Galerkin's Method



    To start we need some background regarding an the inner product.  

Definition (Inner Product).  Consider the vector space of real functions whose domain is the closed interval  [Graphics:Images/GalerkinMod_gr_1.gif].  We define the inner product of two functions  [Graphics:Images/GalerkinMod_gr_2.gif] as follows  


Remark.  The inner product is a continuous infinite dimensional analog to the ordinary dot product that is studied in linear algebra.  If the inner product is zero then [Graphics:Images/GalerkinMod_gr_6.gif] are said to be orthogonal to each other on [Graphics:Images/GalerkinMod_gr_7.gif].  All the functions we use are assumed to be square-integrable, i. e.  [Graphics:Images/GalerkinMod_gr_8.gif].  

Mathematica Function (Inner Product). To compute the inner product of two real functions over [Graphics:Images/GalerkinMod_gr_4.gif].  



Example 1 (a).  Find the inner product of   [Graphics:Images/GalerkinMod_gr_9.gif]  and  [Graphics:Images/GalerkinMod_gr_10.gif]  over [Graphics:Images/GalerkinMod_gr_11.gif].  
1 (b).  Find the inner product of   [Graphics:Images/GalerkinMod_gr_12.gif]  and  [Graphics:Images/GalerkinMod_gr_13.gif]  over  [Graphics:Images/GalerkinMod_gr_14.gif].   
1 (c).  Find the inner product of   [Graphics:Images/GalerkinMod_gr_15.gif]  and  [Graphics:Images/GalerkinMod_gr_16.gif]  over  [Graphics:Images/GalerkinMod_gr_17.gif].   
Solution 1 (a).
Solution 1 (b).
Solution 1 (c).


Lemma.  If  [Graphics:Images/GalerkinMod_gr_30.gif]  for any function  [Graphics:Images/GalerkinMod_gr_31.gif],  then  [Graphics:Images/GalerkinMod_gr_32.gif].  

Basis for a Vector Space

    A complete basis for a vector space V of functions is a set of linear independent functions [Graphics:Images/GalerkinMod_gr_33.gif]  which has the property that any [Graphics:Images/GalerkinMod_gr_34.gif] can be written uniquely as a linear combination  

For example, if V the set of all polynomials and power series, then a complete basis is  [Graphics:Images/GalerkinMod_gr_36.gif].  

Property.  If  [Graphics:Images/GalerkinMod_gr_37.gif]  and  [Graphics:Images/GalerkinMod_gr_38.gif]  all  [Graphics:Images/GalerkinMod_gr_39.gif]  then  [Graphics:Images/GalerkinMod_gr_40.gif].  

We mention these concepts without proof so as to provide a little background.  


Weighted Residual Methods

    A weighted residual method uses a finite number of functions [Graphics:Images/GalerkinMod_gr_41.gif].  Consider the differential equation  

(1)        [Graphics:Images/GalerkinMod_gr_42.gif]     over the interval  [Graphics:Images/GalerkinMod_gr_43.gif].  

The term [Graphics:Images/GalerkinMod_gr_44.gif] denotes a linear differential operator.  

    Multiplying (1) by any arbitrary weight function [Graphics:Images/GalerkinMod_gr_45.gif] and integrating over the interval [Graphics:Images/GalerkinMod_gr_46.gif] one obtains

(2)        [Graphics:Images/GalerkinMod_gr_47.gif]    for any arbitrary  [Graphics:Images/GalerkinMod_gr_48.gif].    

Equations (1) and (2) are equivalent, because [Graphics:Images/GalerkinMod_gr_49.gif] is any arbitrary function.  

    We introduce a trial solution [Graphics:Images/GalerkinMod_gr_50.gif] to (1) of the form

(3)        [Graphics:Images/GalerkinMod_gr_51.gif],

and replace [Graphics:Images/GalerkinMod_gr_52.gif] with [Graphics:Images/GalerkinMod_gr_53.gif] on the left side of  (1).  

    The residual is defined as follows

(4)        [Graphics:Images/GalerkinMod_gr_54.gif]     

    The goal is to construct [Graphics:Images/GalerkinMod_gr_55.gif] so that the integral of the residual will be zero for some choices of weight functions.  That is, [Graphics:Images/GalerkinMod_gr_56.gif] will partially satisfy (2) in the sense that

(5)        [Graphics:Images/GalerkinMod_gr_57.gif]    for some choices of  [Graphics:Images/GalerkinMod_gr_58.gif].  


