Finding the Harmonic Conjugate Which Satisfies $v(0,0)=1$
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Define a function $u:mathbbR^2rightarrowmathbbR$ by $u(x,y,)=e^xsin(y)+4xy$. Find the harmonic conjugate $v$ for $u$ that satisfies $v(0,0)=1$.
As $mathbbR^2$ is simply connected and $u$ is harmonic, then $exists v:mathbbR^2rightarrowmathbbR$ such that $f(x+iy)=u(x,y)+iv(x,y)$ is holomorphic in $mathbbR^2$. Hence CRE hold.
$$fracpartial vpartial y=e^xsin(y)+4y, fracpartial vpartial x=-e^xcos(y)-4x$$
Now consider
beginalign
fracpartial vpartial y&=e^xsin(y)+4y \
v(x,y)&=-e^xcos(y)+2y^2+f(x) \
fracpartial vpartial x&=-e^xcos(y)+f'(x) \
endalign
Hence $$f(x)=-2x^2+C$$
So $$v(x,y)=-e^xcos(y)+2y^2-2x^2+C$$ and using $v(0,0)=1$, $C=2$. But the solution I have found states $C=0$. I do not see how this can be.
complex-analysis proof-verification harmonic-functions
asked Aug 16 at 2:39
Bell
691313
add a comment |Â
up vote
0
down vote
favorite
Define a function $u:mathbbR^2rightarrowmathbbR$ by $u(x,y,)=e^xsin(y)+4xy$. Find the harmonic conjugate $v$ for $u$ that satisfies $v(0,0)=1$.
As $mathbbR^2$ is simply connected and $u$ is harmonic, then $exists v:mathbbR^2rightarrowmathbbR$ such that $f(x+iy)=u(x,y)+iv(x,y)$ is holomorphic in $mathbbR^2$. Hence CRE hold.
$$fracpartial vpartial y=e^xsin(y)+4y, fracpartial vpartial x=-e^xcos(y)-4x$$
Now consider
beginalign
fracpartial vpartial y&=e^xsin(y)+4y \
v(x,y)&=-e^xcos(y)+2y^2+f(x) \
fracpartial vpartial x&=-e^xcos(y)+f'(x) \
endalign
Hence $$f(x)=-2x^2+C$$
So $$v(x,y)=-e^xcos(y)+2y^2-2x^2+C$$ and using $v(0,0)=1$, $C=2$. But the solution I have found states $C=0$. I do not see how this can be.
complex-analysis proof-verification harmonic-functions
asked Aug 16 at 2:39
Bell
691313
2
I think this is correct and $C=2$.
– Nosrati
Aug 16 at 2:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Define a function $u:mathbbR^2rightarrowmathbbR$ by $u(x,y,)=e^xsin(y)+4xy$. Find the harmonic conjugate $v$ for $u$ that satisfies $v(0,0)=1$.
As $mathbbR^2$ is simply connected and $u$ is harmonic, then $exists v:mathbbR^2rightarrowmathbbR$ such that $f(x+iy)=u(x,y)+iv(x,y)$ is holomorphic in $mathbbR^2$. Hence CRE hold.
$$fracpartial vpartial y=e^xsin(y)+4y, fracpartial vpartial x=-e^xcos(y)-4x$$
Now consider
beginalign
fracpartial vpartial y&=e^xsin(y)+4y \
v(x,y)&=-e^xcos(y)+2y^2+f(x) \
fracpartial vpartial x&=-e^xcos(y)+f'(x) \
endalign
Hence $$f(x)=-2x^2+C$$
So $$v(x,y)=-e^xcos(y)+2y^2-2x^2+C$$ and using $v(0,0)=1$, $C=2$. But the solution I have found states $C=0$. I do not see how this can be.
complex-analysis proof-verification harmonic-functions
asked Aug 16 at 2:39
Bell
691313
Define a function $u:mathbbR^2rightarrowmathbbR$ by $u(x,y,)=e^xsin(y)+4xy$. Find the harmonic conjugate $v$ for $u$ that satisfies $v(0,0)=1$.
As $mathbbR^2$ is simply connected and $u$ is harmonic, then $exists v:mathbbR^2rightarrowmathbbR$ such that $f(x+iy)=u(x,y)+iv(x,y)$ is holomorphic in $mathbbR^2$. Hence CRE hold.
$$fracpartial vpartial y=e^xsin(y)+4y, fracpartial vpartial x=-e^xcos(y)-4x$$
Now consider
beginalign
fracpartial vpartial y&=e^xsin(y)+4y \
v(x,y)&=-e^xcos(y)+2y^2+f(x) \
fracpartial vpartial x&=-e^xcos(y)+f'(x) \
endalign
Hence $$f(x)=-2x^2+C$$
So $$v(x,y)=-e^xcos(y)+2y^2-2x^2+C$$ and using $v(0,0)=1$, $C=2$. But the solution I have found states $C=0$. I do not see how this can be.
complex-analysis proof-verification harmonic-functions
asked Aug 16 at 2:39
Bell
691313
asked Aug 16 at 2:39
Bell
691313
asked Aug 16 at 2:39
Bell
691313
asked Aug 16 at 2:39
Bell
691313
691313
2
I think this is correct and $C=2$.
– Nosrati
Aug 16 at 2:49
add a comment |Â
2
I think this is correct and $C=2$.
– Nosrati
Aug 16 at 2:49
2
2
I think this is correct and $C=2$.
– Nosrati
Aug 16 at 2:49
I think this is correct and $C=2$.
– Nosrati
Aug 16 at 2:49
add a comment |Â
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2
I think this is correct and $C=2$.
– Nosrati
Aug 16 at 2:49