Tensor Method to the Weak Solution of the Second Order System of Ordinary Differential Equations with Boundary Value Problems (I)

2016 International Conference on Mathematical, Computational and Statistical Sciences and Engineering (MCSSE 2016) ISBN: 978-1-60595-396-0

Tensor Method to the Weak Solution of the Second-Order System of

Ordinary Differential Equations with Boundary Value Problems (I)

Bing ZHANG

1

, Wei DENG

2

, Bo LIU

1

and Hai-chun LI

1,*

1_{College of Science, Shenyang Agricultural University, Shenyang, Liaoning, 110866, P. R. China} 2_{College of Automation, Shenyang Institute of Engineering, Shenyang, Liaoning, 110136, P. R.}

China

*Corresponding author

Keywords: Tenary valusor method, Bounde problem, System of ODE, Lax-Milgram theorem, Weak solution, Tensor operator, Inverse operator.

Abstract. In this paper, a tensor method for variational approach is established based on Lax-Milgram theorem. Some new results are obtained for the existence of weak solution of ODE system with homogeneous boundary value problems. By seeking the approximate solutions using Matlab, it is proved that the proposed method is effective to the system of ODE with boundary value problems in practice.

Introduction

Lax-Milgram theorem [1] is an efficient way to studying on ODE and its system with boundary value problems. By left multiplying for all W=(w1,w2), we can define the bilinear functional a(U,V), where U,V is in the same Hilbert space. Hence, the existence problem of weak solution of ODE can be proved depending on Lax-Milgram theorem. The purpose of this paper is to develop an algorithm by right multiplying for all based on Lax-Milgram theorem. For this motivate, we establish a new method for variational method by Lax-Milgram theorem, which is called tensor method. We obtain some new results for the existence of weak solution of the system of ODE with boundary value problems in theory.

Preliminaries and Notes

In the paper, the following preliminaries and notes are used for our results.

Theorem 2.1 Lax-Milgram theorem [1]

Let H be a real Hilbert space with the inner product ( , )  and the norm  . Assume that

:

B H H R is a bilinear operator, for which there exist constants  , 0 such that

( , ) , ,

B u v  u  v u vH; and  u 2 B u v( , ), u H. Finally, let f H: R be a bounded

linear functional on H. Then, there exists a unique element uH _{such that}

( , ) ( , )

B u v  f v vH_.

Theorem 2.1' Lax-Milgram theorem [2]

Let X be a Hilbert space. Let B(x,y) be a complex-valued functional defined on the product Hilbert space XX which satisfies the condiction:

(1) Bilinear, i.e,

1 1 2 2 1 1 2 2

( , ) ( , ) ( , )

B a x a x y B x y  B x y and B x a y( , _{1 1}a y_{2 2})₁B x y( , ₁)₂B x y( , ₂) (2) Boundedness, i.e., there exists a positive constant  such that B x y( , )  x  y

. (3) Ellipticity, i.e., there exists a positive constant  such that B x y( , )  x 2.

Definition 2.1 Sobolev space [1]





1 2

0( , ) , ( , ), ( ) ( ) 0

H a b  u u uL a b u a u b 

.

Definition 2.2 Sobolve space

1 2,1 1 1 2

0 0 0 0

( )

( , ) ( , ) ( , ) ( , ) , , , ( , )

( )

u t

HL a b HL a b H a b H a b u v u v L a b

v t

  _{ } 

   _{ }  _

 

 

With u a( )u b( )v a( )v b( )0.

2,2 2 2 2

0 0 0

( )

( , ) ( , ) ( , ) , , , , , ( , )

( )

u t

HL a b H a b H a b u v u v u v L a b

v t

  _{   } 

  _{ }  _

 

 

With u a( )u b( )v a( )v b( )u a'( )u b'( )v a'( )v b'( )0.

Definition 2.3 Tensor derivative

The vector ( ( ),g t g t_{1 2}( ))HL a b2₀( , ) is said to be the derivative of ( ( ),f t₁ f t₂( ))HL a b2₀( , ),

if, for any ( , )u v HL a b2₀( , ), the following formula holds:





1 2





1 2 1 2

1 2

b b

b _{a a} b

b b

a a

a a

ug dt ug dt

u u

g g dt f f dt

v _{vg dt vg dt} v

  _

  _{ }  

  

  _{ }  _

     

 













Definition 2.4 Tensor operator

Letting ( , ) , (u v T w w₁, ₂)THL a b2₀( , ), we define tensor operator

1 1

0 0

2 0

1 1

0 0

1 _{( , )} 2 _{( , )}

( , )

1 ( , ) 2 ( , )

( , ) ( , )

( ) ,

( , ) ( , )

