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This is not a duplicate. I realize similar questions to this have been asked, but what I am asking is slightly different:

I need to prove the following, arguing by complex differentiability only, and NOT by calculating $\displaystyle \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}$:

Let $u$, $v$ be conjugate harmonic functions on a domain $G$ (notice it says nothing about $G$ being simply connected). For any function $g = s + i t$ analytic on the image of $G$ under $u+iv$, the functions $s(u(x,y), \,v(x,y))$ and $t(u(x,y),\, v(x,y))$ are harmonic, and the latter is a harmonic conjugate of the former.

I just proved a little while ago that if $u$ and $v$ are harmonic, and $f = u+iv$ is analytic, then $u^{2}-v^{2}$ is harmonic by considering $f^{2}$, and was told that what I am asking about here is a generalization of that result. However, thus far I have been unable to use that to help me.

The other times questions similar to this have been asked on this site, either it has been asked in the other direction, or OPs have not been specific about what they wanted, and so there have been all kinds of answers given, either doing exactly what I don't want to do here, which is to calculate the 2nd order partials and show that Laplace's Equation is satisfied (I have been specifically told not to do it that way), or referring to such esoteric things as Riemannian manifolds, which are completely useless to me in my current coursework.

Could someone please help me out in proving this result the way I have been asked to prove it, using complex differentiability properties and results only, and not by either calculating 2nd order partials or anything too advanced (also no series)? I am at a loss as to what to do...thank you.

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The composition of holomorphic functions is holomorphic. This is nothing more than the chain rule, and is proved the same way you would in a normal calculus class.

Then your result is just that the real and imaginary parts of the holomorphic function $g \circ f$ are harmonic and indeed harmonic conjugates, which you verify with the Cauchy-Riemann equations. (That is, if $f = u+iv$ is holomorphic, then $u_x = v_y$ and $-u_y = v_x$, so $$u_{xx} + u_{yy} = u_{xx} - v_{xy} = u_{xx} - v_{yx} = u_{xx} - u_{xx} = 0.$$

This is absolutely fundamental. If you're showing something is harmonic, you need to do some work eventually, and this is about as little work as you can get away with doing. I suspect what they mean by "do not calculate the Laplacian" is to not take partials of the function $s(u,v)$ with the chain rule and plugging everything together, but rather to use the observation in the first paragraph. After all, that's what you did in noticing that $u^2-v^2$ is harmonic.

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  • $\begingroup$ but don't you have to calculate the 2nd order partials for that, which I was asked not to do? $\endgroup$ – ALannister Mar 30 '16 at 3:53
  • $\begingroup$ @JessyCat For what, the last paragraph? It's an immediate corollary of the Cauchy-Riemann equations, but yes, there are second order partials involved. This should not be surprising given the definition of harmonic uses second order partials... $\endgroup$ – user98602 Mar 30 '16 at 3:54
  • $\begingroup$ right...but we were instructed not to verify Laplace's Equation. $\endgroup$ – ALannister Mar 30 '16 at 4:16
  • $\begingroup$ @JessyCat Presumably they mean not to explicitly do so to the harmonic functions you produced, as opposed to the general process outlined above that makes no such calculation. If you're not allowed to say "real parts of holomorphic functions are harmonic" somehow, then how do you expect to verify that any function, ever, is harmonic? $\endgroup$ – user98602 Mar 30 '16 at 4:17
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    $\begingroup$ Yes, we want to show that that limit exists. But if you've seen the proof in real analysis, the proof is exactly the same here. $\endgroup$ – user98602 Mar 30 '16 at 17:29
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In fact, you can deduce complex Laplacian operator first. Then apply china rule you will get there. Hint: $\Delta_{\mathbb{R}^2}=4\frac{\partial^2}{\partial z\partial\bar{z}}$. And we know that $u(z)$ is analytic if and only if $\partial_{\bar{z}}u=0$

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