Question

# An important derivative operation in many applications is called the Laplacian; in Cartesian coordinates, for z=f(x, y), the Laplacian is$z _ { x x } + z _ { y y }$. Determine the Laplacian in polar coordinates using the following steps. Use the Chain Rule to find$z _ { x x } = \frac { \partial } { \partial x } \left( z _ { x } \right)$. There should be two major terms, which, when expanded and simplified, result in five terms.

Solution

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Step 1
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Using that

\begin{align*}r_x=\dfrac{x}{\sqrt{x^2+y^2}}=\dfrac{x}{r},\theta_x=-\dfrac{y}{x^2+y^2}=-\dfrac{y}{r^2}\end{align*}

we obtain

\begin{align*}z_{xx}&=\dfrac{\partial}{\partial x}(z_x)=\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}z_r-\dfrac{y}{r^2}\right)=\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}z_r\right)-\dfrac{\partial}{\partial x}\left(\dfrac{y}{r^2}z_{\theta}\right)\\&= \underbrace{\left(\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}\right)z_r+\dfrac{x}{r}\dfrac{\partial}{\partial x}(z_r)\right)}_{\color{#4257b2}\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}z_r\right)}-\underbrace{\left(\dfrac{\partial}{\partial x}\left(\dfrac{y}{r^2}\right)r_{\theta}+\dfrac{y}{r^2}\dfrac{\partial}{\partial x}(r_{\theta})\right)}_{\color{#4257b2}\dfrac{\partial}{\partial x}\left(\dfrac{y}{r^2}z_{\theta}\right)}\hspace{2cm}\text{\color{#4257b2}(Product Rule)}\\& =\dfrac{\partial}{\partial x}\left(\dfrac{x}{\sqrt{x^2+y^2}}\right)z_r+\dfrac{x}{r}\left(\dfrac{\partial}{\partial r}(z_r)\dfrac{\partial r}{\partial x}+\dfrac{\partial}{\partial \theta}(z_r)\dfrac{\partial\theta}{\partial x}\right)\\&\hspace{5cm}-\dfrac{\partial}{\partial x}\left(\dfrac{y}{x^2+y^2}\right)r_{\theta}-\dfrac{y}{r^2}\left(\dfrac{\partial}{\partial r}(r_{\theta})\dfrac{\partial r}{\partial x}+\dfrac{\partial}{\partial \theta}(r_{\theta})\dfrac{\partial\theta}{\partial x}\right)\\&= \dfrac{\sqrt{x^2+y^2}-\dfrac{x^2}{\sqrt{x^2+y^2}}}{x^2+y^2}z_r+\dfrac{x}{r}\left(z_{rr}\dfrac{x}{r}-z_{\theta r}\dfrac{y}{r^2}\right)+\dfrac{2xy}{(x^2+y^2)^2}r_{\theta}-\dfrac{y}{r^2}\left(z_{r\theta}\dfrac{x}{r}-r_{\theta\theta}\dfrac{y}{r^2}\right)\\&=\dfrac{x^2}{r^2}z_{rr}+\dfrac{y^2}{z^4}r_{\theta\theta}-\dfrac{2xy}{r^3}z_{r\theta}+\dfrac{y^2}{r^3}z_r+\dfrac{2xy}{r^4}r_{\theta}\end{align*}

Hence

\begin{align*}\color{#c34632}\boxed{z_{xx}=\dfrac{x^2}{r^2}z_{rr}+\dfrac{y^2}{z^4}r_{\theta\theta}-\dfrac{2xy}{r^3}z_{r\theta}+\dfrac{y^2}{r^3}z_r+\dfrac{2xy}{r^4}r_{\theta}}\end{align*}

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