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Euler-Lagrange-ligning

Euler-Lagrange-ligning

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Euler-Lagrange-ligning
LigningenUdledningEndimensionelt eksempelKildehenvisninger

En Euler-Lagrange-ligning er en partiel differentialligning, for hvilken det gælder, at løsningen er en mængde af funktioner, som opfylder at den første afledte for en given funktional (se funktional-afledte) er lig nul. Euler-Lagrange-ligningen optræder bl.a. inden for analytisk mekanik som betingelsen for stationering af virkningsfunktionalen for et givent mekanisk system.

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Som illustreret er der et vejintegrale for enhver vej. Med Euler-Lagrange-ligningen kan vejen, der minimerer integralet, findes.

Givet et funktional på formen

J [ q 1 , . . . , q n ] = ∫ a b L ( x , q 1 ( x ) , . . . , q n ( x ) , q 1 ′ ( x ) , . . . , q n ′ ( x ) ) d x {\displaystyle J[q_{1},...,q_{n}]=\int _{a}^{b}L(x,q_{1}(x),...,q_{n}(x),q'_{1}(x),...,q_{n}'(x))dx} {\displaystyle J[q_{1},...,q_{n}]=\int _{a}^{b}L(x,q_{1}(x),...,q_{n}(x),q'_{1}(x),...,q_{n}'(x))dx}

da er den første funktional-afledte mht. q i {\displaystyle q_{i}} {\displaystyle q_{i}} ved x {\displaystyle x} {\displaystyle x} givet ved:

δ J δ q i ( x ) = ∂ L ∂ q i − d d ⁡ x ∂ L ∂ q i ′ {\displaystyle {\delta J \over \delta q_{i}(x)}={\partial L \over \partial q_{i}}-{\operatorname {d} \over \operatorname {d} x}{\partial L \over \partial q'_{i}}} {\displaystyle {\delta J \over \delta q_{i}(x)}={\partial L \over \partial q_{i}}-{\operatorname {d} \over \operatorname {d} x}{\partial L \over \partial q'_{i}}}

J {\displaystyle J} {\displaystyle J} er stationær, når den funktional-afledte er lig nul, hvorved Euler-Lagrange-ligningerne fås:

∂ L ∂ q i − d d ⁡ x ∂ L ∂ q i ′ = 0 ∀ i {\displaystyle {\partial L \over \partial q_{i}}-{\operatorname {d} \over \operatorname {d} x}{\partial L \over \partial q'_{i}}=0\quad \forall i} {\displaystyle {\partial L \over \partial q_{i}}-{\operatorname {d} \over \operatorname {d} x}{\partial L \over \partial q'_{i}}=0\quad \forall i}

Hvis L {\displaystyle L} {\displaystyle L} ikke eksplicit afhænger af x {\displaystyle x} {\displaystyle x}, reduceres ligningen til Beltrami-identiteten:

L − q i ′ ∂ L ∂ q i ′ = C i ∀ i {\displaystyle L-q'_{i}{\partial L \over \partial q'_{i}}=C_{i}\quad \forall i} {\displaystyle L-q'_{i}{\partial L \over \partial q'_{i}}=C_{i}\quad \forall i}

hvor C i {\displaystyle C_{i}} {\displaystyle C_{i}} er konstant.[1]

Et funktional på formen

J = ∫ a b L ( x , q , q ′ ) d x {\displaystyle J=\int _{a}^{b}L(x,q,q')dx} {\displaystyle J=\int _{a}^{b}L(x,q,q')dx}

skal stationeres:

δ J = δ ∫ a b L ( x , q , q ′ ) d x = 0 {\displaystyle \delta J=\delta \int _{a}^{b}L(x,q,q')dx=0} {\displaystyle \delta J=\delta \int _{a}^{b}L(x,q,q')dx=0}

Variationen i J {\displaystyle J} {\displaystyle J} kan skrives ved variationerne i q {\displaystyle q} {\displaystyle q} og q ′ {\displaystyle q'} {\displaystyle q'}:

∫ a b [ ∂ L ∂ q δ q + ∂ L ∂ q ′ δ q ′ ] d x = 0 {\displaystyle \int _{a}^{b}\left[{\partial L \over \partial q}\delta q+{\partial L \over \partial q'}\delta q'\right]dx=0} {\displaystyle \int _{a}^{b}\left[{\partial L \over \partial q}\delta q+{\partial L \over \partial q'}\delta q'\right]dx=0}

