Can You Solve the Challenging Math Problem from POTW #168?

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Here is this week's POTW:

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Let $X$ be a domain in $\Bbb R^2$. Suppose $u,v \in C^2(X\to\Bbb R)$ such that

$$\oint_c uv\frac{\partial v}{\partial \mathbf{n}}\, ds = -\frac{1}{2}\oint_c v^2\frac{\partial u}{\partial \mathbf{n}}\, ds$$

for every simple closed curve $c$ in $X$. Prove that to every $\varepsilon > 0$, there corresponds a subharmonic function $h_\varepsilon$ on $X$ such that

$$u(x,y)v(x,y)^2 + h_\varepsilon(x,y) = \varepsilon(x^2 + y^2)$$

for all $(x,y)\in X$.
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Hi MHB,

I've made a correction to the problem by placing missing $v$'s in the integrals above. Now the problem is doable. My apologies for the error.

 

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No one solved this week's problem. You can read my solution below.

Spoiler

First I'll show $uv^2$ is harmonic. Suppose to the contrary that $uv^2$ is not harmonic on $X$. Without loss of generality, assume $\Delta (uv^2)(x_0) > 0$ for some $x_0 \in X$. Then there is an open disc $D(x_0)$ in $X$ containing $x_0$ such that $\Delta (uv^2) > 0$ on $D(x_0)$. If $c$ is the circle which bounds $D(x_0)$, then

$$ \oint_c \left(uv\frac{\partial v}{\partial \mathbf{n}}\, ds + \frac{1}{2}v^2\frac{\partial u}{\partial \mathbf{n}}\right)\, ds = \oint_c \left(\frac{1}{2}u\frac{\partial v^2}{\partial \mathbf{n}} + \frac{1}{2}v^2\frac{\partial u}{\partial \mathbf{n}}\right)\, ds = \frac{1}{2}\oint_c \frac{\partial (uv^2)}{\partial \mathbf{n}}\, ds = \frac{1}{2}\iint_{D(x_0)} \Delta (uv^2) > 0,$$

a contradiction.

Since $uv^2$ is harmonic on $X$, for a given $\epsilon > 0$, the function $h_\varepsilon : X \to \Bbb R$ defined by $h_\varepsilon(x,y) = \varepsilon(x^2 + y^2) - u(x,y)v(x,y)^2$ is $C^2(X\to \Bbb R)$ such that $\Delta h_\varepsilon = 4\varepsilon > 0$ on $X$. Hence, $h_\varepsilon$ is subharmonic on $X$, and satisfies the equation $u(x,y)v(x,y)^2 + h_\varepsilon(x,y) = \varepsilon(x^2 + y^2)$ in $X$.