2.1.73 \(u_t + t u_x = 0\) with \(u(x,0)=e^x\)

problem number 73

Added May 23, 2019.

From Math 5587 midterm I, Fall 2016, practice exam, problem 4.

Solve for \(u(x,t)\) \begin {align*} u_t + t u_x = 0 \end {align*}

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

Mathematica ✓

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

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

Maple ✓

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

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

Hand solution

Solve \(u_{t}+xu_{x}=0\) with \(u\left ( x,0\right ) =e^{x}\). 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} & =t\tag {2}\\ \frac {du}{ds} & =0\tag {3} \end {align}

With initial conditions at \(s=0\)\[ t\left ( 0\right ) =0,x\left ( 0\right ) =\xi ,u\left ( 0\right ) =e^{\xi }\] Equation (1) gives\begin {align} t & =s+t\left ( 0\right ) \nonumber \\ & =s\tag {5} \end {align}

Equation (2) now becomes \(\frac {dx}{ds}=s\), whose solution is \begin {align} x & =\frac {s^{2}}{2}+x\left ( 0\right ) \nonumber \\ x & =\frac {s^{2}}{2}+\xi \tag {6} \end {align}

From (5,6) solving for \(\xi \) gives\begin {align} \xi & =x-\frac {s^{2}}{2}\nonumber \\ & =x-\frac {t^{2}}{2}\tag {7} \end {align}

Equation (3) gives\begin {align*} u & =u\left ( 0\right ) \\ & =e^{\xi }\\ & =e^{\left ( x-\frac {t^{2}}{2}\right ) } \end {align*}

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