 

 

 

 

(1)

hvor

δ q ′ ≡ d d t ( δ q ) {\displaystyle \delta q'\equiv {\frac {d}{dt}}\left(\delta q\right)} {\displaystyle \delta q'\equiv {\frac {d}{dt}}\left(\delta q\right)}

Det kræves desuden, at variationen i hver ende er nul:

δ q ( a ) = δ q ( b ) = 0 {\displaystyle \delta q(a)=\delta q(b)=0} {\displaystyle \delta q(a)=\delta q(b)=0}

 

 

 

 

(2)

Pga. produktreglen gælder:

d d t ( ∂ L ∂ q ′ δ q ) = d d t ( ∂ L ∂ q ′ ) δ q + ∂ L ∂ q ′ d d t ( δ q ) {\displaystyle {\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)={\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q+{\partial L \over \partial q'}{\frac {d}{dt}}\left(\delta q\right)} {\displaystyle {\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)={\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q+{\partial L \over \partial q'}{\frac {d}{dt}}\left(\delta q\right)}

Og dermed:

∂ L ∂ q ′ d d t ( δ q ) = − d d t ( ∂ L ∂ q ′ ) δ q + d d t ( ∂ L ∂ q ′ δ q ) {\displaystyle {\partial L \over \partial q'}{\frac {d}{dt}}\left(\delta q\right)=-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q+{\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)} {\displaystyle {\partial L \over \partial q'}{\frac {d}{dt}}\left(\delta q\right)=-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q+{\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)}

Dette indsættes i lign. 1, og integralet splittes op:

∫ a b [ ∂ L ∂ q δ q − d d t ( ∂ L ∂ q ′ ) δ q + d d t ( ∂ L ∂ q ′ δ q ) ] d x = 0 ∫ a b [ ∂ L ∂ q δ q − d d t ( ∂ L ∂ q ′ ) δ q ] d x + ∫ a b d d t ( ∂ L ∂ q ′ δ q ) d x = 0 ∫ a b [ ∂ L ∂ q δ q − d d t ( ∂ L ∂ q ′ ) δ q ] d x + [ ∂ L ∂ q ′ δ q ] a b = 0 {\displaystyle {\begin{aligned}\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q+{\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)\right]dx&=0\\\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q\right]dx+\int _{a}^{b}{\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)dx&=0\\\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q\right]dx+\left[{\partial L \over \partial q'}\delta q\right]_{a}^{b}&=0\end{aligned}}} {\displaystyle {\begin{aligned}\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q+{\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)\right]dx&=0\\\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q\right]dx+\int _{a}^{b}{\frac {d}{dt}}\left({\partial L \over \partial q'}\delta q\right)dx&=0\\\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q\right]dx+\left[{\partial L \over \partial q'}\delta q\right]_{a}^{b}&=0\end{aligned}}}

Pga. lign. 2 gælder:

[ ∂ L ∂ q ′ δ q ] a b = 0 {\displaystyle \left[{\partial L \over \partial q'}\delta q\right]_{a}^{b}=0} {\displaystyle \left[{\partial L \over \partial q'}\delta q\right]_{a}^{b}=0}

Derfor:

∫ a b [ ∂ L ∂ q δ q − d d t ( ∂ L ∂ q ′ ) δ q ] d x = 0 ∫ a b [ ∂ L ∂ q − d d t ( ∂ L ∂ q ′ ) ] ( δ q ) d x = 0 {\displaystyle {\begin{aligned}\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q\right]dx&=0\\\int _{a}^{b}\left[{\partial L \over \partial q}-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\right]\left(\delta q\right)dx&=0\end{aligned}}} {\displaystyle {\begin{aligned}\int _{a}^{b}\left[{\partial L \over \partial q}\delta q-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\delta q\right]dx&=0\\\int _{a}^{b}\left[{\partial L \over \partial q}-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)\right]\left(\delta q\right)dx&=0\end{aligned}}}

Integralet består nu af to faktorer. Siden integralet altid skal være nul, og det skal gælde for alle variationer af q {\displaystyle q} {\displaystyle q}, må den første faktor være nul:

∂ L ∂ q − d d t ( ∂ L ∂ q ′ ) = 0 {\displaystyle {\partial L \over \partial q}-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)=0} {\displaystyle {\partial L \over \partial q}-{\frac {d}{dt}}\left({\partial L \over \partial q'}\right)=0}

