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

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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 (II)

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: Tensor method, Boundary value 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.

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

1 1 2 2

( ) ( ) ( ) ( ) ( ) ( )

b b

T T T T

aA U W  B U W dx a A U W  B 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 0

1 2 _{( , )} 1 2

(( ) ) b( ) T

HL a b a

F F W F F W dx

   

_

  2 2

0 0

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

( ) ( )

HL a b HL a b

F W a F W

 

   

So that the operator 2 0( , )

(FW)_{HL a b} is a linear operator. This is complete.

Theorem 3.1 Extend Riesz Representation Theorem

If a b_ij, _ij 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 K_i (1, 2,3, 4) have inverse operator.

Proof. Due to a bij, ij 0,then, the bilinear operators

















11 11 12 12

21 21 21 21

, ,

( 1, 2)

, ,

b b

i i i i

a a

b b

i i i i

a a

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

i

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

 _{ }_ _{ }_

 _



    





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are systemical, bounded and elliptic. According to Lax-Milgram theorem, then there exists linear inverse operatorsK K K₁, ₂, ₃and K₄such that

















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 inH a b1₀( , ). Therefore

2 0

1 1 2 1 1 2 2 2

( , )

3 1 4 1 3 2 4 2

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

( )

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

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

     __ _{ } __

 _ _

 

  ₍₈₎

Noting that

 





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.

Lemma 3.1

The operator K; defined by (9) is a linear operator.

Proof: Let ₁ ( , )₁ ₁ T

U  u v and ₂ ( ,₂ ₂)T

U  u v . Since K_i (1, 2,3, 4) are linear operators, then, for all  , R we have

1 1 1 2 2 1 2 2

1 2

3 1 3 2 4 1 4 2

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

K u K u K v K v

K U U

K u K u K v K v

   

 

   

  

 

 _{ } _

  

 

1 2

( ) ( )

K U K U

 

 

Hence, the operator K is a linear operator. This is complete.

Definition 3.2 Inverse operator

The operator K is called inverse, if there exists K1 such that

1 1

( ( )) ( ( ))

K K U K K U IU

i.e., K K1 KK1I, where I is the identity operator.

According to (7), we suppose that Ki are inverse operators. If K has left inverse 1

L

K , letting

1 2 1

3 4 ( )

L

K K u u

K K U

K K v v

   

 _     _

 

     

     

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1 3 2 4

( )( ) ( )

K K u K K u

K K u K K v

   

   

   



 _ _ _ _



i.e.,

1 3 2 4

, 0

) 0, )

K K I K K

K K K K I

   

   

   



 _ _ _ _

 ₍₁₀₎

where I is the identity operator and 0 is the zero operator. By right multiplying K₃1, we get

1 1 1

1 3 3 3 3

1 1

2 4 4 4 0

K K K K K

K K K K

 

  

 

  



 



i.e.,

1 1 1

1 3 2 4 3 (K K K K ) K

  _  _  If K K1 31 K K2 41 0

 _  _

, which is inverse, then

1

1 1 1 1 1 1 1 1

3 ( 1 3 2 4 ) ( 1 3 2 4 ) 3 ( 1 2 4 3)

K K K K K K K K K K K K K K

 _   _   __  _  _ _ _  

 

In a similar way, by right multiplying K₂1,K₃1, andK₄1, whenK K₃ ₁1K K₄ ₂10,

1 1

2 4 1 3 0

K K K K  ,K K4 21 K K3 11 0

 _  _ , we obtain

1

1 1 1 1 1 1 1 1

1 3 1 4 2 3 1 4 2 1 3 4 2 1

1

1 1 1 1 1 1 1 1

4 2 4 1 3 2 4 1 3 4 2 1 3 4

1

1 1 1 1 1 1 1 1

2 4 2 3 1 4 2 3 1 2 4 3 1 2

( ) ( ) ( )

K K K K K K K K K K K K K K









       



       



       

 _ _ __ _ _ _ _

 



 _ _ __ _ _ _ _

 _ _



 

   _  _  



If K has right inverse K_R1, letting

1 2 1 1 1

3 4 1 1 ( )

R

K K u u

KK U

K K v v

   

 _     _

         

 

then

1 1 2 1 1 1 2 1 3 1 4 1 3 1 4 1

( )( ) ( )

K K u K K u

K K u K K v

   

   

   



 _ _ _ _



So that

1 1 2 1 1 1 2 1 3 1 4 1 3 1 4 1

, 0,

0, ,

K K I K K

K K K K I

   

   

   



 _ _ _ _



where I is the identity operator and 0 is the zero operator.

