First and second $r$-derivatives in terms of $x,y$

Question

For a differentiable function $f(x, y)$ with continuous second partial derivatives, $x=r\cos\theta$ and $y=r\sin\theta$, show that $f_r=f_x\cos\theta+f_y\sin\theta$ and $f_{rr}=f_{xx}\cos^2\theta+2f_{xy}\cos\theta\sin\theta+f_{yy}\sin^2\theta$. [Note that $f_x=\frac{\partial f}{\partial x}$ and $f_{xx}=\left(f_x\right)_x=\frac{\partial^2f}{\partial x^2}$]

My attempt:

I was able to prove the first part of the question. Here's what I did:

$$\begin{align} f_x&=\frac{\partial f}{\partial r}\times\frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\times\frac{\partial \theta}{\partial x}\\ &=f_r\times\frac1{\cos\theta}-f_{\theta}\times r\sin\theta\\ f_y&=\frac{\partial f}{\partial r}\times\frac{\partial r}{\partial y}+\frac{\partial f}{\partial \theta}\times\frac{\partial \theta}{\partial y}\\ &=f_r\times\frac1{\sin\theta}+f_{\theta}\times r\sin\theta \end{align}$$ I eliminated $f_\theta$ from these two equations and got the first part. But I am stuck on the second part. Any hints will be appreciated.

Answer 1

You have done the harder part already. Here's an outline to the second part:

• Apply the first part to $g=f_r$. You will get an expression for $f_{rr} = g_r$ in terms of $f_{rx} = g_x$ and $f_{ry} = g_y$.

• Since $f$ has continuous second derivatives, $f_{rx} = f_{xr}$ and $f_{ry} = f_{yr}$.

• Use the first part again on $f_x$ and $f_y$.

Answer 2

Denote by $\mathbf{J}= \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & + r \cos \theta \end{pmatrix}$ the Jacobian matrix so that $d\mathbf{x}=\mathbf{J} d\mathbf{u}$.

Here $ d\mathbf{x}= \begin{pmatrix} dx \\ dy \end{pmatrix}$ and $ d\mathbf{u}= \begin{pmatrix} dr \\ d\theta \end{pmatrix} $.

It is easy to see that gradients in both coordinates are linked by the relation $\mathbf{g}_u = \mathbf{J}^T \mathbf{g}_x$.

Here $ \mathbf{g}_x= \begin{pmatrix} f_x \\ f_y \end{pmatrix}$ and $ \mathbf{g}_u= \begin{pmatrix} f_r \\ f_\theta \end{pmatrix} $.

Differentiating this expression, yields \begin{eqnarray} d\mathbf{g}_u &=& (d\mathbf{J})^T \mathbf{g}_x + \mathbf{J}^T d\mathbf{g}_x = \left[ \mathbf{A} + \mathbf{J}^T \mathbf{H}_x \mathbf{J} \right] d\mathbf{u} \\ \end{eqnarray}

I leave you to check that the upper-left term of $\mathbf{A}$ is null so to compute $f_{rr}$ we are left with the upper-left term of $\mathbf{J}^T \mathbf{H}_x \mathbf{J}$ which is equal to $f_{xx} \cos^2 \theta+2f_{xy} \cos \theta \sin \theta+f_{yy} \sin^2 \theta$.

Answer 3

Here's what I did: $f_x=\ldots$

Why do that? You have to show that $f_r$ equals an expression on $f_x$ and $f_y$ and have $x$ and $y$ as expressions of $r$ and $\theta$. So instead start by applying the chain rule to $f_r$.

$\qquad\begin{align}f_r &= f_x~x_r+f_y~y_r&&\text{chain rule} \\[2ex] f_{rr} &=(f_x)_r~x_r+f_x~x_{rr}+(f_y)_r ~y_r+f_y~y_{rr}&&\text{product rule}\\&= (f_{xx}~x_r+f_{xy}~y_r)~x_r+f_x~x_{rr}+(f_{yx}~x_r+f_{yy}~y_r)~y_r+f_y~y_{rr}&&\text{chain rule}\end{align}$

The rest is just finding $x_r, x_{rr}, y_r,$ and $y_{rr}$.