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Let $u(x,y)$ by the integral surface of the equation:


Where $a,b$ are positives differential function in the hole plane.

Let $D=\{(x,y)||x|<1 ,|y|<1\}$

How do I prove that the proyection of characteristics in the plane $(x,y)$ that intersect D must intersect the boundary $D$ in two point?

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Hint: think of the slope of those projections, particularly the sign of the slope. Also: projection, whole, positive differentiable functions. –  user31373 Oct 1 '12 at 16:41
I was thinking in that, but I also think of this example $f(t)=(-e^{-t},-e^{-t})$, It has a positive gradient but it only touch the boundary of D in a single point. –  Porufes Oct 2 '12 at 4:24

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