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: Collections: Mathematics