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Cauchy-Riemann conditions

  1. Mar 4, 2007 #1
    1. The problem statement, all variables and given/known data
    The functions u(x,y) and v(x,y) are the real and imaginary parts, respectively, of an analytic function w(z).
    Assuming that the required derivatives exist, show that

    [tex]\bigtriangledown^2 u=\bigtriangledown^2 v=0[/tex]

    Solutions of Laplace's equation such as u(x,y) and v(x,y) are called harmonic functions.

    2. Relevant equations
    Cauchy-Riemann conditions:

    [tex]\frac{\delta u}{\delta x} = \frac{\delta v}{\delta y}[/tex]
    [tex]\frac{\delta u}{\delta y} = -\frac{\delta v}{\delta x}[/tex]

    3. The attempt at a solution
    I expanded [tex]\bigtriangledown^2 u = \frac{\delta u}{\delta x}\frac{\delta u}{\delta x} + \frac{\delta u}{\delta y}\frac{\delta u}{\delta y}[/tex] and using the Cauchy-Riemann conditions I found

    [tex]\bigtriangledown^2 u = \frac{\delta v}{\delta y}\frac{\delta v}{\delta y} + \frac{\delta v}{\delta x}\frac{\delta v}{\delta x}=\bigtriangledown^2 v[/tex]

    What I can't figure out how to do is prove that this is equal to zero.
  2. jcsd
  3. Mar 4, 2007 #2

    Meir Achuz

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    Your eqs. for del^2 are wrong.
    [tex]\nabla^2 u=\partial_x\partial_x u+\partial_y\partial_y u.[/tex]
  4. Mar 4, 2007 #3
    Dang, you're right. Can I dot it into an element of length like this?

    [tex]\bigtriangledown^2 u \bullet d\vec{r}^2 = \frac{\delta}{\delta x}\frac{\delta u}{\delta x} dx^2 + \frac{\delta}{\delta y}\frac{\delta u}{\delta y} dy^2[/tex]
  5. Mar 4, 2007 #4
    There is a hint in the problem that says I need to construct vectors normal to the curves [tex]u(x,y)=c_i[/tex] and [tex]v(x,y)=c_j[/tex]. Wow, I'm pretty lost.
  6. Mar 4, 2007 #5


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    The Cauchy-Riemann equations are
    [tex]\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}[/tex]
    [tex]\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}[/tex]
    which is what you have, allowing for your peculiar use of [itex]\delta[/itex] rather than [itex]\partial[/itex]!

    Now just do the obvious: differentiate both sides of the first equation with respect to x and differentiate both sides of the second equation with respect to y and compare them.

    Are you sure that the hint is for this particular problem? A normal vector to u(x,y)= c is
    [tex]\frac{\partial u}{\partial x}\vec{i}+ \frac{\partial u}{\partial y}\vec{j}[/tex]
    and a normal vector to v(x,y)= c is
    [tex]\frac{\partial v}{\partial x}\vec{i}+ \frac{\partial v}{\partial y}\vec{j}[/tex].
    Using the Cauchy-Riemann equations, that second equation is
    [tex]-\frac{\partial u}{\partial y}\vec{i}+ \frac{\partial u}{\partial x}\vec{j}[/tex]
    which tells us the the two families of curves are orthogonal but that does not directly tell us about [itex]\nabla^2 u[/itex] and [itex]\nabla^2 v[/itex].
  7. Mar 4, 2007 #6
    Thanks for your reply.

    There is a part b) to the problem, and it is this:

    b) Show that

    [tex]\frac{\partial u}{\partial x}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\frac{\partial v}{\partial y} = 0[/tex]

    I solved it easily using the Cauchy-Riemann equations, so I figured that the hint was for the first part.
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