Radial derivative of the characteristic function of the unit disc in $\mathbb R^2$.

Question

I need help to answer this question from Hörmander, Let $u$ be the characteristic function of the unit disc in $\mathbb R^2$. Calculate $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$.

The answer in the book is: minus the arc length measure on the unit circle.

I am not sure if my calculations are right.

Let $\phi \in D(\mathbb R^2)$,

$$\left<x \frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}, \phi\right>= - \left<u,\frac{\partial {(x \phi(x,y))}}{\partial x} + \frac{\partial (y \phi(x,y))}{\partial y} \right>= - \left<u, \operatorname{div} \begin{bmatrix}x \phi(x,y)\\y \phi(x,y)\end{bmatrix}\right>. $$

Now I think I have to use the Gauss-Green formula to get the length measure, but I don't know how to finish my answer if the above calculations are right.

Any help please would be perfect.

Answer 1

\begin{align} & \iint\limits_\text{unit disk} \left<u, \operatorname{div} \begin{bmatrix}x \varphi(x,y) \\ y \varphi(x,y) \end{bmatrix}\right> \, dA \\[10pt] = {} & \int\limits_\text{boundary} \Big( (x\varphi(x,y), y\varphi(x,y)) \cdot (\text{outward unit normal}) \, d(\text{arc length}) \Big) \\ & \qquad\qquad\text{by the divergence theorem} \\[15pt] = {} & \int\limits_\text{boundary} (x\varphi(x,y),y\varphi(x,y)) \cdot (x,y)\, d(\text{arc length}) \\[10pt] = {} & \int\limits_\text{boundary} (x^2 + y^2)\varphi(x,y) \, d(\text{arc length}) \\[10pt] = {} & \int_\text{boundary} \varphi(x,y)\,d(\text{arc length}) \end{align}

Answer 2

There are (at least) three ways of performing this computation. In what follows, $D$ is the unit disk and $\mathbf 1_A$ is the function that equals $1$ on $A$ and $0$ otherwise.

1. Consider the formal integral $$\iint_{\mathbb R^2} (x\partial_x+y\partial_y)(\mathbf 1_D)\,f(x, y)\, dx\,dy, $$ where $f\in C^\infty_c(\mathbb R^2)$ and integrate by parts. (This formal integral is often written as a braket, like in the post by the OP). This is the method of the OP and of Micheal's answer.
2. Consider a smoothed version of $\mathbf 1_D$, such as the one obtained by convolution against the heat kernel $K(t, x,y)=e^{-(x^2+y^2)/2t}/(4\pi t)$, and observe that $$(x\partial_x+y\partial_y)\mathbf 1_D \ast K(t, x,y) = \mathbf 1_D \ast (x\partial_x+y\partial_y)K(t, x, y),$$ then compute $(x\partial_x+y\partial_y)K(t, x, y)$ and let $t\to 0^+$ in the convolution in the right-hand side.
3. Observe that $x\partial_x+y\partial_y$ is the generator of dilations, in the sense that for any distribution $T$ one has that $$\partial_\lambda T(\lambda x , \lambda y)|_{\lambda=1} =(x\partial_x+y\partial_y)T(x, y),$$ and then compute the derivative in the left-hand side.

The method 3 can be carried out as follows: taking an arbitrary $f\in C^\infty(\mathbb R^2)$, and observing that $\mathbf 1_D(\lambda x, \lambda y)= \mathbf 1_{\lambda^{-1}D}(x, y)$, we compute $$\partial_\lambda \iint_{\mathbb R^2} \mathbf 1_{\lambda^{-1}D}f(x, y)\, dx\,dy = \partial_\lambda \iint_{\lambda^{-1}D} f\, dx\,dy = -\lambda^{-2}\int_{\partial\lambda^{-1}D} f\, ds.$$ The last formula is the derivative of an integral over a moving region, and can be found, e.g. in Evans book on PDEs, in the Appendix (actually, this formula is a special case of Reynolds transport theorem). Here $ds$ denotes the arc-length measure on the circle. Notice that the $-\lambda^{-2}$ factor comes from the differentiation of $\lambda^{-1}$. Evaluating this expression at $\lambda=1$ yields the desired conclusion. $\Box$