Suppose I have the PDE:
$u_x + u_y = 1 - u, u(x, x+x^2) = \sin(x), x > 0$
The Lagrange Charpit equations are
$\frac{dx}{dt} = 1, x(0, s) = s$,
$\frac{dy}{dt} = 1, y(0, s) = s + s^2$
$\frac{dz}{dt} = 1-z, z(0, s) = \sin(s)$
Now I know how to solve these and I obtain $z(s, t) = 1 + (\sin s - 1)e^{-t}$.
I can then find $u$.
My question is:
How are these equations derived?
Do I set $u(x, t) = z(t(x, y), s(x, y)$ initially or what?
We proved this in class in the form of a very general Cauchy problem, and the expectation in that module is to just write out the Lagrange Charpit equations and solve. However, in classes for other lecturers, I'll probably need some justification before I just write out the equations.
So I would like to know how I should start so that I can obtain the equations, e.g. 'let $u(x, t) = z(t(x, y), s(x, y))$, then equations are...' (or whatever the correct starting point is)
$$u_x+u_y=1-u$$ $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{1-u}=dt$$ A first characteristic equation comes from $\frac{dx}{1}=\frac{dy}{1}$ : $$y-x=c_1$$ A second characteristic equation comes from $\frac{dx}{1}=\frac{du}{1-u}$ : $$(u-1)e^x=c_2$$ General solution of the PDE on implicit form $c_2=F(c_1)$ : $$(u-1)e^x=F(y-x)$$ $$u(x,y)=1+e^{-x}F(y-x)$$ $F$ is an arbitrary function to be determined according to the boundary condition.
CONDITION :
$$u(x+x^2)=\sin(x)=1+e^{-x}F((x^2-x)-x)$$ $$F(x^2)=e^x(\sin(x)-1)$$ The function $F$ is determined : $$F(X)=e^{\sqrt{|X|} }(\sin(\sqrt{|X|} )-1)$$ We put it into the above general solution where $X=y-x$ : $$u(x,y)=1+e^{-x}e^{\sqrt{|y-x|} }\left(\sin\left(\sqrt{|y-x|}\right)-1\right)\quad ,\quad x>0$$