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Contradictory (complex) integral transformation

  1. May 27, 2013 #1
    The Schwarz-Christoffel mapping (a Riemann-mapping) from the unit disk (z-plane) to a twice-symmtric area (a cross, ζ-plane)

    $$ \zeta : \mathbf C \to \mathbf C $$

    is given by:

    $$\frac{ \mathrm{d}\zeta }{ \mathrm{d} z} = \left( \frac{ ( z^2-b^2 ) ( z^2-\frac 1 {b^2} ) }{ ( z^2-a^2 ) ( z^2-\frac 1 {a^2} ) ( z^2-c^2 ) ( z^2-\frac 1 {c^2} ) } \right)^\frac12 $$

    attachment.php?attachmentid=59045&stc=1&d=1369655156.png

    where a, b, and c are points in counter-clockwise order in the first quadrant on the unit circle. If the integration constant is chosen such that

    $$\zeta ( z ) = \int_0^z \mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y)$$

    the cross-shape in the ζ-plane to which the unit circle in the z-plane is mapped has a symmetry about the real and the imaginary axis, i.e. ∀ |z| = 1:

    $$ \zeta \left( \frac 1 z \right) = \bar \zeta ( z ) $$

    $$ \zeta \left( - \frac 1 z \right) = -\bar \zeta ( z ) $$

    attachment.php?attachmentid=59046&stc=1&d=1369655156.png

    This is true and evident from numeric evaluation but it is contradicted by:

    $$ \zeta \left( \frac 1 z \right) = \int_0^1 \mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y) + \int_1^{\frac 1 z }\mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y) $$

    Transformation of the second integral with x := 1/y

    $$\begin{aligned}\int_1^{\frac 1 z }\mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y) &= -\int_1^z \mathrm{d}x \frac 1 {x^2} \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}\left(y = \frac 1 x \right) \\
    &= -\int_1^z \mathrm{d}x \frac{ \mathrm{d}\zeta }{ \mathrm{d} x} ( x )
    \end{aligned}$$

    Where the second line follows from the first by simply inserting dζ/dz and distributing the x² factor on the fraction and pulling out pairwise terms of the forms b and 1/b. So we have

    $$ \zeta \left( \frac 1 z \right) = \int_0^1 \mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y) - \int_1^z\mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y) $$

    but ∫01… is purely real and ∫1z… is mixed real and imaginary, so this contradicts the symmetry about the real axis from above because

    $$ \zeta ( z ) = \int_0^1 \mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y) + \int_1^z\mathrm{d}y \frac{ \mathrm{d}\zeta }{ \mathrm{d} y}(y) $$

    In fact, the transformation turns out to be wrong. ∫11/z… is indeed the complex conjucate of ∫1z…, not the negative, so there must be a problem with the derivation above.

    Note: The cross in the ζ-plane is mapped such that 0 maps to the center, 1 maps to the tip of the left arm (i.e. negative real part, 0 imaginary part), -1 maps to the tip of the right arm and i, -i respectively, map to tips of the bottom and top arm. Therefore ∫01… points from the center outwards to the left arm and is purealy real while, say, ∫1i… points from the tip of the left arm to the tip of the bottom arm and is mixed real and imaginary.

    PS: I'm afraid this is again related to choosing the correct branches on the exponentiation in the integral, so that by the (correct) choice of branches, I get the correct result, that is, the complex conjucate. But this seems messed up: Is there a tool which allows me to treat the transformation consistently without thinking through in detail which branch goes where and when so these be equivalent?

    PS II: Even despite that vague suspicion, I can't really see how choosing branches could yield the complex conjugate, because that would only yield a sign for the expression (unless I switch branches inside the interval)

    PS III: I'm aware of alternative methods to prove the relation, but that doesn't explain the above contradiction.
     

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    Last edited: May 27, 2013
  2. jcsd
  3. May 27, 2013 #2
    I might be able to help you, but...your notation is a little confusing. Do you mean ##\displaystyle \zeta(z) = \int_0^z \left. dy \frac{d\zeta}{dy}\right|_y## or ##\displaystyle \zeta(z)=\int_0^z \left. \frac{d\zeta}{dy}\right|_y \, dy##?
     
  4. May 27, 2013 #3

    micromass

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    Physicists tend to put the ##dy## before the function that is integrated.
     
  5. May 27, 2013 #4
    Here is where I can't follow you. If the modulus of z is equal to 1, then that appears to be correct, because this basically inserts the value for which zeta equals 1/z, and if |z|=1, its complex conjugate is identically it's multiplicative inverse. Thus, I see no apparent contradiction.

    Check limits of integration below that point.
     
    Last edited: May 27, 2013
  6. May 27, 2013 #5
    I suspect this is a misunderstanding: Yes, the last line you quoted is indeed obtained by just inserting z* = 1/z into the definition, so to speak, of ζ. The actual contradiction is shown in the following (after where you stoped quoting me) and is concluded where I say

    Did this clarify it?
     
