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Show that if z=u(x,y)\, is an integral surface of V=<a,b,c>\, 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)\,


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)\,.

Main Page : Partial Differential Equations : Method of Characteristics