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Nonlinear PDE

ryo0071

New member
Mar 6, 2013
12
Okay so I am working on this problem:

Solve \(\displaystyle xu_t + uu_x = 0\) with \(\displaystyle u(x, 0) = x.\)(Hint: Change variables \(\displaystyle x \rightarrow x^2\).)

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.
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Okay so I am working on this problem:

Solve \(\displaystyle xu_t + uu_x = 0\) with \(\displaystyle u(x, 0) = x.\)(Hint: Change variables \(\displaystyle x \rightarrow x^2\).)

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

$\chi$ $\sigma$
 

ryo0071

New member
Mar 6, 2013
12
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 \(\displaystyle x \rightarrow \eta^2\)). Anyway, I was able to solve it by method of characteristics without using the coordinate transform. Again, thank you both for the assistance.