- #1
daudaudaudau
- 302
- 0
If we have a functional
[tex]
J(y)=\int L(y,y',x)dx
[/tex]
then the first variation is
[tex]
\delta J=\int\left(\frac{\partial L}{\partial y}\eta(x)+\frac{\partial L}{\partial y'}\eta'(x)\right)dx,
[/tex]
where [itex]\eta(x)[/itex] is the variation of the stationary solution. Now, if [itex]L[/itex] is independent of [itex]y(x)[/itex], then [itex]\frac{\partial L}{\partial y}=0[/itex] and I have
[tex]
\delta J=\int\left(\frac{\partial L}{\partial y'}\eta'(x)\right)dx.
[/tex]
At this point, why can't I simply say that [itex]\eta'(x)[/itex] is arbitrary and hence the E-L equation is [itex]\frac{\partial L}{\partial y'}=0[/itex] ?
[tex]
J(y)=\int L(y,y',x)dx
[/tex]
then the first variation is
[tex]
\delta J=\int\left(\frac{\partial L}{\partial y}\eta(x)+\frac{\partial L}{\partial y'}\eta'(x)\right)dx,
[/tex]
where [itex]\eta(x)[/itex] is the variation of the stationary solution. Now, if [itex]L[/itex] is independent of [itex]y(x)[/itex], then [itex]\frac{\partial L}{\partial y}=0[/itex] and I have
[tex]
\delta J=\int\left(\frac{\partial L}{\partial y'}\eta'(x)\right)dx.
[/tex]
At this point, why can't I simply say that [itex]\eta'(x)[/itex] is arbitrary and hence the E-L equation is [itex]\frac{\partial L}{\partial y'}=0[/itex] ?