MHB Solving xu_t + uu_x = 0: Need Help with Change of Variables

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The discussion centers on solving the nonlinear partial differential equation xu_t + uu_x = 0 with the initial condition u(x, 0) = x, where a change of variables hint (x → x^2) is provided. Participants clarify that while the hint may be useful, the method of characteristics is a valid approach to find the solution. The general procedure for solving such equations involves setting up a system of ordinary differential equations, which leads to a solution expressed in terms of the original variables. Ultimately, one participant successfully solves the equation using the method of characteristics without applying the suggested change of variables. The conversation highlights the flexibility in methods for tackling nonlinear PDEs.
ryo0071
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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.
 
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Yes, I'm not sure why the hint is even suggested unless maybe you solved

$u_t + u u _x = 0,\;\; u(x,0)=f(x)$.

already. The method of characteristic will work fine. Are you familiar with this method?
 
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

$\chi$ $\sigma$
 
Thank you both for the help. I was trying to use the hint but what was confusing me was that they used the same variable x for both parts of the change of coordinates (rather than doing something like $$x \rightarrow \eta^2$$). Anyway, I was able to solve it by method of characteristics without using the coordinate transform. Again, thank you both for the assistance.
 

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