Results 1 to 9 of 9

Math Help - Harmonic Function

  1. #1
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Maccaman View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85
    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??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Maccaman View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85
    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)....
    Last edited by Maccaman; October 18th 2008 at 10:38 PM. Reason: Latex Error
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2008
    Posts
    903
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85
    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....
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85
    Quote Originally Posted by shawsend View Post
    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)..
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member Maccaman's Avatar
    Joined
    Sep 2008
    Posts
    85
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: August 25th 2011, 02:33 PM
  2. Is this function harmonic?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 8th 2009, 03:19 PM
  3. some help about harmonic function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 28th 2009, 12:10 PM
  4. ln Harmonic function
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 27th 2009, 08:47 PM
  5. Harmonic function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 27th 2009, 10:54 AM

Search Tags


/mathhelpforum @mathhelpforum