Calculus of Variations: Functional is product of 2 integrals

[anf3]

Homework Statement

Minimize the functional: ∫₀¹ dx y'²⋅ ∫₀¹ dx(y(x)+1) with y(0)=0, y(1)=a

Homework Equations

(1) δI=∫ dx [∂f/∂y δy +∂f/∂y' δy']
(2) δy'=d/dx(δy)
(3) ∫ dx ∂f/∂y' δy' = δy ∂f/∂y' |₀¹ - ∫ dx d/dx(∂f/∂y') δy
where the first term goes to zero since there is no variation at the endpoints

The Attempt at a Solution

Using (1) to get
δI= ∫dx 2y'δy' ∫dx(y+1) + ∫dx y'² ∫dx δy
Using (2) and (3) to get
δI= ∫dx y'² ∫dx δy - ∫dx d/dx(2y')δy ∫dx(y+1) = 0
I then tried various ways of manipulating this equation to something that I could minimize trying to isolate δy but was unable to do so since it was always within an integral. I tried to simplify d/dx(2 y') to 2y'' to see if that helped but could not find any benefit to doing so. I also thought about using an integral by parts again but couldn't see where to do so. Using the values at the end points seems like something I should be doing but I don't know how I would. Any advice on how to proceed?

 

[stevendaryl]

You're on the right track. You have:

\int_0^1 dx (y')^2 \int_0^1 \delta y dx - \int_0^1 dx 2y'' \delta y\int_0^1 (y+1) dx = 0

(Yes, it is better to write \frac{d}{dx} 2 y' as 2 y''.)

At this point, let's just name two integrals:
I_1 = \int_0^1 dx (y')^2

I_2 = \int_0^1 dx (y+1)

So we have:
I_1 \int_0^1 \delta y dx - I_2 \int_0^1 dx 2y'' \delta y = 0

Now, the important thing to recognize is that I_1 and I_2 are not functions of x (since x has been integrated out). So you can treat them as if they were constants, and you can combine the two integrals:

\int_0^1 \delta y (I_1 - I_2 2y'') dx = 0

 

[anf3]

Now, the important thing to recognize is that I_1 and I_2 are not functions of x (since x has been integrated out). So you can treat them as if they were constants, and you can combine the two integrals:

\int_0^1 \delta y (I_1 - I_2 2y'') dx = 0

Thank you! Yes, the integrals becoming constants after being evaluated was what I was failing to think about.