Holla, everybody!
I'm studying first order PDE and I have difficulties figuring out a few points. Will appreciate any help!
So, say, we have a first order inhomogeneous equation: $$X_1(x_1,...,x_n,u)\frac{\partial u}{\partial x_1}+...+X_n(x_1,...,x_n,u)\frac{\partial u}{\partial x_n}= T(x_1, ...,x_n,u)$$ We write down a system of ODEs in order to find characteristics: $$\frac{dx_1}{ds} = X_1,...,\frac{dx_n}{ds} = X_n, \frac{du}{ds}=T$$ Then we write a symmetric form of a system down: $$\frac{dx_1}{X_1(x_1,...,x_n,u)} =...=\frac{dx_2}{X_n(x_1,...,x_n,u)} = \frac{du}{T} = ds$$ The first question: where did $ds$ come away? All the books use symmetric form without $ds$: $$\frac{dx_1}{X_1(x_1,...,x_n,u)} =...=\frac{dx_2}{X_n(x_1,...,x_n,u)} = \frac{du}{T}$$ Then we found n first integrals $\psi_1,...,\psi_n$ of the system - it's curves along which solution $u$ has constant value, and the problem in constructing the solution $u$ is that we don't know how does $u$ change between these curves.
And there without any derivation books say that a general solution of PDE is $\Phi(\psi_1,...,\psi_n)=0$, where $\Phi$ - arbitrary function. The second question: where did it come from?
Lets consider our equation to be homogeneous, so $T \equiv 0$, then we have system $$\frac{dx_1}{X_1(x_1,...,x_n,u)} =...=\frac{dx_2}{X_n(x_1,...,x_n,u)} = \frac{du}{0}$$ One of its first integrals is $u=const$, so our solution shall be $$\Phi(\psi_1,...,\psi_{n-1},u)=0$$ And then it is rewritten, saying that we resolve $u$, $u = \Phi(\psi_1,...,\psi_{n-1}) $ The third question: why we can resolve $u$ in such a way?