Harmonic Conjugates

Complex Analysis
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Tsakanikas Nickos
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Harmonic Conjugates

#1

Post by Tsakanikas Nickos »

(1) Prove that if \( \displaystyle f=u+iv \) is defined and is analytic on an open subset \( \displaystyle A \) of \( \mathbb{C} \) , then \( \displaystyle u = Re(f) \) and \( \displaystyle v = Im(f) \) are harmonic on \( \displaystyle A \).

(2) Prove that the opposite in not generally true. Specifically, prove that if \( \displaystyle u \) and \( \displaystyle v \) are harmonic on an open subset \( \displaystyle A \) of \( \mathbb{C} \) , then the function \( \displaystyle f : A \longrightarrow \mathbb{C} , f = u+iv \) need not be analytic.

(3) Prove now that if \( \displaystyle u \) and \( \displaystyle v \) are harmonic on an open subset \( \displaystyle A \) of \( \mathbb{C} \) and additionally the satisfy the Cauchy-Riemann equations, then the function \( \displaystyle f : A \longrightarrow \mathbb{C} , f = u+iv \) is analytic on \( \displaystyle A \). (This illustrates the case that the converse of (1) is true)

[The proofs of (1),(2),(3) are quite easy. The question following is much more difficult to deal with]

Consequently, if a complex-valued function is analytic, then its real and imaginary parts are harmonic functions. So, it is natural to ask whether every harmonic function is the real part of an analytic function. Answer to this question· that is to the question:

" Given a harmonic function \( \displaystyle u \), which is defined on an open subset \( \displaystyle A \) of \( \mathbb{C} \), need there be a harmonic function \( \displaystyle v \), which is also defined on \( \displaystyle A \), such that the function \( \displaystyle f : A \longrightarrow \mathbb{C} , f = u+iv \) is analytic on \( \displaystyle A \) ? "
Papapetros Vaggelis
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Re: Harmonic Conjugates

#2

Post by Papapetros Vaggelis »

Hello Nickos. Interesting questions.

\(\displaystyle{(1).}\) Let

\(\displaystyle{f:A\longrightarrow \mathbb{C}\,,f(z)=u(x,y)+i\,v(x,y)\,,z=x+i\,y}\) ,

\(\displaystyle{\left(x,y\right)\in D=\left\{\left(x,y\right)\in\mathbb{R}^2: x+i\,y\in A\right\}\subset \mathbb{R}^2}\) ,

where \(\displaystyle{A}\) is an open subset of \(\displaystyle{\mathbb{C}}\) , be an analytic function on \(\displaystyle{A}\) .

Note : The set \(\displaystyle{D}\) is an open set of \(\displaystyle{\mathbb{R}^2}\) : Indeed, let \(\displaystyle{\left(x,y\right)\in D}\) . Then,

\(\displaystyle{z=x+i\,y\in A}\) and since \(\displaystyle{A}\) is an open subset of \(\displaystyle{\mathbb{C}}\) , there is \(\displaystyle{\epsilon>0}\)

such that \(\displaystyle{B_{\mathbb{C}}\,(z,\epsilon)=\left\{w\in\mathbb{C}:\left|w-z\right|<\epsilon\right\}\subseteq A}\) . So,

if \(\displaystyle{w=(w_1,w_2)\in B_{\mathbb{R}^2}\,((x,y),\epsilon)}\) ,

then :

\(\displaystyle{||(w_1,w_2)-(x,y)||<\epsilon\implies \sqrt{\left(w_1-x\right)^2+\left(w_2-y\right)^2}<\epsilon}\) ,

which means that the complex number \(\displaystyle{w=w_1+i\,w_2}\) is an element of \(\displaystyle{B_{\mathbb{C}}\,(z,\epsilon)}\).

Therefore, \(\displaystyle{w=w_1+i\,w_2\in A\implies \left(w_1,w_2\right)\in D}\) and thus

\(\displaystyle{B_{\mathbb{R}^2}\,(z,\epsilon)\subseteq D}\) .

The real and the imaginary part of the fynction \(\displaystyle{f}\) have continuous partial derivatives of any order.

Also, by Cauchy-Riemann equations we have that :

\(\displaystyle{u_{x}(x,y)=v_{y}(x,y)\,,u_{y}(x.y)=-v_{x}(x.y)\,,\forall\,\left(x,y\right)\in D}\) . So,

\(\displaystyle{u_{xx}=v_{yx}\,,u_{xy}=v_{yy}\,,u_{yx}=-v_{xx}\,,u_{yy}=-v_{xy}}\) .

