Why does $log|(x,y) - (ξ,η)| $ satisfy the laplace equation for all $(x,y) ne (ξ,η)$
$begingroup$
Why does $G = log|(x,y) - (ξ,η)| $ satisfy the laplace equation for all $(x,y) ne (ξ,η)$
Could someone explain why this is true?
I thought $frac{partial G}{partial x} = frac{1}{(x,y) - (ξ,η)}$ and $frac{partial G}{partial y} = frac{1}{(x,y) - (ξ,η)}$
and so
$frac{partial^2 G}{partial x^2}+ frac{partial^2 G}{partial y^2} = frac{-1}{[(x,y) - (ξ,η)]^2} + frac{-1}{([(x,y) - (ξ,η)]^2}$
pde partial-derivative
edited Dec 18 '18 at 13:56
Harry49
7,07131239
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
Why does $G = log|(x,y) - (ξ,η)| $ satisfy the laplace equation for all $(x,y) ne (ξ,η)$
Could someone explain why this is true?
I thought $frac{partial G}{partial x} = frac{1}{(x,y) - (ξ,η)}$ and $frac{partial G}{partial y} = frac{1}{(x,y) - (ξ,η)}$
and so
$frac{partial^2 G}{partial x^2}+ frac{partial^2 G}{partial y^2} = frac{-1}{[(x,y) - (ξ,η)]^2} + frac{-1}{([(x,y) - (ξ,η)]^2}$
pde partial-derivative
edited Dec 18 '18 at 13:56
Harry49
7,07131239
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
Why does $G = log|(x,y) - (ξ,η)| $ satisfy the laplace equation for all $(x,y) ne (ξ,η)$
Could someone explain why this is true?
I thought $frac{partial G}{partial x} = frac{1}{(x,y) - (ξ,η)}$ and $frac{partial G}{partial y} = frac{1}{(x,y) - (ξ,η)}$
and so
$frac{partial^2 G}{partial x^2}+ frac{partial^2 G}{partial y^2} = frac{-1}{[(x,y) - (ξ,η)]^2} + frac{-1}{([(x,y) - (ξ,η)]^2}$
pde partial-derivative
edited Dec 18 '18 at 13:56
Harry49
7,07131239
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
$endgroup$
Why does $G = log|(x,y) - (ξ,η)| $ satisfy the laplace equation for all $(x,y) ne (ξ,η)$
Could someone explain why this is true?
I thought $frac{partial G}{partial x} = frac{1}{(x,y) - (ξ,η)}$ and $frac{partial G}{partial y} = frac{1}{(x,y) - (ξ,η)}$
and so
$frac{partial^2 G}{partial x^2}+ frac{partial^2 G}{partial y^2} = frac{-1}{[(x,y) - (ξ,η)]^2} + frac{-1}{([(x,y) - (ξ,η)]^2}$
pde partial-derivative
pde partial-derivative
edited Dec 18 '18 at 13:56
Harry49
7,07131239
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
edited Dec 18 '18 at 13:56
Harry49
7,07131239
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
edited Dec 18 '18 at 13:56
Harry49
7,07131239
edited Dec 18 '18 at 13:56
Harry49
7,07131239
edited Dec 18 '18 at 13:56
Harry49
7,07131239
7,07131239
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
asked Dec 14 '18 at 17:03
pablo_mathscobarpablo_mathscobar
996
996
add a comment |
add a comment |
1 Answer
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$begingroup$
Your calculation of the partial derivatives is wrong. The expression $(x,y) - (xi,eta)$ is a vector and doesn't make sense in a denominator. Instead you should write
$$G(x,y) = log sqrt{(x-xi)^2 + (y - eta)^2} = frac 12 log left( (x-xi)^2 + (y - eta)^2right)$$ and work from there. For instance,
$$frac{partial G}{partial x}(x,y) = frac{x-xi}{(x-xi)^2 + (y - eta)^2}$$ so that
$$ frac{partial^2 G}{partial x ^2}(x,y) = frac{ -(x-xi)^2 + (y-eta)^2}{(x-xi)^2 + (y - eta)^2}.$$
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
$endgroup$
1
$begingroup$
Cheers mate, i was confused about the notation
