Canonical coordinates of first order differential equation.

Question

Consider the first order differential equation \begin{align} \frac{\text{d}y}{\text{d}x}=\frac{\text{d}f(x)}{\text{d}x}y+g(x) \tag{1} \end{align} for some arbitrary functions $f$ and $g$. This DE is invariant under the one parameter transformation group \begin{align} x_1=x+\mathcal{O}(\varepsilon^2),~~~~y_1=y+\varepsilon e^{f(x)}+\mathcal{O}(\varepsilon^2). \tag{2} \end{align} To solve (1) I am trying to use (2) to obtain the canonical coordinates $(u(x,y),v(x,y))$ which will allow me to write (1) in the simpler form \begin{align} \frac{\text{d}v}{\text{d}u}=G(u) \end{align} for some function $G(u)$. By demanding transnational invariance in $v$, the quasi-linear equations satisfied by the canonical coordinates are \begin{align} X\frac{\partial u}{\partial x}+Y\frac{\partial u}{\partial y}=0 ~~~~\text{ and }~~~~ X\frac{\partial v}{\partial x}+Y\frac{\partial v}{\partial y}=1. \end{align} From (2) we have that $X=0$ and $Y=e^{f(x)}$; \begin{align} e^{f(x)}\frac{\partial u}{\partial y}=0,~~~~~~~e^{f(x)}\frac{\partial v}{\partial y}=1 \tag{3a, 3b}. \end{align} Equation (3a) says that $u=h(x)$ for some arbitrary function $h$ and equation (3b) says that $v=ye^{-f(x)}+k(x)$ for some arbitrary function $k$. If I now try to write (1) in terms of the canonical coordinates I get \begin{align} \frac{\text{d}v}{\text{d}u}=\frac{\text{d}v/\text{d}x}{\text{d}u/\text{d}x}&=\frac{\frac{\text{d}y}{\text{d}x}e^{-f}-y \frac{\text{d}f}{\text{d}x}e^{-f}+\frac{\text{d}k}{\text{d}x}}{\frac{\text{d}h}{\text{d}x}}\\[6pt] &=\frac{\bigg(\frac{\text{d}f}{\text{d}x}y+g\bigg)e^{-f}-y \frac{\text{d}f}{\text{d}x}e^{-f}+\frac{\text{d}k}{\text{d}x}}{\frac{\text{d}h}{\text{d}x}}\\[6pt] &=\frac{ge^{-f}+\frac{\text{d}k}{\text{d}x}}{\frac{\text{d}h}{\text{d}x}} \tag{4}. \end{align} Now I am not sure how I can write (4) as $\text{d}v/\text{d}u=G(h(x))$ for some function $G$. I think I am probably allowed to choose $k(x)=0$, but I still don't see how to get the results I am looking for.

Answer 1

I have not checked your whole procedure, but you would do the following, by assuming the most simple forms for $h$ and $k$:

$$u=h(x)=x \implies x(u,v)=u$$ $$v=y\exp(-f(x))+k(x)=y\exp(-f(x)) \implies y(u,v)=v\exp(f(u)).$$

Now, determine:

$$\dfrac{dy}{dx}=\dfrac{\dfrac{\partial y}{\partial u}du+\dfrac{\partial y}{\partial v}dv}{\dfrac{\partial x}{\partial u}du+\dfrac{\partial x}{\partial v}dv}=\dfrac{\dfrac{\partial y}{\partial u}+\dfrac{\partial y}{\partial v}\dfrac{dv}{du}}{\dfrac{\partial x}{\partial u}+\dfrac{\partial x}{\partial v}\dfrac{dv}{du}}=\dfrac{v\exp(f(u))\dfrac{df}{du}+\exp(f(u))\dfrac{dv}{du}}{1+0\cdot\dfrac{dv}{du}}$$ $$=v\exp(f(u))\dfrac{df}{du}+\exp(f(u))\dfrac{dv}{du}$$

Using the ODE will give:

$$v\exp(f(u))\dfrac{df}{du}+\exp(f(u))\dfrac{dv}{du}=\dfrac{df}{du}v\exp(f(u))+g(u)$$

$$\implies\dfrac{dv}{du}=\exp(f(u))g(u)\implies v=c+\int\exp(f(u))g(u)du$$