harmonic conjugate for the $u(x,y)=x^3 +2xy-4xy^2$
If $$u(x,y)=x^3 +2xy-4xy^2$$ find the harmonic conjugate $v(x,y)$ and explain why the function is entire!
So I tried to solve it and that's what I got
$$ux=3x^2+2y-4y^2$$
$$uy=2x-8xy$$
after applying Cauchy-Riemann Equations
$$vy=ux=3x^2+2y-4y^2$$
and after integration
$$v(x,y)=3x^2y+y^2-(4/3) y^3+beta(X)$$
and after trying to solve for $beta$
I found it equal to
$$beta=x^2y-x^2$$
and after applying it to the
$$v(x,y)= 4x^2+y^2=(4/3)y^3-x^2$$
which in obviously is not applying C-R equations if we want to prove the solution is correct and ux is not equal to vy !!!
So does this function is not analytic or entire at all or I am doing something wrong in somewhere!
regards .
calculus complex-analysis complex-numbers systems-of-equations
edited Nov 28 at 20:17
dantopa
6,40932042
asked Nov 27 at 0:10
Smb Youz
265
add a comment |
If $$u(x,y)=x^3 +2xy-4xy^2$$ find the harmonic conjugate $v(x,y)$ and explain why the function is entire!
So I tried to solve it and that's what I got
$$ux=3x^2+2y-4y^2$$
$$uy=2x-8xy$$
after applying Cauchy-Riemann Equations
$$vy=ux=3x^2+2y-4y^2$$
and after integration
$$v(x,y)=3x^2y+y^2-(4/3) y^3+beta(X)$$
and after trying to solve for $beta$
I found it equal to
$$beta=x^2y-x^2$$
and after applying it to the
$$v(x,y)= 4x^2+y^2=(4/3)y^3-x^2$$
which in obviously is not applying C-R equations if we want to prove the solution is correct and ux is not equal to vy !!!
So does this function is not analytic or entire at all or I am doing something wrong in somewhere!
regards .
calculus complex-analysis complex-numbers systems-of-equations
edited Nov 28 at 20:17
dantopa
6,40932042
asked Nov 27 at 0:10
Smb Youz
265
There are several mistakes in your integration step. You should have had $3x^2 y + y^2 - frac 4 3 y^3 + beta(x)$, so your first and third terms are incorrect.
– T. Bongers
Nov 27 at 0:22
ok,,,i wrote it wrongly , so let me fix it please ,thanks
– Smb Youz
Nov 27 at 0:24
No answer yet guys ?
– Smb Youz
Nov 27 at 22:03
add a comment |
If $$u(x,y)=x^3 +2xy-4xy^2$$ find the harmonic conjugate $v(x,y)$ and explain why the function is entire!
So I tried to solve it and that's what I got
$$ux=3x^2+2y-4y^2$$
$$uy=2x-8xy$$
after applying Cauchy-Riemann Equations
$$vy=ux=3x^2+2y-4y^2$$
and after integration
$$v(x,y)=3x^2y+y^2-(4/3) y^3+beta(X)$$
and after trying to solve for $beta$
I found it equal to
$$beta=x^2y-x^2$$
and after applying it to the
$$v(x,y)= 4x^2+y^2=(4/3)y^3-x^2$$
which in obviously is not applying C-R equations if we want to prove the solution is correct and ux is not equal to vy !!!
So does this function is not analytic or entire at all or I am doing something wrong in somewhere!
regards .
calculus complex-analysis complex-numbers systems-of-equations
edited Nov 28 at 20:17
dantopa
6,40932042
asked Nov 27 at 0:10
Smb Youz
265
If $$u(x,y)=x^3 +2xy-4xy^2$$ find the harmonic conjugate $v(x,y)$ and explain why the function is entire!
So I tried to solve it and that's what I got
$$ux=3x^2+2y-4y^2$$
$$uy=2x-8xy$$
after applying Cauchy-Riemann Equations
$$vy=ux=3x^2+2y-4y^2$$
and after integration
$$v(x,y)=3x^2y+y^2-(4/3) y^3+beta(X)$$
and after trying to solve for $beta$
I found it equal to
$$beta=x^2y-x^2$$
and after applying it to the
$$v(x,y)= 4x^2+y^2=(4/3)y^3-x^2$$
which in obviously is not applying C-R equations if we want to prove the solution is correct and ux is not equal to vy !!!
