Let $v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$. Show that for $x in B(0,R)^c$ we have $v(x) =...











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Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$



Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.






integration pde harmonic-functions potential-theory







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    up vote
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    Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$



    Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.






    integration pde harmonic-functions potential-theory







    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$



      Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.






      integration pde harmonic-functions potential-theory







      share|cite|improve this question















      Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$



      Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.






      integration pde harmonic-functions potential-theory




      integration pde harmonic-functions potential-theory






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      edited Nov 23 at 19:53

























      asked Nov 20 at 8:18









      h3h325

      18910




      18910






















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          Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:



          $$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$



          Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$



          You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$






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            Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:



            $$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$



            Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$



            You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$






            share|cite|improve this answer

























              up vote
              0
              down vote













              Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:



              $$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$



              Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$



              You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:



                $$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$



                Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$



                You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$






                share|cite|improve this answer












                Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:



                $$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$



                Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$



                You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 at 20:13









                Will M.

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                2,257313






























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