# Does having harmonic functions $u(x,y)$ and $v(x,y)$ guarantee having an analytic function?

Verify that each given function $$u$$ is harmonic (in the region where it is defined) and then find a harmonic conjugate of $$u$$.

(a) $$u=y$$

I was able to verify that $$u$$ is harmonic pretty easily. But I found that the harmonic conjugate of $$u$$ is $$-x+a$$. Thus, I get that $$f(z)= y + i(-x+a)$$. Using that $$y = \frac{z-\bar{z}}{2i}$$ and that $$x=\frac{z+\bar{z}}{2}$$, I get that $$f(z) = -i(\bar{z}+a)$$, which as I understand is not an analytic function. I would like to know if $$f(z)$$ is indeed analytic, and whether or not having a harmonic function $$u$$ along with its conjugate $$v$$ in $$f(z)$$ guarantees having an analytic function $$f(z)$$?

• I think there's an error in your computation. In your example, $f(z) = -iz + ia$, which is affine and hence analytic. – Travis Willse Feb 10 '19 at 21:45
• In any case, one can always find a harmonic conjugate locally but not always globally, at least on domains that are not simply connected. The standard example is $z \mapsto \log |z|$ on $\Bbb C - {0}$. – Travis Willse Feb 10 '19 at 21:52
• I was wondering what is meant by locally? Taking an example from the harmonic function $u=y$, does that mean that the domain of $v$ (harmonic conjugate) can be found in some neighborhood of the domain of $u$? – K.M Feb 10 '19 at 21:58
• On a domain that is not simply connected, one is actually guaranteed more than local existence but one is not guaranteed existence of a harmonic function on all of its domain. (Here, local existence would just mean that for any point $z_0$ in the domain of $u$ that there is a neighborhood $A$ of $z_0$ and a harmonic conjugate $v$ of $u\vert_A$ on $A$.) I've written up an answer that explains more precisely what is guaranteed. – Travis Willse Feb 11 '19 at 4:03

Suppose $$D \subset \Bbb C$$ is a domain and $$u : D \to \Bbb R$$ is a harmonic function.
If $$D$$ is simply connected, then $$u$$ admits a harmonic conjugate (globally). Explicitly, up to an overall constant it is (for any choice of base point $$z_0 \in D$$) $$v(z) := \int_{z_0}^z -u_y \,dx + u_x \,dy .$$ Computing the exterior derivative of the integrated $$1$$-form gives $$d(-u_y \,dx + u_x \,dy) = (u_{xx} + u_{yy}) \, dx \wedge dy = 0 ,$$ where the second equality just uses that $$u$$ is harmonic, so $$-u_y \,dx + u_x \,dy$$ is closed and hence (since $$D$$ is simply connected) exact, and thus the integral is well-defined. If $$-u_y \,dx + u_x \,dy = df$$, then $$v(z) = \int_{z_0}^z df = f(z) - f(z_0)$$.
Now, differentiating shows that the pair $$(u, v)$$ satisfies the Cauchy-Riemann equations---in fact the definition of $$v$$ is rigged precisely so that this happens---or equivalently that the function $$f : D \to \Bbb C$$, $$f := u + i v$$, is analytic on $$D$$.
On the other hand, if $$D$$ is not simply connected, $$u$$ may not admit a global harmonic conjugate, i.e., one on all of $$D$$, whereas the above argument shows that for any simply connected subset $$A \subset D$$, $$u\vert_A$$ does admit a harmonic conjugate $$v : A \to \Bbb R$$. To find a counterexample, inspecting the argument shows that we certainly need a $$1$$-form that is closed but not exact (any such $$1$$-form must have domain that is not simply connected). The standard example is $$\frac{-y \,dx + x \,dy}{x^2 + y^2}.$$ On any simply connected subset of $$\Bbb C - \{ 0 \}$$ this is the exterior derivative of some function, and the Cauchy-Riemann equations imply that (the negative of) its harmonic conjugate has exterior derivative $$\frac{x \,dx + y \,dy}{x^2 + y^2} .$$ But this expression defines a $$1$$-form on all of $$\Bbb C - \{ 0 \}$$, and this $$1$$-form is exact: An easy integration shows that it's just the exterior derivative of $$u(z) := \frac{1}{2} \log (x^2 + y^2) = \log |z| .$$ By construction, $$u$$ does not have a harmonic conjugate on all of $$\Bbb C - \{ 0 \}$$. Up to additive constants, the harmonic conjugates of the restrictions of $$u$$ to simply connected subsets are branches of the argument function.