Calculus of Variations: The Euler equation becomes an indentity

Question

I'm trying to solve a problem when the $F$ function depends linearly on $x′$ alone. e.g:

$$\int_0^1 (3+ 2x') \,dt$$

With boundary conditions $x_{(0)} = 0$ and $x_{(1)} = 1$. Since:

$$F= (3+ 2x')$$ and: $F_{x}=0$ ; $F_{x'} = 2$

The Euler equation: $F_{x} = \frac{d}{dt}F_{x'}$ Can be written as: $0=0$

My question is How do I find a solution if the Euler equation becomes an indentity? I am interesed in a definite solution for $x$ in a general case e.g. $F=a+bx'$

Thanks.

Answer 1

By FTC the functional can be written $$3+2(x_1-x_0) =5$$ so it makes perfect sense that there is no content to your Euler Lagrange equations. Changing the function does nothing.

Answer 2

Just integrate the objective functional: $$\int_0^1[3+2x'(t)]{\rm d}t=[3t+2x(t)]_0^1=[3+2x(1)]-2x(0)=5.$$ This has to hold for all feasible solutions. Since all of the yield the same value, they are all optimal.