2.1.8 Transport equation $u_t-x{u}_x=0$ IC $u(x,0)=\frac{1}{1+x^2}$. Peter Olver textbook, problem 2.2.17

problem number 8

Added Sept 12, 2019.

Taken from Peter Olver textbook, Introduction to Partial differential equations. problem 2.2.17

Solve $u_t-x{u}_x=0$ with IC $u(x,0)=\frac{1}{1+x^2}$

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x,t], t] -x*D[u[x,t], x]== 0; 
ic = u[x,0]==1/(1+x^2); 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[x,t], {x,t}], 60*10]];
 

$$\left\{\{u(x,t)\to \frac{1}{e^{2t}{x}^2+1}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(x,t), t) -x*diff(u(x,t),x) =0; 
ic:=u(x,0)=1/(1+x^2); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t))),output='realtime'));
 

$$u(x,t)=\frac{1}{x^2{\mathrm{e}}^{2t}+1}$$

Hand solution

Solve the initial value problem $$u_t-x{u}_x=0$$ With initial conditions $u(0,x)=\frac{1}{1+x^2}$

Solution

Let $u=u(x\left(t\right),t)$. Then $$\begin{array}{}\text{(2)}& \frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt}+\frac{\partial u}{\partial t}\end{array}$$ Comparing (1),(2) shows that $$\begin{array}{}\text{(3)}& \frac{du}{dt}& =0\text{(4)}& \frac{dx}{dt}& =-x\end{array}$$

Solving (3) gives$$\begin{array}{rl}u& =u\left(x\left(0\right)\right)\\ \text{(5)}& & =\frac{1}{1+x{\left(0\right)}^2}\end{array}$$

We just need to find $x\left(0\right)$ to finish the solution. From (4)$$\begin{array}{rl}\ln \left|x\right|& =-t+C\\ x& =x\left(0\right)e^{-t}\\ \text{(6)}& x\left(0\right)& =x{e}^t\end{array}$$

Substituting (6) in (5) gives$$u(x\left(t\right),t)=\frac{1}{1+x^2{e}^{2t}}$$ The following is an animation of the solution

Source code used for the above

Figure 2.11: Source code 2D

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