So does this function is not analytic or entire at all or I am doing something wrong in somewhere!
regards .
calculus complex-analysis complex-numbers systems-of-equations
calculus complex-analysis complex-numbers systems-of-equations
edited Nov 28 at 20:17
dantopa
6,40932042
asked Nov 27 at 0:10
Smb Youz
265
edited Nov 28 at 20:17
dantopa
6,40932042
asked Nov 27 at 0:10
Smb Youz
265
edited Nov 28 at 20:17
dantopa
6,40932042
edited Nov 28 at 20:17
dantopa
6,40932042
edited Nov 28 at 20:17
dantopa
6,40932042
6,40932042
asked Nov 27 at 0:10
Smb Youz
265
asked Nov 27 at 0:10
Smb Youz
265
asked Nov 27 at 0:10
Smb Youz
265
265
There are several mistakes in your integration step. You should have had $3x^2 y + y^2 - frac 4 3 y^3 + beta(x)$, so your first and third terms are incorrect.
– T. Bongers
Nov 27 at 0:22
ok,,,i wrote it wrongly , so let me fix it please ,thanks
– Smb Youz
Nov 27 at 0:24
No answer yet guys ?
– Smb Youz
Nov 27 at 22:03
add a comment |
There are several mistakes in your integration step. You should have had $3x^2 y + y^2 - frac 4 3 y^3 + beta(x)$, so your first and third terms are incorrect.
– T. Bongers
Nov 27 at 0:22
ok,,,i wrote it wrongly , so let me fix it please ,thanks
– Smb Youz
Nov 27 at 0:24
No answer yet guys ?
– Smb Youz
Nov 27 at 22:03
There are several mistakes in your integration step. You should have had $3x^2 y + y^2 - frac 4 3 y^3 + beta(x)$, so your first and third terms are incorrect.
– T. Bongers
Nov 27 at 0:22
There are several mistakes in your integration step. You should have had $3x^2 y + y^2 - frac 4 3 y^3 + beta(x)$, so your first and third terms are incorrect.
– T. Bongers
Nov 27 at 0:22
ok,,,i wrote it wrongly , so let me fix it please ,thanks
– Smb Youz
Nov 27 at 0:24
ok,,,i wrote it wrongly , so let me fix it please ,thanks
– Smb Youz
Nov 27 at 0:24
No answer yet guys ?
– Smb Youz
Nov 27 at 22:03
No answer yet guys ?
– Smb Youz
Nov 27 at 22:03
add a comment |
2 Answers
2
active
oldest
votes
$u$ is not harmonic because $u_{xx}+u_{yy}neq0$ so it won't have a harmonic conjugate.
answered Nov 28 at 19:49
Empy2
33.4k12261
that's what i found sir , but will the professor accept it ? that is the problem . regards
– Smb Youz
Nov 28 at 20:33
Why do you think we know what your professor will accept? Also, why should we care?
– zhw.
Nov 28 at 22:37
add a comment |
The function is not harmonic, so it cannot have a conjugate.
On the other hand, if we fix the example as
$$
u(x,y)=x^3+2xy-color{red}{3}xy^2,
$$
then
begin{alignat}{2}
u_x&=3x^2+2y-3y^2 &qquadqquad u_{xx}&=6x \[4px]
u_y&=2x-6xy & u_{yy}&=-6x
end{alignat}
and this $u$ is harmonic.
The harmonic conjugate $v$ must satisfy
$$
v_x=-u_y qquadqquad v_y=u_x
$$
Thus
$$
v_x=-2x+6xy qquadqquad v_y=3x^2+2y-3y^2
$$
Hence
$$
v=-x^2+3x^2y+f(y)
$$
and
$$
v_y=3x^2+beta'(y)
$$
that gives $beta'(y)=2y-3y^2$, so $beta(y)=y^2-y^3$ (plus an arbitrary constant).
