Finding an harmonic function

Question

I was doing an exercice that i'm struggling doing.

Consider the function $u$ defined on $\mathbb{R}^2$ by $$ u\left(x,y\right)=e^{x}\left(x\cos\left(y\right)-y\sin\left(y\right)\right) $$

1] Prove that $u$ is harmonic on $\mathbb{R}^2$

2] Find a function $v$ so that $f=u+iv$ be holomorphic on $\mathbb{C}$. Explicite $f$ using the variable $z$ using analytic extension theorem.

1. My attempt : $$ \frac{\partial u }{\partial x}\left(x,y\right)=e^{x}\left(x\cos\left(y\right)-y\sin\left(y\right)+\cos\left(y\right)\right) $$ So I found $$ \frac{\partial^2u}{\partial x^2}\left(x,y\right)=e^{x}\left(x\cos\left(y\right)-y\sin\left(y\right)+2\cos\left(y\right)\right) $$ And for $y$ $$ \frac{\partial u }{\partial y}\left(x,y\right)=e^{x}\left(-x\sin\left(y\right)-\sin\left(y\right)-y\cos\left(y\right)\right) $$ hence $$ \frac{\partial^2 u }{\partial y^2}\left(x,y\right)=e^{x}\left(-x\cos\left(y\right)-\cos\left(y\right)+y\sin(y)-\cos(y)\right) $$ Finally I happily found that

$$ \frac{\partial^2u}{\partial x^2}\left(x,y\right)+\frac{\partial^2u}{\partial y^2}\left(x,y\right)=0 $$

Is it all good for the first one ?

2. I was wondering how I could link this to being holomorphic, how should I proceed ?

Answer 1

You want to find a particular $v(x,y)$ that satisfies

$$ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = e^x(x\sin y + y\cos y + \sin y) $$ $$ \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = e^x(x\cos y - y\sin y + \cos y) $$

The process is straightforward. Start by integrating one of the partials, let's say the second one $$ v(x,y) = \int e^x(x\cos y - y\sin y + \cos y) dy = e^x(x\sin y + y\cos y) + g(x) $$

Then differentiate the other way $$ \frac{\partial v}{\partial x} = e^x(x\sin y + y\cos y + \sin y) + f'(x) $$

By comparison, $f'(x) = 0$, so $f(x) = C = 0$. Therefore $$ v(x,y) = e^x(x\sin y + y\cos y) $$

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Let's simplify the complex function $$ \begin{align} f(z) = u(x,y) + iv(x,y) &= e^x(x\cos y - y\sin y + ix\sin y + iy\cos y) \\ &= e^x\big((x+iy)\cos y + i(x+iy)\sin y\big) \\ &= e^x(x+iy)(\cos y + i\sin y) \\ &= (x+iy)e^{x+iy} \\ &= ze^z \end{align} $$

As for the last question, if $u$ is harmonic (which you've proven), then $v$ is guaranteed to exist. Since they're both continuous and twice-differentiable on $\Bbb R^2$, $f(z)$ must be holomorphic on $\Bbb C$

Answer 2

If $f:\Bbb R^2\to\Bbb R^2$ defined as $f(x,y)=u(x,y)+v(x,y)$ is harmonic, then $\tilde f:\Bbb C\to\Bbb C$ defined as $\tilde f(x+iy)=u(x+iy)+iv(x+iy)$ is holomorphic.

Answer 3

Hint: The form of the function $u(x,y)$ suggests looking at $e^z\cdot z,$ for both parts 1. and 2.

Answer 4

The real part of the holomorphic function $$ e^{x+iy}(x+iy)=e^x[(x\cos(y)-y\sin(y))+i(y\cos(y)+x\sin(y))] $$ is $$ e^x(x\cos(y)-y\sin(y)) $$ and the real part of a holomorphic function is harmonic.