$d^2u/dx^2 - d^2u/dy^2 = f(x,y)$ $rightarrow$ $d^2u/dξdη =...
$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$
Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$
I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$
My working is:
$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
Where am I going wrong?
Could someone please help?
linear-algebra pde partial-derivative
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
add a comment |
$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$
Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$
I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$
My working is:
$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
Where am I going wrong?
Could someone please help?
linear-algebra pde partial-derivative
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08
Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08
As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27
add a comment |
$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$
Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$
I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$
My working is:
$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
Where am I going wrong?
Could someone please help?
linear-algebra pde partial-derivative
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
$$frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = f(x,y) implies frac{partial^2u}{partial ξpartial η} = frac{1}{4}fleft(frac{1}{2}(ξ+η),frac{1}{2}(η-ξ)right)$$
Setting $ξ = x − y, η = x + y$ I get $x = frac{1}{2}(ξ+η)$ and $y=frac{1}{2}(η-ξ)$
I cant seem to show that $frac{partial^2u}{partial x^2} - frac{partial^2u}{partial y^2} = 4times frac{partial^2u}{partial ξpartial η}$
My working is:
$$partial u/partial x = frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$partial u/partial x = -frac{1}{2}frac{partial u} {partial xi} + frac{1}{2}frac{partial u}{partial eta}$$
$$frac{partial^2u}{partial x^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}+ frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
$$frac{partial^2u}{partial y^2} = frac{1}{4}frac{partial^2u}{partial xi^2} + frac{1}{4}frac{partial^2u}{partial eta^2}- frac{1}{2}frac{partial^2u}{partial ξpartial η}$$
Where am I going wrong?
Could someone please help?
linear-algebra pde partial-derivative
linear-algebra pde partial-derivative
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
edited Nov 26 '18 at 19:14
B. Mehta
11.8k22144
11.8k22144
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
asked Nov 26 '18 at 19:04
pablo_mathscobar
836
836
Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08
Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08
As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27
add a comment |
Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08
Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08
As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27
Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08
Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08
Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08
Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08
As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27
As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27
add a comment |
1 Answer
1
active
oldest
votes
You applied the chain rule incorrectly.
Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17
add a comment |
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1 Answer
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1 Answer
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You applied the chain rule incorrectly.
Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17
add a comment |
You applied the chain rule incorrectly.
Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17
add a comment |
You applied the chain rule incorrectly.
Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
You applied the chain rule incorrectly.
Applying the chain rule twice, we get
begin{align}
frac{partial^2u}{partial x^2} &= frac{partial}{partial x} left( frac{partial u}{partial xi}frac{partial xi}{partial x}+frac{partial u}{partial eta}frac{partial eta}{partial x} right)
= frac{partial}{partial x} left( frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= frac{partial^2 u}{partial xi^2}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial x}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial x}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial x}\
&= frac{partial^2 u}{partial xi^2} + 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial x} = frac{partial eta}{partial x} = 1. $$
Analogously,
begin{align}
frac{partial^2u}{partial y^2} &= frac{partial}{partial y} left( frac{partial u}{partial xi}frac{partial xi}{partial y}+frac{partial u}{partial eta}frac{partial eta}{partial y} right)
= frac{partial}{partial x} left( -frac{partial u}{partial xi}+frac{partial u}{partial eta} right)\
&= -frac{partial^2 u}{partial xi^2}frac{partial xi}{partial y}
- frac{partial^2 u}{partial xi partial eta}frac{partial eta}{partial y}
+ frac{partial^2 u}{partial eta partial xi}frac{partial xi}{partial y}
+ frac{partial^2 u}{partial eta^2}frac{partial eta}{partial y}\
&= frac{partial^2 u}{partial xi^2} - 2frac{partial^2 u}{partial xi partial eta} + frac{partial^2 u}{partial eta^2},
end{align}
since
$$ frac{partial xi}{partial y} = -1, quadtext{and}quad = frac{partial eta}{partial y} = 1. $$
Subtracting the two gives the result.
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
answered Nov 26 '18 at 19:13
MisterRiemann
5,7791624
5,7791624
thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17
add a comment |
thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17
thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17
thank you, i was making a stupid mistake
– pablo_mathscobar
Nov 26 '18 at 19:17
add a comment |
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Are the $d$s in your question intended to be $partial$?
– B. Mehta
Nov 26 '18 at 19:08
Yes they are, i dont know how to get the partial symbol
– pablo_mathscobar
Nov 26 '18 at 19:08
As you can see in my edit, you can use partial to get that symbol.
– B. Mehta
Nov 26 '18 at 19:27