Harmonic holomorphic function in Ω
$Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$
$$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$
Prove that there exists a holomorphic function $u+iv$ in $Ω$
I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.
Photo/solution
holomorphic-functions
asked Dec 4 '18 at 10:29
malilini
44
add a comment |
$Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$
$$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$
Prove that there exists a holomorphic function $u+iv$ in $Ω$
I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.
Photo/solution
holomorphic-functions
asked Dec 4 '18 at 10:29
malilini
44
add a comment |
$Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$
$$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$
Prove that there exists a holomorphic function $u+iv$ in $Ω$
I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.
Photo/solution
holomorphic-functions
asked Dec 4 '18 at 10:29
malilini
44
$Ω$ is simply connected in $C$, $u$ is a harmonic function in $Ω$ , $v$ in $Ω$
$$v(x,y) = int_0^1 (yu{Tiny x} (tx,ty)-xu{Tiny y} (tx,ty)) dt$$
Prove that there exists a holomorphic function $u+iv$ in $Ω$
I have the solution - can someone explain me the equalities to where i put the orange question marks? I dont understand why the first part of the term disappears.
Photo/solution
holomorphic-functions
holomorphic-functions
asked Dec 4 '18 at 10:29
malilini
44
asked Dec 4 '18 at 10:29
malilini
44
asked Dec 4 '18 at 10:29
malilini
44
asked Dec 4 '18 at 10:29
malilini
44
asked Dec 4 '18 at 10:29
malilini
44
44
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Let $g(t):= tu_y(tx,ty)$, then
$ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$
answered Dec 4 '18 at 10:40
Fred
44.2k1845
thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
– malilini
Dec 4 '18 at 10:53
I didnt think about tagging you @Fred
– malilini
Dec 4 '18 at 19:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $g(t):= tu_y(tx,ty)$, then
$ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$
answered Dec 4 '18 at 10:40
Fred
44.2k1845
thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
– malilini
Dec 4 '18 at 10:53
I didnt think about tagging you @Fred
– malilini
Dec 4 '18 at 19:45
add a comment |
Let $g(t):= tu_y(tx,ty)$, then
$ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$
answered Dec 4 '18 at 10:40
Fred
44.2k1845
thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
– malilini
Dec 4 '18 at 10:53
I didnt think about tagging you @Fred
– malilini
Dec 4 '18 at 19:45
add a comment |
Let $g(t):= tu_y(tx,ty)$, then
$ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$
answered Dec 4 '18 at 10:40
Fred
44.2k1845
Let $g(t):= tu_y(tx,ty)$, then
$ int_0^1 frac{d}{dt}(tu_y(tx,ty)) dt = int_0^1 g'(t) dt = g(1)-g(0) = u_y(x,y).$
answered Dec 4 '18 at 10:40
Fred
44.2k1845
answered Dec 4 '18 at 10:40
Fred
44.2k1845
answered Dec 4 '18 at 10:40
Fred
44.2k1845
answered Dec 4 '18 at 10:40
Fred
44.2k1845
44.2k1845
thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
– malilini
Dec 4 '18 at 10:53
I didnt think about tagging you @Fred
– malilini
Dec 4 '18 at 19:45
add a comment |
thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
– malilini
Dec 4 '18 at 10:53
I didnt think about tagging you @Fred
– malilini
Dec 4 '18 at 19:45
thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
– malilini
Dec 4 '18 at 10:53
thanks, this i can understand. but what happened with $$-u{Tiny y} (tx,ty) $$
– malilini
Dec 4 '18 at 10:53
I didnt think about tagging you @Fred
– malilini
Dec 4 '18 at 19:45
I didnt think about tagging you @Fred
– malilini
Dec 4 '18 at 19:45
add a comment |
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