\ x & =\xi e^{s}\tag {6}\end{align}

From (5,6) solving for \(\xi \) gives

\begin{align} \xi & =xe^{-s}\nonumber \\ & =xe^{-t}\tag {7}\end{align}

Equation (3) gives

\begin{align*} u & =u\left ( 0\right ) \\ & =\xi ^{2}\\ & =x^{2}e^{-2t}\end{align*}

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