Originally Posted by

**Maccaman** Here is what I have done...

$\displaystyle u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos(2x(y - 2)) $

so

$\displaystyle u_x = 3x^2 - 3y^2 + 2x e^{x^2-(y-2)^2} \ cos(2x(y-2))-e^{x^2-(y-2)^2} \ sin(2x(y-2))(2y-4) $

$\displaystyle u_{xx} = 6x+2e^{x^2-(y-2)^2} \ cos(2x(y-2)) + 4x^2e^{x^2-(y-2)^2} \ cos(2x(y-2))-4xe^{x^2-(y-2)^2}$ $\displaystyle \ sin(2x(y-2))(2y-4) - e^{x^2-(y-2)^2} \ cos(2x(y-2))(2y-4)^2 $

and

$\displaystyle u_y = -6xy + (-2y+4)e^{x^2-(y-2)^2} \ cos(2x(y-2))- 2e^{x^2-(y-2)^2} \ sin(2x(y-2))x $

$\displaystyle u_{yy} =-6x-2e^{x^2-(y-2)^2} \ cos(2x(y-2)) + (-2y+4)^2 e^{x^2-(y-2)^2} \ cos(2x(y-2)) - $$\displaystyle (4(-2y+4)) e^{x^2-(y-2)^2} sin(2x(y-2)) \ x-4x^2 e^{x^2-(y-2)^2} \ cos(2x(y-2)) $

therefore

$\displaystyle u_{xx} + u_{yy} = 0 $ so $\displaystyle u(x,y) $ is a harmonic function.

Now to find a conjugate harmonic function $\displaystyle v(x,y) $

$\displaystyle v_y = u_x $

This is where I get lost.....am I now supposed to differentiate $\displaystyle v_y $ w.r.t x or y?? I've tried both and keep getting bizarre values. How do I get the conjugate function??

As far as the third part goes (with out v)

$\displaystyle \bigtriangledown u = u_xi + u_yj $

$\displaystyle \bigtriangledown v = v_xi + v_yj $

then I think im supposed to do something like

$\displaystyle \bigtriangledown u *\bigtriangledown v = ??A \ something + - * B \ something?? = 0 $

which enables me to solve for some values A and B??

(Thinking)