# PDEMOC9

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

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

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

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

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

$=u_{x}a+u_{y}b-c(x(t),y(t),z(t))=c(x(t),y(t),u)-c(x(t),y(t),z(t))\,$

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

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

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

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