# Calculus of variations

• May 22nd 2010, 01:56 PM
bigdoggy
Calculus of variations
I have a problem relating to the extremal function for a surface of revolution.

The Euler-Lagrange equation is ${d \over dx}({\partial F \over \partial y'})- {\partial F \over \partial y}=0.$

Surface area of revolution = $\int {2\pi y \sqrt{1+(y')^2}}dx$

So $F(x,y,y') = {2\pi y \sqrt{1+(y')^2}}dx$ [eqn. (1)] is a function of y and y' only and so the Euler-Lagrange equation is $F-y'{\partial f \over \partial y'}=0$ because x is absent from F.

But in a similar problem $F(x,y,y')= (y')^2+yy'+y^2$ [eqn. (2)] and the E-L equation is stated as ${d \over dx}({\partial F \over \partial y'})- {\partial F \over \partial y}=0$ because x is not absent from F....

why is eqn. (1) independent of x but eqn. (2) isn't !?
• May 22nd 2010, 03:01 PM
mr fantastic
Quote:

Originally Posted by bigdoggy
I have a problem relating to the extremal function for a surface of revolution.

The Euler-Lagrange equation is ${d \over dx}({\partial F \over \partial y'})- {\partial F \over \partial y}=0.$

Surface area of revolution = $\int {2\pi y \sqrt{1+(y')^2}}dx$

So $F(x,y,y') = {2\pi y \sqrt{1+(y')^2}}dx$ [eqn. (1)] is a function of y and y' only and so the Euler-Lagrange equation is $F-y'{\partial f \over \partial y'}=0$ because x is absent from F. Mr F says: This is wrong. It should be ${\color{red} F-y'{\partial {\color{blue}F} \over \partial y'}= {\color{blue}C}}$ where C is a constant. This is a first integral of the Euler-Lagrange equation.

But in a similar problem $F(x,y,y')= (y')^2+yy'+y^2$ [eqn. (2)] and the E-L equation is stated as ${d \over dx}({\partial F \over \partial y'})- {\partial F \over \partial y}=0$ because x is not absent from F....

why is eqn. (1) independent of x but eqn. (2) isn't !?

As posted, both equations obviously are explicitly independent of x. Perhaps there is a typo in the second equation ....?
• May 22nd 2010, 03:13 PM
bigdoggy
Thanks for the reply Mr Fantastic.

Thanks for pointing out the error in red, but there is no typo with regards to the second equation:
the boundary conditions are y(0)=0 and y(1)=1 and the extremal function is $y(x)={sinhx \over sinh1}$ am I missing something or making an error?
• May 22nd 2010, 03:48 PM
mr fantastic
Quote:

Originally Posted by bigdoggy
Thanks for the reply Mr Fantastic.

Thanks for pointing out the error in red, but there is no typo with regards to the second equation:
the boundary conditions are y(0)=0 and y(1)=1 and the extremal function is $y(x)={sinhx \over sinh1}$ am I missing something or making an error?

Clearly both equations can be used.

Using ${d \over dx}({\partial F \over \partial y'})- {\partial F \over \partial y}=0$ leads to solving $\frac{d^2 y}{dx^2} - y = 0$ subject to the given boundary conditions.

Using $F-y'{\partial F \over \partial y'}=C$ leads to solving $y^2 - \left( \frac{dy}{dx}\right)^2 = C$ subject to the given boundary conditions.

Both differential equations should lead to the same answer. You can decide which one you'd rather solve .... (Clearly the book decided the former was easier to solve but have not explained this).
• May 22nd 2010, 04:31 PM
bigdoggy
Thanks, but how could I solve
$

y^2 - \left( \frac{dy}{dx}\right)^2 = C
$
• May 22nd 2010, 07:51 PM
mr fantastic
Quote:

Originally Posted by bigdoggy
Thanks, but how could I solve
$

y^2 - \left( \frac{dy}{dx}\right)^2 = C
$

Make dy/dx the subject etc.