I have the linear PDE
$$yu_x - xu_y = 0 \qquad x^2 + y^2 < a^2 \\ u(0,y) = (a^2-y^2)^{\frac{1}{2}} \qquad y \in(-a,a)$$
where $a > 0$ is a constant.
So what I have done is to say that
$$a_1 = y \quad a_2=-x \quad b=0 \\ \gamma(x_0(s), y_0(s)) = \gamma(0,s) = (a^2-s^2)^{\frac{1}{2}}$$
with $\gamma(0,s)$ being the Cauchy curve.
From this I say that
$$\widetilde{x}_\tau = \widetilde{y} \qquad \widetilde{x}(0,s) = 0 \\ \widetilde{y}_\tau = -\widetilde{x} \qquad \widetilde{y}(0,s) = s \\ \widetilde{z}_\tau = 0 \qquad \widetilde{z}(0,s) = (a^2-s^2)^{\frac{1}{2}}$$
Now I see that
$$ \widetilde{x}_{\tau\tau} = \widetilde{y}_\tau = -\widetilde{x}$$
this gives
$$\widetilde{x} = A\,cos(\tau) + B\,sin(\tau)$$
and from the initial boundary condition it can be shown that $A=0$. Similarly,
$$ \widetilde{y}_{\tau\tau} = \widetilde{x}_\tau = -\widetilde{y}$$
giving
$$\widetilde{y} = C\,e^{\tau} + B\,e^{\tau}$$
which using the boundary condition gives
$$s = C + D$$
now I know I have made a mistake somewhere because I get stuck here. I don't know how to find the values of $B,\, C,\,D$. But I don't know what I ahve done wrong.
I know that there is a second method where you multiply $x_\tau$ and $y_\tau$ together but I don't want to do it that way.
$$yu_x-xu_y=0$$ $$\frac{\partial u}{x\partial x}-\frac{\partial u}{y\partial y}=0$$ With $X=x^2$ and $Y=y^2$ : $$\frac{\partial u}{\partial X}-\frac{\partial u}{\partial Y}=0$$ The general solution is well known : $$u=F(X+Y)$$ $$u(x,y)=F(x^2+y^2) \tag 1$$ $F$ is an arbitrary differentiable function, to be determined according to the boundary condition.
Condition : $\quad u(0,y)=\sqrt{a^2-y^2}$ $$F(0^2+y^2)=\sqrt{a^2-y^2}$$ Let $X=y^2$ $$F(X)=\sqrt{a^2-X}$$ Now, the function $F(X)$ is determined. We put it into the above general solution Eq.$(1)$, where $X=x^2+y^2$. Thus $F(x^2+y^2)=\sqrt{a^2-(x^2+y^2)}$ $$u(x,y)=\sqrt{a^2-(x^2+y^2)}$$