Solving $nabla K(x) = delta(x)$
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I'm learning about the 'PDE'
$$nabla K(x) = delta(x)$$
In a book it says that it is symmetric in all variables $x_1,cdots,x_n$ and is also radially symmetric (its value depends only on $r = |x|$. Since $delta(x) = 0$ for $xneq 0$, then the 'PDE' requires $K$ to be harmonic for $r>0$. Thus he tries to solve the 'PDE' with a radially symmetric function $K(x)$ and end up with:
$$fracpartial^2 Kpartial r^2 + fracn-1rfracpartial Kpartial r = 0$$
Why $K$ is symmetric and radially symmetric? How did he arrive at this PDE exactly?
pde
asked Aug 25 at 3:01
Guerlando OCs
29321244
add a comment |Â
up vote
1
down vote
favorite
I'm learning about the 'PDE'
$$nabla K(x) = delta(x)$$
In a book it says that it is symmetric in all variables $x_1,cdots,x_n$ and is also radially symmetric (its value depends only on $r = |x|$. Since $delta(x) = 0$ for $xneq 0$, then the 'PDE' requires $K$ to be harmonic for $r>0$. Thus he tries to solve the 'PDE' with a radially symmetric function $K(x)$ and end up with:
$$fracpartial^2 Kpartial r^2 + fracn-1rfracpartial Kpartial r = 0$$
Why $K$ is symmetric and radially symmetric? How did he arrive at this PDE exactly?
pde
asked Aug 25 at 3:01
Guerlando OCs
29321244
What book are you citing?
– Brian Borchers
Aug 25 at 3:03
@BrianBorchers math.ucla.edu/~yanovsky/handbooks/PDEs.pdf
– Guerlando OCs
Aug 25 at 3:13
Shouldn't it be $$nabla^2 K = delta (x)$$ ??
– Mattos
Aug 25 at 3:21
The operator is the Laplacian, which some authors write as $nabla^2 K$ and others write as $Delta K$. The special cases $n=2$ and $n=3$ should be easy if you know the Laplacian operator in polar and spherical coordinates.
– Brian Borchers
Aug 25 at 3:42
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm learning about the 'PDE'
$$nabla K(x) = delta(x)$$
In a book it says that it is symmetric in all variables $x_1,cdots,x_n$ and is also radially symmetric (its value depends only on $r = |x|$. Since $delta(x) = 0$ for $xneq 0$, then the 'PDE' requires $K$ to be harmonic for $r>0$. Thus he tries to solve the 'PDE' with a radially symmetric function $K(x)$ and end up with:
$$fracpartial^2 Kpartial r^2 + fracn-1rfracpartial Kpartial r = 0$$
Why $K$ is symmetric and radially symmetric? How did he arrive at this PDE exactly?
pde
asked Aug 25 at 3:01
Guerlando OCs
29321244
I'm learning about the 'PDE'
$$nabla K(x) = delta(x)$$
In a book it says that it is symmetric in all variables $x_1,cdots,x_n$ and is also radially symmetric (its value depends only on $r = |x|$. Since $delta(x) = 0$ for $xneq 0$, then the 'PDE' requires $K$ to be harmonic for $r>0$. Thus he tries to solve the 'PDE' with a radially symmetric function $K(x)$ and end up with:
$$fracpartial^2 Kpartial r^2 + fracn-1rfracpartial Kpartial r = 0$$
Why $K$ is symmetric and radially symmetric? How did he arrive at this PDE exactly?
pde
asked Aug 25 at 3:01
Guerlando OCs
29321244
asked Aug 25 at 3:01
Guerlando OCs
29321244
asked Aug 25 at 3:01
Guerlando OCs
29321244
asked Aug 25 at 3:01
Guerlando OCs
29321244
29321244
What book are you citing?
– Brian Borchers
Aug 25 at 3:03
@BrianBorchers math.ucla.edu/~yanovsky/handbooks/PDEs.pdf
– Guerlando OCs
Aug 25 at 3:13
Shouldn't it be $$nabla^2 K = delta (x)$$ ??
– Mattos
Aug 25 at 3:21
The operator is the Laplacian, which some authors write as $nabla^2 K$ and others write as $Delta K$. The special cases $n=2$ and $n=3$ should be easy if you know the Laplacian operator in polar and spherical coordinates.
– Brian Borchers
Aug 25 at 3:42
add a comment |Â
What book are you citing?
– Brian Borchers
Aug 25 at 3:03
@BrianBorchers math.ucla.edu/~yanovsky/handbooks/PDEs.pdf
– Guerlando OCs
Aug 25 at 3:13
Shouldn't it be $$nabla^2 K = delta (x)$$ ??
– Mattos
Aug 25 at 3:21
The operator is the Laplacian, which some authors write as $nabla^2 K$ and others write as $Delta K$. The special cases $n=2$ and $n=3$ should be easy if you know the Laplacian operator in polar and spherical coordinates.
– Brian Borchers
Aug 25 at 3:42
What book are you citing?
– Brian Borchers
Aug 25 at 3:03
What book are you citing?
– Brian Borchers
Aug 25 at 3:03
@BrianBorchers math.ucla.edu/~yanovsky/handbooks/PDEs.pdf
– Guerlando OCs
Aug 25 at 3:13
@BrianBorchers math.ucla.edu/~yanovsky/handbooks/PDEs.pdf
– Guerlando OCs
Aug 25 at 3:13
Shouldn't it be $$nabla^2 K = delta (x)$$ ??
– Mattos
Aug 25 at 3:21
Shouldn't it be $$nabla^2 K = delta (x)$$ ??
– Mattos
Aug 25 at 3:21
The operator is the Laplacian, which some authors write as $nabla^2 K$ and others write as $Delta K$. The special cases $n=2$ and $n=3$ should be easy if you know the Laplacian operator in polar and spherical coordinates.
– Brian Borchers
Aug 25 at 3:42
The operator is the Laplacian, which some authors write as $nabla^2 K$ and others write as $Delta K$. The special cases $n=2$ and $n=3$ should be easy if you know the Laplacian operator in polar and spherical coordinates.
– Brian Borchers
Aug 25 at 3:42
add a comment |Â
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What book are you citing?
– Brian Borchers
Aug 25 at 3:03
@BrianBorchers math.ucla.edu/~yanovsky/handbooks/PDEs.pdf
– Guerlando OCs
Aug 25 at 3:13
Shouldn't it be $$nabla^2 K = delta (x)$$ ??
– Mattos
Aug 25 at 3:21
The operator is the Laplacian, which some authors write as $nabla^2 K$ and others write as $Delta K$. The special cases $n=2$ and $n=3$ should be easy if you know the Laplacian operator in polar and spherical coordinates.
– Brian Borchers
Aug 25 at 3:42