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Double integral/change of variable

  1. Mar 11, 2009 #1

    hnh

    User Avatar

    Hello, it may look long but if written down-
    is really quite short, I have only included a detailed explanation to try and be more clear.
    My question is; I am doing a change of variable which doesn't seem to 'fit' the theory of finding the jacobian to make
    the transformation. It seems to me that I make a straightforward substitution:

    The integral is \int\int_A f'(t) ln( f(x) - f(t) + t - x)dxdt In my area, x <= t so the order cannot be switched.

    I make the following change of variable:

    Let r=x-f(x) and h(r) = x and let s=t-f(t) and h(s)=t
    notice that this gives f(x)=h(r)-r and f(t)=h(s)-s, hence f'(t)dxdt = dxf'(t)dt = h'(r)dr(h'(s) - 1)ds=h'(r)(h'(s)-1)drds.

    Now the integral appears to be

    \int\int_A' ln(r-s)h'(r)(h'(s)-1)drds

    Is this correct-can I do this change of variable without actually doing the jacobian. It seems that I do not actually
    have the integral \int\int f(x,y) dxdt in this case to make the transformation with the jacobian so I am wondering
    if there is anything wrong with the way I have done this?
     
  2. jcsd
  3. Mar 12, 2009 #2
    I think it would work because your r and s variables are not each one dependent on x and t at the same time, so it's like you're doing two one-variable changes, so you really are doing the Jacobian thingy even if you don't realise that you are, if you know what I mean. You have:


    [tex] r(x) \equiv x -f(x) \ \ \rightarrow \ \ x=h(r); [/tex]
    [tex] s(t) \equiv t-f(t) \ \ \rightarrow \ \ t=h(s); [/tex]

    [tex] f(x)=x-r(x) \ \ \rightarrow \ \ f(x(r)) \equiv F(r) = h(r)-r; [/tex]
    [tex] f(t)=t-s(t) \ \ \rightarrow \ \ f(t(s)) \equiv F(s) = h(s)-s; [/tex]


    The Jacobian would be:

    [tex]

    J = det \left(
    \begin{array}{cc}
    \frac {\partial x} {\partial r} & \frac {\partial t} {\partial r}\\
    \frac {\partial x} {\partial s} & \frac {\partial t} {\partial s}
    \end{array}
    \right) = det
    \left( \begin{array}{cc}
    h'(r) & 0\\
    0 & h'(s)
    \end{array}
    \right) = h'(r) h'(s)

    [/tex]

    so you have:

    [tex] dxdt = J \ dr ds = h'(r) h'(s) dr ds [/tex]

    and

    [tex] ln|f(x(r))-f(t(s))+t(s)-x(r)| = ln|s-r| [/tex]

    and

    [tex] f'(t) \ = \ \frac {df} {dt} = \left( 1 - \frac {ds} {dt} \right) [/tex]

    And so:

    [tex]

    \iint_\textrm{A} f'(t) \ ln|f(x)-f(t)+t-x|dxdt = \iint_\textrm{A'} \left( 1 - \frac {ds} {dt} \right) ln|s-r|h'(r) h'(s) dr ds= \iint_\textrm{A'} \left( h'(s)- \frac {ds} {dt} \frac {dh} {ds} \right) ln|s-r|h'(r) dr ds [/tex]

    [tex]= \iint_\textrm{A'} \left( h'(s)- \frac {ds} {dt} \frac {dt} {ds} \right) ln|s-r|h'(r) dr ds= \iint_\textrm{A'} \left( h'(s)- 1 \right) ln|s-r|h'(r) dr ds

    [/tex]

    which is what you got so I think it's correct :)
     
  4. Mar 13, 2009 #3

    hnh

    User Avatar

    Thank you very much-that is a huge help!
     
  5. Mar 14, 2009 #4
    You're welcome :smile:
     
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