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Math Help - Laplace's equation

  1. #1
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    Post Laplace's equation

    I understand using Laplace transforms but I think this is something different the question reads.

    Show that the following functions u(x,y) are harmonic (ie satisfy Laplace's equation) and in each case find f(z) such that f(z) = u + iv is regular.
    (a) u = x^4 - 6x^2y^2 + y^4

    (b) u = sinxsinhy
    Any help would be great.

    Jez
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  2. #2
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    Quote Originally Posted by jezzyjez View Post
    I understand using Laplace transforms but I think this is something different the question reads.

    Show that the following functions u(x,y) are harmonic (ie satisfy Laplace's equation) and in each case find f(z) such that f(z) = u + iv is regular.
    (a) u = x^4 - 6x^2y^2 + y^4

    (b) u = sinxsinhy
    Any help would be great.

    Jez
    This is nothing to do with Laplace transforms.

    Laplace's equation is this:

     \nabla^2 \phi = 0

    That is:

     \frac{\partial \phi}{\partial u} + \frac{\partial \phi}{\partial v} = 0
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  3. #3
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    Quote Originally Posted by jezzyjez View Post
    I understand using Laplace transforms but I think this is something different the question reads.

    Show that the following functions u(x,y) are harmonic (ie satisfy Laplace's equation) and in each case find f(z) such that f(z) = u + iv is regular.
    (a) u = x^4 - 6x^2y^2 + y^4

    (b) u = sinxsinhy
    Any help would be great.

    Jez
    If u = x^4 - 6x^2y^2 + y^4

    is harmonic then:

    \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0

    You can take the second partials and check that right? Same dif for the other one. Now, in order to find a complex-analytic function for which the real part is u(x,y), then you need to find the harmonic conjugate of u(x,y) using the Cauchy-Riemann equations and a little bit of integration.
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  4. #4
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    Quote Originally Posted by AnonymitySquared View Post
    This is nothing to do with Laplace transforms.

    Laplace's equation is this:

     \nabla^2 \phi = 0

    That is:

     \frac{\partial \phi}{\partial u} + \frac{\partial \phi}{\partial v} = 0
    You've missed the "second order"! Laplace's equation is
    \nabla^2u= \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2u}{\partial y^2}= 0
    (I have changed AnonimitySquared's " \phi" to "u" since the functions are given as "u(x,y)".)
    Also, jezzyjez, a function, u(x,y)+ iv(x,y), is "regular" if it satifies the Cauchy-Riemann equations:
    \frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}
    and
    \frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}
    Since you are given u is both cases, you can find the right hand sides of these two equations and use them to find v.
    Last edited by Jester; December 23rd 2009 at 06:43 AM. Reason: fixed latex
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  5. #5
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    Cheers guys got the answer to the first part now.

    Not to difficult once you get your head round what all the words mean.
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  6. #6
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    You can find the harmonic conjugate by the following expression:

    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

    Yeah, I know it looks intimidating. Start from the inside and work out and when you do an integration, drop the constant of integration. I'll do the first part for u(x,y)=u = x^4 - 6x^2y^2 + y^4 then:

    \frac{\partial u}{\partial x}=4 x^3-12 x y^2

    see, I'm starting from the most inner level. Now, I want to partially integrate that with respect to y:

    \int 4x^3-12xy^2 \partial y=4x^3 y-12/3 xy^3

    Now, I want to take the partial of that with respect to x. See, that's the first three steps inside the parantheses. Now get the partial of u with respect to y, negate them, then integrate the bunch with respect to x. See, got that whole parentheses thing now. Do that last integral, bingo-bango.

    Edit: Oh yeah, I forgot to mention don't use that expression unless you can go through the steps in deriving it. Any Complex Analysis text book goes over finding the complex conjugate. I just combined all the steps into one expression for fun.
    Last edited by shawsend; December 24th 2009 at 03:33 AM.
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