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Analytical functions, proof

  1. Jun 11, 2007 #1

    malawi_glenn

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    1. The problem statement, all variables and given/known data

    See pdf - file



    2. Relevant equations

    See pdf - file

    3. The attempt at a solution

    See pdf - file

    Is it right so far? How can I proceed?
     

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  3. Jun 11, 2007 #2

    StatusX

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    Can you relate [itex]\partial u(\bar z)/\partial x[/itex] and [itex]\partial u(z)/\partial x[/itex], or [itex]\partial u(\bar z)/\partial y[/itex] and [itex]\partial u(z)/\partial y[/itex]?
     
  4. Jun 11, 2007 #3

    malawi_glenn

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    yeah, that is what I am trying to figure out HOW :P

    [itex]\partial u(\bar z)/\partial x[/itex] = [itex]\partial u(z)/\partial x[/itex]

    I think

    and

    [itex]\partial u(\bar z)/\partial y[/itex] = [itex]- \partial u(z)/\partial y[/itex]



    ?? =)
     
  5. Jun 11, 2007 #4

    StatusX

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    Not quite. The notation is a little confusing here, since the same symbol is used to show the variable being differentiated and the place the derivative is evaluated. To be clear, you should specify where the derivative is evaluated separately.

    In this case we have:

    [tex]\frac{\partial u(\bar z)}{\partial x} =\frac{\partial u(x-iy)}{\partial x}[/tex]

    Now we should rewrite this as:

    [tex]= \frac{\partial u(x'-iy')}{\partial x'} \left|_{x'=x, y'=y}[/tex]

    This might seem stupid, but it allows us to get what you need as follows:

    [tex]= \frac{\partial u(x'+iy')}{\partial x'} \left|_{x'=x, y'=-y}[/tex]

    [tex]= \frac{\partial u(z')}{\partial x'} \left|_{z'=\bar z}[/tex]

    and similarly for the derivative with respect to y, although there's one more step there.
     
  6. Jun 12, 2007 #5

    malawi_glenn

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    I am sorry, the notation really confused me :S
     
  7. Jun 12, 2007 #6

    Dick

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    Try think of it this way. If f(z)=f(x+iy)=u(x,y)+i*v(x,y). So g(z)=u(x,-y)-i*v(x,-y). So using the CR equations for u and v, prove the CR equations for U(x,y)=u(x,-y) and V(x,y)=-v(x,-y).
     
  8. Jun 12, 2007 #7

    malawi_glenn

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    I still don't get it, what is worng? :S

    [tex]U(x,y)=u(x,-y) and V(x,y)=-v(x,-y)[/tex]

    [tex]\dfrac{\partial U(x,y)}{\partial x} =\dfrac{\partial u(x,-y)}{\partial x} = \dfrac{\partial u(x,y)}{\partial x}[/tex]

    [tex]\dfrac{\partial U(x,y)}{\partial y} =\dfrac{\partial u(x,-y)}{\partial y} = - \dfrac{\partial u(x,y)}{\partial y} [/tex]

    [tex]\dfrac{\partial V(x,y)}{\partial x} =\dfrac{\partial (-v(x,-y))}{\partial x} = -\dfrac{\partial v(x,y)}{\partial x} [/tex]

    [tex]\dfrac{\partial V(x,y)}{\partial y} =\dfrac{\partial (-v(x,-y))}{\partial y} = \dfrac{\partial v(x,y)}{\partial y}[/tex]

    Is apparently wrong, how would you do this Dick? Dont show me all, just show for one of them, then I try more myself.

    thanks.
     
  9. Jun 12, 2007 #8

    Dick

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    Is not wrong. Is right! I'll do one. Since du(x,y)/dy=-dv(x,y)/dx or du(x,-y)/dy=dv(x,-y)/dx (CR for f), from what you've sent, dU(x,y)/dy=du(x,-y)/dy=dv(x,-y)/dx=-dV(x,y)/dx which is the second CR for g.
     
