Proving that the solution to $yu_x-xu_y=0$ containing $x^2+y^2=a^2,u=y$ doesn't exist. This is basically a Cauchy problem.Parametrizing the given curve(Initial curve):$x=a\cos s,y=a\sin s,u=a\sin s$
$y'(s)y(s)+x(s)x'(s)=a\cos s(a\sin s)-(a\cos s)(a\sin s)=0 \implies$ Characteristic curve and Initial curve are same so we can't apply Method of characteristics to solve this problem.
My question is: why can't we apply Method of characteristics if Initial curve and Characteristic curves are same?In the present problem ,HOW TO PROVE THAT :
the solution to $yu_x-xu_y=0$ containing $x^2+y^2=a^2,u=y$ doesn't exist (because  in this case Method of characteristics is inconclusive due to the fact that characteristic curve and Initial curve are same in this case)?

 A: The problem in this case is not that the method of characteristics is inconclusive (which would have meant you have infinitely many solutions) but rather that there is a contradiction between the initial conditions between the equation and the initial conditions.
Problems with the method of characteristics occur whenever the projection onto the $xy$ plane of the initial curve coincides with the characteristics. Why does this happen? because the equation determines the rate of change in $u$ as you move along each of the characteristics. However, if the initial condition is specified along one of these curves then it can either:
1. Agree with the equation, but then you don't have any information about what happens as you move away away from that curve.
2. Contradict the equation and then the equation cannot be satisfied.
In you particular example the characteristic curves are circles with their center at the origin, and the derivative along these curves is zero. That means that if you would have had the initial condition $x^2+y^2=a^2, u=c$ then you could have infinitely many solutions since the directional derivative along the initial curve would have been zero as required by the equation. Unfortunately in this case you are given an initial condition with non-vanishing directional derivative along the curve ($u$ is not constant).
Let's see this explicitly: a paramatrization of the initial condition can be $u(a\cos t,a\sin t)=a\sin t$, and differentiating it with respect to $t$ leads to $-a\sin t u_x+a\cos t u_y = a\cos t\rightarrow -yu_x+xu_y=x\neq 0$.
EDIT: An attempt at an intuitive explanation of the characteristics method.
Let's start with a linear PDE with constant coefficients, that is, the LHS is now $au_x+bu_y$, this can be rewritten as $(a,b)\cdot(u_x,u_y)=(a,b)\cdot\nabla u$ which is, up to a normalizing constant the directional derivative of $u$ along the $(a,b)$ direction, that means that one you change the variables you will be left with a simple ODE along the direction of the characteristic lines. Now, if your initial condition intersects each of these line at a single point you can imagine yourself solving the ODE with each of these intersection points as the initial condition. An important point is to note that the characteristic lines in this case cover the entire plane and do not intersect each other (for parallel lines this is obvious).
For more complicated cases such as yours, despite the curves not being parallel lines the same idea still holds. Specifically, in your case these curves are the circles mentioned above.
Problems arise when the initial condition coincides with a characteristic curve instead of intersecting it. In such cases you get two different inputs about the behavior of the function along this curve, one from solving the ODE and one from the value of the function being specified along the curve in the initial condition. If these two inputs agree then the problem is that there are many ways to extend the function beyond the curve (the initial condition does not add any information not in the equation), and if they contradict each other it means the equation cannot be satisfied.
