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I have the following question:

Let $u(x, y)$ be a function with continuous second order partial derivatives. Use the chain rule to transform the expression: $$ x^2\frac{\partial^2u}{\partial y^2}-xy\frac{\partial^2u}{\partial y\partial x}+x\frac{\partial u}{\partial y} $$ into polar coordinates.

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This is what I did, but I'm not sure if this is right.

First I find the second-order partial derivatives, by using the chain rule:

$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\Rightarrow\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}=\frac{\partial u}{\partial \theta}-\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\Rightarrow\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}-\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\frac{\partial \theta}{\partial x}$

$\frac{\partial u}{\partial x}=-\frac{\partial u}{\partial \theta}\cdot\frac{y}{x^2+y^2}+\frac{\partial u}{\partial y}r\cos(\theta)\cdot\frac{y}{x^2+y^2}$

$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}\Rightarrow\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial u}{\partial r}-\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}\Rightarrow\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}-\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}\frac{\partial r}{\partial y}$

$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\cdot\frac{y}{(x^2+y^2)^\frac{1}{2}}-\frac{\partial u}{\partial x}\cos(\theta)\cdot\frac{y}{(x^2+y^2)^\frac{1}{2}}$

$\frac{\partial^2 u}{\partial y \partial x}=\frac{\partial}{\partial y}(\frac{\partial u}{\partial x})=-\frac{\partial u}{\partial \theta}\frac{\partial}{\partial y}(\frac{y}{x^2+y^2})+\frac{\partial u}{\partial y}r \cos \theta\frac{\partial}{\partial y}(\frac{y}{x^2+y^2})$

$\frac{\partial^2 u}{\partial y \partial x}=-\frac{\partial u}{\partial \theta}\cdot\frac{x^2-y^2}{(x^2+y^2)^2}+\frac{\partial u}{\partial y}r\cos(\theta)\cdot\frac{x^2-y^2}{(x^2+y^2)^2}$

$\frac{\partial^2 u}{\partial y \partial x}=-\frac{\partial u}{\partial \theta}\cdot\frac{r^2cos(2\theta)}{r^4}+\frac{\partial u}{\partial y}r\cos(\theta)\cdot\frac{r^2cos(2\theta)}{r^4}$

$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial r}\frac{\partial}{\partial y}(y\cdot(x^2+y^2)^{-\frac{1}{2}})-\frac{\partial u}{\partial x}\cos(\theta)\frac{\partial}{\partial y}(y\cdot(x^2+y^2)^{-\frac{1}{2}})$

$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial r}(\frac{1}{\sqrt{x^2+y^2}}-\frac{3}{2}2y^2(x^2+y^2)^{-\frac{3}{2}})-\frac{\partial u}{\partial x}\cos(\theta)(\frac{1}{\sqrt{x^2+y^2}}-\frac{3}{2}2y^2(x^2+y^2)^{-\frac{3}{2}})$

$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial r}(\frac{1}{r}-\frac{3}{2}2r^2\sin^2(\theta)r^\frac{1}{2})-\frac{\partial u}{\partial x}\cos(\theta)(\frac{1}{r}-\frac{3}{2}2r^2\sin^2(\theta)r^\frac{1}{2})$

Now using the fact that $x=r \cos(\theta)$ and $y=r \sin(\theta)$, I find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ in terms of $r$ and $\theta$:

$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}\Rightarrow\frac{\partial u}{\partial x}\cos (\theta)+\frac{\partial u}{\partial y}\sin(\theta)$

$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}\Rightarrow\frac{\partial u}{\partial x}\left(-r\right)\sin(\theta)+\frac{\partial u}{\partial y}r \cos(\theta)$

Writing in terms of a matrix:

$\begin{bmatrix} \frac{\partial u}{\partial r}\\ \frac{\partial u}{\partial \theta} \end{bmatrix}=\begin{bmatrix} \cos \theta & \sin \theta\\ -r \sin \theta & r \cos \theta \end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x}\\ \frac{\partial u}{\partial y} \end{bmatrix}$

$\begin{bmatrix} \frac{\partial u}{\partial x}\\ \frac{\partial u}{\partial y} \end{bmatrix}=\begin{bmatrix} \cos \theta & \sin \theta\\ -r \sin \theta & r \cos \theta \end{bmatrix}^{-1}\begin{bmatrix} \frac{\partial u}{\partial r}\\ \frac{\partial u}{\partial \theta} \end{bmatrix}$

$\begin{bmatrix} \frac{\partial u}{\partial x}\\ \frac{\partial u}{\partial y} \end{bmatrix}=\begin{bmatrix} \cos \theta & -\frac{\sin \theta}{r}\\ \sin \theta & \frac{\cos \theta}{r} \end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial r}\\ \frac{\partial u}{\partial \theta} \end{bmatrix}$

From here it's just a simple case of plugging in terms into the expression, which I am too lazy to do. Can anyone confirm that this is indeed correct?

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