Given <mi mathvariant="normal">&#x03A9;<!-- Ω --> := [ x a </msub> ,

istremage8o

istremage8o

Answered question

2022-05-25

Given Ω := [ x a , x b ] × [ y a , y b ] ⊂ R 2 , consider the problem
max f ∈ C 2 ( Ω , R ) ∬ Ω [ f y ( x , y ) ∫ x a x f ( z , y ) d z ] d x d y
subject to a constraint of the form
∬ Ω [ ∫ x a x f y ( z , y ) d z − ∫ x a x b z f y ( z , y ) d z ] 2 d x d y = c ∈ R
where the subscript y denotes the partial derivative w.r.t. the second variable, i.e., y.

I thought the first step would have been to write down the Euler-Lagrange equations, i.e., given the Langragian L ,
∂ ∂ y ( ∂ L ∂ f y ) = ∂ L ∂ f
but the Lagrangian is pretty involved and I got stuck. Do you have any hint?

Answer & Explanation

bgu999dq

bgu999dq

Beginner2022-05-26Added 9 answers

You can re-write:
F ( x , y ) = ∫ x a x f ( z , y ) d z
It is easy to check that F x ( x , y ) = f ( x , y ). Your problem becomes:
∫ ∫ Ω F x y ( x , y ) F ( x , y ) d x d y
It seems much harder to deal with the restriction when there is one. To demonstrate that, you can utilize differentiation under the integral sign.
∫ x a x f y ( z , y ) d z = F y ( x , y )
As for the second term, you can use the integration by parts formula, which is (for F ′ = f):
∫ a b x f ( x ) d x = x F ( x ) | a b − ∫ a b F ( x ) d x
Now we can get rid of that "annoying" z in front of f y :
∫ x a x b z f y ( z , y ) d z = ∫ x a x b z F z y ( z , y ) d z = z F y ( z , y ) | x a x b − ∫ x a x b F y ( z , y ) d z
Now that everything appears to be going well, your constraint is:
∫ ∫ Ω ( F y ( x , y ) − z F y ( z , y ) | x a x b + ∫ x a x b F y ( z , y ) d z ) 2 d x d y = c
It might be possible to simplify the constrain even more by using Fubini, since all of the terms seem to involve a ∂ y , but that is a really lengthy calculation.

Now use that your function f ∈ C 2 , which means that F ∈ C 2 . It also makes exchanging the order of derivatives possible. Now you may discover the appropriate PDE by using higher-order Euler-Lagrange equations.
Your Euler-Lagrange equation is now (I purposefully used different notations):
F x y − ∂ x y F = 0
which is a Null-Lagrangian. That means it holds trivially for a all smooth functions. Now you just have to deal with the constraint. Note, however, you have only found a a candidate for a critical point so far.

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