$endgroup$
– pablo_mathscobar
Dec 14 '18 at 17:59
add a comment |
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1 Answer
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$begingroup$
Your calculation of the partial derivatives is wrong. The expression $(x,y) - (xi,eta)$ is a vector and doesn't make sense in a denominator. Instead you should write
$$G(x,y) = log sqrt{(x-xi)^2 + (y - eta)^2} = frac 12 log left( (x-xi)^2 + (y - eta)^2right)$$ and work from there. For instance,
$$frac{partial G}{partial x}(x,y) = frac{x-xi}{(x-xi)^2 + (y - eta)^2}$$ so that
$$ frac{partial^2 G}{partial x ^2}(x,y) = frac{ -(x-xi)^2 + (y-eta)^2}{(x-xi)^2 + (y - eta)^2}.$$
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
$endgroup$
1
$begingroup$
Cheers mate, i was confused about the notation
$endgroup$
– pablo_mathscobar
Dec 14 '18 at 17:59
add a comment |
$begingroup$
Your calculation of the partial derivatives is wrong. The expression $(x,y) - (xi,eta)$ is a vector and doesn't make sense in a denominator. Instead you should write
$$G(x,y) = log sqrt{(x-xi)^2 + (y - eta)^2} = frac 12 log left( (x-xi)^2 + (y - eta)^2right)$$ and work from there. For instance,
$$frac{partial G}{partial x}(x,y) = frac{x-xi}{(x-xi)^2 + (y - eta)^2}$$ so that
$$ frac{partial^2 G}{partial x ^2}(x,y) = frac{ -(x-xi)^2 + (y-eta)^2}{(x-xi)^2 + (y - eta)^2}.$$
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
$endgroup$
1
$begingroup$
Cheers mate, i was confused about the notation
$endgroup$
– pablo_mathscobar
Dec 14 '18 at 17:59
add a comment |
$begingroup$
Your calculation of the partial derivatives is wrong. The expression $(x,y) - (xi,eta)$ is a vector and doesn't make sense in a denominator. Instead you should write
$$G(x,y) = log sqrt{(x-xi)^2 + (y - eta)^2} = frac 12 log left( (x-xi)^2 + (y - eta)^2right)$$ and work from there. For instance,
$$frac{partial G}{partial x}(x,y) = frac{x-xi}{(x-xi)^2 + (y - eta)^2}$$ so that
$$ frac{partial^2 G}{partial x ^2}(x,y) = frac{ -(x-xi)^2 + (y-eta)^2}{(x-xi)^2 + (y - eta)^2}.$$
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
$endgroup$
Your calculation of the partial derivatives is wrong. The expression $(x,y) - (xi,eta)$ is a vector and doesn't make sense in a denominator. Instead you should write
$$G(x,y) = log sqrt{(x-xi)^2 + (y - eta)^2} = frac 12 log left( (x-xi)^2 + (y - eta)^2right)$$ and work from there. For instance,
$$frac{partial G}{partial x}(x,y) = frac{x-xi}{(x-xi)^2 + (y - eta)^2}$$ so that
$$ frac{partial^2 G}{partial x ^2}(x,y) = frac{ -(x-xi)^2 + (y-eta)^2}{(x-xi)^2 + (y - eta)^2}.$$
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
answered Dec 14 '18 at 17:58
Umberto P.Umberto P.
39.3k13166
39.3k13166
1
$begingroup$
Cheers mate, i was confused about the notation
$endgroup$
– pablo_mathscobar
Dec 14 '18 at 17:59
add a comment |
1
$begingroup$
Cheers mate, i was confused about the notation
$endgroup$
– pablo_mathscobar
Dec 14 '18 at 17:59
1
1
$begingroup$
Cheers mate, i was confused about the notation
$endgroup$
– pablo_mathscobar
Dec 14 '18 at 17:59
$begingroup$
Cheers mate, i was confused about the notation
$endgroup$
– pablo_mathscobar
Dec 14 '18 at 17:59
add a comment |
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