H a b H a b

HL a b

H a b H a b

u w u w

U W

v w v w

 

 

 

 

  ₍₁₎

1 1 1 1

0 0 0 0

2 0

1 1 1 1

0 0 0 0

1 ( , ) 1 ( , ) 2 ( , ) 2 ( , )

( , )

1 _{( , )} 1 _{( , )} 2 _{( , )} 2 _{( , )}

( , ) ( , ) ( , ) ( , )

( )

( , ) ( , ) ( , ) ( , )

H a b H a b H a b H a b

HL a b

H a b H a b H a b H a b

a u w b v w a u w b v w

A U W

c u w d v w c u w d v w

 

 

 

 

   

  ₍₂₎

where a( , ), ( , ), ( , ), ( , )  b   c   d   H a b₀1( , ) are the bilinear functional. It is clearly that tensor operator (UV) is systemical.

Lemma 2.1 The continuous functions fi(t) (i=1,2) is bounded in [a,b]. For all

2

1 2 0

( , )T ( , ),

W  w w HL a b ( ) 0 0 0

0 0

T

FW _ _

  . Then( ( ),1 2( )) (0, 0) .

T T

f t f t 

Theorem 2.2 Extend Riesz Representation Theorem

There exists a linear operator K such that 2 2

0( , ) 0( , )

( ) ( ( ) ) ,

HL a b HL a b

a UW  K U W where the

tensor operator 2

0( , )

( )

HL a b

a UW defined by (2).

Proof: Since a( , ), ( , ), ( , ), ( , )  b   c   d   H a b₀1( , ) are the bilinear functional, then, by Lax-Milgram theorem, there exists linear operatorsK1, K2, K3, K4 such that

1 1 1 2 1

2 2 1 2 2

1 1 3 4 1

1 1 3 4 2

( , ) ( , ) ( ( ) ( ), )

( , ) ( , ) ( ( ) ( ), )

( , ) ( , ) ( ( ) ( ), )

( , ) ( , ) ( ( ) ( ), )

a u w b v w K u K v w

a u w b v w K u K v w

c u w d v w K u K v w

c u w d v w K u K v w

  



 _{  }



 _{  }



 _{  }

2 0

2 0

1 2 1 1 2 2 1 2 1

( , )

3 4 1 3 4 2 3 4 2 _{( , )}

( ( ) ( ), ) ( ( ) ( ), )

( )

( ( ) ( ), ) ( ( ) ( ), )

HL a b

hL a b

K u K v w K u K v w K K u w

a U W

K u K v w K u K v w K K v w

   

      

 _{ }_{  }_{ }

  _{   }

    

Letting operator 1 2

3 4

,

K K

K

K K

 

  

  such that

1 2 1 2

3 4 3 4

( ) ( )

( ) ( )

K K u K u K v

K K v K u K v



    



    _ 

 

   ,

then, there exists linear operator K such that 2 2

0( , ) 0( , )

( ) ( ( ) ) ,

HL a b HL a b

a UW  K U W This is

complete.

Main Results for the Second-order System of ODE with Boundary Value Problem

In the section, we, firstly, establish the tensor operators associated with the system

 

( ) , ,

( ) ( ) 0

AU BU F x a b

U a U b

 

   



 

 ₍₃₎ where A(a x_ij( ))_{2 2} ,B( ( ))b x_{ij 2 2} ,U ( , ) ,u v T F ( ( ),f t₁ f t₂( )), a x b x f t_ij( ), _ij( ), ( )_i are continues.

ChooseW ( ,w w_{1 2}) Right multiplying the first equation of the system (3) by WT, and integrating over [a,b],we get

( ( ) )

b _T b _T

a  AU BU W dx a FW dx





According to the integration by parts and the boundary conditions, we have

( ( ) )

b _{T T} b _T

a AU W BUW dx aFW dx

   





Now, we define tensor operator

2 0

2 2

0 0

( , )

( )_{HL a b} : ( , ) ( , ) ,

a UW HL a b HL a b M

given by

2 0( , )

( )_{HL a b}

a UW

=

1 2

3 4

( ( ) )

b _{T T}

a

M M

AU W BUW dx

M M

 

   _{  }

 



(4)

11 1 11 1 12 1 12 1 11 2 11 2 12 2 12 2

21 1 21 1 22 1 22 1 21 2 21 2 22 2 22 2

b

a

a u w b uw a v w b vw a u w b uw a v w b vw

a u w b uw a v w b vw a u w b uw a v w b vw

          

 

  _{          } 

 



and the linear tensor operator 2 0

2 0 ( , )

(FW)_{HL a b} :HL a b( , )M, described as

2 0

1 1 1 2

( , )

2 1 2 2

( )

b b

b _{a a}

T

HL a b a b b

a a

f w dx f w dx

F W FW dx

f w dx f w dx

 

 

  _{  }

 

 











₍₅₎

According to (4) and (5), then

2 2

0( , ) 0( , )

( ) ( ) .

