I started learning PDE on my Own. I was doing Example 0.14 in the book (1) p. 32 but I stuck at one step. I do not understand how the Author come at the conclusion about strip condition.
Example 0.14 Find the characteristics of the PDE $$ p^2+q^2=2 $$ and determine the integral surface which passes through $x=0$, $z=y$.
Here, $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$ (see book (1) p. 24).
I understand everything except strip condition equation 2
Any Help will be appreciated
(1) K. Sankara Rao, Introduction to Partial Differential Equations, PHI Learning Pvt. Ltd., 2010. GBooks Preview
One has $$z_0(s) = u(x_0(s),y_0(s))$$ so differentiating with respect to $s$ leads to the so-called strip condition $$ \frac{d z_0}{ds} = p_0 \frac{d x_0}{ds} + q_0 \frac{d y_0}{ds}$$ In your case, $$ \frac{d z_0}{ds} = 1; \quad \frac{d x_0}{ds} = 0; \quad \frac{d y_0}{ds} = 1$$ so one gets $$1 = p_0(0) + q_0(1)$$ (notice that $p_0(0)$ and $q_0(1)$ mean $p_0\times(0)$ and $q_0\times(1)$ here... I don't think it is a good notation). So one gets $q_0 = 1$.
Now you can get $p_0$ from $q_0$ using the fact that $$p_0^2+q_0^2-2=0$$ that gives $p_0^2 = 1$, i.e. $p_0 = \pm 1$.
Remark. The strip condition does not depend on $F$ (here $F(x,y,z,p,q) = p^2+q^2-2$).