Analytic functions having harmonic real and imaginary parts.
$begingroup$
I've bee set the following question in a homework assignment for my complex analysis class, but have literally no idea what it means by sufficiently regular.
Let $f : mathbb{C} to mathbb{C}$ be an analytic function, with $f = u + iv$. Prove that if $f$ is sufficiently regular the real
part and the imaginary part of $f$ (i.e. u and v) are harmonic functions in $mathbb{R}^2$
. That is, if we consider
$u$ as a function of $x$ and $y$ (where $z = x + iy$) we have
$u_{xx} + u_{yy} = 0$ ; $v_{xx} + v_{yy} = 0$.
Using $u_{xx}$ to denote the 2nd derivative of $u$ with respect to $x$.
We haven't mentioned anything about regularity conditions in lectures and I can't find anything about it in the online notes for the course.
Could someone please point me in the direction as to what it might mean by that? My first thought is that it might be something to do with considering it as a function from $mathbb{R}^2$ to $mathbb{R}^2$ with $u$ and $v$ as the component functions. But not sure what it means from there.
Thanks in advance
complex-analysis analysis harmonic-functions analytic-functions
edited Nov 25 '18 at 22:19
Mason
1,9551530
asked Nov 25 '18 at 21:58
user601175user601175
114
$endgroup$
|
show 3 more comments
$begingroup$
I've bee set the following question in a homework assignment for my complex analysis class, but have literally no idea what it means by sufficiently regular.
Let $f : mathbb{C} to mathbb{C}$ be an analytic function, with $f = u + iv$. Prove that if $f$ is sufficiently regular the real
part and the imaginary part of $f$ (i.e. u and v) are harmonic functions in $mathbb{R}^2$
. That is, if we consider
$u$ as a function of $x$ and $y$ (where $z = x + iy$) we have
$u_{xx} + u_{yy} = 0$ ; $v_{xx} + v_{yy} = 0$.
Using $u_{xx}$ to denote the 2nd derivative of $u$ with respect to $x$.
We haven't mentioned anything about regularity conditions in lectures and I can't find anything about it in the online notes for the course.
Could someone please point me in the direction as to what it might mean by that? My first thought is that it might be something to do with considering it as a function from $mathbb{R}^2$ to $mathbb{R}^2$ with $u$ and $v$ as the component functions. But not sure what it means from there.
Thanks in advance
complex-analysis analysis harmonic-functions analytic-functions
edited Nov 25 '18 at 22:19
Mason
1,9551530
asked Nov 25 '18 at 21:58
user601175user601175
114
$endgroup$
$begingroup$
I have no idea what is meant by "sufficiently regular" in that context. An analytic function is already as regular as can be!
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:04
$begingroup$
@Mason Isn't OP explaining what he means by harmonic there?
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:06
$begingroup$
@OlivierMoschetta. You must be right. Hard to prove something sufficiently regular if the MSE community and the OP doesn't know what the expression means... Actually.... Maybe there is a way. What additional conditions are required for an analytic function to be shown to be harmonic in this way?
$endgroup$
– Mason
Nov 25 '18 at 22:09
$begingroup$
The real part of an analytic function is always harmonic. No extra conditions required. This is an immediate consequence of the Cauchy-Riemann equations.
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:16
$begingroup$
Are you using the Cauchy Riemann equations in the following way? ux=vy => uxx=vyx : uy=-vx => uyy= -vxy. Then adding them to get 0? In this case how do we know vyx=vxy
$endgroup$
– user601175
Nov 25 '18 at 22:27
|
show 3 more comments
$begingroup$
I've bee set the following question in a homework assignment for my complex analysis class, but have literally no idea what it means by sufficiently regular.
