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Directional Derivative of Complex Function

  1. Oct 20, 2014 #1
    1. The problem statement, all variables and given/known data
    We are given that ##f(z) = u(x,y) + iv(x,y)## and that the function is differentiable at the point ##z_0 = x_0 + iy_0##. We are asked to determine the directional derivative of ##f##

    1. along the line ##x=x_0##, and

    2. along the line ##y=y_0##.

    in terms of ##u## and ##v##.

    2. Relevant equations

    Definition of directional derivative: ##\lim\limits_{h \rightarrow 0} \frac{f(z+hw)-f(z)}{h}##

    where ##w## is a vector designating the direction along which we are differentiating.

    3. The attempt at a solution

    I am first going to deal with part 1. The direction along which we are differentiating is ##w = x##. Thus,

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f(z_0+hw)-f(z_0)}{h} = ##

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f((x_0+iy_0)+hx_0)-f(z_0)}{h} =##

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f((x_0+ hx_0) + iy_0)-f(z_0)}{h} =##

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f(x_0(1+ h) + iy_0)-f(z_0)}{h} =##

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) + iv(x_0(1+h),y_0 - (u(x_0,y_0) + iv(x_0,y_0))}{h} =## (*)

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) - u(x_0,y_0) + iv(x_0(1+h),y_0) - iv(x_0,y_0)}{h} =##

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) - u(x_0,y_0)}{h} + \lim\limits_{h \rightarrow 0} \frac{iv(x_0(1+h),y_0) - iv(x_0,y_0)}{h} =##

    ##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) - u(x_0,y_0)}{h} + i \lim\limits_{h \rightarrow 0} \frac{v(x_0(1+h),y_0) - v(x_0,y_0)}{h} =##

    I have two issues: the first is, did I properly substitute at step (*); second, assuming I did properly substitute, is the last line simply the partial derivatives of u and v, with respect x? The reason I ask is, that I am accustomed to seeing the change of a function, as x changes, represented as ##f(x+\delta x, y) - f(x,y)##.
     
  2. jcsd
  3. Oct 20, 2014 #2

    HallsofIvy

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    As far as finding directional derivatives is concerned, the fact that these are complex numbers is really irrelevant- you want to treat the real and imaginary parts as independent variables.

    I'm not sure what you are doing with the "difference quotients". If you do not know what "u" and "v" are the only thing you can do is write the derivatives of f in terms or [itex]u_x[/itex], [itex]u_y[/itex], [itex]v_x[/itex], and [itex]v_y[/itex].

    .
     
  4. Oct 20, 2014 #3
    The reason why I am using the difference quotient is to calculate the directional derivative, which is the method my book presents, as far as I can understand.

    But I don't know such a thing, that f can be written in terms of the partials of u and v, until I have reached the last line of my work.
     
  5. Oct 20, 2014 #4

    vela

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    Why are you multiplying ##h## by ##x_0##? ##h## multiplies ##w##, which in the first case is equal to ##w=1+0i##, right?
     
  6. Oct 20, 2014 #5
    As far as I understand, the ##w = x + 0i##.
     
  7. Oct 21, 2014 #6
    Sorry, I should have written ##w = x_0 + 0i##.
     
  8. Oct 21, 2014 #7

    BvU

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    Funny. "Along the line x = x0" to me would mean the direction vector is ##(0, i)##
    Am I way off? I understand the exercise asks for the derivatives in the point ##z_0## and that's also the way you go to work...
     
  9. Oct 21, 2014 #8

    Ray Vickson

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    No: the directional derivative of a function ##f(x,y)## in direction ##p = (u,v)## at the point ##(x_0,y_0)## is defined to be
    [tex] D_pf(x_0,y_0) = \lim_{t \to 0} \frac{f(x_0 + t u,y_0 + tv) - f(x_0,y_0)}{t} [/tex]
    (see, eg., http://en.wikipedia.org/wiki/Directional_derivative ). Sometimes authors will use the vector of length 1 in direction ##p##, so would define the directional derivative as ##\frac{1}{|p|} D_p f ##, where ##D_p f## is as above.
     
    Last edited: Oct 21, 2014
  10. Oct 21, 2014 #9

    vela

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    You're right. From his work, I was thinking the OP was looking for the partial derivative in the x-direction, but the first part of the problem wasn't asking for that.
     
  11. Oct 21, 2014 #10
    Okay, well I am quite confused. What do we all agree "along the line x=x_0 means?
     
  12. Oct 21, 2014 #11

    Ray Vickson

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    It is the vertical line through the point ##(x_0,0)##. In the complex plane it is the set of points ##\{ x_0 + i t | t \in \mathbb{R} \}##. Its direction is ##(0,1)## in ##\mathbb{R}^2##, or just ##i## in the complex plane. However, in terms of ##u(x,y), v(x,y)## it should be given as ##(0,1)##.
     
    Last edited: Oct 21, 2014
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