Hence
$$
v(x,y)=-x^2+3x^2y+y^2-y^3.
$$
You can show that
$$
f(z)=f(x+iy)=u(x,y)+iv(x,y)=z^3-iz^2
$$
answered Nov 28 at 21:20
egreg
178k1484200
yes i know if the example is $x^3+2xy-3xy^2$ is harmonic because the professor give us both of them and i solved this one but the one that i am asking for is my problem ! thanks
– Smb Youz
Nov 28 at 21:37
@SmbYouz Since $u_1=x^3+2xy-3xy^2$ is harmonic, if also $u_2=x^3+2xy-4xy^2$ is harmonic, then $u_3=xy^2$ would also be, which clearly isn't, because its Laplacian is $2x$. There's no way you can find a harmonic conjugate to a non harmonic function.
– egreg
Nov 28 at 21:46
You are absolutely Right . Special Thanks For you .
– Smb Youz
Nov 28 at 22:09
add a comment |
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2 Answers
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2 Answers
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votes
$u$ is not harmonic because $u_{xx}+u_{yy}neq0$ so it won't have a harmonic conjugate.
answered Nov 28 at 19:49
Empy2
33.4k12261
that's what i found sir , but will the professor accept it ? that is the problem . regards
– Smb Youz
Nov 28 at 20:33
Why do you think we know what your professor will accept? Also, why should we care?
– zhw.
Nov 28 at 22:37
add a comment |
$u$ is not harmonic because $u_{xx}+u_{yy}neq0$ so it won't have a harmonic conjugate.
answered Nov 28 at 19:49
Empy2
33.4k12261
that's what i found sir , but will the professor accept it ? that is the problem . regards
– Smb Youz
Nov 28 at 20:33
Why do you think we know what your professor will accept? Also, why should we care?
– zhw.
Nov 28 at 22:37
add a comment |
$u$ is not harmonic because $u_{xx}+u_{yy}neq0$ so it won't have a harmonic conjugate.
answered Nov 28 at 19:49
Empy2
33.4k12261
$u$ is not harmonic because $u_{xx}+u_{yy}neq0$ so it won't have a harmonic conjugate.
answered Nov 28 at 19:49
Empy2
33.4k12261
answered Nov 28 at 19:49
Empy2
33.4k12261
answered Nov 28 at 19:49
Empy2
33.4k12261
answered Nov 28 at 19:49
Empy2
33.4k12261
33.4k12261
that's what i found sir , but will the professor accept it ? that is the problem . regards
– Smb Youz
Nov 28 at 20:33
Why do you think we know what your professor will accept? Also, why should we care?
– zhw.
Nov 28 at 22:37
add a comment |
that's what i found sir , but will the professor accept it ? that is the problem . regards
– Smb Youz
Nov 28 at 20:33
Why do you think we know what your professor will accept? Also, why should we care?
– zhw.
Nov 28 at 22:37
that's what i found sir , but will the professor accept it ? that is the problem . regards
– Smb Youz
Nov 28 at 20:33
that's what i found sir , but will the professor accept it ? that is the problem . regards
– Smb Youz
Nov 28 at 20:33
Why do you think we know what your professor will accept? Also, why should we care?
– zhw.
Nov 28 at 22:37
Why do you think we know what your professor will accept? Also, why should we care?
– zhw.
Nov 28 at 22:37
add a comment |
The function is not harmonic, so it cannot have a conjugate.
On the other hand, if we fix the example as
$$
u(x,y)=x^3+2xy-color{red}{3}xy^2,
$$
then
begin{alignat}{2}
u_x&=3x^2+2y-3y^2 &qquadqquad u_{xx}&=6x \[4px]
u_y&=2x-6xy & u_{yy}&=-6x
end{alignat}
and this $u$ is harmonic.
The harmonic conjugate $v$ must satisfy
$$
v_x=-u_y qquadqquad v_y=u_x
$$
Thus
$$
v_x=-2x+6xy qquadqquad v_y=3x^2+2y-3y^2
$$
Hence
$$
v=-x^2+3x^2y+f(y)
$$
and
$$
v_y=3x^2+beta'(y)
$$
that gives $beta'(y)=2y-3y^2$, so $beta(y)=y^2-y^3$ (plus an arbitrary constant).
