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Question of Galerkins method of weighted residual- the physical meaning

  1. Jun 12, 2012 #1
    I have been reading on Galerkins method of weighted residuals according to which an approximate solution to a differential equation is expressed as:

    [itex]y^*(x)=Ʃc_iN_i(x)[/itex] where i lies between 1 and n

    where y* is the approximate solution expressed as the product of ci unknown,
    constant parameters to be determined and Ni (x ) trial functions.

    The major requirement placed on the trial functions is that they be admissible functions;
    that is, the trial functions are continuous over the domain of interest and satisfy
    the specified boundary conditions exactly. In addition, the trial functions should
    be selected to satisfy the “physics” of the problem in a general sense.

    Given these somewhat lax conditions, it is highly unlikely that the solution represented by
    above equation is exact. Instead, on substitution of the assumed solution into the
    main differential equation a residual error results.

    The method of weighted residuals requires that the unknown parameters ci be evaluated such that:

    [itex]∫w_i(x)R(x)dx = 0[/itex]

    Can anyone please explain the physical interpretation of the above equation?The physical significance of weights?

    What are these weights physically signify?


    What is the maigic in these weights such that error (residue) is 0?

    Vishal
     
  2. jcsd
  3. Jun 16, 2012 #2
    Please can anyone help?
     
  4. Jun 16, 2012 #3
    You are sampling the domain of the differential equation at discrete points. You are then writing a linear matrix equation to replace the differential equation. The zero represents the error (difference between the approximate sampled solution and the exact solution.)

    The big picture is that you are replacing a general linear operator system (differential equation, integral equation, whatever) with a finite order matrix equation which you then solve numerically to get an approximate answer.
     
  5. Jun 18, 2012 #4
    Dear Antiphon,

    Thank you very much for the reply.

    Now,as you said:

    Yes- this in effect is the interpretation of:

    [itex]y^*(x)=Ʃc_iN_ix[/itex]

    I would like to know,that, when we say:

    The method of weighted residuals requires that the unknown parameters ci be evaluated such that:

    [itex]∫w_i(x)R(x)=0[/itex]

    What are we exactly doing ? what is the significance of weight? Can you express an interpretation for this as you did for [itex]y^*(x)=Ʃc_iN_ix[/itex]

    Thanks again!

    Vishal
     
  6. Jun 18, 2012 #5
    Ok.

    The residual is the difference between the known right-hand side of the original operator equation and the unknown function but approximated by the basis functions and acted upon by the operator. Thats the magic step that turns it from an operator equation into an algebraic one. Because you know the form of the basis functions. There are i=1..N basis functions whose analytic form is given. There are j=1..N weight functions.

    The weighted residuals (the definite integral of R and W) evaluated for all i and j form an NxN matrix who's inverse can be multiplied into the coefficients for the known excitation to yield the unknown weights on the basis functions. Once you have the weights, you have the approximate numerical solution to the operator equation.
     
  7. Jun 18, 2012 #6
    Thanks again..few questions...as below

    Which equation depeicts this statement?- especially when you say;

    then,

    Can you give an example for this?
     
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