According to \(\displaystyle{\rm{Weierstrass}}\) theorem, we get :

\(\displaystyle{u_{xy}=u_{yx}\implies v_{yy}=-v_{xx}\implies v_{xx}+v_{yy}=0}\) .

\(\displaystyle{v_{xy}=v_{yx}\implies-u_{yy}=u_{xx}\implies u_{xx}+u_{yy}=0}\) .

Consequently, the functions \(\displaystyle{u\,,v}\) are harmonic on \(\displaystyle{D}\) .

\(\displaystyle{(2).}\) If \(\displaystyle{u\,,v:\mathbb{R}^2\longrightarrow \mathbb{R}\,,u(x,y)=x\,,v(x,y)=-y}\) , then

\(\displaystyle{u_{x}(x,y)=1\,,u_{xx}(x,y)=0\,,u_{y}(x,y)=0=u_{yy}(x,y)\,,v_{x}(x,y)=0=v_{xx}(x,y)\,,v_{y}(x,y)=-1\,,v_{yy}(x,y)=0}\)

and \(\displaystyle{u_{xx}+u_{yy}=0\,,v_{xx}+v_{yy}=0}\), which means that \(\displaystyle{u\,,v}\) are harmonic on \(\displaystyle{D=\mathbb{R}^2}\) ,

but the complex-valued function \(\displaystyle{f:A=\mathbb{C}\longrightarrow \mathbb{C}\,,f(z)=x-i\,y\,,z=x+i\,y\,,f(z)=\bar{z}}\) is not

analytic on \(\displaystyle{A=\mathbb{C}}\) . This follows form the fact that the limit \(\displaystyle{\lim_{z\to 0}\dfrac{\bar{z}}{z}}\)

does not exist on \(\displaystyle{\mathbb{C}}\) .

Considering the sequences \(\displaystyle{x_{n}=\dfrac{1}{n}+i\,0\,,n\in\mathbb{N}\,,y_{n}=0+i\,\dfrac{1}{n}\,,n\in\mathbb{N}}\) ,

we have that \(\displaystyle{\dfrac{\bar{x_{n}}}{x_{n}}=1\longrightarrow 1\,,\dfrac{\bar{y_{n}}}{y_{n}}=-1\longrightarrow -1}\) .

\(\displaystyle{(3).}\) Since \(\displaystyle{u\,,v}\) satisfy the Cauchy-Riemann equations at each point

\(\displaystyle{\left(x,y\right)\in D=\left\{\left(x,y\right)\in\mathbb{R}^2: x+i\,y\in A\right\}}\) ,

it makes sense to examine if the function \(\displaystyle{f=u+i\,v}\) is analytic on \(\displaystyle{A}\) .

The functions \(\displaystyle{u_{x},u_{y},v_{x},v_{y}}\) are continuous on \(\displaystyle{D}\) and satisfy the Cauchy-Riemann equations,

so the function \(\displaystyle{f}\) is analytic.



Additional question :

It's known that \(\displaystyle{u_{xx}+u_{yy}=0}\) . This equation for \(\displaystyle{P=-u_{y}\,,Q=u_{x}}\) takes the form :

\(\displaystyle{\dfrac{\partial}{\partial y}P(x,y)=\dfrac{\partial}{\partial x}Q(x,y)\,,(x,y)\in D}\) and thus there is \(\displaystyle{v \,: \,D \longrightarrow \mathbb{R}}\) ,

such that

\(\displaystyle{grad v(x,y)=(v_{x}(x,y),v_{y}(x,y))=(P(x,y),Q(x,y))=(-u_{y}(x,y),u_{x}(x,y))\iff }\)

\(\displaystyle{\iff u_{x}=v_{y}\,,u_{y}=-v_{x}}\) .

According to \(\displaystyle{(3)}\) , the function \(\displaystyle{f=u+i\,v}\) is analytic on \(\displaystyle{A}\) .

Nickos, let me ask one more question :

Find all analytic functions \(\displaystyle{f:A\longrightarrow \mathbb{C}}\) whose real part given by

\(\displaystyle{u=u(x,y)=\dfrac{1}{2}\,\ln\,\left(x^2+y^2\right)\,,\left(x,y\right)\in D=\left(0,+\infty\right)\times \mathbb{R}}\) .
Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Harmonic Conjugates

#3

Post by Tsakanikas Nickos »

Thank you for your answer, mr.Papapetros!

