After the definition of the fundamental solution of Laplace equation (page 22)

$\Phi(x) = \begin{cases} -\frac{1}{2\pi} \, \log(|x|), \, & n=2, \\ \frac{1}{n \, (n-2) \, \omega_n} \, \frac{1}{|x|^{n-2}}, \, & n\geq 3, \end{cases}$

Evans gives two estimates:

(i) $|D \Phi(x)| \leq \frac{C}{|x|^{n-1}}$

(ii) $|D^2 \Phi(x)|\leq \frac{C}{|x|^n}$

for some constant $C>0$

I have been able to prove (i) with no difficulties, but I'm really stuck on (ii), for the case n \leq 3.

I know that $D^2 \Phi(x)$ denotes the hessian matrix of $\Phi$, but I don't understand how to compute its norm. As far as I know (I looked up on the appendix), given a matrix $A$, Evans writes \begin{align} |A|=(\sum_{i=1}^{n} \sum_{j=1}^n a_{ij}^2)^{1/2} \quad \star \end{align}

So I write in the following the $(i,j)$-th element of $D\Phi(x)$

\begin{align} \frac{\partial}{\partial x_j} [\frac{\partial \Phi(x)}{\partial x_i}]=\frac{\partial}{\partial x_j}[v'(|x|) \frac{x_i}{|x|}]=v''(|x|) \frac{x_i x_j}{|x|^2} - v'(|x|) \frac{x_i x_j}{|x|^3} \end{align}

Now I plug in the definition of $v(|x|)=C |x|^{1-N}$, $C$ constant, hence I have that

\begin{align} (D \Phi(x))_{ij}=C(1-N) \frac{x_i x_j}{|x|^{2+N}}-C \frac{x_i x_j}{|x|^{2+N}} = -N \frac{x_i x_j}{|x|^{2+N}} \end{align}

Now I should just apply $\star$, but I don't know how to move from here


I have that $(D \Phi(x))_{ij}=-N \frac{x_i x_j}{|x|^{2+N}}\leq -N \frac{|x|^2}{|x|^{2+N}}=-N \frac{1}{|x|^N}$

and since this bound does not depend on $(i,j)$ anymore I apply straightforward $\star$ and get

\begin{align} \frac{N}{|x|^N} \sqrt{\sum_{i=1}^N \sum_{j=1}^{N}}1 =\frac{N^2}{|x|^N} \end{align}

and hence I have the desired bound

EDIT$^2$ [17/5/2019]

After the comments with @SeverinSchraven, differentiating I end up with

\begin{align} \frac{\partial}{\partial x_i}[v'(|x|) \frac{x_i}{|x|}]= \ldots=\frac{C}{|x|^N}-N \frac{x_i^2}{|x|^{N+2}} \end{align}

and so I have, in the Hessian matrix:

\begin{equation} a_{ij}= \begin{cases} \frac{N}{|x|^N} \quad i \ne j \\ \frac{(C-N)}{|x|^N} \quad i =j \end{cases} \end{equation}

So, using $\star$ I have:

$\sqrt{\sum_{i=1}^{N} \sum_{j=1}^{N} a_{ij}^2}= \sqrt{\sum_{i} \sum_{i} a_{ii}^2 + \frac{N^2}{|x|^N}}=\sqrt{\frac{1}{|x|^{2N}}(N^2+(N^2(C-N)^2)}=\frac{N}{|x|^N} \cdot \tilde{C}$, where $\tilde{C}=\sqrt{1+(C-N)^2}$

  • $\begingroup$ Note that $x_i x_j \leq x_i^2 + x_j^2 \leq \vert x \vert^2$. $\endgroup$ Commented May 16, 2019 at 22:33
  • $\begingroup$ @SeverinSchraven could it be okay now? $\endgroup$
    – VoB
    Commented May 17, 2019 at 16:13
  • $\begingroup$ You should take absolute value. Otherwise you are in trouble due to the minus sign. $\endgroup$ Commented May 17, 2019 at 16:46
  • $\begingroup$ @SeverinSchraven You're right! I took it, but since $N>0$ I did not wrote it! Finally, I can say that $|D^2 \Phi| \leq \frac{N^2}{|x|^N}$, and the constant depends on $N$. Do you think it's right? I mean, the book just writes $C$ $\endgroup$
    – VoB
    Commented May 17, 2019 at 17:24
  • $\begingroup$ I would guess that it depends on $N$. What is is more important, you need also to consider the case $i=j$, which you omitted so far. $\endgroup$ Commented May 17, 2019 at 18:29

1 Answer 1


After the comments, I'll write an answer.

Evans defines $|A|$, where $A$ is a matrix, as

\begin{align} |A|=(\sum_{i=1}^{n} \sum_{j=1}^n a_{ij}^2)^{1/2} \quad \star \end{align}

so, in order to compute the norm of $D^2 \Phi$ (Hessian matrix), I need to compute each component $a_{ij}=\frac{\partial}{\partial x_i}(\frac{\partial \Phi(x)}{\partial x_j})$

As pointed out in comments, I need to distinguish the case $i=j$ and $i \ne j$. [Recall that in Evans $v'(|x|)=v'(r)=C r^{1-N}$ ]

So, for $i \ne j$:

$ \frac{\partial}{\partial x_i} [\frac{\partial \Phi(x)}{\partial x_j}]=\frac{\partial}{\partial x_i}[v'(|x|) \frac{x_j}{|x|}]=v''(|x|) \frac{x_i x_j}{|x|^2} - v'(|x|) \frac{x_i x_j}{|x|^3} $

Pluggin in the definition of $v(|x|)$:

\begin{align} a_{ij}= -N \frac{x_i x_j}{|x|^{2+N}}, \quad i \ne j \end{align}

In the case $i=j$ I have:

$\frac{\partial}{\partial x_i}[v'(|x|) \frac{x_i}{|x|}]=v''(|x|) \frac{x_i^2}{|x|^2} + v'(|x|) [ \frac{1}{|x|^2} - \frac{x_i^2}{|x|^3}]=C(1-N) \frac{x_i^2}{|x|^{N+2}}+\frac{C}{|x|^N}-C \frac{x_i^2}{|x|^{N+2}}$


\begin{align} a_{ij}= \frac{C}{|x|^N}-NC \frac{x_i^2}{|x|^{N+2}}, \quad i =j \end{align}

In order to compute $\star$, I use that $x_i^2 \leq |x|^2$ and so I can bound $a_{ii}$ in the following way

$a_{ii} \leq \frac{C}{|x|^N}+NC \frac{|x|^2}{|x|^{N+2}}=C(1+N) \frac{1}{|x|^N}$

So, \begin{align} \sum_i \sum _i a_{ii}^2 \leq N^2 C^2(1+N)^2 \frac{1}{|x|^{2N}} \end{align}

Now, bounding again $a_{ij}$ ($i \ne j$) by using the fact that $x_i x_j \leq |x|^2$, I have

\begin{align} \sum_i \sum_j a_{ij}^2 \leq \sum_i \sum_j (\frac{N}{|x|^{N}})^2=N(N-1)N^2 \frac{1}{|x|^{2N}} \end{align}

Now I put together the two sums and get

\begin{align} \frac{N^2}{|x|^{2N}} [N(N-1)+C^2(1+N)^2] \end{align}

hence, by taking the square root:

$|D^2 \Phi(x)| \leq \frac{N}{|x|^N} \sqrt{N(N-1)+C^2(1+N)^2}=\tilde{C} \frac{N}{|x|^N}$

where $\tilde{C}=\sqrt{N(N-1)+C^2(N+1)}$

and the bound is proved.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .