Polar Coordinates tranformation for Linear Homogeneous Differential Equations (1st order)

Question

While studying a book of Differential Equations I found this problem so interesting.

Suppose \begin{equation} M(x,y)dx+N(x,y)dy=0 \tag 1 \end{equation}

is a homogeneous ODE. Show that the transformation $x=r \cos (\theta) $ and $y= r \sin (\theta) $ reduces the equation to a separable equation in the variables $r$ and $\theta$

Is from the book Diff. Eq's, Shepley L. Ross.

So starting from the hypothesis that the equation is homogenous then $(1)$ is equivalent to

\begin{equation} \frac{dy}{dx}=g\left(\frac{y}{x}\right) \tag 2 \end{equation} So the thing is thatI don't know how to relate $x$ and $y$ , or more precisely how to find the relation $\dfrac{dr}{d\theta}$ (or maybe the other way around)

The first thing that came to mind was $r^2=x^2+y^2$ but how do I differentiate it ? I mean, I don´t see clearly how to use the chain rule

I have seen this $2rr'=2xx'+2yy'$. Although, still not clear how did they do it.

Later, a silly approach (I think so) was to take the differentials of either $x$ and $y$ with respect to $r$ and $\theta$, respectively. Therefore:

$$x=r \cos (\theta) \Rightarrow dx=\cos (\theta) dr $$ and $$y= r \sin (\theta) \Rightarrow dy= r \cos (\theta) d\theta .$$ Later

$$\frac{dy}{dx}=\frac{r \cos (\theta) d\theta} { \cos (\theta) dr} = \frac{r d\theta} {dr}$$

So after substituting in $(2)$

\begin{equation} \frac{r d\theta} {dr}=g\left(\frac{\sin \theta }{ \cos \theta }\right) \end{equation}

which reduces it to a separable equation

\begin{equation} \frac{dr } {r }=\frac{d\theta} { g(\tan \theta )} \end{equation}

But.... come on! At least I tried...

Later the book has also as an exercise to prove that the same equations is invariant under the tranformations $x=k\alpha$ and $y=k\beta$ with $k$ constant. But I think that the previous one seems more approachable.

Could someone help me with this kind of problems? Thanks. :)

Answer 1

A differential equation of the form

$M(x, y)dx + N(x, y)dy = 0 \tag{1}$

is homogeneous of degree $n$ provided that there is an $n \in \Bbb Z$ such that

$M(tx, ty) = t^nM(x, y) \tag{2}$

and

$N(tx, ty) = t^nN(x, y); \tag{3}$

if (1) is such an equation, setting

$x = r\cos \theta \tag{4}$

and

$y = r\sin \theta, \tag{5}$

we find

$M(r\cos \theta, r\sin \theta)dx + N(r\cos \theta, r\sin \theta)dy = 0 \tag{6}$

becomes

$r^nM(\cos \theta, \sin \theta)dx + r^nN(\cos \theta, \sin \theta) dy = 0. \tag{7}$

Now (4) and (5) yield

$dx = dr\cos \theta - rd\theta \sin \theta \tag{8}$

and

$dy = dr \sin \theta + rd\theta \cos \theta; \tag{9}$

inserting these two equations into (7) we obtain

$r^nM(\cos \theta, \sin \theta)(dr\cos \theta - rd\theta \sin \theta)$ $+ r^nN(\cos \theta, \sin \theta) (dr \sin \theta + rd\theta \cos \theta) = 0. \tag{10}$

We gather like terms (in $dr$ and $d\theta$):

$r^n(M(\cos \theta, \sin \theta)\cos \theta + N(\cos \theta, \sin \theta)\sin \theta)dr$ $- r^{n + 1}(M(\cos \theta, \sin \theta)\sin \theta - N(\cos\theta, \sin \theta)\cos \theta) d\theta = 0. \tag{11}$

Some minor algebraic fiddling yields (note we can cancel $r^n$ as long as $r \ne 0$, where polars are in any event undefined):

$\dfrac{dr}{r} = \dfrac{M(\cos \theta, \sin \theta)\sin \theta - N(\cos\theta, \sin \theta)\cos \theta}{M(\cos \theta, \sin \theta)\cos \theta + N(\cos \theta, \sin \theta)\sin \theta} d\theta, \tag{12}$

and voila!!! seperated variables.

Now try integrating it.

Answer 2

This is what I've got:

Given this fancy thing (At least I think so..)

$$ \frac{dy}{dx}=\frac{dy}{d\theta} \times \frac{d\theta}{d x} = \frac{ \frac{dy}{d\theta} } { \frac{dx}{d\theta}} $$

So $x=r\cos(\theta)$ and $y=r \sin(\theta)$ ,yield: $$\frac{dx}{d\theta}= \cos (\theta) \frac{dr}{d\theta}-r \sin(\theta)$$ and $$\frac{dy}{d\theta}= \sin (\theta) \frac{dr}{d\theta}+r \cos(\theta)$$

Therefore:

$$ \frac{dy}{dx}=\frac{ \sin (\theta) \frac{dr}{d\theta}+r \cos(\theta) } { \cos(\theta) \frac{dr}{d\theta}-r \sin(\theta)} $$ (*)

Knowing that (1) is equivalent to

$$ \frac{dy}{dx} = g(\frac{y}{x}) $$ , thus plugging (*)...

$$\frac{ \sin (\theta) \frac{dr}{d\theta}+r \cos(\theta) } { \cos (\theta) \frac{dr}{d\theta}-r \sin(\theta)} = g(\frac{y}{x}) $$

Doing a little of algebra, (Hopefully I did it right, tho)...

$$ \frac{dr}{r} = \frac{ \sin (\theta)+ \cos(\theta) } { \sin(\theta) - \cos (\theta) } \frac{d\theta}{g( \tan \theta )} $$

What do you think? :s