"Binomial" partial derivatives/PDE

Question

I have encountered the following expression,

$$x^2 \frac{\partial^2 u}{\partial x^2} + 2 x y \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2}=C,\tag{1}$$

where $u=f(x,y)$ (or even possibly $u=f(g(x,y),h(x,y))$) and everything is sufficiently differentiable.

Is there a name for this seemingly quasibinomial combination of partial derivatives? I'm well aware this is not literally $\left( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right)^2$, but it seems like a mistake most of us made once.

I've seen asserted that using the substitution $x=r \cos \theta$ and $y=r\sin\theta$ yields

$$x^2 \frac{\partial^2 u}{\partial x^2} + 2 x y \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2}=r^2\frac{\partial^2 u}{\partial r}. \tag{2}$$

Given that the usual way to show this would include several pages of laborious calculation, I've tried to be clever:

$$\begin{aligned} x^2 \frac{\partial^2 u}{\partial x^2} + 2 x y \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} &= x^2 \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} + 2 x y \frac{\partial^2 u}{\partial x \partial y} \\ &= r^2 \nabla^2 u + 2 x y \frac{\partial^2 u}{\partial x \partial y} \\ &= r^2 \left( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} \right) + 2 x y \frac{\partial^2 u}{\partial x \partial y} \\ &=r^2 \frac{\partial^2 u}{\partial r^2} + r \frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial \theta^2} + 2 x y \frac{\partial^2 u}{\partial x \partial y} . \end{aligned}$$

Unfortunately, being clever ends there (if it ever began - I'm not sure I can even do this $\nabla^2$ trick), since I can't find a way to show that

$$r \frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial \theta^2} + 2 x y \frac{\partial^2 u}{\partial x \partial y} = 0$$

without having to do the tedious calculations I wanted to avoid in the first place.

So, to review, my questions are:

1. Is there a name for this expression/PDE in $(1)$?
2. Is there a way to show $(2)$ is true without brute force calculation?

UPDATE:

Since Mathematica confirms that $(2)$ (case $n=2$) is correct, I'd like to conjecture that for sufficiently smooth $u$

$$\sum_{k=0}^{n} {n\choose k} x^k y^{n-k} \frac{\partial^n u}{\partial x^k \partial y^{n-k}} = r^n \frac{\partial^n u}{\partial r^n},$$ under the substitution $x=r\cos\theta$ and $y=r\sin\theta$.

Mathematica 13.1 confirms this is true for $n\leq7$, for example,

n = 7;
DSolveChangeVariables[
    Inactive[DSolve][
   Sum[Binomial[n, k] x^k y^(n - k) D[
      u[x, y], {x, k}, {y, n - k}], {k, 0, n}], u, {x, y}], 
    u, {r, \[Theta]}, "Cartesian" -> "Polar"] // Simplify

outputs

DSolve[r^7 D[u[r, \[Theta]],{r,7}], u, {r, \[Theta]}]

Answer 1

If you perform the change of variable $(u,v)\gets(\ln x,\ln y)$ then you can express: $$\left(x^2\frac{\partial^2}{\partial x^2}+2xy\frac{\partial^2}{\partial x\partial y}+y^2\frac{\partial^2}{\partial y^2}\right)f=\left(\frac\partial{\partial u}+\frac\partial{\partial v}\right)^2f$$

You can get there as: $$\frac\partial{\partial u}=\frac{\partial x}{\partial u}\frac\partial{\partial x}$$ and $$\frac{\partial x}{\partial u}=x$$

Answer 2

If you have coordinates $(x_1, \dots, x_n)$ on an open set $O$ and you change to coordinates $(y_1, \dots, y_n)$ on an open set $\Omega$ (meaning $x = F(y)$ with $F : \Omega \to O$ a diffeomorphism), then $$\partial_{x_j}\partial_{x_k}u = \frac{1}{\sqrt{g}}\partial_{y_\ell}(\sqrt{g}\frac{\partial y_{\ell}}{\partial x_{j}}\frac{\partial y_{m}}{\partial x_k}\partial_{y_m}u),$$ where $g(y) = \det(\frac{\partial x_i}{\partial y_j})^2 = \det(DF(y))^2$. The above formula can be proved by integrating by parts and using the change of variables theorem of integration to show that for any $\phi \in C_c^{\infty}(O)$, $$\int_{O}\partial_{x_j}\partial_{x_k}u\phi\,dx = \int_{O}\frac{1}{\sqrt{g}}\partial_{y_\ell}(\sqrt{g}\frac{\partial y_{\ell}}{\partial x_{j}}\frac{\partial y_{m}}{\partial x_k}\partial_{y_m}u)\phi\,dx.$$

In principle now you can plug in the formula and see what happens.