Hence
$$
v(x,y)=-x^2+3x^2y+y^2-y^3.
$$
You can show that
$$
f(z)=f(x+iy)=u(x,y)+iv(x,y)=z^3-iz^2
$$
answered Nov 28 at 21:20
egreg
178k1484200
yes i know if the example is $x^3+2xy-3xy^2$ is harmonic because the professor give us both of them and i solved this one but the one that i am asking for is my problem ! thanks
– Smb Youz
Nov 28 at 21:37
@SmbYouz Since $u_1=x^3+2xy-3xy^2$ is harmonic, if also $u_2=x^3+2xy-4xy^2$ is harmonic, then $u_3=xy^2$ would also be, which clearly isn't, because its Laplacian is $2x$. There's no way you can find a harmonic conjugate to a non harmonic function.
– egreg
Nov 28 at 21:46
You are absolutely Right . Special Thanks For you .
– Smb Youz
Nov 28 at 22:09
add a comment |
The function is not harmonic, so it cannot have a conjugate.
On the other hand, if we fix the example as
$$
u(x,y)=x^3+2xy-color{red}{3}xy^2,
$$
then
begin{alignat}{2}
u_x&=3x^2+2y-3y^2 &qquadqquad u_{xx}&=6x \[4px]
u_y&=2x-6xy & u_{yy}&=-6x
end{alignat}
and this $u$ is harmonic.
The harmonic conjugate $v$ must satisfy
$$
v_x=-u_y qquadqquad v_y=u_x
$$
Thus
$$
v_x=-2x+6xy qquadqquad v_y=3x^2+2y-3y^2
$$
Hence
$$
v=-x^2+3x^2y+f(y)
$$
and
$$
v_y=3x^2+beta'(y)
$$
that gives $beta'(y)=2y-3y^2$, so $beta(y)=y^2-y^3$ (plus an arbitrary constant).
Hence
$$
v(x,y)=-x^2+3x^2y+y^2-y^3.
$$
You can show that
$$
f(z)=f(x+iy)=u(x,y)+iv(x,y)=z^3-iz^2
$$
answered Nov 28 at 21:20
egreg
178k1484200
yes i know if the example is $x^3+2xy-3xy^2$ is harmonic because the professor give us both of them and i solved this one but the one that i am asking for is my problem ! thanks
– Smb Youz
Nov 28 at 21:37
@SmbYouz Since $u_1=x^3+2xy-3xy^2$ is harmonic, if also $u_2=x^3+2xy-4xy^2$ is harmonic, then $u_3=xy^2$ would also be, which clearly isn't, because its Laplacian is $2x$. There's no way you can find a harmonic conjugate to a non harmonic function.
– egreg
Nov 28 at 21:46
You are absolutely Right . Special Thanks For you .
– Smb Youz
Nov 28 at 22:09
add a comment |
The function is not harmonic, so it cannot have a conjugate.
On the other hand, if we fix the example as
$$
u(x,y)=x^3+2xy-color{red}{3}xy^2,
$$
then
begin{alignat}{2}
u_x&=3x^2+2y-3y^2 &qquadqquad u_{xx}&=6x \[4px]
u_y&=2x-6xy & u_{yy}&=-6x
end{alignat}
and this $u$ is harmonic.
The harmonic conjugate $v$ must satisfy
$$
v_x=-u_y qquadqquad v_y=u_x
$$
Thus
$$
v_x=-2x+6xy qquadqquad v_y=3x^2+2y-3y^2
$$
Hence
$$
v=-x^2+3x^2y+f(y)
$$
and
$$
v_y=3x^2+beta'(y)
$$
that gives $beta'(y)=2y-3y^2$, so $beta(y)=y^2-y^3$ (plus an arbitrary constant).
Hence
$$
v(x,y)=-x^2+3x^2y+y^2-y^3.
$$
You can show that
$$
f(z)=f(x+iy)=u(x,y)+iv(x,y)=z^3-iz^2
$$
answered Nov 28 at 21:20
egreg
178k1484200
The function is not harmonic, so it cannot have a conjugate.
On the other hand, if we fix the example as
$$
u(x,y)=x^3+2xy-color{red}{3}xy^2,
$$
then
begin{alignat}{2}
u_x&=3x^2+2y-3y^2 &qquadqquad u_{xx}&=6x \[4px]
u_y&=2x-6xy & u_{yy}&=-6x
end{alignat}
and this $u$ is harmonic.
