Harmonic Function

• Oct 18th 2008, 10:00 AM
Maccaman
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.
• Oct 18th 2008, 02:57 PM
mr fantastic
Quote:

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?
• Oct 18th 2008, 07:54 PM
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??
(Thinking)
• Oct 18th 2008, 08:40 PM
mr fantastic
Quote:

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??
(Thinking)

To get the harmonic conjugate of u, read this: PlanetMath: harmonic conjugate function
• Oct 18th 2008, 10:37 PM
Maccaman
Here is what I have now.....(Wondering)

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

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)....(Crying)
• Oct 19th 2008, 05:19 AM
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. 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.
• Oct 19th 2008, 05:36 AM
Maccaman
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....
• Oct 19th 2008, 05:57 AM
Maccaman
Quote:

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)..
• Oct 19th 2008, 06:39 AM
Maccaman
Shawsend, this is what I was trying to do with the gradient expression.....this answers the question, does it not? (Wondering)

$\bigtriangledown u = u_x i + u_y j$

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

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.