Question of Galerkins method of weighted residual- the physical meaning

In summary, Galerkin's method of weighted residuals involves using an approximate solution to a differential equation expressed as a product of unknown parameters and trial functions. The trial functions must be continuous and satisfy the boundary conditions, but are otherwise chosen based on the "physics" of the problem. The method then requires evaluating the unknown parameters in such a way that the definite integral of the residual error is equal to zero. This approach essentially replaces a general linear operator equation with a finite order matrix equation, which can then be numerically solved to obtain an approximate solution. The weights in this method represent the difference between the known right-hand side of the equation and the unknown function, approximated by the basis functions and acted upon by the operator. By evaluating the
  • #1
svishal03
129
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
 
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  • #2
Please can anyone help?
 
  • #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.
 
  • #4
Dear Antiphon,

Thank you very much for the reply.

Now,as you said:

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.

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
 
  • #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.
 
  • #6
Thanks again..few questions...as below

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.

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

approximated by the basis functions and acted upon by the operator.

then,

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

Can you give an example for this?
 
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1. What is the physical meaning of Galerkin's method?

Galerkin's method is a numerical technique used in solving differential equations by approximating the solution with a finite set of functions. The physical meaning of this method is to find the best possible approximation of the solution by minimizing the error between the approximate solution and the actual solution.

2. How does Galerkin's method work?

Galerkin's method works by multiplying both sides of a differential equation by a weight function and then integrating over the domain. This weight function is chosen to satisfy certain boundary conditions and minimize the error between the approximate and actual solution. The resulting equations are then solved using algebraic methods.

3. What is the difference between Galerkin's method and other numerical methods?

Galerkin's method differs from other numerical methods in that it uses a weighted residual approach instead of a collocation or least squares approach. This means that it aims to minimize the error globally over the entire domain, rather than at specific points or regions.

4. What are the advantages of using Galerkin's method?

One advantage of Galerkin's method is its flexibility in choosing the weight function, which allows for different types of boundary conditions to be satisfied. It also has a wider convergence range compared to other numerical methods, making it suitable for a variety of differential equations.

5. Are there any limitations to Galerkin's method?

One limitation of Galerkin's method is its sensitivity to the choice of weight function, which can affect the accuracy and convergence of the solution. It is also computationally intensive and may require a large number of terms in the approximate solution to achieve a desired level of accuracy.

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