MHB Partial derivatives-polar coordinates

[evinda]

Hello! :)

From the relations:
$$\partial_{r}=\cos \theta \cdot \partial_{x}+ \sin \theta \cdot \partial_y$$
$$\partial_{\theta}=-r \sin \theta \cdot \partial_x+ r \cos \theta \cdot \partial_y$$
we get:

$$\partial_y=\sin \theta \cdot \partial{r}+\frac{\cos \theta}{r} \cdot \partial_{ \theta}$$
$$\partial_x=\cos \theta \cdot \partial{r}-\frac{\sin \theta}{r} \cdot \partial_{\theta}$$

Now,I want to calculate $ \partial_{xx}$.

That's what I have tried:

$$\partial_{xx}=\frac{\partial}{\partial{x}}\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right)=\frac{\partial}{\partial{x}}\left(\cos \theta\right)\partial_r+ \cos \theta \cdot \partial_x \partial_r-\frac{\partial}{\partial{x}}\left(\frac{\sin\left(\theta\right)}{r}\right) \partial_{\theta}-\frac{\sin \theta}{r} \cdot \frac{\partial}{\partial{x}}\left(\frac{\partial}{\partial_{\theta}}\right)= \\ \left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \cos \theta \cdot \partial_r+\cos \theta \cdot \left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right)\partial_r-\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \left(\frac{\sin \theta}{r}\right) \cdot \partial_{\theta}-\frac{\sin \theta}{r}\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \frac{\partial}{\partial_{\theta}}= \\ \cos^2 \theta \cdot \partial_{rr}-\frac{\sin \theta}{r} \cdot \left(- \sin \theta \cdot \partial_r + \cos \theta \cdot \partial_{r \theta}\right)+\cos \theta\left(\cos \theta \partial_{rr}-\frac{\sin \theta}{r} \cdot \partial_{\theta r}\right)- \cos \theta\left(-\frac{\sin \theta}{r^2} \partial_{\theta}+\frac{\sin \theta}{r} \cdot \partial_{r \theta}\right)+ \frac{\sin \theta}{r^2} \cdot \cos \theta \partial_{\theta}+ \frac{\sin^2 \theta}{r^2} \cdot \partial_{\theta \theta}-\frac{\sin \theta}{r} \cos \theta \cdot \partial_{r \theta}+\frac{\sin^2 \theta}{r^2} \partial_{\theta \theta }$$
$\displaystyle{\partial_{*}=\frac{\partial}{\partial{*}}}$

I collected the terms and simplified them and I got:
$$2 \cos^2 \theta \partial_{rr}+\frac{2 \sin^2\theta}{r^2} \partial_{\theta \theta}+\frac{\sin^2\theta}{r} \partial_r-\frac{4 \cos \theta \sin \theta}{r} \partial_{r \theta}+\frac{2 \sin \theta \cos \theta}{r^2} \partial_{\theta}$$But...in my notes the result is:
$$\cos^2 \theta \partial_{rr}+\frac{\sin^2 \theta}{r^2} \partial_{\theta \theta}+\frac{\sin^2 \theta}{r} \partial_r-\frac{2 \cos \theta \sin \theta}{r} \partial_{r \theta}+\frac{2 \sin \theta \cos \theta}{r^2} \partial_{\theta}$$

Have I done something wrong? (Thinking)

 

[I like Serena]

Hi! (Blush)

I'm trying to follow what you did.

How did you get from:
$$\frac{\partial}{\partial{x}}\left(\cos \theta\right)\partial_r+ \cos \theta \cdot \partial_x \partial_r-\frac{\partial}{\partial{x}}\left(\frac{\sin\left(\theta\right)}{r}\right) \partial_{\theta}-\frac{\sin \theta}{r} \cdot \frac{\partial}{\partial{x}}\left(\frac{\partial}{\partial_{\theta}}\right)$$
to this:
$$\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \cos \theta \cdot \partial_r+\cos \theta \cdot \left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right)\partial_r-\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \left(\frac{\sin \theta}{r}\right) \cdot \partial_{\theta}-\frac{\sin \theta}{r}\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \frac{\partial}{\partial_{\theta}}$$
(Thinking)(Wondering)

 

[evinda]

Hi! (Blush)

I'm trying to follow what you did.

