- #1
phi1123
- 9
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Simply put, can you find the function which extremizes the integral
[tex]J[f]=\iint L\left(x,y,f(x),f(y),f'(x),f'(y)\right) \,dx \,dy[/tex]
Where ##f## is the function to be extremized, and ##x## and ##y## are independent variables? A result seems possible by using the usual calculus of variation technique of finding
[tex]\left.\frac{J[f+\epsilon \eta]}{d\epsilon}\right |_{\epsilon=0}[/tex]
treating ##f(x)## and ##f(y)## as two different functions, and setting this to zero since ##J## is extremized. My (rather sketchy) calculations seem to give the differential equations:
[itex]\frac{\partial L}{\partial f(x)}-\frac{\partial}{\partial x}\left(\frac{\partial L}{\partial f'(x)}\right)=0\text{ and}[/itex]
[itex]\frac{\partial L}{\partial f(y)}-\frac{\partial}{\partial y}\left(\frac{\partial L}{\partial f'(y)}\right)=0[/itex]
Can anyone with more experience than me in calculus of variations (or has maybe encountered this scenario before) confirm my result?
Also maybe help me out with how Lagrange multipliers would work under such a situation? Basically the same as with two independent variables?
[tex]J[f]=\iint L\left(x,y,f(x),f(y),f'(x),f'(y)\right) \,dx \,dy[/tex]
Where ##f## is the function to be extremized, and ##x## and ##y## are independent variables? A result seems possible by using the usual calculus of variation technique of finding
[tex]\left.\frac{J[f+\epsilon \eta]}{d\epsilon}\right |_{\epsilon=0}[/tex]
treating ##f(x)## and ##f(y)## as two different functions, and setting this to zero since ##J## is extremized. My (rather sketchy) calculations seem to give the differential equations:
[itex]\frac{\partial L}{\partial f(x)}-\frac{\partial}{\partial x}\left(\frac{\partial L}{\partial f'(x)}\right)=0\text{ and}[/itex]
[itex]\frac{\partial L}{\partial f(y)}-\frac{\partial}{\partial y}\left(\frac{\partial L}{\partial f'(y)}\right)=0[/itex]
Can anyone with more experience than me in calculus of variations (or has maybe encountered this scenario before) confirm my result?
Also maybe help me out with how Lagrange multipliers would work under such a situation? Basically the same as with two independent variables?