Let $f : mathbb{C} to mathbb{C}$ be an analytic function, with $f = u + iv$. Prove that if $f$ is sufficiently regular the real
part and the imaginary part of $f$ (i.e. u and v) are harmonic functions in $mathbb{R}^2$
. That is, if we consider
$u$ as a function of $x$ and $y$ (where $z = x + iy$) we have
$u_{xx} + u_{yy} = 0$ ; $v_{xx} + v_{yy} = 0$.
Using $u_{xx}$ to denote the 2nd derivative of $u$ with respect to $x$.
We haven't mentioned anything about regularity conditions in lectures and I can't find anything about it in the online notes for the course.
Could someone please point me in the direction as to what it might mean by that? My first thought is that it might be something to do with considering it as a function from $mathbb{R}^2$ to $mathbb{R}^2$ with $u$ and $v$ as the component functions. But not sure what it means from there.
Thanks in advance
complex-analysis analysis harmonic-functions analytic-functions
edited Nov 25 '18 at 22:19
Mason
1,9551530
asked Nov 25 '18 at 21:58
user601175user601175
114
$endgroup$
I've bee set the following question in a homework assignment for my complex analysis class, but have literally no idea what it means by sufficiently regular.
Let $f : mathbb{C} to mathbb{C}$ be an analytic function, with $f = u + iv$. Prove that if $f$ is sufficiently regular the real
part and the imaginary part of $f$ (i.e. u and v) are harmonic functions in $mathbb{R}^2$
. That is, if we consider
$u$ as a function of $x$ and $y$ (where $z = x + iy$) we have
$u_{xx} + u_{yy} = 0$ ; $v_{xx} + v_{yy} = 0$.
Using $u_{xx}$ to denote the 2nd derivative of $u$ with respect to $x$.
We haven't mentioned anything about regularity conditions in lectures and I can't find anything about it in the online notes for the course.
Could someone please point me in the direction as to what it might mean by that? My first thought is that it might be something to do with considering it as a function from $mathbb{R}^2$ to $mathbb{R}^2$ with $u$ and $v$ as the component functions. But not sure what it means from there.
Thanks in advance
complex-analysis analysis harmonic-functions analytic-functions
complex-analysis analysis harmonic-functions analytic-functions
edited Nov 25 '18 at 22:19
Mason
1,9551530
asked Nov 25 '18 at 21:58
user601175user601175
114
edited Nov 25 '18 at 22:19
Mason
1,9551530
asked Nov 25 '18 at 21:58
user601175user601175
114
edited Nov 25 '18 at 22:19
Mason
1,9551530
edited Nov 25 '18 at 22:19
Mason
1,9551530
edited Nov 25 '18 at 22:19
Mason
1,9551530
1,9551530
asked Nov 25 '18 at 21:58
user601175user601175
114
asked Nov 25 '18 at 21:58
user601175user601175
114
asked Nov 25 '18 at 21:58
user601175user601175
114
114
$begingroup$
I have no idea what is meant by "sufficiently regular" in that context. An analytic function is already as regular as can be!
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:04
$begingroup$
@Mason Isn't OP explaining what he means by harmonic there?
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:06
$begingroup$
@OlivierMoschetta. You must be right. Hard to prove something sufficiently regular if the MSE community and the OP doesn't know what the expression means... Actually.... Maybe there is a way. What additional conditions are required for an analytic function to be shown to be harmonic in this way?
$endgroup$
– Mason
Nov 25 '18 at 22:09
$begingroup$
The real part of an analytic function is always harmonic. No extra conditions required. This is an immediate consequence of the Cauchy-Riemann equations.
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:16
$begingroup$
Are you using the Cauchy Riemann equations in the following way? ux=vy => uxx=vyx : uy=-vx => uyy= -vxy. Then adding them to get 0? In this case how do we know vyx=vxy
$endgroup$
– user601175
Nov 25 '18 at 22:27
|
show 3 more comments
$begingroup$
I have no idea what is meant by "sufficiently regular" in that context. An analytic function is already as regular as can be!
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:04
$begingroup$
@Mason Isn't OP explaining what he means by harmonic there?
