Not able to understand the simplification done here.

Question

How do we get from $(1)$ to $(2)$ ? $$\frac{\partial}{\partial x} \frac{f(x,y)}{G(y)}+\frac{\partial}{\partial y} \frac{f(x,y)}{G(y)}=0 \tag{1}$$ $$ \frac{f(x,y+z)}{G(y+z)}=\frac{h(x-z,y)}{G(y)}, \text{for } x \geq z\tag{2}$$

Answer 1

First of all, we have to recognize that we are treating $\frac{h(t,x)}{N(x)}$ as a single function of two variables. Let's write this explicitly:

$$ f(t,x) = \frac{h(t,x)}{N(x)}. $$

Now the fact that $$\frac{\partial}{\partial t} f(t,x)+\frac{\partial}{\partial x} f(t,x) = 0 \tag1$$ tells us that the directional derivative of $f$ in the direction $(1,1)$ is zero. That is, if you increase each of the two input variables of $f$ always at the exact same rate, the value of $f$ never changes. After doing some amount of increase to each variable in this way, you will have increasing each of them by the exact same amount without changing the value of $f$.

So for some values of $t,$ $x,$ and $z,$ we start with $f(t - z, x)$, which has some value, and we increase the first and second input variables by the same amount, $z$, arriving at $f(t, x+z)$, which produces the same value. For this to work, however, $f$ must be defined at each of the input pairs $(t-z,x)$ and $(t,x+z)$ and Equation$~(1)$ must be true at every point on the straight line segment between those two points (interpreted as Cartesian coordinates), that is, Equation$~(1)$ must be true whenever we replace $(t,x)$ with $(t - z + \eta, x + \eta)$ for $0 < \eta < z.$

The condition $t \geq z$ does not come directly from Equation $(1)$ but it may come from some limitation on the domain over which Equation $(1)$ is true.

For a more mechanical proof, for a given $t$, $x$, and $z$ you can define $u(\eta) = t - z + \eta,$ $v(\eta) = x + \eta$, and $$ g(\eta) = f(u(\eta), v(\eta)). $$ Then $$ \frac d{d\eta} g(\eta) = \frac{\partial f(u,v)}{\partial u} \frac{du}{d\eta} + \frac{\partial f(u,v)}{\partial v} \frac{dv}{d\eta} = \frac{\partial f(u,v)}{\partial u} + \frac{\partial f(u,v)}{\partial v} = 0.$$ Now simply integrate $$ g(z) - g(0) = \int_0^z \frac d{d\eta} g(\eta)\,d\eta = 0. $$

Hence $f(t,x+z) = g(z) = g(0) = f(t-z,x).$