Question

A (nonconstant) harmonic function takes its maximum value and its minimum value

on the boundary of any region (not at an interior point). Thus, for example, the
electrostatic potential V in a region containing no free charge takes on its largest
and smallest values on the boundary of the region; similarly, the temperature T of a
body containing no sources of heat takes its largest and smallest values on the surface
of the body. Prove this fact (for two-dimensional regions) as follows: Suppose that
it is claimed that u(x, y) takes its maximum value at some interior point a; this
means that, at all points of some small disk about a, the values of u(x, y) are no
larger than at a. Show by Problem 36 that such a claim leads to a contradiction
(unless u = const.). Similarly prove that u(x, y) cannot take its minimum value at
an interior point

Answer 1

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Step-by-step explanation:

Answer 2

Step-by-step explanation:

Consider that F (any function) <0 .

u(x,y) is a coontinuous function in the closed interval or region R.

Let us consider a point (p,q) that is inside the region and it is a maximum point.

Then it should be must

uxx (p,q) <0 where uxx means double differentiation

and uy(p,q) >0

Since ux(p,q) = 0 = uy(p,q) where ux and uy means single differentiation with respect to x and y respectively.

Say, Maximum limits of the region is T

therefore q<T

then uy (p,q) = 0 if q<T

if q = T then

point (p,q) = (p,T) will be on the boundary of R then we claim that

uy(p,q) >0

Similarly for the minimum also it will work