I would like to make some comments,as well.

Comment on (2): It is known that the Cauchy-Riemann equations
\[ \displaystyle {\partial u \over \partial x} = {\partial v \over \partial y} , {\partial u \over \partial y} = - {\partial v \over \partial x} \] [ at a point \( \displaystyle (x_{0},y_{0}) = z_{0} \) ] are equivalent to the equation
\[ \displaystyle {\partial f\over \partial \bar{z}} = 0 \] [ at \( \displaystyle z=z_{0} \) ].

Another way to prove (2) is the following one: Since, in (2), \( \displaystyle f(z) = \bar{z} \) , it is obvious that \( \displaystyle {\partial f \over \partial \bar{z}} = 1 \neq 0 \) , which implies that the C-R equations are not satisfied. Therefore, \( \displaystyle f \) cannot be analytic.


Comment on additional question: Another way to answer to this question, locally though, is to prove the following: " Let \( \displaystyle A \) be an open subset of \( \mathbb{C} \) and suppose that \( \displaystyle u : A \longrightarrow \mathbb{R} \) is a harmonic function. Show that there is a disk \( \displaystyle D(z_{0},r) \), where \( r>0 \text{ and } z_{0}=x_{0} + iy_{0} \), such that the equation
\[ \displaystyle v(x_{1},y_{1}) = c + \int_{y_{0}}^{y_{1}} {\partial u \over \partial x}( x_{1},y ) \mathrm{d}y - \int_{x_{0}}^{x_{1}} {\partial u \over \partial y}( x,y_{1} ) \mathrm{d}x \; , \; (x_{1},y_{1}) \in D(z_{0},r) \]
defines a harmonic conjugate for \( \displaystyle u \) on \( \displaystyle D(z_{0},r) \) with \( \displaystyle v(x_{0},y_{0})=c\)."

Having proved the above (which is quite easy to prove), we know that, at least locally, every harmonic function is the real part of an analytic function.
Tsakanikas Nickos
Community Team
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Harmonic Conjugates

#4

Post by Tsakanikas Nickos »

Papapetros Vaggelis wrote:Nickos, let me ask one more question :

Find all analytic functions \(\displaystyle{f:A\longrightarrow \mathbb{C}}\) whose real part given by

\(\displaystyle{u=u(x,y)=\dfrac{1}{2}\,\ln\,\left(x^2+y^2\right)\,,\left(x,y\right)\in D=\left(0,+\infty\right)\times \mathbb{R}}\) .
Let me, now, try to answer your question, Vaggelis!

Let \( \displaystyle u : \left( 0, +\infty \right) \times \mathbb{R} \longrightarrow \mathbb{R} \; , \; u(x,y) = \frac{1}{2}\log\left(x^2+y^2\right) \). We have that
\[ \displaystyle {\partial u \over \partial x} = \frac{x}{x^2+y^2} \]
\[ \displaystyle {\partial u \over \partial y} = \frac{y}{x^2+y^2} \]
\[ \displaystyle {\partial^2 u \over \partial x^2} = \frac{y^2-x^2}{(x^2+y^2)^2} \]
\[ \displaystyle {\partial^2 u \over \partial x^2} = \frac{-(y^2-x^2)}{(x^2+y^2)^2} \]
Hence, \[ \displaystyle \nabla^2 u = 0, \] which means that the given function \( \displaystyle u \) is harmonic on the open subset \( \displaystyle \left( 0 , +\infty \right) \times \mathbb{R} \) of \( \mathbb{C} \).

Suppose that \( \displaystyle v \) is the harmonic conjugate of \( \displaystyle u \). Then the functions \( \displaystyle u,v \) must satisfy the Cauchy-Riemann equations, so we have that
\[\displaystyle {\partial v \over \partial y} = {\partial u \over \partial x} = \frac{x}{x^2+y^2} \implies
v= \int \frac{x}{x^2+y^2} \mathrm{d}y \, + \, g_{1}(y) \implies
v(x,y)= \arctan\left( \frac{y}{x} \right) \,+ \, g_{1}(y) \, + \, c_{1} \]

\[ \displaystyle {\partial v \over \partial x} = -{\partial u \over \partial y} = -\frac{y}{x^2+y^2} \implies
v= \int \frac{y}{x^2+y^2} \mathrm{d}x \, + \, g_{2}(x) \implies
v(x,y)= -\arctan\left( \frac{x}{y} \right) \,+ \, g_{2}(x) \, + \, c_{2} \]

