For a quasi-linear partial differential equation of order 1 of $u:\mathbb{R}^2\rightarrow\mathbb{R}$, the characteristic method requires, for a solution surface with parametrization $(x(t,s),y(t,s),u(x,y))$ that $\det \frac{\partial (x,y)}{\partial (t,s)} \neq 0$. It was mentioned that this is to ensure that its a surface defined, and not degenerated form as a path or a dot.
I'm wish to make sure I understand the reason for this: a non-zero Jacobian ensures us that the function is reversible, thus $\dim (x,y,u)(t,s)=2$ (and for a continuous $u$ this type of representation ensures us it's a 2-dimensional manifold, as well).
Am I correct?
Another question, regarding the derivation of the method. Let $F(x,y,u(x,y))=0$ be the solution surface for a quasi-linear PDE. Then $\nabla F=(u_x,u_y, -1)$. Thus, if $a(x,y,y),b(x,y,u),c(x,y,u)$ satisfy: $(a, b, c)\cdot\nabla F$ it means that $(a,b,c)$ belongs to the tangent space of the manifold at $(x,y,u(x,y))$ which induces for a parametrization $(x(t),y(t),u(x,y)), t\in I\subset \mathbb{R}$:
$\frac{dx}{dt}=a(x,y,u) \\
\frac{dy}{dt}=b(x,y,u) \\
\frac{du}{dt}=c(x,y,u)$
I'm not sure though why being in the tangent space implies the following equations?
$(a,b,c)$ belong to the tangent space because $(a, b, c)\cdot\nabla F=0$. The gradient is orthogonal to the tangent space and the inner product is zero, so, the vector is in the tangent space. Now, we consider a curve $(x(t),y(t),z(t)$ in the surface solution, then its derivative, so is, its tangent is in the tangent space. Particulary, and it is the point, we have a curve into the surface if we make to coincide its tangent with $(a,b,c)$. In fact, we obtain a whole family of curves all of them into the surface, the characteristic curves.
Intuitively, we obediently follow the tangent vector $(a,b,c)$ and then we see what curves we draw into the surface. The bonus is that the family of curves this way earned reconstruct itself the surface solution. Summarizing, you equate the tangent of the curve to a vector tangent to the surface; you integrate to get a curve into the surface.