Harmonic Function

• Oct 18th 2008, 10:00 AM
Maccaman
Harmonic Function
Show that $\displaystyle u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos[2x(y - 2)]$ is an harmonic function, and find a conjugate harmonic function $\displaystyle v(x,y)$.

Check that $\displaystyle \vec{\bigtriangledown} u * \vec{\bigtriangledown} v = 0$ holds for all x and y.
• Oct 18th 2008, 02:57 PM
mr fantastic
Quote:

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

Check that $\displaystyle \vec{\bigtriangledown} u * \vec{\bigtriangledown} v = 0$ 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...

$\displaystyle u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos(2x(y - 2))$
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)) - $$\displaystyle (4(-2y+4)) e^{x^2-(y-2)^2} sin(2x(y-2)) \ x-4x^2 e^{x^2-(y-2)^2} \ cos(2x(y-2)) therefore \displaystyle u_{xx} + u_{yy} = 0 so \displaystyle u(x,y) is a harmonic function. Now to find a conjugate harmonic function \displaystyle v(x,y) \displaystyle v_y = u_x This is where I get lost.....am I now supposed to differentiate \displaystyle v_y w.r.t x or y?? I've tried both and keep getting bizarre values. How do I get the conjugate function?? As far as the third part goes (with out v) \displaystyle \bigtriangledown u = u_xi + u_yj \displaystyle \bigtriangledown v = v_xi + v_yj then I think im supposed to do something like \displaystyle \bigtriangledown u *\bigtriangledown v = ??A \ something + - * B \ something?? = 0 which enables me to solve for some values A and B?? (Thinking) • Oct 18th 2008, 08:40 PM mr fantastic Quote: Originally Posted by Maccaman Here is what I have done... \displaystyle u(x,y) = x^3 - 3xy^2 + e^{[x^2-(y-2)^2]} \ cos(2x(y - 2)) so \displaystyle u_x = 3x^2 - 3y^2 + 2x e^{x^2-(y-2)^2} \ cos(2x(y-2))-e^{x^2-(y-2)^2} \ sin(2x(y-2))(2y-4) \displaystyle u_{xx} = 6x+2e^{x^2-(y-2)^2} \ cos(2x(y-2)) + 4x^2e^{x^2-(y-2)^2} \ cos(2x(y-2))-4xe^{x^2-(y-2)^2} \displaystyle \ sin(2x(y-2))(2y-4) - e^{x^2-(y-2)^2} \ cos(2x(y-2))(2y-4)^2 and \displaystyle u_y = -6xy + (-2y+4)e^{x^2-(y-2)^2} \ cos(2x(y-2))- 2e^{x^2-(y-2)^2} \ sin(2x(y-2))x \displaystyle u_{yy} =-6x-2e^{x^2-(y-2)^2} \ cos(2x(y-2)) + (-2y+4)^2 e^{x^2-(y-2)^2} \ cos(2x(y-2)) -$$\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)

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)

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

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

====>

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

so

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

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

so

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

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:

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

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

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

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

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

which satisfies the Cauchy-Riemann equations with $\displaystyle u(x,y)$ 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

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

( 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 $\displaystyle v_x = -u_y$

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)

$\displaystyle \bigtriangledown u = u_x i + u_y j$

$\displaystyle \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 + $$\displaystyle [-6xy + (-2y+4)e^{x^2-(y-2)^2} \ cos(2x(y-2))- 2e^{x^2-(y-2)^2} \ sin(2x(y-2))x] j similarly, \displaystyle \bigtriangledown v = v_xi +v_yj \displaystyle \therefore \ \bigtriangledown u = [ 6yx+2xe^{x^2-(y-2)^2} \ sin(2x(y-2))+e^{x^2-(y-2)^2} \ cos(2x(y-2))(2y-4)]i +$$\displaystyle [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

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

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