# Prove that a function $u: u= \ln\|x\|{_{2}}$ has $\Delta u = 0$.

I had a similar case some time ago and following the advices there I tried to solve this one too. I tried to find its first partial derivative and I got: $$\frac{\partial}{\partial x_{i}}=\frac{1}{2\cdot \|x\|_{2}^{1/2}}$$ Now i have to find the second derivative of this and I got $$\frac{1+4\cdot (\sum x_{i}^{2})^{5}}{8\cdot(\sum x_{i}^{2})^{5/4} }$$.

And now I am stuck,I have no idea if my calculations are okay or how can I continue to solve this. I would be incredibly thankful for some help.

Annalisa

• Are you sure about your first derivative? Commented Jun 9, 2020 at 21:59
• connected, but on a different viewpoint : math.stackexchange.com/q/2316637 Commented Jun 9, 2020 at 22:17
• If you want to be more precise, $\Delta \ln (||x||_2) = 0$ for $x\in \mathbb{R}^2 - \{(0,0)\}$, but in fact one has $\Delta \ln (||x||_2) = 2\pi \delta(x)$ as a distribution on $\mathbb{R}^2$. Commented Jun 9, 2020 at 22:33

You have, for $$x\in\mathbb{R}^d$$, $$u(x) = \ln \lVert x\rVert_2 = \ln \sqrt{\sum_{i=1}^d x_i^2} = \frac{1}{2}\ln \sum_{i=1}^d x_i^2$$ from which, for $$x\in\mathbb{R}^d$$ and $$1\leq i\leq d$$, $$\frac{\partial}{\partial x_i}u(x) = \frac{1}{2}\frac{\partial}{\partial x_i}\ln \sum_{i=1}^d x_i^2= \frac{1}{2}\cdot\frac{2x_i}{\sum_{i=1}^d x_i^2} = \frac{x_i}{\lVert x\rVert_2^2}\,.$$ From there, $$\frac{\partial^2}{\partial x_i^2}u(x) = \frac{\partial}{\partial x_i}\frac{x_i}{\lVert x\rVert_2^2} = \frac{1\cdot\lVert x\rVert_2^2-x_i\cdot 2x_i}{\lVert x\rVert_2^4}= \frac{\lVert x\rVert_2^2-2x_i^2}{\lVert x\rVert_2^4}$$ so that $$\Delta u(x) =\sum_{i=1}^d\frac{\partial^2}{\partial x_i^2}u(x) = \frac{d\lVert x\rVert_2^2-2\lVert x\rVert_2^2}{\lVert x\rVert_2^4}= \frac{d - 2}{\lVert x\rVert_2^4}$$ This is only 0 if $$d=2$$.
Let $$\|\mathbf{x}\|=\sqrt{x^2+y^2}$$. Then \begin{align*} u & =\ln \|\mathbf{x}\|\\ \frac{\partial u}{\partial x}&=\frac{1}{\|\mathbf{x}\|}\frac{\partial}{\partial x}\sqrt{x^2+y^2}\\ &=\frac{1}{\|\mathbf{x}\|}\frac{x}{\sqrt{x^2+y^2}}\\ &=\frac{x}{\|\mathbf{x}\|^2} \end{align*} Likewise $$\frac{\partial u}{\partial y}=\frac{y}{\|\mathbf{x}\|^2}.$$ Now \begin{align*} \frac{\partial ^2u}{\partial x^2}&=\frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right)\\ &=\frac{(x^2+y^2)\frac{\partial x}{\partial x}-x\frac{\partial (x^2+y^2)}{\partial x}}{(x^2+y^2)^2}\\ &=\frac{\|\mathbf{x}\|^2-2x^2}{\|\mathbf{x}\|^4}. \end{align*}
Likewise $$\frac{\partial ^2u}{\partial \color{red}{y}^2}=\frac{\|\mathbf{x}\|^2-2\color{red}{y}^2}{\|\mathbf{x}\|^4}.$$ Hopefully you can take it from here and generalize it to higher dimensions.