$Finding Harmonic function$

Question

$Finding Harmonic function$

Find harmonic function in cartesian coordinates u(x,y) which satisfies in polar coordinates the next condition:

u(rcos(0),rsin(0))=u(rcos(pi/4),rsin(pi/4))=a
given that:
u(x0,y0)=2a for 0<y0<x0

Complex Analysis

Danielatia123

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Let's construct the function as a sum u = u1 + u2 where u1 = a everywhere and u2 = 0 on the x-axis {y=0} and on the bisector {y=x}, and then adjust u2 (multiply by a constant) so that its value is a at (x0,y0).

The simplest try for u2 could be y (x - y), but that doesn't work (is not harmonic). Actually there is a "reflection principle" which implies that if a harmonic function is zero on {y=0} and {y=x}, then it's zero also on {x=0} and {y=-x} (Cf. comment).

Another way to guess the correct form of u2 is to try to make it "antisymmetric" in (x,y) so that u_xx = - u_yy.

Anyways, both approaches lead to u2(x,y) = x y (x-y)(x+y) = x y (x² - y²), which is indeed harmonic
(u'' = 6 x y = - u¨ where I denote by primes ' the derivative with respect to x and by (two) dots ¨ the (second) derivative w.r.t. y),
and which is zero on (x,y) = (r,0) and on (x,y) = (r/√2, r/√2), as desired.

It remains to adjust the coefficient of u2 to get the required value in (x0, y0).

So, u(x,y) = a + b x y (x² - y²) with b to be determined.
Now, u(x0,y0) = 2a <=> b x0 y0 (x0² - y0²) = a <=> b = a/(x0 y0 (x0² - y0²)),
which is well defined in view of the assumption 0 < y0 < x0.

So, in conclusion, the function u(x,y) = a + a x y (x² - y²)/(x0 y0 (x0² - y0²))
satisfies all of the requirements.

M F H

363

Danielatia123

0

Can you maybe let me in on how you came up with this function?
- M F H
  
  0
  
  We need a harmonic function that is zero on the x-axis and on {y=x}. When harmonic functions are zero on two lines, there is some "reflection principle" that implies that they must be zero also on further lines that are found by reflecting one line on the other one (and this can be iterated). So in our case it must also be zero on the y axis and on {y=-x}. So I tried x y (x-y)(x+y) = x y (x²-y²) = x³ y - x y³ which is easily seen to be harmonic (because of the "antisymmetry" in (x,y)).
M F H

0

("We need" => rather: "I wanted"...)
- M F H
  
  0
  
  ( I started from the idea to write the function as a sum u1 + u2 where u1 = a everywhere and u2 = 0 on the axes, and equal to a (so that u1+u2 = 2a) in (x0,y0). Sorry if I was a bit brief in the solution.)
M F H

0

I edited the solution to add these explanations. I can try to find a reference for the "reflection principle" if you are interested (but you can probably find it through Google.)

$Finding Harmonic function$

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