Maximum principle (elliptic)

Prove that a function $u(x,y)$ that is harmonic in a bounded closed domain and is not constant in this domain cannot attain its maximum or minimum at an interior point of the domain (the maximum principle).

Prove that a function $u=u(x,y,z)$ that is continuous on a bounded closed domain $V$ and harmonic in its interior, if it is not constant, cannot attain its maximum and minimum values at interior points of the domain (the maximum principle).

Consider the regular pentagram centered at the origin in the complex plane. Let $u$ be the harmonic function in the interior of the pentagram which has boundary values $1$ on the two segments shown and $0$ on the rest of the boundary. What is the value of $u$ at the origin? Justify your claim.

Let $\Omega \subset \mathbb {R}^{n}$, $n > 1$, be a bounded domain with smooth boundary. Let $u \in C^1(\overline{\Omega })$ be harmonic in $\Omega$.

Let $\Omega$ be an open subset of $\mathbb {R}^{n}$. Suppose $u \in C^2(\overline{\Omega })$ is a solution of the equation $\Delta u = u^3$ with the property that $\left|\nabla u(x)\right| \leq 1$ for each $x \in \partial \Omega$. Prove that $\left|\nabla u(x)\right| \leq 1$ for all $x \in \Omega$.

Consider (a nonlinear) equation

-u = u^2(1-u) in , u = 0 on , 1

where $\lambda > 0$ is a parameter, and $\partial \Omega$ sufficiently smooth.

Let $\Omega$ be a bounded domain in $\mathbb {R}^{n}$ with smooth boundary $\partial \Omega$ and assume $u \in C^2(\overline{\Omega })$ satisfies the following

where $f \in C^0(\overline{\Omega })$ and $\phi \in C^0(\partial \overline{\Omega })$. Let $B_R(O)$ be the smallest ball centered at the origin $O$ such that $\Omega \subset B_R(O)$. Show that

Also give an example to demonstrate that the estimate is sharp, i.e., the equality sign can be attained in your example.

We suppose that the functions $\varphi (z)$ and $\psi (z)$ are regular in the closed disk $\left|z\right|\leq 1$ and non-zero in the open disk $\left|z\right|<1$ and that, besides, $\varphi (0)$ and $\psi (0)$ are real and positive. If $\varphi (z)$ and $\psi (z)$ have the same modulus on the circle $\left|z\right|=1$ then $\varphi (z)=\psi (z)$ identically.

In three dimensional space, the $n$ points $P_{1},P_{2},\ldots ,P_{n}$ are given and $P$ denotes a variable point. The function

($\overline{PP_{\nu }}$ is the distance between $P$ and $P_{\nu }$) of the point $P$ assumes its maximum in any domain on the boundary. (Generalization of problem 137.)

An analytic function that is regular in the closed disk $\mathfrak {R}$ cannot assume real values at all the boundary points of $\mathfrak {R}$ except if it is a real constant.

A harmonic function is supposed to be regular and single-valued in a domain $\mathfrak {D}$. Then it attains its maximum and its minimum on the boundary and only on the boundary unless it is a constant.

A harmonic function that is regular in the domain $\mathfrak {D}$ and vanishes at all the boundary points of $\mathfrak {D}$ is identically zero.

The equilibrium described in solution 31 is not stable.

A harmonic function is assumed to be single-valued in the domain $\mathfrak {D}$ and regular with the exception of finitely many points at which it becomes $-\infty$ (i.e.the function converges to $-\infty$ as $z$ approaches such a point). Then it assumes its maximum on the boundary of $\mathfrak {D}$.

Let $u_{1}(x,y),u_{2}(x,y),\ldots ,u_{n}(x,y)$, $z=x+iy$, be regular harmonic functions in a domain $\mathfrak {D}$. The continuous function

assumes its maximum on the boundary of $\mathfrak {D}$.

The closed curve $\mathfrak {C}$ without double points is the boundary of the region $\mathfrak {R}$, a part of the $x,y$-plane. The function $f(x,y)$ is continuous in the closure of $\mathfrak {R}$ (the union of $\mathfrak {R}$ and $\mathfrak {C}$) and has continuous partial derivatives of the second order in $\mathfrak {R}$. If $f(x,y)$ vanishes along $\mathfrak {C}$ and also at some interior point of $\mathfrak {R}$, then there exists a point $\xi ,\eta$ in $\mathfrak {R}$ at which

Prove that a function, harmonic in some domain and not identically constant, can attain neither its greatest nor its least value at an interior point of this domain. (The maximum principle for harmonic functions.)

Let the function $u(z)$ be harmonic in a domain $G$. Denote $M = \sup \left\{ u(z): z \in G\right\} , m = \inf \left\{ u(z): z \in G\right\}$. Prove that if $m < M$, then at every point of the domain $G$ the inequalities $m < u(z) < M$ hold. (The maximum and minimum principle for harmonic functions.)

Prove the maximum principle for harmonic functions. This says that if $h$ is harmonic in a bounded domain $D$, then