Harmonic Functions
[11/100%]An analytic function that is regular in the closed disk cannot assume real values at all the boundary points of except if it is a real constant.
We denote by , , a harmonic function that is regular in the disk . The value of at the center is
consequently
If a harmonic function is regular in a closed disk its absolute value at the center does not exceed the arithmetic mean of its absolute values on the boundary. Under what conditions is there equality?
A harmonic function is supposed to be regular and single-valued in a domain . 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 and vanishes at all the boundary points of is identically zero.
The equilibrium described in solution 31 is not stable.
A harmonic function is assumed to be single-valued in the domain and regular with the exception of finitely many points at which it becomes (i.e.the function converges to as approaches such a point). Then it assumes its maximum on the boundary of .
Let the function be regular in the disk . The function maps the disk onto a certain domain of the -plane that is assumed to be star-shaped with respect to . Let . Then the images of the concentric circles , , are also star-shaped with respect to the origin.
Suppose that the function is regular in the disk and that it maps the disk onto a convex domain of the -plane. Then the image of an arbitrary circle in the open disk is also convex.
Let , , be regular harmonic functions in a domain . The continuous function
assumes its maximum on the boundary of .
Consider a regular harmonic function in the disk . We denote by its maximum on the circle , . When we have
i.e. is a convex function of .
Deduce Hadamard's three circle theorem 304 from problem 320 and vice versa problem 320 from Hadamard's three circle theorem.