How to determine strip condition of nonlinear PDE?

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

• Dear Sir It is form Introduction of Partial differential equation By K S Rao. Page number 32.Please Help me Aug 25, 2019 at 15:39

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 $$F(x,y,z,p,q) = p^2+q^2-2$$).