Second Derivative of Polar Coordinate Equation

Question

I am trying to convert the following equation to polar coordinates: $$ x'' - x' + x^2x'+x'^3 + x =0 $$

I know that $x' = r'\cos(\theta)-r\theta \sin(\theta)$, however I am not too sure where to go from here?

I know it should be $x'' = \frac{d}{dt}(r'\cos(\theta)-r\theta \sin(\theta))$ but not entirely sure how to evaluate this.

Thanks.

Answer 1

As you did, I'll denote with a prime the time derivative. Although you did not mention it explicitly, I assume that you are considering $$x = r\cos\theta\,.$$ Then everything is just a matter of differentiating compositions and products of functions.

You have $$x' = (r\cos\theta)' = r'\cos\theta + r(\cos\theta)' = r'\cos\theta -r\theta'\sin\theta\,.$$ Then $$(r'\cos\theta)' = r''\cos\theta + r'(\cos\theta)' = r''\cos\theta -r'\theta'\sin\theta$$ and $$(r\theta'\sin\theta)' = r'\theta'\sin\theta + r\theta''\sin\theta + r\theta'(\sin\theta)' = r'\theta'\sin\theta + r\theta''\sin\theta + r(\theta')^2\cos\theta\,.$$ Putting all together you have $$x'' = r''\cos\theta -2r'\theta'\sin\theta -r\theta''\sin\theta -r(\theta')^2\cos\theta\,.$$ Now you have the expressions for $x''$ and $x'$ and you have just to plug them in your initial ODE.

Answer 2

Another way to polar coordinates, that is also more adapted to the equation $$ x''+(x^2+x'^2-1)x'+x=0, $$ is to compute the radius and angle as polar coordinates of the pair $(x,y)=(x,x')$, so that $x=r\cosθ$, $y=x'=r\sinθ$ purely from the geometry. Then \begin{alignat}{3} rr'&=xx'+yy'&&=(x''+x)x'\\ r^2θ'&=xy'-x'y&&=-r^2+x(x''+x)\\ \end{alignat} and insert that $x''+x=(1-r^2)x'$, so that you end up with \begin{alignat}{3} r'&=&r(1-r^2)\sin^2θ&=\frac12r(1-r^2)(1-\cos(2θ))\\ θ'&=&-1+(1-r^2)\sinθ\cosθ&=-1+\frac12(1-r^2)\sin(2θ) \end{alignat} On average, this gives a clockwise rotation with constant speed $1$, the radius moving towards $r=1$ from all points except $r=0$.