Studying the method of characteristics, the argument goes as follows:
We are interested in the equation: $a(x, y)u_x+b(x, y) u_y=f(x, y, u)$;
$(a(x, y), b(x, y), f(x, y, u))(u_x, u_y, – 1) =(a,b,f)\nabla{F}=0$, where $F =u(x, y) – u=0$.
Hence, $(a, b, f)$ is orthogonal to $\nabla F$.
Then, they say that $\nabla F$ is orthogonal to the solution surface, and so we get that $(a, b, f) $ lies in the tangential plane to the solution surface. Therefore, we can build a characteristic curve, starting from a point on the boundary. We parametrise our unknown characteristic curve as ${x(t), y(t), z(t)} $, and find the tangent vector to it at every point – $v=(x_t (t), y_t (t), z_t (t)) $.
Then, we find $x, y, z$ from the condition $v=c(a, b, f)$. This is a system of ODEs, comprising the method of Characteristics.
However, I don't understand why they say that $\nabla F=(u_x, u_y, – 1)$ is orthogonal to the solution surface.
Best Answer
The tangent plane of the surface parametrized by $$\Gamma =\{\sigma (x,y):=(x,y,u(x,y))\mid (x,y)\in I\},$$ at the point $(a,b)$ is given by $$u(a,b)+\text{Span}\Big\{\sigma _x(a,b),\sigma _y(a,b)\Big\},$$(whenever $\{\sigma _x(a,b),\sigma _y(a,b)\}$ is free) and thus, has normal vector $$\sigma _x(a,b)\times \sigma _y(a,b),$$ which is collinear to $(u_x,u_y,-1)$.