Calculus of Variations problem

Sep 2006
782
100
The raggedy edge.
Find solutions of the Euler-Lagrange equations for critical points of the following functionals with the given endpoint conditions.

\(\displaystyle I(y)=\int_1^2 x(y')^2~dx\) where \(\displaystyle y(1)=1\) and \(\displaystyle y(2)=2\)
So I decided to use the Euler Lagrange equations:

\(\displaystyle \frac{\partial f}{\partial y}=0\) and \(\displaystyle \frac{\partial f}{\partial y'}=2(y')x\)

Inputting this into the Euler-Lagrange equations gives:

\(\displaystyle -\frac{d}{dx}(2y'x)=0 \ \Rightarrow -2y'=0 \Rightarrow y'=0\)


So integrating both sides w.r.t x gives \(\displaystyle y=A\) where \(\displaystyle A\) is a positive constant.

From here I was expecting to use the conditions on y to work out what A is. Unfortunately this doesn't work! y can't equal a constant and yet change from 1 to 2 as described in the conditions.

What's going wrong?
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
No , we should take total differential not partial ...

It is

\(\displaystyle \frac{d}{dx}( 2xy' ) = 0 \)

Actually , we obtain \(\displaystyle 2xy' = c \) immediately but in other situations , we should be careful that what we are doing is taking total differential .

\(\displaystyle \frac{d}{dx}( 2xy' ) = 2 \frac{d}{dx}( xy' ) \)

\(\displaystyle = 2 [\frac{\partial }{\partial x}(xy') \frac{dx}{dx} + \frac{\partial }{\partial y'}(xy') \frac{dy'}{dx}] \)

\(\displaystyle = 2 ( y' + x y'' ) \)
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
So I decided to use the Euler Lagrange equations:

\(\displaystyle \frac{\partial f}{\partial y}=0\) and \(\displaystyle \frac{\partial f}{\partial y'}=2(y')x\)

Inputting this into the Euler-Lagrange equations gives:

\(\displaystyle -\frac{d}{dx}(2y'x)=0 \ \Rightarrow -2y'=0 \Rightarrow y'=0\)


So integrating both sides w.r.t x gives \(\displaystyle y=A\) where \(\displaystyle A\) is a positive constant.

From here I was expecting to use the conditions on y to work out what A is. Unfortunately this doesn't work! y can't equal a constant and yet change from 1 to 2 as described in the conditions.

What's going wrong?
Just because a functional has an extremal does not mean that it is actually a minimum or maximum, just like how the first derivative equals zero does not imply you have a maximum there. In fact, the E-L equations can be thought very informally as an infinite dimensional analogue of a gradient.
 
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