[SOLVED]Euler-Lagrange equation first integral question

dwsmith

Well-known member
$\int_0^1yy'dx$
where $$y(0) = 0$$ and $$y(1) = 0$$.
The first integral is
$f - y'\frac{\partial f}{\partial y'} = c.$
Using this, I get $$yy' - y'y = 0 = c$$ so ofcourse $$y(0)$$ and $$y(1)$$ equal $$0$$ then but is this correct?

It just seems odd.

M R

Active member
Re: E-L eq first integral question

$\int_0^1yy'dx$
where $$y(0) = 0$$ and $$y(1) = 0$$.
The first integral is
$f - y'\frac{\partial f}{\partial y'} = c.$
Using this, I get $$yy' - y'y = 0 = c$$ so ofcourse $$y(0)$$ and $$y(1)$$ equal $$0$$ then but is this correct?

It just seems odd.
Big edit: I'm not sure...

(edited again, as mentioned in post 10)

$$\displaystyle \int_0^1 (yy')dx=\frac{1}{2}\int_0^1 (y^2)' dx=\frac{y(1)^2-y(0)^2}{2}$$

So, with the conditions that $$\displaystyle y(0)=y(1)=0$$ you have $$\displaystyle \int_0^1yy'dx=0$$ and there isn't a unique solution.

Last edited:

dwsmith

Well-known member
Re: E-L eq first integral question

Big edit: I'm not sure...

$$\displaystyle \int_0^1 (yy')dx=\frac{1}{2}\int_0^1 (y^2)' dx=\frac{y(1)^2}{2}$$

So, with the condition that $$\displaystyle y(1)=0$$ you have $$\displaystyle \int_0^1yy'dx=0$$ and there isn't a unique solution.
I am not integrating the function per se. I am using the E-L eqs to determine the answer.

M R

Active member
Re: E-L eq first integral question

I am not integrating the function per se. I am using the E-L eqs to determine the answer.

The answer to what question though?

I had assumed you were trying to find a function for which the functional has a stationary value but in this case, $$\displaystyle \int yy' dx$$, the integral is zero for every function that satisfies the boundary conditions.

May I ask, what is the source of the question?

dwsmith

Well-known member
Re: E-L eq first integral question

The answer to what question though?

I had assumed you were trying to find a function for which the functional has a stationary value but in this case, $$\displaystyle \int yy' dx$$, the integral is zero for every function that satisfies the boundary conditions.

May I ask, what is the source of the question?
Shouldn't I be able to use the E-L first integral rule here? From class.

M R

Active member
Re: E-L eq first integral question

Shouldn't I be able to use the E-L first integral rule here? From class.
In this example the first integral tells you nothing about y.

Maybe that's why you were given this question to think about.

dwsmith

Well-known member
Re: E-L eq first integral question

In this example the first integral tells you nothing about y.

Maybe that's why you were given this question to think about.
So what I did was correct but it leads no where then, correct?

M R

Active member
Re: E-L eq first integral question

So what I did was correct but it leads no where then, correct?
I guess so but see my PM.

dwsmith

Well-known member
Re: E-L eq first integral question

I guess so but see my PM.
So the minimum and maximum of that problem would be both 0 then correct?

M R

Active member
Re: E-L eq first integral question

So the minimum and maximum of that problem would be both 0 then correct?
Yes. For every y(x) that satisfies y(0)=y(1)=0 the value of that integral is 0.

Going back to my first post it should have said $$\displaystyle \int_0^1 yy' dx=\frac{y(1)^2-y(0)^2}{2}$$

but since y(0)=0 it didn't really matter, but then y(1)=0 too so that didn't matter either.