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

ryo0071
Messages
12
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top