$U\in \mathbb{R}^2$ bounded, smooth edge, $0\in U$, $f\in C_0^{\infty}(U)$. $V(x)=\frac{x}{\|x\|^2}$. Prove:$\int_U(V,\nabla f)d\lambda_2=-2 \pi f(0)$

Question

$U \in \mathbb{R}^2$ is a bounded region with a smooth edge, such that $0 \in U$, $f \in C_0^{\infty}(U)$ $$ V(x) = \begin{cases} \dfrac{x}{\|x\|^2} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases} $$ Prove that: $$\int_U (V, \nabla f) \ d\lambda_2 = -2 \pi f(0)$$

Hint: think about the region $U_t = U \ \setminus \ \{ \|x\| \leq \epsilon \}$

I found such exercise in the set of exercises preparing for an exam. I don't know how to start since I don't understand the idea of integrating over function $V$ on one dimension and gradient of $f$ on the other. Any help would be much appreciated.

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As I understand it, $$\nabla f = \left( \frac{\partial f}{\partial x_1} \ , \ \frac{\partial f}{\partial x_2} \right)$$

We have: $$V(x) = \frac{x}{\|x\|^2} = \left( \frac{x_1}{x_1^2 + x_2^2} \ , \ \frac{x_2}{x_1^2 + x_2^2} \right)$$

The dot product is given by: $$(V, \nabla f) = \frac{\partial f}{\partial x_1} \frac{x_1}{x_1^2 + x_2^2} + \frac{\partial f}{\partial x_2} \frac{x_2}{x_1^2 + x_2^2} = \frac{\dfrac{\partial f}{\partial x_1} x_1 + \dfrac{\partial f}{\partial x_2} x_2}{x_1^2 + x_2^2}$$

Therefore we have: $$ (V, \nabla f) = \begin{cases} \dfrac{\dfrac{\partial f}{\partial x_1} x_1 + \dfrac{\partial f}{\partial x_2} x_2}{x_1^2 + x_2^2} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases} $$ Right now, as I understand it, I would like to show that: $$\int_D \frac{\frac{\partial f}{\partial x_1} x_1 + \frac{\partial f}{\partial x_2} x_2}{x_1^2 + x_2^2} \lambda_2$$

Is equal to: $$ −\int_D f \nabla \cdot V \ d \lambda_2 + \int_{\partial D} fV \cdot ndS $$

But while calculating $\int_D V \cdot \nabla f \ d \lambda_2$ I don't get why: $$\int_U (V, \nabla f) \ d\lambda_2 = \int_U (\nabla f, V) \ d\lambda_2$$

I guess dot product is commutative...

Answer 1

Let $ U \subset \mathbb{R}^2 $ be a bounded region with a smooth edge, such that $ 0 \in U $. Given $ f \in C_0^{\infty}(U) $ and the vector field $$ V(x) = \begin{cases} \frac{x}{\|x\|^2}, & \text{for } x \neq 0, \\ 0, & \text{for } x = 0, \end{cases} $$ prove that: $$ \int_U (V, \nabla f) \, d\lambda_2 = -2\pi f(0). $$

Proof:

We will use the hint provided and consider the region $ U_\epsilon = U \setminus \{ \|x\| \leq \epsilon \} $. Applying the Divergence Theorem on $ U_\epsilon $, we start by computing the divergence of $ V $.