Since \[ \displaystyle \arctan(\xi) + \arctan\left(\frac{1}{\xi}\right) = \frac{\pi}{2}, \] we have that
\[ \displaystyle \arctan\left(\frac{y}{x}\right) = - \arctan\left(\frac{x}{y}\right) + \frac{\pi}{2} \]
Therefore the functions \( \displaystyle g_{1},g_{2} \) are constant and, respectively, equal to \( \displaystyle c_{3},c_{4} \). Taking
\[ \displaystyle c_{1} + c_{3} = c \text{ and } c_{2} + c_{4} = c + \frac{\pi}{2} \] we conclude that the function
\[ \displaystyle v : \left( 0, +\infty \right) \times \mathbb{R} \longrightarrow \mathbb{R} \; , \; v(x,y) = \arctan\left(\frac{y}{x}\right) + c, \] where c is an arbitrary constant, is the harmonic conjugate of
\[ \displaystyle u : \left( 0, +\infty \right) \times \mathbb{R} \longrightarrow \mathbb{R} \; , \; u(x,y) = \frac{1}{2}\log\left(x^2+y^2\right) \]
Hence, the requested analytic function (modulo constant) is
\[ \displaystyle f : \left( 0, +\infty \right) \times \mathbb{R} \longrightarrow \mathbb{C} \; , \; f(x,y) = \frac{1}{2}\log\left(x^2+y^2\right) + i\arctan\left(\frac{y}{x}\right) \]
Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Harmonic Conjugates

#5

Post by Tsakanikas Nickos »

Two additional and quite interesting exercises, related to the initial one, are the following:

(1) Suppose that \( \displaystyle f \) is analytic on the set \( \displaystyle A=\left\{ z \in \mathbb{C} \big| Re(z)>1 \right\} \) and that \( \displaystyle {\partial u \over \partial x} + {\partial v \over \partial y} = 0 \) on \( \displaystyle A \). Show that there are a real constant \( \displaystyle c \) and a complex constant \( \displaystyle d \) such that \( \displaystyle f(z) = -icz +d \) on \( \displaystyle A \).


(2) (a) Show that \( \displaystyle u(x,y)={e}^{x}\cos(y) \) is harmonic on \( \mathbb{C} \).
(b) Find a harmonic conjugate \( \displaystyle v=v(x,y) \) for \( \displaystyle u \) on \( \mathbb{C} \) such that \( \displaystyle v(0,0)=0 \).
(c) Show that \( \displaystyle f(z)={e}^{z} \) is analytic on \( \mathbb{C} \).
Papapetros Vaggelis
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Re: Harmonic Conjugates

#6

Post by Papapetros Vaggelis »

Hello Nickos, I'' ll make a try.

\(\displaystyle{\bullet\,\,(1)}\)

Firstly, the set \(\displaystyle{A}\) is an open set of \(\displaystyle{\mathbb{C}}\) . Indeed, let \(\displaystyle{z\in A}\)

and \(\displaystyle{r\in\left(0,Re(z)-1\right)}\) . If \(\displaystyle{y\in B\,\left(z,r\right)}\) , then :

\(\displaystyle{\left|Re(y)-Re(z)\right|\leq \sqrt{\left(Re(y)-Re(z)\right)^2+\left(Im(y)-Im(z)\right)^2}=\left|y-z\right|<r\implies }\)

\(\displaystyle{\implies Re(y)>Re(z)-r>Re(z)-Re(z)+1=1\implies y\in A}\) .

Now, \(\displaystyle{f(z)=u(x,y)+i\,v(x,y)\,,z=x+i\,y\in A}\) .

Due to the fact that the function \(\displaystyle{f}\) is analytic on \(\displaystyle{A}\) , we have the following :

\(\displaystyle{u_{x}=v_{y}\,,u_{y}=-v_{x}\,,u_{xx}+u_{yy}=0\,,v_{xx}+v_{yy}=0}\) and according to the hypothesis :


\(\displaystyle{u_{x}=-v_{y}}\) . So, \(\displaystyle{v_{y}=0}\), which means that

\(\displaystyle{v(x,y)=c_1+g(x)\,,(x,y)\in\left(1,+\infty\right)\times \mathbb{R}}\) ,whereas \(\displaystyle{c_1\in\mathbb{R}}\) .

