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For the real valued function $u(x,y)=xy^3-x^3y$, decide whether there is a real valued function $v(x,y)$ such that $f(x+iy)=u(x,y)+iv(x,y)$ is analytic with $f(0)=2i$.

My Approach: I used Cauchy-Reumann equations to find $v(x,y)$. I get $$v(x,y)=\frac{y^4}{4}-\frac{3}{4}x^2y^2+\phi(x)$$

I then take $$\frac{\partial v}{\partial x}=-\frac{3}{2}xy^2+\phi'(x)$$ Solving for $\phi'(x)$ using Cauchy Reumann, I obtain $$\phi'(x)=-\frac{\partial u}{\partial y}+\frac{3}{2}xy^2=\frac{9}{2}xy^2-x^3$$ Integrating to obtain $\phi(x)$, $$\phi(x)=\frac {9}{4}x^2y^2-\frac{x^4}{4}+C$$ Thus $$v(x,y)=\frac{y^4}{4}-\frac{3}{4}x^2y^2+\frac {9}{4}x^2y^2-\frac{x^4}{4}+C$$ Given the above condition, I get $C=2$. My question is, how can we tell if there's a real valued function $v(x,y)$? Where would the conditions of a real valued function fail - if Cauchy Reumann equations is not satisfied? How do I explain there is no such function?

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possible duplicate of Finding imaginary part of a function $f(x+iy)=u(x,y)+iv(x,y)$ –  kahen Sep 17 '13 at 8:23
    
Not exactly the same question. My question is how would I know such a function will not exist? I know how to find $v(x,y)$ - that was not my question as it is in this link –  User69127 Sep 17 '13 at 8:29

2 Answers 2

up vote 4 down vote accepted

My question is, how can we tell if there's a real valued function $v(x,y)$?

If there is a real-valued function $v$ such that $f(x+iy) = u(x,y) + i v(x,y)$ is holomorphic, then the Cauchy-Riemann equations

$$\begin{align} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} \end{align}$$

yield upon further differentiation

$$\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 v}{\partial x\partial y} = \frac{\partial^2v}{\partial y\partial x} = - \frac{\partial^2 u}{\partial y^2},$$

so $u$ must be a harmonic function, $\Delta u = 0$. The same holds of course for imaginary parts of holomorphic functions.

That is a necessary criterion, but not always a sufficient criterion.

However, it is locally sufficient: For every real-valued harmonic function $u$ on an open set $\Omega \subset \mathbb{C}$, and every $z \in \Omega$, there is an open neighbourhood $W\subset \Omega$ of $z$, and a harmonic function $v\colon W \to\mathbb{R}$, such that $f(x+iy) = u(x,y) + iv(x,y) \colon W\to\mathbb{C}$ is holomorphic.

It can happen - as for example for $u (x,y) = \frac12\log (x^2+y^2)$ on $\mathbb{C}\setminus\{0\}$ - that the local solutions cannot be glued together to a global solution, but for simply connected $\Omega$ - in particular $\Omega = \mathbb{C}$ - there is always a globally defined conjugate harmonic function $v$ of a harmonic $u\colon \Omega \to\mathbb{R}$.

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Great. This is the part I was missing. I know that such a function $v(x,y)$ will not exist for $u(x,y)=xy^3+x^3y$ since $u$ is not harmonic. –  User69127 Sep 17 '13 at 19:59

I think you are reading too much into the 'real-valued function' bit; all it means is that we are assuming that $u(x,y)$ is the real part of our putative analytic function $f$. Otherwise there would be trivial solutions like $v(x,y) = i\,u(x,y) + 2$.

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Does this really answer all your questions, User69127? –  Gerry Myerson Sep 17 '13 at 8:49

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