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Query on indexing to determine coefficients

  1. Aug 27, 2012 #1
    Folks,
    I am interested to know what the author is doing in the following

    ##\displaystyle B_{ij}=EL ij (L)^{(i+j-1)} \left[ \frac{(i-1)(j-1)}{i+j-3} -\frac{2(ij-1)}{i+j-2}+\frac{(i+1)(j+1)}{i+j-1}\right]##

    he states that this expression is not valid for ##B_{ij}## when ##i=1## and ##j=1,2,...N##

    ......yet he goes on to actually calculate

    ##B_{11}=4EIL##, ##B_{1j}=B_{j1}=2EIL^j##, ##(j>1)##

    I understand the the numerator in the first 2 terms inside the big brakets are both 0 when i=j=1 but we still yield a value from the third term...
    Any insight will be appreciated
    Regards

    PS:I notice there is some editing problem with the 3 terms inside the big brackets. There should be a minus and plus separating the terms.
     
  2. jcsd
  3. Aug 27, 2012 #2

    Mark44

    Staff: Mentor

    That's not at all obvious. The only restrictions I see are that
    1. i + j ≠ 3 (would make the first denominator vanish)
    2. i + j ≠ 2 (would make the second denominator vanish)
    3. i + j ≠ 1 (would make the third denominator vanish)
    I don't see how. With i = 1, j = 1, the second term in the brackets is 0/0.
     
  4. Aug 27, 2012 #3

    micromass

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    Staff Emeritus
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    2016 Award

    Could you give a link to where you found this question?
     
  5. Aug 28, 2012 #4
    Are you saying because one of the terms is indeterminate then the whole equation is invalid and thus cannot be usedt o calcualte ##B_{ij}## for i=j=1?

    See attached jpeg of question. The answer involves converting the DE into a weak form using a weight function w and splitting the differentiation between the weight function and the dependent variable u.
    Would the choice of the approximation functions affect the outcome?

    regards
     

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