  7. May 27, 2013 #6
    Kind of. Just so I'm clear, what have you defined y to be? What is it in terms of z?
     
  8. May 27, 2013 #7
    This is meant in functional notation, not system notation. Two functions f(x) and f(y) have the same form (actually, they are the same function). There are no globally bound variables or any such hocus-pocus that is popular among physicists :-p

    $$ \frac { \mathrm{d}\zeta } { \mathrm{d} y } (y) \equiv \left. \frac { \mathrm{d}\zeta } { \mathrm{d} z } \right|_{z=y}$$
    (meaning y is just bound by the integral, i.e. a placeholder name)
     
    Last edited: May 27, 2013
  9. May 29, 2013 #8
    Your notation is awkward and not easy to follow. Try and come up with something better like doing away entirely with the notation [itex]\frac{d\zeta}{dz}[/itex] and just calling it f(z), then you could write the integrand simply as [itex]\frac{1}{x^2} f(\frac{1}{x})dx[/itex]. Now ain't that easier to read?

    Also, I do not agree that:

    [tex]\frac{d\zeta}{dz}\biggr|_{z=x}=\frac{1}{x^2} \frac{d\zeta}{dz}\biggr|_{z=1/x}[/tex]
     
  10. May 30, 2013 #9
    You seem to conveniently forget that there is a derivative involved. How would that be written with your "easy" notation?
    $$\frac{\mathrm{d}f}{\mathrm{d}\left[\frac1x\right]}\left(\frac1x\right)$$
    [tex]\begin{aligned}
    \frac1{x^2}\frac{\mathrm{d}\zeta}{\mathrm{d}z}\left(z=\frac1x\right) &= \frac1{x^2} \left( \frac{ ( \left(\frac1x\right)^2-b^2 ) ( \left(\frac1x\right)^2-\frac 1 {b^2} ) }{ ( \left(\frac1x\right)^2-a^2 ) ( \left(\frac1x\right)^2-\frac 1 {a^2} ) ( \left(\frac1x\right)^2-c^2 ) ( \left(\frac1x\right)^2-\frac 1 {c^2} ) } \right)^\frac12 \\
    &= \frac1{x^2} \left( \frac{ x^4 ( 1-x^2 b^2 ) ( 1-x^2 \frac 1 {b^2} ) }{ ( 1-x^2a^2 ) ( 1-x^2 \frac 1 {a^2} ) ( 1-x^2 c^2 ) ( 1-x^2\frac 1 {c^2} ) } \right)^\frac12 \\
    &= \left( \frac{ ( \frac 1 {b^2}-x^2 ) ( b^2-x^2 ) }{ ( \frac 1 {a^2} -x^2 ) ( a^2-x^2 ) ( \frac 1 {c^2} -x^2 ) ( c^2-x^2) } \right)^\frac12 \\
    &= \frac{\mathrm{d}\zeta}{\mathrm{d}x}(x)
    \end{aligned}
    [/tex]

    (this is contradicted by a similar calculation where you take the complex vonjugate of the whole term, instead, which yields the correct result)
     
    Last edited: May 30, 2013
  11. May 30, 2013 #10
    Ok, I see it now. Thanks for clearing that up for me. I'll continue working with it as it's an interesting problem for me. I suspect it's a branching issue you're having but don't see it yet.
     
  12. May 30, 2013 #11
    I do not get mixed real and imaginary for the second integral when I check your integrals with [itex]z=e^{\pi i/12}[/itex] and integrate over a path which follows the unit circle: Let's take [itex]z=e^{\pi i/12}[/itex] so that, numerically, [itex]\zeta(z)\approx 0.9978+0.2422i[/itex] and thus [itex]\overline{\zeta}(z)\approx 0.9978-0.2422i[/itex]. Now letting the derivative be u(z):
    [tex]\int_0^1 u(w)dw\approx 0.9978[/tex]
    [tex]\int_1^z u(w)dw=\int_0^{\pi/12} u(e^{it}) i e^{it} dt\approx 0.2422i[/tex]
    and therefore:

    [tex]\zeta(1/z)\approx 0.9978-0.2422i[/tex]
    which agrees with your derivation.
     
    Last edited: May 30, 2013
  13. Jun 4, 2013 #12
    I dont know your particular choice for a, b, and c, but while ∫01 lies on the end of the left, horizontal arm, ∫0exp(iπ/12) is likely to lie on the same edge, which means they have the same real part.

    In general, however, ∫1exp(iπr) with r ∈ ℝ, does not point strictly vertically and thus has mixed real- and imaginary parts.
     
  14. Jun 4, 2013 #13
    I chose your choices:

    [tex]a=e^{i/2}, b=e^i, c=e^{3/2 i}[/tex]

    I belive I've determined the cause of your problem, at least to my satisfaction, and if this was a final exam question, this is how I would risk answering it: You have

    [tex]\int_1^{1/z} u(y)dy[/tex]

    and you propose making the substitution [itex]x=1/y[/itex]. That substitution introduces the function [itex]u(1/x)=v(x)[/itex] and v(x) between the endpoints x=1 and x=z has a principal branch cut which if I just integrated numerically without regards to analytic continuitity, would integrate (discontinuously) over a branch-cut. However if I integrated over an analytically-continuous sheet of v(x), I could get the correct answer. But I'm not sure I can determine which sheet of v(x) to integrate over without knowing the answer beforehand.

    Conclusions: The substitution x=1/y is not a recommended technique to evaluate the integral above.

    Here's mine:
     

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    Last edited: Jun 4, 2013
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