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Show that a harmonic function u(x, y) is equal at every point a to its average value on any circle centered at a [and lying in the region where f (z) = u(x, y) + iv(x, y) is analytic]. Hint: In (3.9), let z = a + reiθ (that is, C is a circle with center at a), and show that the average value of f(z) on the circle is f(a) (see Chapter 7, Section 4 for discussion of the average of a function). Take real and imaginary parts of f(a) = [u(x, y) + iv(x, y)]z=a.

Respuesta :

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Answer:

u(ax,ay)=1/2π∫2π0u(ax+rcosθ, ay+rsinθ)dθ

Step-by-step explanation:

For any holomorphic function (f), we have that,

f(a)=1/2πi∮Cf(z)z−adz = 1/2πi∫2π0f(a+reiθ)a+re^iθ−aire^iθdθ = 1/2π∫2π0f(a+re^iθ)dθ

Now we know that for any real harmonic function u(x,y), there is another real harmonic function v(x,y) such that f(x+iy)=u(x,y)+iv(x,y) is holomorphic.

Letting ax=Re[a] and ay=Im[a] and applying the above gives

u(ax,ay)+iv(ax,ay)=1/2π∫2π0u(ax+rcosθ,ay+rsinθ)dθ+i/2π∫2π0v(ax+rcosθ,ay+rsinθ)dθ

Taking real parts then gives us the desired result

u(ax,ay)=1/2π∫2π0u(ax+rcosθ, ay+rsinθ)dθ

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