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:06
$begingroup$
@OlivierMoschetta. You must be right. Hard to prove something sufficiently regular if the MSE community and the OP doesn't know what the expression means... Actually.... Maybe there is a way. What additional conditions are required for an analytic function to be shown to be harmonic in this way?
$endgroup$
– Mason
Nov 25 '18 at 22:09
$begingroup$
The real part of an analytic function is always harmonic. No extra conditions required. This is an immediate consequence of the Cauchy-Riemann equations.
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:16
$begingroup$
Are you using the Cauchy Riemann equations in the following way? ux=vy => uxx=vyx : uy=-vx => uyy= -vxy. Then adding them to get 0? In this case how do we know vyx=vxy
$endgroup$
– user601175
Nov 25 '18 at 22:27
$begingroup$
I have no idea what is meant by "sufficiently regular" in that context. An analytic function is already as regular as can be!
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:04
$begingroup$
I have no idea what is meant by "sufficiently regular" in that context. An analytic function is already as regular as can be!
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:04
$begingroup$
@Mason Isn't OP explaining what he means by harmonic there?
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:06
$begingroup$
@Mason Isn't OP explaining what he means by harmonic there?
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:06
$begingroup$
@OlivierMoschetta. You must be right. Hard to prove something sufficiently regular if the MSE community and the OP doesn't know what the expression means... Actually.... Maybe there is a way. What additional conditions are required for an analytic function to be shown to be harmonic in this way?
$endgroup$
– Mason
Nov 25 '18 at 22:09
$begingroup$
@OlivierMoschetta. You must be right. Hard to prove something sufficiently regular if the MSE community and the OP doesn't know what the expression means... Actually.... Maybe there is a way. What additional conditions are required for an analytic function to be shown to be harmonic in this way?
$endgroup$
– Mason
Nov 25 '18 at 22:09
$begingroup$
The real part of an analytic function is always harmonic. No extra conditions required. This is an immediate consequence of the Cauchy-Riemann equations.
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:16
$begingroup$
The real part of an analytic function is always harmonic. No extra conditions required. This is an immediate consequence of the Cauchy-Riemann equations.
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:16
$begingroup$
Are you using the Cauchy Riemann equations in the following way? ux=vy => uxx=vyx : uy=-vx => uyy= -vxy. Then adding them to get 0? In this case how do we know vyx=vxy
$endgroup$
– user601175
Nov 25 '18 at 22:27
$begingroup$
Are you using the Cauchy Riemann equations in the following way? ux=vy => uxx=vyx : uy=-vx => uyy= -vxy. Then adding them to get 0? In this case how do we know vyx=vxy
$endgroup$
– user601175
Nov 25 '18 at 22:27
|
show 3 more comments
1 Answer
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active
oldest
votes
$begingroup$
If a function is analytic on a given domain, it doesn't need to be anything other than that to satisfy Cauchy-Riemann equations.
By then taking second-order partial derivatives of u(x,y) and v(x,y) you easily arrive at Laplace's equation, which defines harmonic functions. Therefore, the only additional requirement is for the second-order partial derivatives of u(x,y) and v(x,y) to exist and be continuous (on the same domain as u(x,y) and v(x,y)). Continuity is needed to apply equality of mixed partials (Schwarz's theorem) when deriving Laplace's equation.
Calculation details (now that I have learnt some MathJax basics):
Cauchy-Riemann equations:
$frac{partial u}{partial x} = frac{partial v}{partial y}$
$frac{partial u}{partial y} = -frac{partial v}{partial x}$.
If the second-order partial derivatives of u(x,y) and v(x,y) exist on the domain considered, we have:
$frac{partial^2 u}{partial x^2} = frac{partial^2 v}{partial y partial x}$
$frac{partial^2 u}{partial y^2} = - frac{partial^2 v}{partial x partial y}$.
