Let $ \alpha$ be real number and $h=h(x)$ be a continuous function in $\mathbb{R}$.Consider following initial value problem: $$yu_x + xu_y=\alpha u, u(x, 0) =h(x) $$ Then
a) Find all points on ${(y=0)} $ where $ {(y=0)}$ is characteristic. What is the compatibility condition on h at these points?
b) Find the solution of the initial value problem.what is the domain of this solution in general?
c) For each case h(x) =x and $\alpha$ =3, check whether this solution can be extended over the points in (a)
(d) For each points in (a) find all the characteristics curves containing it. What is the relation of these curves and the domain in (b)?
Further hints are appreciated
I suppose that the main difficulty encountered is to fit the general solution to the bourdary condition. So, the solving below is only a summary without much details, except at the end for the particular solution.
$$yu_x+xu_y=\alpha u$$ $$\frac{dx}{y}=\frac{dy}{x}=\frac{du}{\alpha u}$$ First characteristic equation, from $\frac{dx}{y}=\frac{dy}{x} \quad\to\quad x^2-y^2=c_1$
Second characteristic equation, from $\frac{dx}{y}=\frac{dy}{x}=\frac{dx+dy}{x+y}=\frac{du}{\alpha u} \quad\to\quad \ln|x+y|=\frac{1}{\alpha}\ln|u|$+C $$\frac{u}{(x+y)^\alpha}=c_2$$ General solution on implicit form, any differentiable function $\Phi$ : $$\Phi\left((x^2-y^2)\:,\:\frac{u}{(x+y)^\alpha}\right)=0$$ General solution on explicit form, any differentiable function $F$ : $$u(x,y)=(x+y)^\alpha F\left(x^2-y^2\right)$$ Given condition : $\quad u(x,0)=h(x) \quad\to\quad (x+0)^\alpha F\left(x^2+0\right)=h(x) $
Let $x=\pm\sqrt{|X|} \quad\to\quad |X|^{\alpha/2} F\left(X\right)=h(\pm\sqrt{|X|}) \quad\to\quad F(X)=|X|^{-\alpha/2}h(\pm\sqrt{|X|}) $
Thus $ F(x^2-y^2)=|x^2-y^2|^{-\alpha/2}h(\pm\sqrt{|x^2-y^2|})$ $$u(x,y)=|x+y|^\alpha |x^2-y^2|^{-\alpha/2}h\left(\pm\sqrt{|x^2-y^2|}\right)$$ $$u(x,y)=\left|\frac{x+y}{x-y}\right|^{\alpha/2}h\left(\pm\sqrt{|x^2-y^2|}\right) $$