Final answer:
To show that uV+vU is harmonic, we use the properties of harmonic functions and their conjugates, apply the Laplacian operator, and utilize the Cauchy-Riemann equations that lead to the cancelation of cross-terms in the sum of second partial derivatives.
Step-by-step explanation:
To show that uV+vU is harmonic when u, v, U, and V are harmonic functions and v is a harmonic conjugate of u, and V is a harmonic conjugate of U, we will use the properties of harmonic functions and their conjugates. A harmonic function f(x, y) satisfies Laplace's equation, which in two dimensions is ∇2f/∇2x + ∇2f/∇2y = 0. If u and v are harmonic conjugates, then they are related through the Cauchy-Riemann equations u_x = v_y and u_y = -v_x, where subscripts denote partial derivatives.
Applying the Laplacian operator to uV+vU and considering the product rule of differentiation, we get:
- ∇2(uV)/∇2x = u_{xx}V + 2u_xV_x + uV_{xx}
- ∇2(uV)/∇2y = u_{yy}V + 2u_yV_y + uV_{yy}
- ∇2(vU)/∇2x = v_{xx}U + 2v_xU_x + vU_{xx}
- ∇2(vU)/∇2y = v_{yy}U + 2v_yU_y + vU_{yy}
Since u, v, U, and V are harmonic, their second partial derivatives sum to zero. Hence, we're left with the cross-terms which, due to the Cauchy-Riemann equations, cancel out each other. Thus:
∇2(uV+vU)/∇2x + ∇2(uV+vU)/∇2y = 0
This implies that uV+vU is a harmonic function. Its harmonic conjugate can then be found using the Cauchy-Riemann equations, which relate the partial derivatives of harmonic conjugates. If we denote w as the conjugate of uV+vU, we know that w_x = (uV+vU)_y and w_y = -(uV+vU)_x. Integrating these, using the properties of u, v, U, and V, can then yield w.