vector field $X= \frac{\partial }{\partial \rho} + \frac{\partial }{\partial \theta}$ corresponding differential equation in Cartesian coordinates

Question

Given the vector field in the plane that has the following expression in polar coordinate $(\rho,\theta)$ : $$X= \frac{\partial }{\partial \rho} + \frac{\partial }{\partial \theta}$$ what is the corresponding differential equation in Cartesian coordinates. I'm looking for a solution in the form of: $$\begin{cases} \dot{x}=f(x,y) \\ \dot{y}=g(x,y) \end{cases}$$ The problem is that I'm not used to the representation of vector fields as above.

Answer 1

By the chain rule, \begin{align*} \frac{\partial}{\partial\rho} &= \frac{\partial x}{\partial\rho}\frac{\partial}{\partial x} + \frac{\partial y}{\partial\rho}\frac{\partial}{\partial y} \\ &= \cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y} = \frac x{\sqrt{x^2+y^2}}\frac{\partial}{\partial x} + \frac y{\sqrt{x^2+y^2}}\frac{\partial}{\partial y} \quad\text{and} \\ \frac{\partial}{\partial\theta} &= \frac{\partial x}{\partial\theta}\frac{\partial}{\partial x} + \frac{\partial y}{\partial\theta}\frac{\partial}{\partial y} \\ &= -\rho\sin\theta \frac{\partial}{\partial x} + \rho\cos\theta \frac{\partial}{\partial y} = -y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}. \end{align*} So we have $$X = \frac{\partial}{\partial\rho} + \frac{\partial}{\partial\theta} = \left(\frac x{\sqrt{x^2+y^2}} - y\right) \frac{\partial}{\partial x} + \left(\frac y{\sqrt{x^2+y^2}} +x\right) \frac{\partial}{\partial y}.$$ Can you take it from here?

Answer 2

In $xy$ coordinates your vector field is \begin{align} X &= \frac{\partial }{\partial \rho} + \frac{\partial }{\partial \theta}\\ &=\frac{\partial x}{\partial \rho}\frac{\partial }{\partial x}+\frac{\partial y}{\partial \rho}\frac{\partial }{\partial y}+\frac{\partial x}{\partial \theta}\frac{\partial }{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial }{\partial y}\\ &=\cos\theta\frac{\partial }{\partial x}+\sin\theta\frac{\partial }{\partial y}-\rho\sin\theta\frac{\partial }{\partial x}+\rho\cos\theta\frac{\partial }{\partial y}\\ &=\left(\frac{x}{\sqrt{x^2+y^2}}-y\right)\frac{\partial }{\partial x}+\left(\frac{y}{\sqrt{x^2+y^2}}+x\right)\frac{\partial }{\partial y}. \end{align} So you get the equation $$ \dot x=\left(\frac{x}{\sqrt{x^2+y^2}}-y\right),\quad \dot y=\left(\frac{y}{\sqrt{x^2+y^2}}+x\right). $$ But in case you want to solve the equation, you really should use polar coordinates.