Introduction
In this problem we will derive the Laplacian in polar coordinates using the identities

At the end of the problem you will find a complete list of the notations that you will need to enter and how to enter them.

Part 1 - Find $\displaystyle\frac{\partial r}{\partial x},\frac{\partial r}{\partial y}, \frac{\partial \theta}{\partial x}, \frac{\partial \theta}{\partial y}$

To find $\displaystyle\frac{\partial r}{\partial x},\frac{\partial r}{\partial y}, \frac{\partial \theta}{\partial x}$, and $\displaystyle\frac{\partial \theta}{\partial y}$ we will implicitly differentiate the above equations. Unlike calc I, x and y are both independent variables, so that $\displaystyle\frac{dy}{dx} = 0$.
Implicitly differentiating both sides of (iii) with respect to x, without simplifying, gives us

Solving for $\displaystyle \frac{\partial r}{\partial x}$ we get

Similarly, implicitly differentiating both sides of (iii) with respect to y, without simplifying, gives us

Solving for $\displaystyle \frac{\partial r}{\partial y}$ we get

Now we need to find $\displaystyle \frac{\partial \theta}{\partial x}$ and $\displaystyle \frac{\partial \theta}{\partial y},$ which take a bit more work. Implicitly differentiating both sides of (iv) with respect to x gives

Substitute (i) and (ii) into the right hand side of this equation for x and y respectively, and solve for $\displaystyle \frac{\partial \theta}{\partial x}$

Implicitly differentiating both sides of (iv) with respect to y gives

Substitute (i) and (ii) into the right hand side of this equation for x and y respectively, and solve for $\displaystyle \frac{\partial \theta}{\partial y}$

Part 2 - Find $\displaystyle\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}$

By the chain rule, we know that

$\displaystyle\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}$   and   $\displaystyle\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial y}$

Substituting the results from part 1 into these equations we get

$\displaystyle\frac{\partial u}{\partial x} =$

$\displaystyle\frac{\partial u}{\partial y} =$

Part 3 - Find $\displaystyle\frac{\partial^2 u}{\partial x^2}$

$\displaystyle\frac{\partial^2 u}{\partial x^2} = \frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right) = \left(\cos(\theta)\frac{\partial }{\partial r} - \frac{\sin(\theta)}{r}\frac{\partial }{\partial \theta}\right)\left(\cos(\theta)\frac{\partial u}{\partial r} - \frac{\sin(\theta)}{r}\frac{\partial u}{\partial \theta}\right)$

$\displaystyle\hskip 85pt = \cos(\theta)\frac{\partial}{\partial r}\left(\cos(\theta)\frac{\partial u}{\partial r}\right) + \cos(\theta)\frac{\partial}{\partial r}\left(-\frac{\sin(\theta)}{r}\frac{\partial u}{\partial \theta}\right) -\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\left(\cos(\theta)\frac{\partial u}{\partial r}\right)+ \frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\left(\frac{\sin(\theta)}{r}\frac{\partial u}{\partial \theta}\right)$

Remembering to use the product rule when it's appropriate

$\displaystyle\cos(\theta)\frac{\partial}{\partial r}\left(\cos(\theta)\frac{\partial u}{\partial r}\right) =$

$\displaystyle\cos(\theta)\frac{\partial }{\partial r}\left( -\frac{\sin(\theta)}{r}\frac{\partial u}{\partial \theta}\right) =$

$\displaystyle-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\left(\cos(\theta)\frac{\partial u}{\partial r}\right) =$

$\displaystyle\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\left(\frac{\sin(\theta)}{r}\frac{\partial u}{\partial \theta}\right) =$

Part 4 - Find $\displaystyle\frac{\partial^2 u}{\partial y^2}$

$\displaystyle\frac{\partial^2 u}{\partial y^2} = \frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y}\right) = \left(\sin(\theta)\frac{\partial }{\partial r} + \frac{\cos(\theta)}{r}\frac{\partial }{\partial \theta}\right)\left(\sin(\theta)\frac{\partial u}{\partial r} + \frac{\cos(\theta)}{r}\frac{\partial u}{\partial \theta}\right)$

$\displaystyle\hskip 85pt = \sin(\theta)\frac{\partial}{\partial r}\left(\sin(\theta)\frac{\partial u}{\partial r}\right) + \sin(\theta)\frac{\partial}{\partial r}\left(\frac{\cos(\theta)}{r}\frac{\partial u}{\partial \theta}\right) +\frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial u}{\partial r}\right)+ \frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta}\left(\frac{\cos(\theta)}{r}\frac{\partial u}{\partial \theta}\right)$

$\displaystyle\sin(\theta)\frac{\partial}{\partial r}\left(\sin(\theta)\frac{\partial u}{\partial r}\right) =$

$\displaystyle\sin(\theta)\frac{\partial}{\partial r}\left(\frac{\cos(\theta)}{r}\frac{\partial u}{\partial \theta}\right) =$

$\displaystyle\frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial u}{\partial r}\right) =$

$\displaystyle\frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta}\left(\frac{\cos(\theta)}{r}\frac{\partial u}{\partial \theta}\right) =$

Part 5 - Putting it all together

Using the results from parts 3 and 4 we see that

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

Similarly

$\displaystyle\frac{\partial^2 u}{\partial y^2} = \sin^2(\theta)\frac{\partial^2 u}{\partial r^2} + \frac{2\sin(\theta)\cos(\theta)}{r}\frac{\partial^2 u}{\partial r \partial\theta}+\frac{\cos^2(\theta)}{r^2}\frac{\partial^2 u}{\partial \theta^2}+\frac{\cos^2(\theta)}{r}\frac{\partial u}{\partial r}-\frac{2\sin(\theta)\cos(\theta)}{r^2}\frac{\partial u}{\partial\theta}$
Therefore, making use of some trig identities,

$\displaystyle\nabla^2 u = \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ =

Notation Here is a list of the notations you will need to input your answers

$\displaystyle \theta = \text{theta} \qquad \frac{\partial \theta}{\partial x} = \text{thetax} \qquad \frac{\partial \theta}{\partial y} = \text{thetay}$

$\displaystyle \theta = \text{theta} \qquad \frac{\partial \theta}{\partial x} = \text{thetax} \qquad \frac{\partial \theta}{\partial y} = \text{thetay}$

$\displaystyle \frac{\partial r}{\partial x} = \text{rx} \qquad \frac{\partial r}{\partial y} = \text{ry}$

$\displaystyle \frac{\partial u}{\partial r} = \text{ur} \qquad \frac{\partial u}{\partial \theta} = \text{utheta}$

$\displaystyle \frac{\partial^2 u}{\partial r^2} = \text{urr} \qquad \frac{\partial^2 u}{\partial \theta^2} = \text{uthetatheta}$

$\displaystyle \frac{\partial^2 u}{\partial r\partial\theta} = \frac{\partial^2 u}{\partial\theta\partial r}=\text{urtheta}=\text{uthetar}$