For $ x \neq 0 $, $$ V(x) = \left( \frac{x_1}{\|x\|^2}, \frac{x_2}{\|x\|^2} \right). $$ The divergence of $ V $ is given by: $$ \nabla \cdot V = \frac{\partial}{\partial x_1} \left( \frac{x_1}{x_1^2 + x_2^2} \right) + \frac{\partial}{\partial x_2} \left( \frac{x_2}{x_1^2 + x_2^2} \right). $$ Using the quotient rule, we compute: $$ \frac{\partial}{\partial x_1} \left( \frac{x_1}{x_1^2 + x_2^2} \right) = \frac{(x_1^2 + x_2^2) \cdot 1 - x_1 \cdot 2x_1}{(x_1^2 + x_2^2)^2} = \frac{x_2^2 - x_1^2}{(x_1^2 + x_2^2)^2}, $$ and similarly, $$ \frac{\partial}{\partial x_2} \left( \frac{x_2}{x_1^2 + x_2^2} \right) = \frac{(x_1^2 + x_2^2) \cdot 1 - x_2 \cdot 2x_2}{(x_1^2 + x_2^2)^2} = \frac{x_1^2 - x_2^2}{(x_1^2 + x_2^2)^2}. $$ Adding these, we get: $$ \nabla \cdot V = \frac{x_2^2 - x_1^2}{(x_1^2 + x_2^2)^2} + \frac{x_1^2 - x_2^2}{(x_1^2 + x_2^2)^2} = 0. $$ Thus, the divergence of $ V $ is zero for $ x \neq 0 $.

Using the Divergence Theorem on $ U_\epsilon $: $$ \int_{U_\epsilon} (\nabla \cdot (fV)) \, d\lambda_2 = \int_{\partial U_\epsilon} (fV \cdot n) \, dS, $$ where $ n $ is the outward unit normal vector on the boundary $ \partial U_\epsilon $.

Since $ \nabla \cdot V = 0 $ and using the product rule for the divergence of $ fV $, we have: $$ \nabla \cdot (fV) = (\nabla f) \cdot V + f (\nabla \cdot V) = (\nabla f) \cdot V. $$ Therefore, $$ \int_{U_\epsilon} (\nabla f) \cdot V \, d\lambda_2 = \int_{\partial U_\epsilon} f (V \cdot n) \, dS. $$

The boundary $ \partial U_\epsilon $ consists of two parts: the outer boundary $ \partial U $ and the inner boundary $ \partial B_\epsilon(0) $ (a circle of radius $ \epsilon $ centered at the origin). Thus, $$ \int_{\partial U_\epsilon} f (V \cdot n) \, dS = \int_{\partial U} f (V \cdot n) \, dS + \int_{\partial B_\epsilon(0)} f (V \cdot n) \, dS. $$

As $ \epsilon \to 0 $, the integral over $ \partial B_\epsilon(0) $ can be approximated. On $ \partial B_\epsilon(0) $, $ V \cdot n = \frac{x}{\|x\|^2} \cdot \frac{-x}{\|x\|} = -\frac{1}{\epsilon} $. Therefore, $$ \int_{\partial B_\epsilon(0)} f (V \cdot n) \, dS = \int_{\partial B_\epsilon(0)} f \left(-\frac{1}{\epsilon}\right) \, dS = -\frac{1}{\epsilon} \int_{\partial B_\epsilon(0)} f \, dS. $$ Since $ f $ is smooth and $ f(0) $ is its value at the origin, $$ \int_{\partial B_\epsilon(0)} f \, dS \approx f(0) \cdot \text{length}(\partial B_\epsilon(0)) = f(0) \cdot 2\pi \epsilon. $$ Thus, $$ -\frac{1}{\epsilon} \int_{\partial B_\epsilon(0)} f \, dS \approx -\frac{1}{\epsilon} f(0) \cdot 2\pi \epsilon = -2\pi f(0). $$

As $ \epsilon \to 0 $, the integral over the outer boundary $ \partial U $ tends to zero because $ f \in C_0^\infty(U) $ vanishes on $ \partial U $. Therefore, $$ \int_{\partial U} f (V \cdot n) \, dS \to 0 \quad \text{as} \quad \epsilon \to 0. $$

So, taking the limit as $ \epsilon \to 0 $, we have: $$ \int_{U} (\nabla f) \cdot V \, d\lambda_2 = -2\pi f(0). $$ Therefore, $$ \int_U (V, \nabla f) \, d\lambda_2 = -2\pi f(0). $$