ryo0071 said:
Okay so I am working on this problem:
Solve $$xu_t + uu_x = 0$$ with $$u(x, 0) = x.$$(Hint: Change variables $$x \rightarrow x^2$$.)
However, I am not sure how to use the change of variables hint that is given or why it is needed. My thinking is that I could just use the method of characteristic as normal to get the solution. Any help is appreciated.
The general solving procedure of a nonlinear PDE of the type... $\displaystyle u_{t} + \varphi(u,x)\ u_{x} = f(x,t),\ u(x,0)= g(x)\ (1)$
... is the following. First we write the system of three ODE...
$\displaystyle \frac{d t}{d s}=1,\ t(0)=0$
$\displaystyle \frac{d x}{d s}= \varphi(u,s),\ x(0)= \xi$
$\displaystyle \frac{d v}{d s} = f(x,t),\ v(0)=g(\xi)\ (2)$... then we solve (2) obtaining $\displaystyle t=t(s, \xi),\ x=x(s,\xi), v=v(s, \xi)$ and then $\displaystyle s=s(x,\ t), \xi=\xi(x,\ t)$. The solution of (1) is...
$\displaystyle u(x,\ t) = v \{s(x,\ t),\ \xi(x,\ t)\}\ (3)$
In Your case is $\displaystyle \varphi (u,\ x) = \frac{u}{x},\ f(x,t) = 0,\ g(x)=x$ so that the (2) becomes...
$\displaystyle \frac{d t}{d s}=1,\ t(0)=0$
$\displaystyle \frac{d x}{d s}= \frac{v}{s},\ x(0)= \xi$
$\displaystyle \frac{d v}{d s} = 0,\ v(0)=\xi\ (4)$
With a little of patience You solve (4) obtaining...
$\displaystyle t=s$
$\displaystyle v=\xi$
$\displaystyle x=\sqrt{2 s \xi + \xi^{2}}\ (5)$
If You invert (5) obtain...
$\displaystyle s=t$
$\displaystyle \xi = -t + \sqrt{t^{2} + x^{2}}\ (6)$
... and (6) permits You to write...
$\displaystyle u(x,\ t)= -t + \sqrt{t^{2} + x^{2}}\ (7)$
Kind regards
$\chi$ $\sigma$