How did you get from:
$$\frac{\partial}{\partial{x}}\left(\cos \theta\right)\partial_r+ \cos \theta \cdot \partial_x \partial_r-\frac{\partial}{\partial{x}}\left(\frac{\sin\left(\theta\right)}{r}\right) \partial_{\theta}-\frac{\sin \theta}{r} \cdot \frac{\partial}{\partial{x}}\left(\frac{\partial}{\partial_{\theta}}\right)$$
to this:
$$\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \cos \theta \cdot \partial_r+\cos \theta \cdot \left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right)\partial_r-\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \left(\frac{\sin \theta}{r}\right) \cdot \partial_{\theta}-\frac{\sin \theta}{r}\left(\cos \theta \cdot \partial_r-\frac{\sin \theta}{r} \cdot \partial_{\theta}\right) \cdot \frac{\partial}{\partial_{\theta}}$$
(Thinking)(Wondering)

I replaced $\displaystyle{\frac{\partial}{\partial_x}}$ with $\displaystyle{\partial_x=\cos \theta \cdot \partial{r}-\frac{\sin \theta}{r} \cdot \partial_{\theta}}$.
Shouldn't I have done it like that? (Thinking)(Thinking)(Thinking)

 

[I like Serena]

I replaced $\displaystyle{\frac{\partial}{\partial_x}}$ with $\displaystyle{\partial_x=\cos \theta \cdot \partial{r}-\frac{\sin \theta}{r} \cdot \partial_{\theta}}$.
Shouldn't I have done it like that? (Thinking)(Thinking)(Thinking)

Aah! Now I see!
You're mixing up products with compositions.

Note that for instance:
$$\frac \partial{\partial x} \cdot \frac \partial{\partial x}=\left(\frac \partial{\partial x}\right)^2$$
is different from:
$$\frac \partial{\partial x} \circ \frac \partial{\partial x}=\frac {\partial^2}{\partial x^2}$$

When you write $\partial_x$ this is short hand notation.
At all times you need to be aware that it "multiplies on the left" and makes a "composition on the right".

 

[evinda]

Aah! Now I see!
And no, you can't do it just like that.
You're mixing up products with compositions.

Note that for instance:
$$\frac \partial{\partial x} \cdot \frac \partial{\partial x}=\left(\frac \partial{\partial x}\right)^2$$
is different from:
$$\frac \partial{\partial x} \circ \frac \partial{\partial x}=\frac {\partial^2}{\partial x^2}$$

When you write $\partial_x$ this is short hand notation.
At all times you need to be aware that it "multiplies on the left" and makes a "composition on the right".

In our case, I can't just find the product: $\displaystyle{\frac \partial{\partial x} \cdot \frac \partial{\partial x}=\left(\frac \partial{\partial x}\right)^2}$ ,because $\displaystyle{\frac{\partial}{\partial_x}}$ is not linear,right?

 

[I like Serena]

In our case, I can't just find the product: $\displaystyle{\frac \partial{\partial x} \cdot \frac \partial{\partial x}=\left(\frac \partial{\partial x}\right)^2}$ ,because $\displaystyle{\frac{\partial}{\partial_x}}$ is not linear,right?

Umm... not quite sure what you're saying here... (Wondering)
It's true enough that $\displaystyle{\frac{\partial}{\partial x}}$ is not linear, but we do have that:

$$\left(\frac \partial{\partial x} \cdot \frac \partial{\partial x}\right)\Big(f(x)\Big)
=\left(\frac \partial{\partial x}\Big(f(x)\Big)\right)^2
=\left(\frac \partial{\partial x}\right)^2\Big(f(x)\Big)$$
Actually, you can replace it like you did, but you have to be very careful to distinguish product from composition.

So for instance:
\begin{aligned}\partial_x(\cos \theta)\partial_r
&= \Big(\partial_x \circ (\cos \theta)\Big) \cdot \partial_r \\
&= \Big(\cos \theta \cdot \partial_r - \frac{\sin \theta}{r} \cdot \partial_\theta\Big) \circ (\cos \theta) \cdot \partial_r \\
&= \Big(\cos \theta \cdot \partial_r(\cos \theta) - \frac{\sin \theta}{r} \cdot \partial_\theta(\cos \theta)\Big)\cdot \partial_r \\
&= \Big(\cos \theta \cdot 0 - \frac{\sin \theta}{r} \cdot -\sin \theta\Big)\cdot \partial_r \\
&=\frac{\sin^2 \theta}{r}\cdot \partial_r
\end{aligned}
(Wasntme)