I am trying to see why

$\big(\partial_{xx} + \partial_{yy}\big) u(r, \theta) = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta}$

I first use the chain rule to say that:

$\frac{\partial u}{\partial x} = u_r r_x + u_{\theta} \theta_x$

And then I calculate:

$r_x = \frac{x}{\sqrt{x^2 + y^2}} = \frac{rcos\theta}{r} = cos\theta$

$\theta_x = \frac{-y}{x^2+y^2} = \frac{-rsin\theta}{r^2} = \frac{-sin\theta}{r}$

Plugging in gives

$\frac{\partial u}{\partial x} = u_r cos\theta - u_{\theta} \frac{sin\theta}{r}$

But I am unsure of how to take the next $x$ derivative and I am wondering if someone can help?


2 Answers 2


Let's start with $$ \begin{cases} x=r\cos(\theta)\\ y=r\sin(\theta). \end{cases} $$

We compute first $u_r:$ $$ u_r=u_xx_r+u_yy_r=\cos\theta u_x+\sin\theta u_y. $$

$$ u_{rr}=\cos\theta u_{xr}+\sin\theta u_{yr}=\cos\theta (u_{xx}x_r+u_{xy}y_r)+sin\theta(u_{xy}x_r +u_{yy}y_r)=\\ =\cos^2\theta u_{xx}+2\cos\theta\sin\theta u_{xy}+\sin^2\theta u_{yy}. $$


$$ u_\theta=u_x x_\theta+u_yy_\theta=-r\sin\theta u_x+r\cos\theta u_y. $$ So $$ u_{\theta\theta}=-r(\cos\theta u_x+\sin\theta u_y)+r^2(\sin^2\theta u_{xx}-2\cos\theta\sin\theta u_{xy}+\cos^2\theta u_{yy}). $$

Dividing both sides by $r^2$, adding $u_{\theta\theta}$ and $u_{rr}$, and rearraging term we obtain: $$ \Delta u=u_{rr}+\frac 1{r^2} u_{\theta\theta}+\frac 1r u_r $$

A bit of calculation omitted at the end, feel free to ask if something is unclear or wrong.


Use the chain rule again on the functions $u_r, u_\theta, r_x, \theta_x$ (and product rule on the products):

$u_{xx}=(u_{rr}r_x+u_{r\theta}\theta_x)r_x+u_r(r_{xr}r_x+r_{x\theta}\theta_x)+(u_{\theta r}r_x+u_{\theta\theta}\theta_x)\theta_x+u_\theta(\theta_{xr}r_x+\theta_{x\theta}\theta_x)$.

I get:


Likewise for the $u_{yy}$'s you get (using $r_y=\sin\theta, \theta_y=\frac{\cos\theta}{r}$):



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