Laplacian of the Euclidean Norm












2












$begingroup$


Let $r:mathbb{R}^n rightarrow mathbb{R}$ be defined by $r(x)=|x|,$ where $large|x|=(x_1^2+dots+x_n^2)^{frac{1}{2}}.$ Compute $nabla^2r,$ the Laplacian of $r.$




I perform the following computations,
$$largepartial_{x_i}r(x)=frac{1}{2} cdot frac{2x_i}{(x_1^2+dots+x_n^2)^{frac{1}{2}}}=frac{x_i}{r(x)},$$
so that we get
$$largepartial_{x_i}partial_{x_i}r(x)=frac{r(x)-frac{x_i^2}{r(x)}}{r(x)^2},$$
and putting these together yields
begin{align*}largenabla^2r(x)
&=largepartial_{x_1}^2r(x)+dots+partial_{x_n}^2r(x) \
&=largefrac{r(x)-x_1^2cdot r(x)^{-1}}{r(x)^2}+dots+frac{r(x)-x_n^2cdot r(x)^{-1}}{r(x)^2}\
&=largefrac{ncdot r(x)-r(x)^{-1}cdot r(x)^2}{r(x)^2}\
&=frac{n-1}{r(x)}.\
end{align*}



Is the above computation correct? I checked the calculation several times, but still have a feeling that I made a mistake somewhere






calculus multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Looks right to me
    $endgroup$
    – Anthony Ter
    Feb 3 at 9:50






  • 1




    $begingroup$
    Take a look to this document which addresses a more general question ; it confirms your result (see the last line !) staff.science.uu.nl/~kolk0101/Books/Analysis/normlinearmap.pdf
    $endgroup$
    – Jean Marie
    Feb 3 at 10:10
















2












$begingroup$


Let $r:mathbb{R}^n rightarrow mathbb{R}$ be defined by $r(x)=|x|,$ where $large|x|=(x_1^2+dots+x_n^2)^{frac{1}{2}}.$ Compute $nabla^2r,$ the Laplacian of $r.$




I perform the following computations,
$$largepartial_{x_i}r(x)=frac{1}{2} cdot frac{2x_i}{(x_1^2+dots+x_n^2)^{frac{1}{2}}}=frac{x_i}{r(x)},$$
so that we get
$$largepartial_{x_i}partial_{x_i}r(x)=frac{r(x)-frac{x_i^2}{r(x)}}{r(x)^2},$$
and putting these together yields
begin{align*}largenabla^2r(x)
&=largepartial_{x_1}^2r(x)+dots+partial_{x_n}^2r(x) \
&=largefrac{r(x)-x_1^2cdot r(x)^{-1}}{r(x)^2}+dots+frac{r(x)-x_n^2cdot r(x)^{-1}}{r(x)^2}\
&=largefrac{ncdot r(x)-r(x)^{-1}cdot r(x)^2}{r(x)^2}\
&=frac{n-1}{r(x)}.\
end{align*}



Is the above computation correct? I checked the calculation several times, but still have a feeling that I made a mistake somewhere






calculus multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Looks right to me
    $endgroup$
    – Anthony Ter
    Feb 3 at 9:50






  • 1




    $begingroup$
    Take a look to this document which addresses a more general question ; it confirms your result (see the last line !) staff.science.uu.nl/~kolk0101/Books/Analysis/normlinearmap.pdf
    $endgroup$
    – Jean Marie
    Feb 3 at 10:10














2












2








2





$begingroup$


Let $r:mathbb{R}^n rightarrow mathbb{R}$ be defined by $r(x)=|x|,$ where $large|x|=(x_1^2+dots+x_n^2)^{frac{1}{2}}.$ Compute $nabla^2r,$ the Laplacian of $r.$




I perform the following computations,
$$largepartial_{x_i}r(x)=frac{1}{2} cdot frac{2x_i}{(x_1^2+dots+x_n^2)^{frac{1}{2}}}=frac{x_i}{r(x)},$$
so that we get
$$largepartial_{x_i}partial_{x_i}r(x)=frac{r(x)-frac{x_i^2}{r(x)}}{r(x)^2},$$
and putting these together yields
begin{align*}largenabla^2r(x)
&=largepartial_{x_1}^2r(x)+dots+partial_{x_n}^2r(x) \
&=largefrac{r(x)-x_1^2cdot r(x)^{-1}}{r(x)^2}+dots+frac{r(x)-x_n^2cdot r(x)^{-1}}{r(x)^2}\
&=largefrac{ncdot r(x)-r(x)^{-1}cdot r(x)^2}{r(x)^2}\
&=frac{n-1}{r(x)}.\
end{align*}



