# PDE with Method of Characteristics and domain of solution

I wanted to solve the following PDE with initial condition

$$\left\{\begin{array}{c} xu_t+u_x=0\\ u(0,x)=f(x) \end{array}\right.$$

Proving that:

(i) if $$f(x) = \sin(x),$$ then it is impossible to find a solution which is valid for all point in $$\mathbb{R}^2.$$

(ii) if $$f(x) = \cos(x),$$ has infinite solutions defined in all $$\mathbb{R}^2.$$

I tried to solve this using the method of characteristics.

First of all the characteristic system is

$$\left\{\begin{array}{c} \frac{d t}{d s} = x\\ \frac{d x}{d s}=1 \\ \frac{d u}{d s}=0 \end{array}\right.$$ with initial conditions $$t(0,\tau)=0$$ $$x(0,\tau)=\tau$$ $$u(0,\tau)=\sin \tau$$

Computing, and proceeding by an analogue method to $$\cos(x),$$ I find $$u(t,x) = \sin (\sqrt{-2t + x^2})$$ and $$u(t,x) = \cos (\sqrt{-2t + x^2}),$$ but both are not defined in $$(1,0).$$ There is a problem with my calculation, or the initial statement are wrong?

• The radicand not defined for some values of the variables is important for the second question. – Rafa Budría Mar 16 at 21:06