HL a b HL a b

a UW  FW

(6) It is called the equivalent integral equation of the system of ODE with boundary value problem (3). It can be developed into Galerkin method for seeking the approximate solutions of the system of ODE with boundary value problem.

Definition 3.1 Weak solution of the system of ODE with boundary value problem

A function U x( )( ( ), ( ))u x v x THL a b2₀( , )is a weak solution of the system of ODE (3) provided

(1) 2 2

0 0

2 0

( , ) ( , )

( )_{HL a b} ( )_{HL a b} , , ( , )

a UW  FW U WHL a b , where 2

0( , )

( )_{HL a b}

a UW defined by (4),

and 2

0( , )

( )

HL a b

FW defined by (5).

(2) U a( )U b( )0, where U a( )( ( ), ( )) , ( )u a v a T U b ( ( ), ( )) .u b v b T

Lemma 3.1

The tensor operator 2 0( , )

( )

HL a b

a UW defined by (4) is a bilinear operator, and the tensor

operator 2

0( , )

(FW)_{HL a b} defined by (5) is a linear operator.

Proof. According to the definition (4), for all  , R, we know

2 0

1 2 _{( , )} 1 2 1 2

(( ) ) b ( ) ( T) ( ) T

HL a b _a

a U U W 



_A U U W B U U W _dx

2 2

0 0

1 _{( , )} 2 _{( , )}

( ) ( )

HL a b HL a b

a U W a U W

 

   

And

2 0

1 2 ( , ) 1 2 1 2

( ( ))_{H a b} b (( ) )T ( )T

a

a U W W 



_AU W W BU W W _dx

2 2

0 0

1 ( , ) 2 ( , )

( )_{HL a b} ( )_{HL a b}

a U W a U W

 

   

According to (5), we get

2 2 2

0 0 0

1 2 ( , ) 1 2 1 ( , ) ( , )

(( ) )_{HL a b} b( ) T ( )_{HL a b} ( )_{HL a b}

a

F F W F F W dx F W a F W

   



     

So that the operator 2 0( , )

( )

HL a b

FW is a linear operator. This is complete.

Theorem 3.1 Extend Riesz Representation Theorem

If aij, bij>0, then there is a linear operator

1 2 3 4 K K K K K     

 , such that

2 2

0( , ) 0( , )

( ) ( ( ) )

HL a b HL a b

a UW  K U W

where Ki (i=1,2,3,4) have inverse operator.

Proof. Due to aij, bij>0, and Lax-Milgram theorem, then, there exists linear inverse operators K1,K2,

K3 and K4, such that

















1 0 1 0 1 0 1 0

11 11 1 ( , )

12 12 2 ( , )

21 21 3 _{( , )}

22 22 4 ( , )

( ( ), ) ,

( ( ), ) ,

( ( ), ) ,

( ( ), ) ,

b

i i _{i H a b}

a b

i i _{i H a b}

a b

i i _{i H a b}

a b

i i _{i H a b}

a

a u w b uw dx K u w

a v w b vw dx K v w

a u w b uw dx K u w

a v w b vw dx K v w

 _{   }                _{   } 









₍₇₎

where ( , )  is the inner product in H a b₀1( , ).

Therefore,

2 0

1 1 2 1 1 2 2 2

( , )

3 1 4 1 3 1 4 1

( ( ), ) ( ( ), ) ( ( ), ) ( ( ), )

( )

( ( ), ) ( ( ), ) ( ( ), ) ( ( ), ) HL a b

K u w K v w K u w K v w

a U W

K u w K v w K u w K v w

     _{  }     2 0

1 2 1

3 4 2 _{( , )}

( ) ( )

( ) ( )

HL a b

K u K v w

K u K v w

 





2

0 2

0

1 2 1

( , )

3 4 2 _{( , )}

, HL a b HL a b

K K u w

K U W

K K v w

    

  

        

 

 

Then, there exists an operator

1 2

3 4

K K

K

K K

 

  

  ₍₉₎

such that 2 2

0( , ) 0( , )

( ) ( ( ) )

HL a b HL a b

a UW  K U W

.This is complete.

Acknowledgements

This work is supported by Shenyang Agricultural University Young Teachers Scientific Research Foundation (20131010); Liaoning Provincial Doctoral Foundation (20081064); Liaoning BaiQian Wan Talents Program (2009921072; 2010921067).

References

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