The harmonic conjugate $v$ must satisfy
$$
v_x=-u_y qquadqquad v_y=u_x
$$
Thus
$$
v_x=-2x+6xy qquadqquad v_y=3x^2+2y-3y^2
$$
Hence
$$
v=-x^2+3x^2y+f(y)
$$
and
$$
v_y=3x^2+beta'(y)
$$
that gives $beta'(y)=2y-3y^2$, so $beta(y)=y^2-y^3$ (plus an arbitrary constant).
Hence
$$
v(x,y)=-x^2+3x^2y+y^2-y^3.
$$
You can show that
$$
f(z)=f(x+iy)=u(x,y)+iv(x,y)=z^3-iz^2
$$
answered Nov 28 at 21:20
egreg
178k1484200
edited Nov 28 at 21:26
answered Nov 28 at 21:20
egreg
178k1484200
answered Nov 28 at 21:20
egreg
178k1484200
answered Nov 28 at 21:20
egreg
178k1484200
178k1484200
yes i know if the example is $x^3+2xy-3xy^2$ is harmonic because the professor give us both of them and i solved this one but the one that i am asking for is my problem ! thanks
– Smb Youz
Nov 28 at 21:37
@SmbYouz Since $u_1=x^3+2xy-3xy^2$ is harmonic, if also $u_2=x^3+2xy-4xy^2$ is harmonic, then $u_3=xy^2$ would also be, which clearly isn't, because its Laplacian is $2x$. There's no way you can find a harmonic conjugate to a non harmonic function.
– egreg
Nov 28 at 21:46
You are absolutely Right . Special Thanks For you .
– Smb Youz
Nov 28 at 22:09
add a comment |
yes i know if the example is $x^3+2xy-3xy^2$ is harmonic because the professor give us both of them and i solved this one but the one that i am asking for is my problem ! thanks
– Smb Youz
Nov 28 at 21:37
@SmbYouz Since $u_1=x^3+2xy-3xy^2$ is harmonic, if also $u_2=x^3+2xy-4xy^2$ is harmonic, then $u_3=xy^2$ would also be, which clearly isn't, because its Laplacian is $2x$. There's no way you can find a harmonic conjugate to a non harmonic function.
– egreg
Nov 28 at 21:46
You are absolutely Right . Special Thanks For you .
– Smb Youz
Nov 28 at 22:09
yes i know if the example is $x^3+2xy-3xy^2$ is harmonic because the professor give us both of them and i solved this one but the one that i am asking for is my problem ! thanks
– Smb Youz
Nov 28 at 21:37
yes i know if the example is $x^3+2xy-3xy^2$ is harmonic because the professor give us both of them and i solved this one but the one that i am asking for is my problem ! thanks
– Smb Youz
Nov 28 at 21:37
@SmbYouz Since $u_1=x^3+2xy-3xy^2$ is harmonic, if also $u_2=x^3+2xy-4xy^2$ is harmonic, then $u_3=xy^2$ would also be, which clearly isn't, because its Laplacian is $2x$. There's no way you can find a harmonic conjugate to a non harmonic function.
– egreg
Nov 28 at 21:46
@SmbYouz Since $u_1=x^3+2xy-3xy^2$ is harmonic, if also $u_2=x^3+2xy-4xy^2$ is harmonic, then $u_3=xy^2$ would also be, which clearly isn't, because its Laplacian is $2x$. There's no way you can find a harmonic conjugate to a non harmonic function.
– egreg
Nov 28 at 21:46
You are absolutely Right . Special Thanks For you .
– Smb Youz
Nov 28 at 22:09
You are absolutely Right . Special Thanks For you .
– Smb Youz
Nov 28 at 22:09
add a comment |
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There are several mistakes in your integration step. You should have had $3x^2 y + y^2 - frac 4 3 y^3 + beta(x)$, so your first and third terms are incorrect.
– T. Bongers
Nov 27 at 0:22
ok,,,i wrote it wrongly , so let me fix it please ,thanks
– Smb Youz
Nov 27 at 0:24
No answer yet guys ?
– Smb Youz
Nov 27 at 22:03