If the second-order partial derivatives are continuous on the domain considered, we have (Schwarz's theorem):
$frac{partial^2 v}{partial y partial x} = frac{partial^2 v}{partial x partial y}$.
On that assumption, if we summate the above equations, we find Laplace's equation for u(x,y):
$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$.
Similarly, we find Laplace's equation for v(x,y):
$frac{partial^2 v}{partial x^2} + frac{partial^2 v}{partial y^2} = 0$.
$endgroup$
add a comment |
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1 Answer
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oldest
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oldest
votes
$begingroup$
If a function is analytic on a given domain, it doesn't need to be anything other than that to satisfy Cauchy-Riemann equations.
By then taking second-order partial derivatives of u(x,y) and v(x,y) you easily arrive at Laplace's equation, which defines harmonic functions. Therefore, the only additional requirement is for the second-order partial derivatives of u(x,y) and v(x,y) to exist and be continuous (on the same domain as u(x,y) and v(x,y)). Continuity is needed to apply equality of mixed partials (Schwarz's theorem) when deriving Laplace's equation.
Calculation details (now that I have learnt some MathJax basics):
Cauchy-Riemann equations:
$frac{partial u}{partial x} = frac{partial v}{partial y}$
$frac{partial u}{partial y} = -frac{partial v}{partial x}$.
If the second-order partial derivatives of u(x,y) and v(x,y) exist on the domain considered, we have:
$frac{partial^2 u}{partial x^2} = frac{partial^2 v}{partial y partial x}$
$frac{partial^2 u}{partial y^2} = - frac{partial^2 v}{partial x partial y}$.
If the second-order partial derivatives are continuous on the domain considered, we have (Schwarz's theorem):
$frac{partial^2 v}{partial y partial x} = frac{partial^2 v}{partial x partial y}$.
On that assumption, if we summate the above equations, we find Laplace's equation for u(x,y):
$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$.
Similarly, we find Laplace's equation for v(x,y):
$frac{partial^2 v}{partial x^2} + frac{partial^2 v}{partial y^2} = 0$.
$endgroup$
add a comment |
$begingroup$
If a function is analytic on a given domain, it doesn't need to be anything other than that to satisfy Cauchy-Riemann equations.
By then taking second-order partial derivatives of u(x,y) and v(x,y) you easily arrive at Laplace's equation, which defines harmonic functions. Therefore, the only additional requirement is for the second-order partial derivatives of u(x,y) and v(x,y) to exist and be continuous (on the same domain as u(x,y) and v(x,y)). Continuity is needed to apply equality of mixed partials (Schwarz's theorem) when deriving Laplace's equation.
Calculation details (now that I have learnt some MathJax basics):
Cauchy-Riemann equations:
$frac{partial u}{partial x} = frac{partial v}{partial y}$
$frac{partial u}{partial y} = -frac{partial v}{partial x}$.
If the second-order partial derivatives of u(x,y) and v(x,y) exist on the domain considered, we have:
$frac{partial^2 u}{partial x^2} = frac{partial^2 v}{partial y partial x}$
$frac{partial^2 u}{partial y^2} = - frac{partial^2 v}{partial x partial y}$.
If the second-order partial derivatives are continuous on the domain considered, we have (Schwarz's theorem):
$frac{partial^2 v}{partial y partial x} = frac{partial^2 v}{partial x partial y}$.
On that assumption, if we summate the above equations, we find Laplace's equation for u(x,y):
$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$.
Similarly, we find Laplace's equation for v(x,y):
$frac{partial^2 v}{partial x^2} + frac{partial^2 v}{partial y^2} = 0$.
$endgroup$
add a comment |
$begingroup$
If a function is analytic on a given domain, it doesn't need to be anything other than that to satisfy Cauchy-Riemann equations.
