I am tasked with the above, converting the 2-dimensional polar form of the Laplace transform:
$$ \frac{\partial}{\partial r}\Bigl(r^2 \frac{\partial z}{\partial r}\Bigr) + \frac{1}{\sin(\phi)} \frac{\partial} {\partial\phi}\Bigl(\sin(\phi)\frac{\partial z}{\partial \phi}\Bigr) = 0 $$
to the Legendre equation.
I was able to proceed, assuming solution $ z=\Phi(\phi)R(r) $ using method of separation acquiring equation $ \Phi $:
$$ \frac{d}{d\phi}\Bigl(\sin(\phi) \frac{d\Phi}{d\phi}\Bigr) + \lambda \sin(\phi)\Phi=0 $$
and equation $ R $:
$$ \frac{d}{dr}\Bigl(r^2 \frac{dR}{dr}\Bigr) - \lambda R=0 $$
Consider only Equation $ \Phi $. I substituted $ x = \cos(\phi) \implies \theta = \arccos(x)$ and applied chain rule twice:
$$ \frac{d(\Phi)}{dx}=\frac{d(\Phi)}{d\phi}\frac{d\phi}{dx}=-\frac{d(\Phi)}{d\phi}\frac{1}{\sqrt{1-x^2}} $$
Resulting in:
$$ \sqrt{1-x^2}\frac{d}{dx}\Bigl(y\sqrt{1-x^2}\frac{d\Phi}{dx}\Bigr) + \lambda y\Phi = 0 $$
However, the answer was supposed to simplify to:
$$ \frac{d}{dx}\Bigl((1-x^2)\frac{d\Phi}{dx}\Bigr) + \lambda\Phi =0 $$
I see the potential for the cancellation of factors which would get the correct answer, but the derivatives won't allow this. What have I done wrong?
OP here. I assumed the solution required the used of polar coordinates. This is incorrect, since the term $ \sin(\phi) $ should have been found in terms of $ x $ given $ x = \cos(\phi) $
Consider:
$$ \sin^2(\phi) + \cos^2(\phi) = 1 $$ $$ \sin(\phi)=\sqrt{1-\cos^2(\phi)} $$
Hence
$$ \sin(\phi) = \sqrt{1-x^2} $$
Substituting this into the equation rather than $ y $ results in the correct result.