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Am I going about this question the right way?

Solve $u_t + x^2tu_x = 0 $ with initial condition $u_0(x) = \cos x$

I first started by finding the vector field for where $u$ is constant which is $(x^2t,1)$ and so I'm looking for a set of curves such that $\frac{d}{d\tau}(x(\tau), t(\tau)) = (x(\tau)^2t(\tau), 1)$ and so I got that $t(\tau) = \tau$ and so after inverting $t(\tau) = \tau$ we're looking for a function $x(t)$ such that $\dot{x}(t) = x(t)^2t$ but I can't think of such a function as this requires solving a non-linear ODE, am I on completely the wrong track?

EDIT

I have thought of a function! It wasn't too complicated after all, the function is $x(t) = -2t^{-2}$. I will attempt the solution now and make another edit.

EDIT 2: On second thoughts I'm missing the constant $c$, but when I make $x(t) = -2t^{-2} + c$ it makes the function really complicated when squaring etc.. does this matter?

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    This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date.2012-09-10

1 Answers 1

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$u_t+x^2tu_x=0$

$\dfrac{u_t}{t}+x^2u_x=0$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=\dfrac{1}{t}$ , letting $t(0)=0$ , we have $\dfrac{t^2}{2}=s$

$\dfrac{dx}{ds}=x^2$ , letting $x(0)=\dfrac{1}{x_0}$ , we have $\dfrac{1}{x}=-s+x_0=-\dfrac{t^2}{2}+x_0$

$\dfrac{du}{ds}=0$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)=f\biggl(\dfrac{t^2}{2}+\dfrac{1}{x}\biggr)$

$u_0(x)=\cos x$ :

$f\left(\dfrac{1}{x}\right)=\cos x$

$f(x)=\cos\dfrac{1}{x}$

$\therefore u(x,t)=\cos\dfrac{1}{\dfrac{t^2}{2}+\dfrac{1}{x}}=\cos\dfrac{2x}{xt^2+2}$