Euler Lagrange formula with higher derivatives

[Somali_Physicist]

I was trying to Extrapolate Eulers formula , after deriing the basic form I wanted to prove:
∂F/∂y - d(∂F/∂y_x)/dx +d[SUP]2[/SUP](∂F/∂y_xx)/dx² = 0

Here is my attempt but I get different answers:
J(y) = ∫_a^bF(x,y_x,y,y_xx)dx
δ(ε) = J(y+εη(x))
y = y_t+εη(x)
∂y/∂ε = η(x)
∂y_x/∂ε = η^⋅(x)
∂y_xx/∂ε=η^⋅⋅(x)(dots are derivatives with respect to x)
dδ(ε)/dε = ∫_a^b∂F/∂y ∂y/∂ε + ∂F/∂y_x ∂y_x/∂ε + ∂F/∂y_xx ∂y_xx/∂ε dx
∴ dδ(ε)/dε = ∫_a^b∂F/∂y η(x) + ∂F/∂y_x η^⋅(x) + ∂F/∂y_xx η^⋅⋅ (x)dx
Using integration by parts :
∫_a^b∂F/∂y_x η^⋅(x)dx = n(x)∂F/∂y_x |_a^b-∫_a^bd(∂F/∂y_x)η(x)dx

n(x)∂F/∂y_x |_a^b = 0
∴
∫_a^b∂F/∂y_x η^⋅(x)dx = -∫_a^bd(∂F/∂y_x)η(x)dx
Again use IBP:
∫_a^b∂F/∂y_xx η^⋅⋅ (x)dx = ∂F/∂y_xxη^⋅(x)|_a^b - ∫_a^bd(∂F/∂y_xx)/dxη^⋅(x)dx

∫_a^bd(∂F/∂y_xx)/dxη^⋅(x)dx = n(x)d(∂F/∂y_xx)/dx |_a^b - ∫_a^bd²(∂F/∂y_xx)/dx²η(x)dx

n(x)d(∂F/∂y_xx)/dx |_a^b = 0
hence
∫_a^b∂F/∂y_xx η^⋅⋅ (x)dx = ∂F/∂y_xxη^⋅(x)|_a^b + ∫_a^bd²(∂F/∂y_xx)/dx²η(x)dx

dδ(ε)/dε |_ε=0 = dδ(ε)/dε = ∫_a^b∂F/∂y_t η(x) -d(∂F/∂y_tx)η(x)dx+ d²(∂F/∂y_txx)/dx²η(x)dx + ∂F/∂y_xxη^⋅(x)|_a^b = 0

Now this is where I am stuck , because I can intuitively derive the next step:
∂F/∂y_xxη^⋅(x)|_a^b =0 to give
(∂F/∂y - d(∂F/∂y_x)/dx +d[SUP]2[/SUP](∂F/∂y_xx)/dx²)η(x) = 0
∴
as η(x) ≠ 0 (contradiction)
∂F/∂y - d(∂F/∂y_x)/dx +d[SUP]2[/SUP](∂F/∂y_xx)/dx² = 0Note assuming y(a,b) , y_x(a,b) are fixed and differentiable on intervel

 

[Orodruin]

I am not sure I understand what your question is ...

 

[Somali_Physicist]

I am not sure I understand what your question is ...

Is
∂F/∂y_xxη^⋅(x)|_a^b =0 ??
and if so why and a mathematical explanation would be the best.

 

[Orodruin]

If you are just considering variations of functions that are necessarily satisfying boundary conditions such as ##y'(a) = k##, your variations must satisfy ##\eta'(a) = 0##. The argument is the same as that leading to ##\eta(a) = 0##.

However, in general, this may not be the case. The point is that this does not affect the EL equations! In order for ##y(x)## to be a stationary function of the functional under consideration, you must have ##\delta J = 0## for all variations, including those that satisfy ##\eta'(a) = 0##. The only thing that changes when you do not already have the boundary conditions is that, in addition to the EL equations required to satisfy ##\delta J = 0## for those variations that satisfy ##\eta'(a) = 0##, you will also need to consider variations for which ##\eta'(a) \neq 0##. This leads you to an additional boundary condition that ##y## has to satisfy.

Unfortunately, this is a point that is often presented in a manner that makes it seem as if we must require ##\eta(a) = 0## in order to derive the EL equations. This is not true. We derive the EL equations by considering those variations in particular, but if other variations are also allowed, they will give us additional constraints by imposing boundary conditions (such boundary conditions are often referred to as natural boundary conditions).

A prime example of this with higher derivatives is the description of an elastic beam under load. If you fix an end point of the beam you impose ##y(a) = y_0##, which only allows variations that satisfy ##\eta(a) = 0##. If you fix the inclination of the beam at the boundary you fix ##y'(a) = k_0## and only allow variations that satisfy ##\eta'(a) = 0##. You can impose one of those, both, or none, depending on your physical situation. If you do not impose them, you will obtain different natural boundary conditions. In the end, these issues are related to the force and bending moment at the beam endpoint. This is discussed in some detail in section 8.2.2 and problem 8.12 of my book.

