Section 4

Harmonic Functions

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Problem 3.311

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

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Problem 3.312

We denote by u(x,y)u(x,y), z=x+iyz=x+iy, a harmonic function that is regular in the disk (x−x0)2+(y−y0)2≤r2(x-x_{0})^{2}+(y-y_{0})^{2}\leq r^{2}. The value of u(x,y)u(x,y) at the center is

u(x0,y0)=12π∫02πu(x0+rcos⁡ϑ,y0+rsin⁡ϑ) dϑ[problem 118], u(x_{0},y_{0}) = \frac{1}{2\pi }\int _{0}^{2\pi } u(x_{0}+r\cos \vartheta ,y_{0}+r\sin \vartheta )\, d\vartheta \quad [\text{problem 118}],

consequently

∣u(x0,y0)∣≤12π∫02π∣u(x0+rcos⁡ϑ,y0+rsin⁡ϑ)∣ dϑ. \left|u(x_{0},y_{0})\right| \leq \frac{1}{2\pi }\int _{0}^{2\pi } \left|u(x_{0}+r\cos \vartheta ,y_{0}+r\sin \vartheta )\right|\, d\vartheta .

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?

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Problem 3.313

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

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Problem 3.314

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

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Problem 3.315

The equilibrium described in solution 31 is not stable.

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Problem 3.316

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

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Problem 3.317

Let the function f(z)f(z) be regular in the disk ∣z∣≤R\left|z\right|\leq R. The function w=f(z)w=f(z) maps the disk onto a certain domain of the ww-plane that is assumed to be star-shaped with respect to w=0w=0. Let f(0)=0f(0)=0. Then the images of the concentric circles ∣z∣=r\left|z\right|=r, r<Rr<R, are also star-shaped with respect to the origin.

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Problem 3.318

Suppose that the function f(z)f(z) is regular in the disk ∣z∣≤R\left|z\right|\leq R and that it maps the disk onto a convex domain of the ww-plane. Then the image of an arbitrary circle in the open disk ∣z∣<R\left|z\right|<R is also convex.

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Problem 3.319

Let u1(x,y),u2(x,y),…,un(x,y)u_{1}(x,y),u_{2}(x,y),\ldots ,u_{n}(x,y), z=x+iyz=x+iy, be regular harmonic functions in a domain D\mathfrak {D}. The continuous function

∣u1(x,y)∣+∣u2(x,y)∣+⋯+∣un(x,y)∣ \left|u_{1}(x,y)\right| + \left|u_{2}(x,y)\right| + \cdots + \left|u_{n}(x,y)\right|

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

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Problem 3.320

Consider a regular harmonic function in the disk ∣z∣<R\left|z\right|<R. We denote by A(r)A(r) its maximum on the circle ∣z∣=r\left|z\right|=r, r<Rr<R. When 0<r1<r2<r3<R0<r_{1}<r_{2}<r_{3}<R we have

A(r2)≤log⁡r2−log⁡r1log⁡r3−log⁡r1A(r3)+log⁡r3−log⁡r2log⁡r3−log⁡r1A(r1), A(r_{2}) \leq \frac{\log r_{2}-\log r_{1}}{\log r_{3}-\log r_{1}}A(r_{3}) + \frac{\log r_{3}-\log r_{2}}{\log r_{3}-\log r_{1}}A(r_{1}),

i.e.A(r)A(r) is a convex function of log⁡r\log r.

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Problem 3.321

Deduce Hadamard's three circle theorem 304 from problem 320 and vice versa problem 320 from Hadamard's three circle theorem.

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