Identity $Delta u(x) = v´´(r) + frac{n-1}{r}v´(r)$ from a function
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Let $Omega := {{x in mathbb{R}^n}:0 leq Vert xVert_2 < R}$ with $R> 0$.
Further, let $u in C^2(Omega)$ be a radial function.
How can one calculate/prove that this identity follows?
$$Delta u(x) = v´´(r) + frac{n-1}{r}v´(r)$$
I know that since $u in C^2(Omega)$ is a radial function it follows that there exists a function $v in C^2((0,R))$, so that $u(x) = v(r)$ for $x in Omega$, where $r := Vert x Vert_2$.
But I still can't follow the identity from this.
analysis functions norm
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up vote
0
down vote
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Let $Omega := {{x in mathbb{R}^n}:0 leq Vert xVert_2 < R}$ with $R> 0$.
Further, let $u in C^2(Omega)$ be a radial function.
How can one calculate/prove that this identity follows?
$$Delta u(x) = v´´(r) + frac{n-1}{r}v´(r)$$
I know that since $u in C^2(Omega)$ is a radial function it follows that there exists a function $v in C^2((0,R))$, so that $u(x) = v(r)$ for $x in Omega$, where $r := Vert x Vert_2$.
But I still can't follow the identity from this.
analysis functions norm
New contributor
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $Omega := {{x in mathbb{R}^n}:0 leq Vert xVert_2 < R}$ with $R> 0$.
Further, let $u in C^2(Omega)$ be a radial function.
How can one calculate/prove that this identity follows?
$$Delta u(x) = v´´(r) + frac{n-1}{r}v´(r)$$
I know that since $u in C^2(Omega)$ is a radial function it follows that there exists a function $v in C^2((0,R))$, so that $u(x) = v(r)$ for $x in Omega$, where $r := Vert x Vert_2$.
But I still can't follow the identity from this.
analysis functions norm
New contributor
Let $Omega := {{x in mathbb{R}^n}:0 leq Vert xVert_2 < R}$ with $R> 0$.
Further, let $u in C^2(Omega)$ be a radial function.
How can one calculate/prove that this identity follows?
$$Delta u(x) = v´´(r) + frac{n-1}{r}v´(r)$$
I know that since $u in C^2(Omega)$ is a radial function it follows that there exists a function $v in C^2((0,R))$, so that $u(x) = v(r)$ for $x in Omega$, where $r := Vert x Vert_2$.
But I still can't follow the identity from this.
analysis functions norm
analysis functions norm
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asked Nov 22 at 21:37
Math Dummy
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