By then taking second-order partial derivatives of u(x,y) and v(x,y) you easily arrive at Laplace's equation, which defines harmonic functions. Therefore, the only additional requirement is for the second-order partial derivatives of u(x,y) and v(x,y) to exist and be continuous (on the same domain as u(x,y) and v(x,y)). Continuity is needed to apply equality of mixed partials (Schwarz's theorem) when deriving Laplace's equation.
Calculation details (now that I have learnt some MathJax basics):
Cauchy-Riemann equations:
$frac{partial u}{partial x} = frac{partial v}{partial y}$
$frac{partial u}{partial y} = -frac{partial v}{partial x}$.
If the second-order partial derivatives of u(x,y) and v(x,y) exist on the domain considered, we have:
$frac{partial^2 u}{partial x^2} = frac{partial^2 v}{partial y partial x}$
$frac{partial^2 u}{partial y^2} = - frac{partial^2 v}{partial x partial y}$.
If the second-order partial derivatives are continuous on the domain considered, we have (Schwarz's theorem):
$frac{partial^2 v}{partial y partial x} = frac{partial^2 v}{partial x partial y}$.
On that assumption, if we summate the above equations, we find Laplace's equation for u(x,y):
$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$.
Similarly, we find Laplace's equation for v(x,y):
$frac{partial^2 v}{partial x^2} + frac{partial^2 v}{partial y^2} = 0$.
$endgroup$
If a function is analytic on a given domain, it doesn't need to be anything other than that to satisfy Cauchy-Riemann equations.
By then taking second-order partial derivatives of u(x,y) and v(x,y) you easily arrive at Laplace's equation, which defines harmonic functions. Therefore, the only additional requirement is for the second-order partial derivatives of u(x,y) and v(x,y) to exist and be continuous (on the same domain as u(x,y) and v(x,y)). Continuity is needed to apply equality of mixed partials (Schwarz's theorem) when deriving Laplace's equation.
Calculation details (now that I have learnt some MathJax basics):
Cauchy-Riemann equations:
$frac{partial u}{partial x} = frac{partial v}{partial y}$
$frac{partial u}{partial y} = -frac{partial v}{partial x}$.
If the second-order partial derivatives of u(x,y) and v(x,y) exist on the domain considered, we have:
$frac{partial^2 u}{partial x^2} = frac{partial^2 v}{partial y partial x}$
$frac{partial^2 u}{partial y^2} = - frac{partial^2 v}{partial x partial y}$.
If the second-order partial derivatives are continuous on the domain considered, we have (Schwarz's theorem):
$frac{partial^2 v}{partial y partial x} = frac{partial^2 v}{partial x partial y}$.
On that assumption, if we summate the above equations, we find Laplace's equation for u(x,y):
$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$.
Similarly, we find Laplace's equation for v(x,y):
$frac{partial^2 v}{partial x^2} + frac{partial^2 v}{partial y^2} = 0$.
edited Dec 1 '18 at 17:27
answered Nov 30 '18 at 19:01
user621367
add a comment |
add a comment |
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$begingroup$
I have no idea what is meant by "sufficiently regular" in that context. An analytic function is already as regular as can be!
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:04
$begingroup$
@Mason Isn't OP explaining what he means by harmonic there?
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:06
$begingroup$
@OlivierMoschetta. You must be right. Hard to prove something sufficiently regular if the MSE community and the OP doesn't know what the expression means... Actually.... Maybe there is a way. What additional conditions are required for an analytic function to be shown to be harmonic in this way?
$endgroup$
– Mason
Nov 25 '18 at 22:09
$begingroup$
The real part of an analytic function is always harmonic. No extra conditions required. This is an immediate consequence of the Cauchy-Riemann equations.
$endgroup$
– Olivier Moschetta
Nov 25 '18 at 22:16
$begingroup$
Are you using the Cauchy Riemann equations in the following way? ux=vy => uxx=vyx : uy=-vx => uyy= -vxy. Then adding them to get 0? In this case how do we know vyx=vxy
$endgroup$
– user601175
Nov 25 '18 at 22:27