I'm trying to read through Symmetry Methods for Differential Equations by Peter Hydon. However, I have become stuck on something which looks like it should be very simple. On p. 12 it says (paraphrasing)
Differentiating $$\omega(x,y)=\omega(x,y+\epsilon)$$ with respect to $\epsilon$ at $\epsilon=0$ leads to the result $$\omega_y(x,y)=0.$$
However, I don't understand how to reach that result!
Please let me know if more context is required.
Edit:
I have tried thinking about it a different way:
Let $z=y+\epsilon$. Then,
\begin{align} \frac{\partial\omega(x,y+\epsilon)}{\partial \epsilon} &= \frac{\partial\omega(x,z)}{\partial \epsilon} \\ &= \frac{\partial\omega(x,z)}{\partial z} \frac{\partial z}{\partial \epsilon} \\ &= \frac{\partial\omega(x,y+\epsilon)}{\partial (y+\epsilon)} \frac{\partial (y+\epsilon)}{\partial \epsilon} \\ &=\frac{\partial\omega(x,y+\epsilon)}{\partial (y+\epsilon)} (1) \\ \end{align}
Now, taking $\epsilon=0$ yields \begin{align} \frac{\partial\omega(x,y+0)}{\partial \epsilon} &= \frac{\partial\omega(x,y+0)}{\partial (y+0)} \\ 0 &= \frac{\partial\omega(x,y)}{\partial y} \\ \end{align}
Is that valid?
Since you have \begin{align} \omega(x, y) = \omega(x, y+\varepsilon) \end{align} which means $\omega$ doesn't change in the $y$ direction. Anyhow, by chain rule, we have \begin{align} \frac{d}{d\epsilon}\omega(x, y+\epsilon) = \frac{\partial\omega}{\partial y}(x, y+\varepsilon) \frac{d}{d\epsilon}(y+\varepsilon) = \frac{\partial\omega}{\partial y}(x, y+\varepsilon). \end{align} Now, set $\varepsilon=0$, we get \begin{align} \frac{d}{d\epsilon}\omega(x, y+\epsilon)\Big|_{\varepsilon=0} = \frac{\partial\omega}{\partial y}(x, y). \end{align} On the other hand, using the assumption, we have \begin{align} \frac{d}{d\varepsilon}\omega(x, y+\varepsilon) = \frac{d}{d\varepsilon}\omega(x, y) =0. \end{align}