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 $u(x,t)$ $$\begin{array}{r}{u}_t+t{u}_x=0\end{array}$$

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]];
 

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

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'));
 

$$u(x,t)={\mathrm{e}}^{-\frac{t^2}{2}+x}$$

Hand solution

Solve $u_t+x{u}_x=0$ with $u(x,0)=e^x$. Using the method of characteristics, the systems of characteristic lines are (from the PDE itself)$$\begin{array}{}\text{(1)}& \frac{dt}{ds}& =1\text{(2)}& \frac{dx}{ds}& =t\text{(3)}& \frac{du}{ds}& =0\end{array}$$

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$$\begin{array}{rl}t& =s+t\left(0\right)\\ \text{(5)}& & =s\end{array}$$

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

From (5,6) solving for $\xi$ gives$$\begin{array}{rl}\xi & =x-\frac{s^2}{2}\\ \text{(7)}& & =x-\frac{t^2}{2}\end{array}$$

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

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