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I want to take first partial derivative w.r.t. $x_i$ (for $i,j,k=1,\ldots,n$) of $$x\mapsto\frac 1{|x-y|^{n-2}},\quad x\neq y.$$ where $y\in\mathbb{R}^n$ is fixed.

Can I ask here if the following calculation is correct? I would like to make sure because I need to take the second partial derivative w.r.t. $x_i$ as well. \begin{align} \frac{\partial}{\partial x_i} \left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_i} |x-y|^{2-n} \\ &= \frac{\partial}{\partial x_i} \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{\frac{2-n}2} \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} \frac d{dx_i} (x_i-y_i)^2 \\ &= (2-n) \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} (x_i-y_i) \\ &= (2-n) \frac{x_i-y_i}{|x-y|^n} \end{align}

Edit: What about the second partial derivative? (Edit: Yes, Jack is right, the $i$'s below should have been $k$'s.) \begin{align} \frac{\partial^2}{\partial^2 x_k} \left(\frac 1{|x-y|^{n-2}} \right) &= (2-n) \frac{\partial}{\partial x_k} \left(\frac{x_i-y_i}{|x-y|^n} \right) \\ &= (2-n) \left(\frac{|x-y|^n \frac{\partial}{\partial x_k}(x_i-y_i)-(x_i-y_i) \frac{\partial}{\partial x_k}|x-y|^{2n}}{|x-y|^{2n}} \right) \\ &= (2-n) \left(\frac{|x-y|^n (1)-(x_i-y_i) 2n|x-y|^{n-1}(x_k-y_k)}{|x-y|^{2n}} \right) \\ &= (2-n)\left(\frac{1}{|x-y|^n}-2n \frac{(x_i-y_i)(x_k-y_k)}{|x-y|^{n+1}} \right) \end{align}

Second edit: Following Jack's updated answer, I now have \begin{align} \frac{\partial^2}{\partial x_j \partial x_k}\left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_j} \left(\frac{x_k-y_k}{|x-y|^n} \right)\\ &= \frac{\partial}{\partial x_j}(x_k-y_k) \frac 1{|x-y|^n} + (x_k-y_k) \frac{\partial}{\partial x_j} \frac 1{|x-y|^n} \\ &= \frac 1{|x-y|^n} + n \frac{(x_j-y_j)(x_k-y_k)}{|x-y|^{n+2}}, \end{align} the last step owing to $(*)$ in Jack's answer with $n \mapsto n+2$.

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  • $\begingroup$ The notation $i$ in the edited question is incorrect. $\endgroup$
    – user9464
    Commented May 25, 2016 at 20:51
  • $\begingroup$ You wrote $j=k$ in $\partial^2/(\partial_j\partial x_k)$ before your editing. The answer would be different when $j\neq k$. $\endgroup$
    – user9464
    Commented May 25, 2016 at 21:04
  • $\begingroup$ Give me a minute to edit my answer. $\endgroup$
    – user9464
    Commented May 25, 2016 at 21:06
  • $\begingroup$ You calculation of $\partial/\partial x_j(x_k-y_k)$ is incorrect. $\endgroup$
    – user9464
    Commented May 25, 2016 at 21:18

1 Answer 1

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You calculation is basically correct except that you need to be careful with the dummy index.

\begin{align} \tag{*} \frac{\partial}{\partial x_k} \left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_k} |x-y|^{2-n} \\ &= \frac{\partial}{\partial x_k} \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{\frac{2-n}2} \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2}\frac {\partial}{\partial x_k} \left(\sum_{i=1}^n (x_i-y_i)^2 \right) \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} \sum_{i=1}^n\frac{\partial}{\partial x_k} \left( (x_i-y_i)^2 \right) \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} 2(x_k-y_k) \\ &= (2-n) \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} (x_k-y_k) \\ &=(2-n)\frac{x_k-y_k}{|x-y|^{n}} \end{align}


Now to find $\frac{\partial^2}{\partial x_k^2} \left(\frac 1{|x-y|^{n-2}} \right) $ it suffices by linearity to find $$ \frac{\partial}{\partial x_k}\left(\frac{x_k-y_k}{|x-y|^{n}}\right). $$ The product rule tells you that the nontrivial part is $$ \frac{\partial}{\partial x_k}\left(\frac{1}{|x-y|^{n}}\right)\tag{**} $$ But you can use $(*)$ to get $(**)$.


Here is another way to simplify the calculation (and notations) to get $(*)$. First notice that $$ \frac{\partial}{\partial x_k}|x-y|^2=2(x_k-y_k). $$ Then \begin{align} \frac{\partial}{\partial x_k} \left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_k} (|x-y|^2)^{\frac{2-n}{2}}\\ &=\frac{2-n}{2} (|x-y|^2)^{\frac{-n}{2}}\cdot 2(x_k-y_k) \end{align} Similarly, you can find $(**)$.


[Added due to the change in OP] Now to find $\frac{\partial^2}{\partial x_m\partial x_k} \left(\frac 1{|x-y|^{n-2}} \right) $ it suffices by linearity to find $$ \frac{\partial}{\partial x_m}\left(\frac{x_k-y_k}{|x-y|^{n}}\right)= (x_k-y_k)\frac{\partial}{\partial x_m}\left(\frac{1}{|x-y|^{n}}\right)+ \delta_{mk}\left(\frac{1}{|x-y|^{n}}\right) $$ where $\delta_{mk}=1$ when $m=k$ and $\delta_{mk}=0$ when $m\neq k$. The nontrivial part is $$ \frac{\partial}{\partial x_m}\left(\frac{1}{|x-y|^{n}}\right) =\frac{\partial}{\partial x_m}(|x-y|^2)^{(-n/2)}\\ =\frac{-n}{2}(|x-y|^2)^{-\frac{n}{2}-1}2(x_m-y_m)\\ =-n(x_m-y_m)|x-y|^{-n-2} $$

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