We now study the characteristic direction in PDE. This idea comes from the geometrical meaning of a first-order quasi-linear equation. The equation gives a direction field in $(x,y,u)$-space. This direction must lie on the integral surface. Consider the first-order quasi-linear equation $$\begin{equation*} a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0. \quad \text{(1)} \end{equation*}$$ If $u=u(x,y)$ is a solution, then its graph is an integral surface in $(x,y,u)$-space. We write this surface as $$\begin{equation*} f(x,y,u)=u(x,y)-u=0. \quad \text{(2)} \end{equation*}$$ The normal vector to this surface is $$\begin{equation*} \nabla f=(u_x,u_y,-1). \quad \text{(3)} \end{equation*}$$ Now equation (1) may be written as $$\begin{equation*} (a,b,c)\cdot(u_x,u_y,-1)=0. \quad \text{(4)} \end{equation*}$$ Using (3), this becomes $$\begin{equation*} (a,b,c)\cdot \nabla f=0. \quad \text{(5)} \end{equation*}$$

The equation $(a,b,c)\cdot\nabla f=0$ means that the vector $(a,b,c)$ lies in the tangent plane of the integral surface. Choose a point on the sample surface and click Update Tangency View to see the surface, its normal vector, and a tangent direction. The normal vector crosses the tangent plane, while the characteristic direction lies along the plane. Notice that the dot product is zero when the direction is tangent to the surface.