Let $x:\Omega \subset %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{R}^2 %EndExpansion \rightarrow %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ harmonic (I do not know if you need) and $\frac{\partial x}{\partial u_{1}},\frac{\partial x}{% \partial u_{2}}:\Omega \subset %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{R}^2 %EndExpansion \rightarrow %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ their partial derivatives, with $\Omega $ simply connected open. Show that $$ x\left( z\right) =\Re \int_{0}^{z}\left( \frac{\partial x}{% \partial u_{1}}\left( z\right) -i\frac{\partial x}{\partial u_{2}}\left( z\right) \right) dz. $$

I have looked in books of complex analysis, but I can not find the property that allows me this equality.

  • $\begingroup$ @saulspatz Yes, I have already edited. $\endgroup$ – Mancala Sep 12 '18 at 14:26

I don't think this has much to do with harmonic functions or complex analysis. Below I use different notation, otherwise I'll get confused.

Assume $\Omega$ is open and connected (I don't think we need simple connectivity.) I'll be writing $w=x+iy$ for $w\in \mathbb C.$ Let $u:\Omega \to \mathbb R$ be continuously differentiable. Suppose $0,z\in \Omega$ and $\gamma:[a,b]\to \Omega$ is a contour with $\gamma(a)=0,\gamma (b)=z.$ Then

$$\tag 1\text { Re }\int_\gamma (u_x(w)-iu_y(w))\, dw = u(z)-u(0).$$

Proof: Write $\gamma(t)= x(t)+iy(t).$ Then the left side of $(1)$ equals

$$\text { Re } \left [\int_a^b (u_x(\gamma(t))-iu_y(\gamma(t)))(x'(t)+iy'(t))\,dt \right ]$$ $$ = \int_a^b [u_x(\gamma(t))x'(t) + u_y(\gamma(t))y'(t)]\,dt $$ $$ = \int_a^b [u(\gamma(t))]'\, dt = u(\gamma(b)) - u(\gamma(a)) = u(z)-u(0).$$

The only things we have used are the chain rule and the FTC.


Ok, some comments:

  • The notation in the integral is a little bit confusing; Let us reinterpret it as being

$$ \text{Re}\left( \int_0^z \left(\frac{\partial x}{\partial u_1}(w) - i \frac{\partial x}{\partial u_2}(w)\right) \mathrm{d}w\right).$$

  • Also, we shall assume that $0 \in \Omega,$ as otherwise the integration does not make sense.

  • Define $y:\Omega \subset \mathbb{R}^2 \to \mathbb{R}$ to be $x'$s harmonic conjugate. This exists, as $\Omega$ is open and simply connected. Therefore, letting $f(z) = x(z) + iy(z),$ we obtain, from the Cauchy-Riemann equation, that the integral (inside the real part) above is

$$ \int_0^z \frac{\partial f}{\partial u_1}(w) \,\mathrm{d}w.$$ (Just use $\partial x/\partial u_2 = - \partial y/\partial u_1$).

  • The last step was only possible because, from the definition, $f$ is holomorphic. Therefore, it does not matter in which direction we take the partial derivative: it shall always yield the (complex) derivative $f'(z)$ of $f.$ Therefore, we rewrite the integral one more time as

$$ \int_0^z f'(w) \, \mathrm{d}w. $$

  • This last is a complex antiderivative to $f'$ on $\Omega,$ and differs from $f$ by only a constant factor. Explicitly, we get to the conclusion that

$$ \int_0^z f'(w) \, \mathrm{d}w = f(z) - f(0).$$

  • We finish by noticing that the harmonic function $x(z) - x(0)$ is the real part of $f(z) - f(0),$ which then yields

$$ \text{Re}\left( \int_0^z \left(\frac{\partial x}{\partial u_1}(w) - i \frac{\partial x}{\partial u_2}(w)\right) \mathrm{d}w\right) = $$ $$ \text{Re} \left(\int_0^z f'(w) \, \mathrm{d}w \right) = \text{Re}(f(z) - f(0)) = x(z) - x(0).$$

(This means: you need a $x(0)$ to be added to the right hand side of your formula, but, aside from that, the result follows.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.