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SOLUTION TO PROBLEM 10651 PROPOSED BY W. K. HAYMAN
 

Summary: SOLUTION TO PROBLEM 10651
PROPOSED BY W. K. HAYMAN
Tewodros Amdeberhan
DeVry Institute, Mathematics
630 US Highway One, North Brunswick, NJ 08902
amdberhan@admin.nj.devry.edu
PROBLEM: P If u1 and u2 are nonconstant real functions of two variables, and if u1;u2;
and u1u2 are all harmonic in a simply connected domain D, prove that u2 = av1 + b; where v1 is a
harmonic conjugate of u1 in D, and a and b are real constants.
PROOF: In R2, we write wx and wy for @w=@x and @w=@y, respectively. Let f = u1 + iv1.
Then f2 is analytic, and hence 2u1v1 = Imf2 is harmonic.
Using the assumptions and the equations,
u1u2 = u1 + u2 + 2ru1 ru2; u1v1 = u1 + v1 + 2ru1 rv1
it follows that both vectors ru2 and rv1 are orthogonal to ru1, in R2. Thus
1 ru2 = arv1;
for some function a = ax;y. Consequently, u2 = av1 + rarv1. Therefore, we obtain
2 rv1 ax;ay = 0:
Rewriting equation 1 as: u2x = av1x; u2y = av1y and di erentiating in y and x, respectively
gives
u2xy = ayv1x + av1xy; u2yx = axv1y + av1yx:

  

Source: Amdeberhan, Tewodros - Department of Mathematics, Tulane University

 

Collections: Mathematics