Additionally,

\(\displaystyle{u_{x}=0\implies \exists\,c_2\in\mathbb{R}: u(x,y)=c_2+h(y)\,,(x,y)\in\left(1,+\infty\right)\times \mathbb{R}}\) .

Therefore :

\(\displaystyle{(x,y)\in\left(1,+\infty\right)\times \mathbb{R}\implies u_{xx}(x,y)+u_{yy}(x,y)=0\implies h^{\prime \prime}(y)=0\implies h(y)=a_1\,y+a_2\,,y\in\mathbb{R}}\)

for some \(\displaystyle{a_1\,,a_2\in\mathbb{R}}\) .

\(\displaystyle{(x,y)\in\left(1,+\infty\right)\times \mathbb{R}\implies v_{xx}(x,y)+v_{yy}(x,y)=0\implies g^{\prime \prime}(x)=0\implies g(x)=b_1\,x+b_2\,,x>1}\)

for some \(\displaystyle{b_1\,,b_2\in\mathbb{R}}\) .

We have that \(\displaystyle{a_1=-b_1}\) cause \(\displaystyle{u_{y}=-v_{x}}\) .

So, if \(\displaystyle{z=x+i\,y\in A}\) , then :

\(\displaystyle{f(z)=u(x,y)+i\,v(x,y)=c_2+a_1\,y+a\,2+i\,\left(c_1-a_1\,x+b_2\right)=-a_1\,i\,(x,+i\,y)+c_2+a_2+i\,(c_1+b_2)}\) .

Setting \(\displaystyle{c=a_1\in\mathbb{R}\,,d=c_2+a_2+i\,\left(c_1+b_2\right)}\) , we have that :

\(\displaystyle{f(z)=-i\,c\,z+d\,,z\in A}\) .

\(\displaystyle{\bullet\,(2)}\)

\(\displaystyle{(a)}\) For each \(\displaystyle{\left(x,y\right)\in\mathbb{R}^2}\) holds :

\(\displaystyle{u_{x}(x,y)=e^{x}\,\cos\,y\,,u_{xx}(x,y)=e^{x}\,\cos\,y\,,u_{y}(x,y)=-e^{x}\,\sin\,y\,,u_{yy}(x,y)=-e^{x}\,\cos\,y}\), so :

\(\displaystyle{u_{xx}(x,y)+u_{yy}(x,y)=e^{x}\,\cos\,y-e^{x}\,\cos\,y=0}\) and \(\displaystyle{u}\) is harmonic on \(\displaystyle{\mathbb{R}^2}\) .

\(\displaystyle{(b)}\) Suppose that \(\displaystyle{v:\mathbb{R}^2\longrightarrow \mathbb{R}}\) is the harmonic conjugate of \(\displaystyle{u}\)

such that \(\displaystyle{v(0,0)=0}\) .

By \(\displaystyle{\rm{Cauchy-Riemann}}\) equations, we have that :

\(\displaystyle{u_{x}=v_{y}\,,u_{y}=-v_{x}}\) and :

\(\displaystyle{v(x,y)=\int u_{x}(x,y)\,\mathrm{d}y+g_1(x)=\int e^{x}\,\cos\,y\,\mathrm{d}y=e^{x}\,\sin\,y+g(x)}\) Also,


\(\displaystyle{u_{y}(x,y)=-v_{x}(x,y)\implies -e^{x}\,\sin\,y=-e^{x}\,\sin\,y-g^\prime(x)\implies g^\prime(x)=0\exists c\in\mathbb{R}: g(x)=c\,,x\in\mathbb{R}}\) .

We conclude that \(\displaystyle{v(x,y)=e^{x}\,\sin\,y\,,\left(x,y\right)\in\mathbb{R}^2}\) cause \(\displaystyle{v(0,0)=0}\) .

\(\displaystyle{(c)}\)

For each \(\displaystyle{z=x+i\,y\in\mathbb{C}}\) holds :

\(\displaystyle{f(z)=e^{z}=e^{x+i\,y}=e^{x}\,e^{i\,y}=e^{x}\,\left(cos\,y+i\,\sin\,y\right)=e^{x}\,\cos\,y+i\,e^{x}\,\sin\,y}\)

We observe that

\(\displaystyle{u(x,y)=e^{x}\,\cos\,y\,,v(x,y)=e^{x}\,\sin\,y\,,\left(x,y\right)\in\mathbb{R}^2}\) and according to

\(\displaystyle{(2)(b)}\) and \(\displaystyle{(3)}\) , we have the desired result.
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