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Question: Determination of the Lagrange multiplier for variational iteration method

Question:Determination of the Lagrange multiplier for variational iteration method

farjad_etg 10 Maple
integration-by-parts lagrange-multipliers
May 25 2017
0
Hi, for determination of the Lagrange multiplier λ(ξ ), I'm trying to use of integration by parts:
                     

To obtain the first integral (first order respet to u) , it's simple.

But for the socond integral (second order respect to u), I could not find a direct solution and I had to solve by multi-steps.
For high orders,this can be  frustrating and time-consuming  
Is there a simpler solution ??

Here is my code:


> 

restart:
> 

with(IntegrationTools):

For firs order:

 

 

> 

Order1_main:=Int(lambda(s)*diff(u(s),s),s)

Int(lambda(s)*(diff(u(s), s)), s)

 (1)
> 

Order1_ans:=Parts(Order1_main, lambda(s))

u(s)*lambda(s)-(Int(u(s)*(diff(lambda(s), s)), s))

 (2)
> 

NULL
 

 

And for second order:

 

 

> 

Order2_main:=Int(lambda(s)*diff(u(s),s,s),s)

Int(lambda(s)*(diff(diff(u(s), s), s)), s)

 (3)
> 

Order2_parts:=Parts(Order2_main, lambda(s))

lambda(s)*(diff(u(s), s))-(Int((diff(u(s), s))*(diff(lambda(s), s)), s))

 (4)

By copy and past the  tow parts of above equaion :

 

> 

Order2_p1:=lambda(s)*(diff(u(s), s));
Order2_p2:=Int((diff(u(s), s))*(diff(lambda(s), s)), s)

lambda(s)*(diff(u(s), s))

  

Int((diff(u(s), s))*(diff(lambda(s), s)), s)

 (5)

 

> 

Order2_p2_ans:=Parts(Order2_p2,diff(lambda(s),s))

u(s)*(diff(lambda(s), s))-(Int(u(s)*(diff(diff(lambda(s), s), s)), s))

 (6)
> 

 

> 

NULLOrder2_ans := Order2_p1-Order2_p2_ans;

lambda(s)*(diff(u(s), s))-u(s)*(diff(lambda(s), s))+Int(u(s)*(diff(diff(lambda(s), s), s)), s)

 (7)
> 

NULL
 

 

> 

 

 

> 

Order3_main:=Int(lambda(s)*diff(u(s),s,s,s),s)

Int(lambda(s)*(diff(diff(diff(u(s), s), s), s)), s)

 (8)
> 

``

 

 

 

> 

``

Download _Integration_by_parts.mw

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Hear is an example :

Is that possible to get equation.4 in the following example by Maple?

 
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