Is the above computation correct? I checked the calculation several times, but still have a feeling that I made a mistake somewhere






calculus multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$




Let $r:mathbb{R}^n rightarrow mathbb{R}$ be defined by $r(x)=|x|,$ where $large|x|=(x_1^2+dots+x_n^2)^{frac{1}{2}}.$ Compute $nabla^2r,$ the Laplacian of $r.$




I perform the following computations,
$$largepartial_{x_i}r(x)=frac{1}{2} cdot frac{2x_i}{(x_1^2+dots+x_n^2)^{frac{1}{2}}}=frac{x_i}{r(x)},$$
so that we get
$$largepartial_{x_i}partial_{x_i}r(x)=frac{r(x)-frac{x_i^2}{r(x)}}{r(x)^2},$$
and putting these together yields
begin{align*}largenabla^2r(x)
&=largepartial_{x_1}^2r(x)+dots+partial_{x_n}^2r(x) \
&=largefrac{r(x)-x_1^2cdot r(x)^{-1}}{r(x)^2}+dots+frac{r(x)-x_n^2cdot r(x)^{-1}}{r(x)^2}\
&=largefrac{ncdot r(x)-r(x)^{-1}cdot r(x)^2}{r(x)^2}\
&=frac{n-1}{r(x)}.\
end{align*}



Is the above computation correct? I checked the calculation several times, but still have a feeling that I made a mistake somewhere






calculus multivariable-calculus partial-derivative




calculus multivariable-calculus partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 3 at 10:11









Jean Marie

31.5k42355




31.5k42355










asked Feb 3 at 9:33









Gaby AlfonsoGaby Alfonso

1,2051418




1,2051418








  • 1




    $begingroup$
    Looks right to me
    $endgroup$
    – Anthony Ter
    Feb 3 at 9:50






  • 1




    $begingroup$
    Take a look to this document which addresses a more general question ; it confirms your result (see the last line !) staff.science.uu.nl/~kolk0101/Books/Analysis/normlinearmap.pdf
    $endgroup$
    – Jean Marie
    Feb 3 at 10:10














  • 1




    $begingroup$
    Looks right to me
    $endgroup$
    – Anthony Ter
    Feb 3 at 9:50






  • 1




    $begingroup$
    Take a look to this document which addresses a more general question ; it confirms your result (see the last line !) staff.science.uu.nl/~kolk0101/Books/Analysis/normlinearmap.pdf
    $endgroup$
    – Jean Marie
    Feb 3 at 10:10








1




1




$begingroup$
Looks right to me
$endgroup$
– Anthony Ter
Feb 3 at 9:50




$begingroup$
Looks right to me
$endgroup$
– Anthony Ter
Feb 3 at 9:50




1




1




$begingroup$
Take a look to this document which addresses a more general question ; it confirms your result (see the last line !) staff.science.uu.nl/~kolk0101/Books/Analysis/normlinearmap.pdf
$endgroup$
– Jean Marie
Feb 3 at 10:10




$begingroup$
Take a look to this document which addresses a more general question ; it confirms your result (see the last line !) staff.science.uu.nl/~kolk0101/Books/Analysis/normlinearmap.pdf
$endgroup$
– Jean Marie
Feb 3 at 10:10










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$begingroup$

If $u(x)=v(|x|)$ then
$$Delta u=frac{n-1}{|x|}v'(|x|)+v''(|x|).$$
See, for example https://web.stanford.edu/class/math220b/handouts/laplace.pdf






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    $begingroup$

    If $u(x)=v(|x|)$ then
    $$Delta u=frac{n-1}{|x|}v'(|x|)+v''(|x|).$$
    See, for example https://web.stanford.edu/class/math220b/handouts/laplace.pdf






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      If $u(x)=v(|x|)$ then
      $$Delta u=frac{n-1}{|x|}v'(|x|)+v''(|x|).$$
      See, for example https://web.stanford.edu/class/math220b/handouts/laplace.pdf






      share|cite|improve this answer









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        2





        $begingroup$

        If $u(x)=v(|x|)$ then
        $$Delta u=frac{n-1}{|x|}v'(|x|)+v''(|x|).$$
        See, for example https://web.stanford.edu/class/math220b/handouts/laplace.pdf






        share|cite|improve this answer









        $endgroup$



        If $u(x)=v(|x|)$ then
        $$Delta u=frac{n-1}{|x|}v'(|x|)+v''(|x|).$$
        See, for example https://web.stanford.edu/class/math220b/handouts/laplace.pdf







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 3 at 11:08









        Aleksas DomarkasAleksas Domarkas

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