2.1.3 Transport equation $u_t+2u_x=0$ IC $u(-1,x)=\frac{x}{1+x^2}$. Peter Olver textbook, 2.2.2 (b)

problem number 3

Added Sept 12, 2019.

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

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

Mathematica ✓

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

$$\left\{\{u(t,x)\to \frac{-2t+x-2}{4t^2-4t(x-2)+x^2-4x+5}\}\right\}$$

Maple ✓

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

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

Hand solution

Solve $$u_t+2u_x=0$$ With initial conditions $u(-1,x)=\frac{x}{1+x^2}$.

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 $$\begin{array}{}\text{(3)}& \frac{du}{dt}& =0\text{(4)}& \frac{dx}{dt}& =2\end{array}$$

Eq (3) says that $u$ is constant on the chataterstic lines, or $u=u\left(x(-1)\right)$. Using the given initial conditions, this becomes $$\begin{array}{}\text{(5)}& u(x\left(t\right),t)=\frac{x(-1)}{1+x{(-1)}^2}\end{array}$$ Eq (4) is now used to find $x(-1)$. Soving (4) gives $x=x\left(0\right)+2t$. Hence $x(-1)=x\left(0\right)-2$ or $x\left(0\right)=x(-1)+2$. Therefore $$\begin{array}{rl}x& =x(-1)+2+2t\\ x(-1)& =x-2-2t\end{array}$$

Now that we found $x(-1)$, we substitute it in (5), giving the solution$$u(x\left(t\right),t)=\frac{x-2-2t}{1+{(x-2-2t)}^2}$$ Alternative method. Using Lagrange-charpit method

$$\frac{dt}{1}=\frac{dx}{2}=\frac{du}{0}$$ Which implies that $du=0$ or $u=C_1$. A constant. Integrating $\frac{dt}{1}=\frac{dx}{2}$ gives $t=\frac{1}{2}x+C_2$ or $C_2=t-\frac{1}{2}x$. But $C_1=F\left(C_2\right)$ always, where $F$ is arbitrary function. Since $C_1=u$ then$$\begin{array}{rl}u& =F\left(C_2\right)\\ \text{(1)}& u& =F(t-\frac{1}{2}x)\end{array}$$

At $t=-1$ the above becomes$$\frac{x}{1+x^2}=F(-1-\frac{1}{2}x)$$ Let $-1-\frac{1}{2}x=z$ which implies $x=-2(1+z)$ The above can be written as$$\begin{array}{rl}\frac{-2(1+z)}{1+{(-2(1+z))}^2}& =F\left(z\right)\\ F\left(z\right)& =-\frac{2(1+z)}{4z^2+8z+5}\end{array}$$

From the above then (1) can be written as$$\begin{array}{rl}u(t,x)& =-\frac{2(1+(t-\frac{1}{2}x))}{4{(t-\frac{1}{2}x)}^2+8(t-\frac{1}{2}x)+5}\\ & =\frac{x-2t-2}{4t^2-4tx+8t+x^2-4x+5}\\ & =\frac{x-2t-2}{1+{(x-2-2t)}^2}\end{array}$$

The following is an animation of the solution

3D	2D

Source code used for the above

Figure 2.3: Source code 3D

Figure 2.4: Source code 2D

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