If 1 1 1 1

2 1 4 3 0, 4 3 2 1 0,

K K K K  K K K K  1 1 1 1

1 2 3 4 0, 3 4 1 2 0,

K K K K  K K K K  by left

multiplying K₁1,K₂1,K₃1,K₄1,we obtain

1 1 1 1 1 1

1 2 1 4 3 2 1 2 4 3

1 1 1 1 1 1

1 4 3 2 1 4 3 4 2 1

1 1 1 1 1 1

1 1 2 3 4 1 2 1 3 4

1 1 1 1 1 1

1 3 4 1 2 3 4 3 1 2

( ) ( )

K K K K K K K K K

 

 

 

 

     

     



    

 

    



     



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1 1 1 1

1 1 1 1 2 4 3 3 4 2 1

1 1 1 1

2 1 3 4 4 3 1 2

( ) ( )

L R

K K K K K K K K

K K K

K K K K K K K K

   

  

   

   

  _{ } _

 

 

According to the above, we have the following theorem.

Theorem 3.2

If the operator K; defined by (9), has an inverse operator K-1; then there is a unique weak solution U such that

2 2

0( , ) 0( , )

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

a UW  K U W

Proof: If the operator K,defined by (9), has an inverse operator K-1; then there is a weak solution

U such that

2 2

0( , ) 0( , )

( ) ( ( ) )

HL a b HL a b

a UW  K U W

. If there exists other K-1 such that

2 2

0( , ) 1 0( , )

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

a UW  K U W

then

2 2

0( , ) 1 0( , )

( ( ) ) ( ( ) )

HL a b HL a b

K U W  K U W

It means that 2

0

1 ( , )

(( ( )K U K U( ))W)_{HL a b} 0.

By lemma (1), we know K U( )K U₁( )0, i.e., K U( )K U₁( ). This is complete.

Theorem 3.3 Existence of weak solutions

Let aij > 0 and bij > 0.

(1) If 11 12 12 22

a a

a a or

11 12 12 22

b b

b b , and functions fi (i = 1; 2) are continues, then there exists a

unique weak solution of system of ODE (3);

(2) If 11 12 11 12 12 22 12 22

a a b b

a  a b b and functions fi (i = 1; 2) are continues, and f1 f2, then there

exists several weak solutions of system of ODE (3);

(3) If 11 12 11 12 12 22 12 22

a a b b

a a b b and f1 f2, then there not exists weak solution of system of ODE

(3).

Proof: We only prove result (1). Since 11 12 12 22

a a

a  a or

11 12 12 22

b b

b b then, for all Ci (i =1; 2; 3; 4);

1 1 2, 1 2 3, 2 3 4, 2 4 4. K C K K C K K C K K C K

It means that

1 1 1 1

1 3 2 4 3 1 4 2

1 1 1 1

2 1 4 3 1 2 3 4 ,

,

K K K K K K K K

K K K K K K K K

   

  

 _ _



According to Lax-Milgram theorem, there exists a unique K such that

2 2

0 0

2 0 ( , ) ( , )

( ) ( ( ) ) , ( , )

HL a b HL a b

a UW  K U W  W HL a b

At the same time, for all linear bounded function ( ,₁ ₂) ,T i

F f f f are linear bounded functions, there exists a unique K such that

2 2

0 0

2 0 ( , ) ( , )

( ( )K u W)_{HL a b} (FW)_{HL a b} , W HL a b( , ),

where UK F1 .

Hence, there exists a unique U K F1 HL a b10( , )



  such that 2 2

0( , ) 0( , )

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

a UW  FW

This is complete.

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1 2

1 4 2 3 3 4

0

K K

K K K K K

K K

 

 _ _   

  .

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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[3] M. E. Taylor. Partial differential equations I: basic theory. New York: Springer-Verlag, 1996.

[4] M. Schetchter. An induction to nonlinear analysis. Cambridge University Press, 2004.