Consider the disk $D:=\{(x,y)\in {\mathbb{R}}^2:\sqrt{x^2+y^2}<1\}$. Let $u(x,y)$ be a function twice differentiable in $D$ and continuous in $\bar{D}$, solving

a) Suppose now that g is any smooth function such that $\mathbf{g}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\ge \mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{)}$. Show that $\mathbf{u}(\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\mathbf{0})\ge \mathbf{1}.$

Let $v:=3x-y$ and $w:=u-v$. Since $v$ harmonic, $w$ also harmonic, and we have

Since $g(x,y)\ge 3x-y$.

Since $w$ harmonic, $D$ bounded, the maximum principle states that

b) Show equality $⟺\mathbf{g}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{y}$.

$⟹$: Assume that $u(\frac{1}{3},0)=1$. Then $w(\frac{1}{3},0)=u(\frac{1}{3},0)-v(\frac{1}{3},0)=1-1=0$

Since we know that $w\ge 0$ in $\bar{D}$, $(\frac{1}{3},0)\in D$ must be a minimum of $w$.

Since $w$ attains its minimum inside $D$, by strong max. principle (since $D$ connected), $w\equiv 0$ in $D$.

Hence,

$⟸$: Assume $g(x,y)=3x-y$. Then $w$ is:

By maximum principle

so

Thus $u\equiv v$, and in particular,