# Show that $\int_{B_1}(\nabla u, \nabla \phi) dx = c_n\phi(0)$

Im given a function $$u:B_1 = \{ x \in \mathbb{R}^n: |x| < 1\} \rightarrow \mathbb{R}$$ such that $$u(x) = |x|^{-n+2}$$ and we assume that $$3 \leq n$$.Then let $$\phi \in C_0^{\infty}(B_1)$$ and show that $$\int_{B_1}(\nabla u, \nabla \phi) dx = c_n\phi(0)$$ where $$c_n > 0$$ is a constant.

Note that $$\partial x_i(u) = (-n+2)x_i|x|^{-n}$$.

I have verified that $$u \in W^{1,p}(B_1)$$ for $$p \leq \frac{n}{n-1}$$.I think this is just a direct calculation, but i get stuck after applying integration by parts for the equation $$\int_{B_1}\sum_{i=1}^n\partial x_i(u)\partial x_i(\phi)dx$$. So i end up with an equation $$(n-2)(\int_{B_1}\sum_{i=1}^n(\phi |x|^{-n}-\phi x_i n|x|^{-n-2})dx)$$. How to proceed?

• Shouldn't it be $x_i^2$ in your last formula?
– F_M_
Commented Oct 12, 2023 at 14:01
• oh i see i made an calculation error :-D yes you are right Commented Oct 12, 2023 at 14:03
• how do i know that $\phi(0) = 0$ ? Commented Oct 12, 2023 at 14:24
• $\int_{B_1}\sum_{i=1}^n(\phi |x|^{-n}-\phi x_i^2 n|x|^{-n-2})dx = \int_{B_1}n\phi |x|^{-n}-n\phi|x|^{-n}dx = 0$ Commented Oct 12, 2023 at 14:26
• In general $\phi(0) \neq 0$ as you only know $\phi \in \mathbb{C}_0^\infty(B_1)$ is vanishing at the boundary of $B_1.$ Note that, since $3 \leq n,$ the function $u$ is not defined in the origin.
– F_M_
Commented Oct 12, 2023 at 14:32

Let us for convenience consider the function $$\gamma(\mathbf{x}) = \frac{1}{n(2-n) \omega_n} |\mathbf{x}|^{2-n} = \frac{1}{n(2-n)\omega_n}u(\mathbf{x}),$$ where $$n\geq 3$$ and $$\omega_n = \frac{1}{n} \int_{\partial B(0,1)}1\,\mathrm{d}\Gamma$$ is the surface area of the unit bal $$B(0,1).$$ As you have shown (in the comments), the function $$\gamma$$ is harmonic for $$\mathbf{x}\neq \mathbf{0}.$$

Let $$\Omega = B_1 = \left\{\mathbf{x} \in \mathbb{R}^n \mid |\mathbf{x}|<1\right\}$$ and $$\Gamma = \partial B_1 = \left\{ \mathbf{x} \in \mathbb{R}^n \mid |\mathbf{x}=1\right\}.$$ Set $$\mathbf{\nu}$$ the outward normal unit vector. Since $$\gamma$$ is not defined in the origin, consider a small ball $$B(\mathbf{0}, \varepsilon) \subset \Omega$$ with radius $$\varepsilon>0$$ around the origin ($$\varepsilon<1$$).

Let $$\varphi \in \mathrm{C}^\infty_0(\Omega)$$ be given. Applying the Green formula (integration by parts) yields

\begin{align*} \int_{\Omega \setminus B(\mathbf{0},\varepsilon)} \gamma(\mathbf{x}) \Delta \varphi(\mathbf{x})\, \mathrm{d}\mathbf{x} &= \int_\Gamma \left(\gamma(\mathbf{x}) \nabla \varphi(\mathbf{x}) \cdot \mathbf{\nu} - \varphi(\mathbf{x})\nabla \gamma(\mathbf{x})\cdot \mathbf{\nu} \right)\,\mathrm{d}\Gamma \\ & \quad + \int_{\partial B(0,\varepsilon)} \left(\gamma(\mathbf{x}) \nabla \varphi(\mathbf{x}) \cdot \mathbf{\nu} - \varphi(\mathbf{x})\nabla \gamma(\mathbf{x})\cdot \mathbf{\nu} \right)\,\mathrm{d}S. \end{align*}

Now, since $$\varphi$$ and all its derivatives vanish at the boundary $$\Gamma$$ of $$\Omega$$, the first integral on the RHS is zero.

Next, since $$\gamma$$ is integrable, we have $$\lim\limits_{\varepsilon\to 0} \int_{\Omega \setminus B(\mathbf{0},\varepsilon)} \gamma(\mathbf{x}) \Delta \varphi(\mathbf{x})\,\mathrm{d}\mathbf{x} = \int_\Omega \gamma(\mathbf{x})\Delta \varphi(\mathbf{x})\, \mathrm{d}\mathbf{x}.$$

We estimate the two remaining integrals.

\begin{align*} \left| \int_{\partial B(\mathbf{0},\varepsilon)} \gamma(\mathbf{x}) \nabla \varphi(\mathbf{x}) \cdot \mathbf{\nu} \,\mathrm{d}S\right| & \leq \gamma(\varepsilon) \int_{\partial B(\mathbf{0},\varepsilon)} \left| \nabla \varphi\right|\,\mathrm{d}S \\ & \leq \gamma(\varepsilon) \varepsilon^{n-1} \sup\limits_{B(\mathbf{0},\varepsilon)}\left| \nabla \varphi\right| \int_{\partial B(\mathbf{0},1)} \,\mathrm{d}S \\&\to 0, \quad \text{ as } \varepsilon \to 0. \end{align*}

Finally,

\begin{align*} - \int_{\partial B(\mathbf{0},\varepsilon)} \varphi(\mathbf{x}) \nabla \gamma(\mathbf{x}) \cdot \mathbf{\nu} \, \mathrm{d}S & = \frac{1}{n\omega_n} \int_{\partial B(\mathbf{0},\varepsilon)} \varphi(\mathbf{x}) \frac{1}{|\mathbf{x}|^{n-1}}\,\mathrm{d}S \\ &= \frac{1}{n\omega_n} \frac{1}{\varepsilon^{n-1}} \int_{\partial B(\mathbf{0},\varepsilon)} \varphi(\mathbf{x}) \,\mathrm{d}S \\ & \to \varphi(\mathbf{0}) \quad \text{ as } \varepsilon \to 0. \end{align*}

Summarizing

$$\varphi(\mathbf{0}) = \int_\Omega \gamma(\mathbf{x}) \Delta \varphi(\mathbf{x}) \, \mathrm{d}\mathbf{x} = \int_\Omega \Delta \gamma(\mathbf{x}) \varphi(\mathbf{x}) \, \mathrm{d}\mathbf{x}.$$

Can you take it from here to your problem?

• i think at the last part you use lebesgue differentation theorem. Commented Oct 12, 2023 at 16:35
• how $\nabla \gamma \cdot v$ becomes $1/|x|^{n-1}$? Commented Oct 12, 2023 at 21:19