Function in calculus

Question

8. If u(x, y) = h(x)e^2y is harmonic, h(0)=0 and h'(0)=1 then find h(x).also find analytic function f =u(x,y)+iv(x,y).

Answer 1

If u is harmonic, then: ∂²u/∂x² + ∂²u/∂y² =0.

By taking the first partial derivatives we obtain: ∂u/∂x = h^'(x) e^2y and ∂u/∂y = 2 h(x) e^2y and by taking the second partial derivatives: ∂²u/∂x² = h^''(x) e^2y and ∂²u/∂y² = 4 h(x) e^2y.

Now the harmonic property: e^2y (h^''(x) + 4 h(x) ) = 0, which means either e^2y=0 (trivial answer) or h^''(x) + 4 h(x) =0, which has the general solution: h(x) = a cos(2x) + b sin(2x). Using the boundary conditions h(0)=0 and h'(0)=1, we obtain: h(x) = sin(2x)/(2cos(2)).

Now v(x,y) is going to be the harmonic conjugate of u(x,y) where it can be found by noting that ∂u/∂x = ∂v/∂y and ∂u/∂y = - ∂v/∂x.