1. ## Harmonic Function

Show that $u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos[2x(y - 2)]$ is an harmonic function, and find a conjugate harmonic function $v(x,y)$.

Check that $\vec{\bigtriangledown} u * \vec{\bigtriangledown} v = 0$ holds for all x and y.

2. Originally Posted by Maccaman
Show that $u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos[2x(y - 2)]$ is an harmonic function, and find a conjugate harmonic function $v(x,y)$.

Check that $\vec{\bigtriangledown} u * \vec{\bigtriangledown} v = 0$ holds for all x and y.
If you know the definition of a harmonic function then you'll realise that all you have to do is a lot of partial differentiation. This is tedious but not intellectually difficult.

I don't help with tedium but can help with intellectual difficulty. Where are you stuck?

3. Here is what I have done...

$u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos(2x(y - 2))$
so
$u_x = 3x^2 - 3y^2 + 2x e^{x^2-(y-2)^2} \ cos(2x(y-2))-e^{x^2-(y-2)^2} \ sin(2x(y-2))(2y-4)$

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

and

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

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

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

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

$v_y = u_x$

This is where I get lost.....am I now supposed to differentiate $v_y$ 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)

$\bigtriangledown u = u_xi + u_yj$

$\bigtriangledown v = v_xi + v_yj$

then I think im supposed to do something like

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

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

4. Originally Posted by Maccaman
Here is what I have done...

$u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos(2x(y - 2))$
so
$u_x = 3x^2 - 3y^2 + 2x e^{x^2-(y-2)^2} \ cos(2x(y-2))-e^{x^2-(y-2)^2} \ sin(2x(y-2))(2y-4)$

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

and

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

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

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

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

$v_y = u_x$

This is where I get lost.....am I now supposed to differentiate $v_y$ 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)

$\bigtriangledown u = u_xi + u_yj$

$\bigtriangledown v = v_xi + v_yj$

then I think im supposed to do something like

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

which enables me to solve for some values A and B??
To get the harmonic conjugate of u, read this: PlanetMath: harmonic conjugate function

5. Here is what I have now.....

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

Therefore
$v(x,y) = 3x^2y + e^{x^2 - (y-2)^2} \ sin (2x(y-2)) + H(x)$

====>

$v_x = 6yx+2xe^{x^2-(y-2)^2} \ sin(2x(y-2))+e^{x^2-(y-2)^2} \ cos(2x(y-2))(2y-4) + H'(x)$

so

$v_x = -u_y + H'(x)$

therefore $H'(x) = 0$ ===> $H(x) = constant$

so

$v(x,y) = 3x^2y + e^{x^2 - (y-2)^2} \ sin (2x(y-2)) + constant$

is that correct??

Also is my procedure for part 3 from above correct?

(I have to go to work now so I wont be able to write a response for another 6 - 7 hrs)....

6. I get:

$v(x,y)=3 x^2 y-y^3-e^{x^2-(-2+y)^2} \sin(4 x-2 x y)$

which satisfies the Cauchy-Riemann equations with $u(x,y)$ given above. Also, I use the standard approach to calculating the complex conjugate of $u(x,y)$ which looks intimidating but is easy to calculate for easy functions:

$v(x,y)=\int\frac{\partial u}{\partial x}dy+\int\left(-\frac{\partial u}{\partial y}-\frac{\partial}{\partial x}\int\frac{\partial u}{\partial x}dy\right)dx$

However it's not easy for your function. I used Mathematica to calculate $v(x,y)$ above and I still had to do some manual processing of the data to arrive at my expression for $v(x,y)$. Is there an easier way to do it?

Also, I don't understand what you're trying to do with that gradient expression. Is that suppose to work for just this particular case? Never seen anything like that for the general case.

7. Could've made a stupid mistake..........I was in a rush since I had to go to work......I'll double check it and see what I get....

8. Originally Posted by shawsend
I get:

$v(x,y)=3 x^2 y-y^3-e^{x^2-(-2+y)^2} \sin(4 x-2 x y)$

which satisfies the Cauchy-Riemann equations with $u(x,y)$ given above.
You are right, and so am I ,almost............I just made a little typo......(or at least i think im almost right....im going to be embarrassed if i claim to be almost right and then i turn out to be wrong)

I should have written

$v(x,y) = 3x^2y -y^3 + e^{x^2 - (y-2)^2} \ sin (2x(y-2))$

( the typo is the - y^3 which I didnt include).

If you differentiate both your answer and mine with respect to x, you end up with $v_x = -u_y$

Now I'll try and show you what I was trying to do with the gradient expression (next post)..

9. Shawsend, this is what I was trying to do with the gradient expression.....this answers the question, does it not?

$\bigtriangledown u = u_x i + u_y j$

$\therefore \ \bigtriangledown u = [3x^2 - 3y^2 + 2x e^{x^2-(y-2)^2} \ cos(2x(y-2))-e^{x^2-(y-2)^2} \ sin(2x(y-2))(2y-4)]i +$ $[-6xy + (-2y+4)e^{x^2-(y-2)^2} \ cos(2x(y-2))- 2e^{x^2-(y-2)^2} \ sin(2x(y-2))x] j$

similarly,

$\bigtriangledown v = v_xi +v_yj$

$\therefore \ \bigtriangledown u = [ 6yx+2xe^{x^2-(y-2)^2} \ sin(2x(y-2))+e^{x^2-(y-2)^2} \ cos(2x(y-2))(2y-4)]i +$ $[3x^2 - 3y^2 + 2x e^{x^2-(y-2)^2} \ cos(2x(y-2))-e^{x^2-(y-2)^2} \ sin(2x(y-2))(2y-4)] j$

So

$\bigtriangledown u * \bigtriangledown v = (Ai + Bj)*(-Bi + Aj) = -AB + AB = 0$

Therefore $\vec{\bigtriangledown} u * \vec{\bigtriangledown} v = 0$ holds for all x and y.