# PDEMOC9

Show that if $\displaystyle z=u(x,y)\,$ is an integral surface of $\displaystyle V=\,$ containing a point $\displaystyle P\,$ , then the surface contains the characteristic curve $\displaystyle \chi\,$ passing through $\displaystyle P\,$ . (Assume the vector field $\displaystyle V\,$ is $\displaystyle C^1\,$ , which means that the first derivative exists everywhere).

It is required to prove that $\displaystyle \chi \subset u\,$ .

Let $\displaystyle \chi:(f(t),g(t),h(t))\,$ and $\displaystyle P:(f(0),g(0),h(0))\,$ .

Now, show that $\displaystyle u(f(t),g(t))=h(t)\,$ .

$\displaystyle \frac{d}{dt} \left[ u(x(t),y(t))-h(t)\right]=u_xx'(t)+u_yy'(t)-h'(t)\,$

$\displaystyle =u_xa+u_yb-c(x(t),y(t),z(t))=c(x(t),y(t),u)-c(x(t),y(t),z(t))\,$

From the mean value theorem, at some point the last equation equals $\displaystyle \frac{\partial c}{\partial z} (u-z)\,$ .

This is true iff $\displaystyle \frac{d(u-z)}{dt} = \frac{\partial c}{\partial z}(u-z)\,$

$\displaystyle \implies (u-z)(t) = (u-z)(0)e^{\int_0^t \frac{\partial c}{\partial z} ds}=0\,$

since $\displaystyle u(x(0),y(0))=z(0)\,$ .