Solve the Cauchy problem $$(x − y − 1)u_x + (y − x − u + 1)u_y = u,$$ if $u=1$ on $x^2+(y+1)^2=1.$
Attempt. $$\frac{dx}{x-y-1}=\frac{dy}{y-x-z+1}=\frac{dz}{z}$$
so $$\frac{dx+dy}{(x-y-1)+(y-x-z+1)}=\frac{dz}{z}\iff d(x+y+z)=0$$ so $g_1(x,y,z)=x+y+z=c_1.$ I have not managed to find the second relation $g_2(x,y,z)=0$, needed in order to get $F(g_1,g_2)=0$ for some $F$. (As far as I am concerned, there are no standard procedures in these cases, one has to work on trial-and -error to find the exact expressions).
Thanks in advance.
$$(x − y − 1)u_x + (y − x − u + 1)u_y = u \tag 1$$ $$\frac{dx}{x-y-1}=\frac{dy}{y-x-u+1}=\frac{du}{u}\quad\text{is correct}$$ You rightly found a first characteristic equation : $$x+y+u=c_1$$ A second characteristic equation comes from $$\frac{dx-dy}{(x-y-1)-(y-x-u+1)}=\frac{du}{u}=\frac{d(x-y-1)}{2(x-y-1)+u}$$ With $v=x-y-1$ $$\frac{du}{u}=\frac{dv}{2v+u}$$ is a separable ODE which solution is $v=-u+c_2u^2$ . Thus $x-y-1=-u+c_2u^2$ . The second characteristic equation is : $$\frac{x-y-1+u}{u^2}=c_2$$ The general solution of the PDE can be expressed on the form of implicit equation : $$\frac{x-y-1+u}{u^2}=F(x+y+u) \tag 2$$ where $F$ is an arbitrary function.
Boundary condition in order to determine the function $F$ :
$u=1$ on $x^2+(y+1)^2=1$ , so $\frac{x-y-1+1}{1^2}=F(x+y+1)$ $$F(x+y+1)=x-y$$ Let $X=x+y+1=\pm\sqrt{1-(y+1)^2}+y+1$
$(X-y-1)^2=1-(y+1)^2.\quad$ To be solved for $y$ which leads to
$y=\frac{X-2\pm\sqrt{2-X^2}}{2}\quad;\quad x=\frac{X\mp\sqrt{2-X^2}}{2}\quad;\quad x-y=\mp\sqrt{2-X^2}+1$ $$F(X)=1\mp\sqrt{2-X^2}$$ So, $F(X)$ is determined. We put it into the above general solution $(2)$ where $X=x+y+u$. $$\frac{x-y-1+u}{u^2}=1\mp\sqrt{2-(x+y+u)^2}$$ The solution which complies to the PDE and the boundary condition is expressed on the form of an implicit equation : $$x-y-1+u-u^2\pm u^2\sqrt{2-(x+y+u)^2}=0 \tag 3$$ To express $u(x,y)$ on explicit form we have to solve a polynomial equation of sixth degree. Thus there is no closed form for $u(x,y)$. The final answer is the above implicit form $(3)$.