Galerkin's Method

    One of the most important weighted residual methods was invented by the Russian mathematician Boris Grigoryevich Galerkin (February 20, 1871 - July 12, 1945).  Galerkin's method selects the weight function functions in a special way:  they are chosen from the basis functions, i.e.  [Graphics:Images/GalerkinMod_gr_59.gif].  It is required that the following [Graphics:Images/GalerkinMod_gr_60.gif] equations hold true
(6)        [Graphics:Images/GalerkinMod_gr_61.gif]    for  [Graphics:Images/GalerkinMod_gr_62.gif].  

To apply the method, all we need to do is solve these [Graphics:Images/GalerkinMod_gr_63.gif] equations for the coefficients [Graphics:Images/GalerkinMod_gr_64.gif].  

Proof  Galerkin Method  


Galerkin's Method for solving an I. V. P.

    Suppose we wish to solve the initial value problem
(i)          [Graphics:Images/GalerkinMod_gr_65.gif],  
             [Graphics:Images/GalerkinMod_gr_66.gif]   over the interval  [Graphics:Images/GalerkinMod_gr_67.gif].  

We use the trial function  

(ii)         [Graphics:Images/GalerkinMod_gr_68.gif].  

There are [Graphics:Images/GalerkinMod_gr_69.gif] equations to solve   [Graphics:Images/GalerkinMod_gr_70.gif]   for  [Graphics:Images/GalerkinMod_gr_71.gif],  i.e.   

(iii)        [Graphics:Images/GalerkinMod_gr_72.gif]    for  [Graphics:Images/GalerkinMod_gr_73.gif].  

Remark.  For the solution of an I. V. P. we choose  [Graphics:Images/GalerkinMod_gr_74.gif].

Proof  Galerkin Method  

Computer Programs  Galerkin Method  


Example 2.  Solve   [Graphics:Images/GalerkinMod_gr_75.gif],   with the initial condition   [Graphics:Images/GalerkinMod_gr_76.gif].  
Solution 2.



Galerkin's Method for solving an a B. V. P.

    Suppose we wish to solve a boundary value problem over the interval  [Graphics:Images/GalerkinMod_gr_142.gif],  
(I)          [Graphics:Images/GalerkinMod_gr_143.gif],  

We define  [Graphics:Images/GalerkinMod_gr_145.gif]  and use the trial function  

(II)         [Graphics:Images/GalerkinMod_gr_146.gif].  

There are [Graphics:Images/GalerkinMod_gr_147.gif] equations to solve   [Graphics:Images/GalerkinMod_gr_148.gif]   for  [Graphics:Images/GalerkinMod_gr_149.gif],  i.e.

(III)        [Graphics:Images/GalerkinMod_gr_150.gif]    for  [Graphics:Images/GalerkinMod_gr_151.gif].  

Remark.  The functions  [Graphics:Images/GalerkinMod_gr_152.gif]  must all be chosen with the boundary properties

               [Graphics:Images/GalerkinMod_gr_153.gif]  and  [Graphics:Images/GalerkinMod_gr_154.gif]    for  [Graphics:Images/GalerkinMod_gr_155.gif].  

Proof  Galerkin Method  

Computer Programs  Galerkin Method  


Example 3.  Solve   [Graphics:Images/GalerkinMod_gr_156.gif].  
3 (a).   Use the boundary values  [Graphics:Images/GalerkinMod_gr_157.gif]  and  [Graphics:Images/GalerkinMod_gr_158.gif].   
3 (b).   Use the boundary values  [Graphics:Images/GalerkinMod_gr_159.gif]  and  [Graphics:Images/GalerkinMod_gr_160.gif].   
Solution 3 (a).
Solution 3 (b).


Example 4.  Solve   [Graphics:Images/GalerkinMod_gr_289.gif].  
4 (a).   Use the boundary values  [Graphics:Images/GalerkinMod_gr_290.gif]  and  [Graphics:Images/GalerkinMod_gr_291.gif].   
4 (b).   Use the boundary values  [Graphics:Images/GalerkinMod_gr_292.gif]  and  [Graphics:Images/GalerkinMod_gr_293.gif].   
Solution 4 (a).
Solution 4 (b).


Research Experience for Undergraduates

Galerkin Method  Internet hyperlinks to web sites and a bibliography of articles.  




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(c) John H. Mathews 2005