MHB Euler-Lagrange equation first integral question

[Dustinsfl]

\[
\int_0^1yy'dx
\]
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]

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)

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

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

 

[Dustinsfl]

Re: E-L eq first integral question

Big edit: I'm not sure...

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

So, with the condition that $$y(1)=0$$ you have $$\int_0^1yy'dx=0$$ 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]

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, $$\int yy' dx$$, the integral is zero for every function that satisfies the boundary conditions.

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

 

[Dustinsfl]

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, $$\int yy' dx$$, 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]

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. :)

 

[Dustinsfl]

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]

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. :)

 

[Dustinsfl]

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]

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

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