# Euler-Lagrange First Integral

1. Jul 26, 2014

### kq6up

1. The problem statement, all variables and given/known data

6.20 ** If you haven't done it, take a look at Problem 6.10. Here is a second situation in which you can find a "first integral" of the Euler—Lagrange equation: Argue that if it happens that the integrand f (y, y', x) does not depend explicitly on x, that is, f = f (y, y'), then

$\frac { df }{ dx } =\frac { \partial f }{ \partial y } y^{ \prime }+\frac { \partial f }{ \partial y^{ \prime } } y^{ \prime \prime }$

Use the Euler—Lagrange equation to replace $\frac { \partial f }{ \partial y }$ on the right, and hence show that
$\frac { df }{ dx } =\frac { d }{ dx } \left( y^{ \prime }\frac { \partial f }{ \partial y^{ \prime } } \right)$
This gives you the first integral

$f-y^{ \prime }\frac { \partial f }{ \partial y^{ \prime } } =const$

This can simplify several calculations. (See Problems 6.21 and 6.22 for examples.) In Lagrangian mechanics, where the independent variable is the time t, the corresponding result is that if the Lagrangian function is independent of t, then energy is conserved. (See Section 7.8.)

For reference here is 6.10:
6.10 * In general the integrand f (y, y', x) whose integral we wish to minimize depends on y, y', and x. There is a considerable simplification if f happens to be independent of y, that is, f = f (y', x). (This happened in both Examples 6.1 and 6.2, though in the latter the roles of x and y were interchanged.) Prove that when this happens, the Euler—Lagrange equation (6.13) reduces to the statement that
$\frac { \partial f }{ \partial y^{ \prime } } =const$ (6.42)
Since this is a first-order differential equation for y(x), while the Euler—Lagrange equation is generally second order, this is an important simplification and the result (6.42) is sometimes called a first integral of the Euler—Lagrange equation. In Lagrangian mechanics we'll see that this simplification arises when a component of momentum is conserved.

2. Relevant equations

Most are above, save the Euler-Lagrange. Here it is: $\frac { \partial f }{ \partial y } -\frac { d }{ dx } \frac { \partial f }{ \partial y^{\prime} }=0$

3. The attempt at a solution

All the steps of this problem are easy save one.

The first equation listed is just a total derivative of f.

When making the suggested substitution with the E-L equation I get:

$\frac { df }{ dx } =\frac { d }{ dx } \left( y^{ \prime }\frac { \partial f }{ \partial y^{ \prime } } \right)+\frac { \partial f }{ \partial y^{ \prime } } y^{ \prime \prime }$. This leads me to believe that last term on the right is equal to zero when compared to the equation I was supposed to get shown above. However, I have no clue how to show that. The last step to the QED is just a straightforward application of an integral. No problem there. Hints would be appreciated.

Thanks,
Chris

2. Jul 26, 2014

### TSny

You didn't quite make the substitution correctly. In the first equation above, you want to make a substitution for just $\frac { \partial f }{ \partial y }$.

3. Jul 27, 2014

### kq6up

I believe I did. I simply inserted the left side of this equation into the first equation in your quote $\frac { \partial f }{ \partial y }=\frac { d }{ dx } \frac { \partial f }{ \partial y^{\prime} }$.

Thanks,
Chris

4. Jul 27, 2014

### TSny

If you substitute $\frac { \partial f }{ \partial y } = \frac { d }{ dx } \frac { \partial f }{ \partial y^{ \prime } }$ into $\frac { df }{ dx } =\left(\frac { \partial f }{ \partial y }\right) y^{ \prime }+\frac { \partial f }{ \partial y^{ \prime } } y^{ \prime \prime }$,

I don't see how you get

$\frac { df }{ dx } =\frac { d }{ dx } \left( y^{ \prime }\frac { \partial f }{ \partial y^{ \prime } } \right)+\frac { \partial f }{ \partial y^{ \prime } } y^{ \prime \prime }$.

In particular, how did you get the $\frac{d}{dx}$ operator to act on the product of $y^{\prime}$ and $\frac { \partial f }{ \partial y^{ \prime } }$?
Shouldn't $\frac{d}{dx}$ act on just $\frac { \partial f }{ \partial y^{ \prime } }$?

5. Jul 27, 2014

### kq6up

I see your point. Let me look at it again.

Chris

6. Jul 28, 2014

### kq6up

Thanks for the tip. Now I see with that parenthesis the book added, it expands into the equation I had using the product rule. Man, I don't know why I missed that.

Thanks,
Chris

7. Jul 28, 2014

OK, good.