PDE with initial value

Question

Problem is that Solve $xu_t+uu_x=0 $ with $u(x,0)=x$

Hint is in the textbook that replacing $x$ into $x^2$

I am stuck with it. Would you help me?

Answer 1

Some version of the characteristics method stabilishes to solve first an auxiliar system of ode's, this:

$\dfrac{dt}{x}=\dfrac{dx}{u}=\dfrac{du}{0}$

The last ratio means that $u=c_1$, so is $u$ is constant along the characteristics. With the first proportion we get the second constant:

$dt=\dfrac{xdx}{c_1}\implies c_1t+c_2=x^2/2\implies ut+c_2=x^2/2$

For the solution we have to impose some relation between the constants, so is, we have to impose that $c_2=f(c_1)$ for some single argument, differentiable function $f$.

$ut+f(c_1)=x^2/2\;;ut+f(u)=x^2/2$ is the general solution.

Imposing the initial condition $u(x,0)=x$ we can get $f$:

$f(x)=x^2/2$ that substituting in the general solution gives (maybe here is where the hint given fits):

$ut+u^2/2=x^2/2$ as the requested solution in implicit form.

Answer 2

$xu_t+uu_x=0$

$u_t+\dfrac{u}{x}u_x=0$

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

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

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{ds}=\dfrac{u}{x}=\dfrac{u_0}{x}$ , we have $x^2=2u_0s+f(u_0)=2ut+f(u)$ , i.e. $u=F(x^2-2ut)$

$u(x,0)=x$ :

$F(x^2)=x$

$F(x)=\pm\sqrt x$

$\therefore u=\pm\sqrt{x^2-2ut}$

$u^2=x^2-2tu$

$u^2+2tu-x^2=0$

$u(x,t)=\dfrac{-2t+\sqrt{4t^2+4x^2}}{2}=-t+\sqrt{x^2+t^2}$