    Last edited: Jun 12, 2007
  10. Jun 12, 2007 #9

    Dick

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    Well, almost correct. Change things like
    [tex]\dfrac{\partial U(x,y)}{\partial x} =\dfrac{\partial u(x,-y)}{\partial x} = \dfrac{\partial u(x,y)}{\partial x}[/tex]
    to:
    [tex]\dfrac{\partial U(x,y)}{\partial x} =\dfrac{\partial u(x,-y)}{\partial x} = \dfrac{\partial u(x,-y)}{\partial x}[/tex]

    You can't just ignore the -y's, the U and V need them.
     
  11. Jun 12, 2007 #10

    malawi_glenn

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    and for the thoughest one: ?

    [tex]\dfrac{\partial V(x,y)}{\partial y} =\dfrac{\partial (-v(x,-y))}{\partial y} = (-1)(-1)\dfrac{\partial (-v(x,-y))}{\partial x}[/tex]
     
  12. Jun 12, 2007 #11

    Dick

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    dV(x,y)/dy=d(-v(x,-y))/dy=-dv(x,-y)/dy=du(x,-y)/dx=dU(x,y)/dx. Since du(x,y)/dx=dv(x,y)/dy implies du(x,-y)/dx=-dv(x,-y)/dy.
     
  13. Jun 12, 2007 #12

    malawi_glenn

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    hmm why is

    dV(x,y)/dy=d(-v(x,-y))/dy=-dv(x,-y)/dy

    ?

    I still think it is:

    [tex]\dfrac{\partial}{\partial y}\left( V(h(x,y),g(x,y))\right) = \dfrac{\partial h(x,y)}{\partial y}\dfrac{\partial V}{\partial y}+ \dfrac{\partial g(x,y)}{\partial y}\dfrac{\partial V}{\partial y}[/tex]

    [tex]
    V(h(x,y),g(x,y)) = -v(x,-y)[/tex]

    [tex]
    h(x,y) = x[/tex]

    [tex]
    g(x,y)=-y
    [/tex]

    gives

    [tex]
    \dfrac{\partial h(x,y)}{\partial y}\dfrac{\partial V}{\partial y}+
    [/tex]

    [tex]\dfrac{\partial g(x,y)}{\partial y}\dfrac{\partial V}{\partial y}=
    [/tex]

    [tex]\dfrac{\partial g(x,y)}{\partial y}\dfrac{\partial V}{\partial y}=\\
    [/tex]

    [tex]
    -1\dfrac{\partial V}{\partial y}= -1\dfrac{\partial (-v(h,g))}{\partial y}\\
    [/tex]

    [tex]
    =(-1)(-1)\dfrac{\partial (v(x,-y))}{\partial y}\\
    [/tex]

    where am I wrong about this? Sorry for being noob :P
     
  14. Jun 12, 2007 #13

    malawi_glenn

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    Since du(x,y)/dx=dv(x,y)/dy implies du(x,-y)/dx=-dv(x,-y)/dy

    is due to:

    d/dy(v(x,h(y)) = (dh/dy)*(dv(x,h)/dy) = -1(dv(x,h)/dy)

    ?

    So why not chain rule when differentiating V(x,y)= -v(x,-y) ?
     
  15. Jun 12, 2007 #14

    malawi_glenn

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    Never mind, my teacher will help me tomorrow. Thanks for all help!
     
  16. Jun 12, 2007 #15

    Dick

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    This is not a chain rule. I'm just substituting equals for equals. V(x,y) is DEFINED to be -v(x,-y). Similarly using:

    du(x,y)/dx=dv(x,y)/dy

    I just SUBSTITUTE -y for y getting du(x,-y)/dx=dv(x,-y)/d(-y). If you want to appeal to the one-dimensional chain rule now dv(x,-y)/d(-y)=(dv(x,-y)/dy)*(dy/d(-y))=-dv(x,-y)/dy. Just a change of variable in y.
    There is no use of the chain rule for partial derivatives here (which I don't think you are stating quite correctly). Does that help? I sympathize here, this can be confusing.
     
  17. Jun 12, 2007 #16

    malawi_glenn

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    I think I got it now, will show my teacher my thougts and work tomorrow. Thanks for all the help guys!

    see ya
     
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