Module

for

Galerkin's Method

 

Background

    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  

                [Graphics:Images/GalerkinMod_gr_3.gif].  

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].  

     [Graphics:Images/GalerkinMod_gr_5.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  
    
            [Graphics:Images/GalerkinMod_gr_35.gif].  

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],  
             with
             [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],  
              with
              [Graphics:Images/GalerkinMod_gr_144.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.  

 

 

 

Download this Mathematica Notebook Galerkin's Method

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2005