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) =...
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
asked Nov 20 at 8:18
h3h325
18910
add a comment |
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
asked Nov 20 at 8:18
h3h325
18910
add a comment |
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
asked Nov 20 at 8:18
h3h325
18910
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
asked Nov 20 at 8:18
h3h325
18910
asked Nov 20 at 8:18
h3h325
18910
edited Nov 23 at 19:53
asked Nov 20 at 8:18
h3h325
18910
asked Nov 20 at 8:18
h3h325
18910
asked Nov 20 at 8:18
h3h325
18910
18910
add a comment |
add a comment |
1 Answer
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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 $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,377314
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1 Answer
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1 Answer
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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 $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,377314
add a comment |
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 $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,377314
add a comment |
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 $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,377314
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 $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,377314
answered Nov 23 at 20:13
Will M.
2,377314
answered Nov 23 at 20:13
Will M.
2,377314
answered Nov 23 at 20:13
Will M.
2,377314
2,377314
add a comment |
add a comment |
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