 

[Somali_Physicist]

If you are just considering variations of functions that are necessarily satisfying boundary conditions such as ##y'(a) = k##, your variations must satisfy ##\eta'(a) = 0##. The argument is the same as that leading to ##\eta(a) = 0##.

However, in general, this may not be the case. The point is that this does not affect the EL equations! In order for ##y(x)## to be a stationary function of the functional under consideration, you must have ##\delta J = 0## for all variations, including those that satisfy ##\eta'(a) = 0##. The only thing that changes when you do not already have the boundary conditions is that, in addition to the EL equations required to satisfy ##\delta J = 0## for those variations that satisfy ##\eta'(a) = 0##, you will also need to consider variations for which ##\eta'(a) \neq 0##. This leads you to an additional boundary condition that ##y## has to satisfy.

Unfortunately, this is a point that is often presented in a manner that makes it seem as if we must require ##\eta(a) = 0## in order to derive the EL equations. This is not true. We derive the EL equations by considering those variations in particular, but if other variations are also allowed, they will give us additional constraints by imposing boundary conditions (such boundary conditions are often referred to as natural boundary conditions).

A prime example of this with higher derivatives is the description of an elastic beam under load. If you fix an end point of the beam you impose ##y(a) = y_0##, which only allows variations that satisfy ##\eta(a) = 0##. If you fix the inclination of the beam at the boundary you fix ##y'(a) = k_0## and only allow variations that satisfy ##\eta'(a) = 0##. You can impose one of those, both, or none, depending on your physical situation. If you do not impose them, you will obtain different natural boundary conditions. In the end, these issues are related to the force and bending moment at the beam endpoint. This is discussed in some detail in section 8.2.2 and problem 8.12 of my book.

Oh so it must work for all situations including η^⋅(x) = 0and hence I use it.Furthermore it being a constant,ie not 0 , is simply because I didn't change the boundary to a specific length.That is Looking at the cases , where η^(m)(x) ≠ 0 : m ∈ R , its because my boundary conditions weren't properly accounted for, correct?

 

[Orodruin]

That is Looking at the cases , where η(m)(x) ≠ 0 : m ∈ R , its because my boundary conditions weren't properly accounted for, correct?

It is unclear what you mean by this. You need to look at those cases when the boundary conditions are not imposed from the beginning, such as by fixing the endpoints. In those cases, finding the stationary functions involve satisfying the requirement ##\delta J = 0## even for variations that have non-zero values at the boundary, which leads to boundary conditions on the stationary function itself. For the purposes of just deriving the EL equations themselves, you can assume that all boundary terms vanish since ##\delta J## must be zero particularly for variations that vanish at the boundaries.

 

[Somali_Physicist]

It is unclear what you mean by this. You need to look at those cases when the boundary conditions are not imposed from the beginning, such as by fixing the endpoints. In those cases, finding the stationary functions involve satisfying the requirement ##\delta J = 0## even for variations that have non-zero values at the boundary, which leads to boundary conditions on the stationary function itself. For the purposes of just deriving the EL equations themselves, you can assume that all boundary terms vanish since ##\delta J## must be zero particularly for variations that vanish at the boundaries.

So its like two doors leading to the same room?It all eventually gets to the correct solution?

 

[Orodruin]

Actually, it is only one door; considering the variations for which ##\eta(a)## (or ##\eta'(a)##) is zero. The question on top of that, if other variations are allowed, is whether or not additionally considering those other variations lead to further requirements - with the result being that they do.

 

[Somali_Physicist]

Actually, it is only one door; considering the variations for which ##\eta(a)## (or ##\eta'(a)##) is zero. The question on top of that, if other variations are allowed, is whether or not additionally considering those other variations lead to further requirements - with the result being that they do.

Ok so to recap.Are you stating that we impose conditions to force it to follow the Euler-Langrange theorem or that regardless of the boundary conditions the theorem works.This is the crux of my question I guess.From what I can garner from your replies I believe you are implying that regardless of the conditions the theorem works to allow ΔJ=0.Im sorry if I'm not understanding it , it just seems counter intuitive as what if you had a function such as e^x if your bounds go on forever ...

 

[Orodruin]

Regardless of whether you put the boundary conditions, the EL equations will need to be satisfied in order for the variation of ##J## to be equal to zero, because ##\delta J## needs to be zero in particular for those variations that do not change the boundary values.

However, if you do not impose any boundary conditions, you will get the boundary conditions that ##y## needs to satisfy out of the variation by also considering those variations that do change the boundary values. In the end, you end up with a differential equation for ##y## and a sufficient set of boundary conditions, whether imposed from the beginning or resulting from the variation itself.

 

[Somali_Physicist]

Regardless of whether you put the boundary conditions, the EL equations will need to be satisfied in order for the variation of ##J## to be equal to zero, because ##\delta J## needs to be zero in particular for those variations that do not change the boundary values.

However, if you do not impose any boundary conditions, you will get the boundary conditions that ##y## needs to satisfy out of the variation by also considering those variations that do change the boundary values. In the end, you end up with a differential equation for ##y## and a sufficient set of boundary conditions, whether imposed from the beginning or resulting from the variation itself.

ok got it! thanks