 

 

 

 

(3)

Dermed er Euler-Lagrange-ligningen udledt.[2]

Et bestemt mekanisk system beskrevet ved én koordinat x {\displaystyle x} {\displaystyle x} og med potentialet V {\displaystyle V} {\displaystyle V} har Lagrangefunktionen:

L ( x , x ˙ ) = T ( x ˙ ) − V ( x ) = 1 2 m x ˙ 2 − V ( x ) {\displaystyle L(x,{\dot {x}})=T({\dot {x}})-V(x)={\frac {1}{2}}m{\dot {x}}^{2}-V(x)} {\displaystyle L(x,{\dot {x}})=T({\dot {x}})-V(x)={\frac {1}{2}}m{\dot {x}}^{2}-V(x)}

Dvs. at den kinetiske energi er givet som for en klassisk punktmasse med massen m {\displaystyle m} {\displaystyle m} og hastigheden:

x ˙ = d e f d ⁡ x d t {\displaystyle {\dot {x}}{\overset {\underset {\mathrm {def} }{}}{=}}{\operatorname {d} x \over \operatorname {d} \!t}} {\displaystyle {\dot {x}}{\overset {\underset {\mathrm {def} }{}}{=}}{\operatorname {d} x \over \operatorname {d} \!t}},

mens den potentielle energi afhænger af positionen (den potentielle energi ville fx i et homogent tyngdefelt være givet på formen:

V ( x ) = m g x {\displaystyle V(x)=mgx} {\displaystyle V(x)=mgx}

Systemet vil udvikle sig således at virkningen I {\displaystyle I} {\displaystyle I} stationeres,[3]

I = ∫ L d t {\displaystyle I=\int Ldt} {\displaystyle I=\int Ldt}

Systemets dynamik er derfor beskrevet ved Euler-Lagrange-ligningen:

d d t ( ∂ L ∂ x ˙ ) − ∂ L ∂ x = 0 {\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {x}}}}\right)-{\frac {\partial L}{\partial x}}=0} {\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {x}}}}\right)-{\frac {\partial L}{\partial x}}=0}
⇓ {\displaystyle \Downarrow } {\displaystyle \Downarrow }
d d t ( ∂ ( 1 2 m x ˙ 2 − V ( x ) ) ∂ x ˙ ) − ∂ ( 1 2 m x ˙ 2 − V ( x ) ) ∂ x = 0 {\displaystyle {\frac {d}{dt}}\left({\frac {\partial \left({\frac {1}{2}}m{\dot {x}}^{2}-V(x)\right)}{\partial {\dot {x}}}}\right)-{\frac {\partial \left({\frac {1}{2}}m{\dot {x}}^{2}-V(x)\right)}{\partial x}}=0} {\displaystyle {\frac {d}{dt}}\left({\frac {\partial \left({\frac {1}{2}}m{\dot {x}}^{2}-V(x)\right)}{\partial {\dot {x}}}}\right)-{\frac {\partial \left({\frac {1}{2}}m{\dot {x}}^{2}-V(x)\right)}{\partial x}}=0}
⇓ {\displaystyle \Downarrow } {\displaystyle \Downarrow }
d d t ( m x ˙ ) + ∂ V ( x ) ∂ x = 0 {\displaystyle {\frac {d}{dt}}\left(m{\dot {x}}\right)+{\frac {\partial V(x)}{\partial x}}=0} {\displaystyle {\frac {d}{dt}}\left(m{\dot {x}}\right)+{\frac {\partial V(x)}{\partial x}}=0}
⇓ {\displaystyle \Downarrow } {\displaystyle \Downarrow }
m x ¨ + ∂ V ( x ) ∂ x = 0 {\displaystyle m{\ddot {x}}+{\frac {\partial V(x)}{\partial x}}=0} {\displaystyle m{\ddot {x}}+{\frac {\partial V(x)}{\partial x}}=0}
⇓ {\displaystyle \Downarrow } {\displaystyle \Downarrow }

m x ¨ = − ∂ V ( x ) ∂ x {\displaystyle m{\ddot {x}}=-{\frac {\partial V(x)}{\partial x}}} {\displaystyle m{\ddot {x}}=-{\frac {\partial V(x)}{\partial x}}}

 

 

 

 

(4)

Eftersom den dobbelte tidsafledte mht. positionen er accelerationen og den positionsafledte mht. det negative potentiale svarer til kraftkomposanten i x {\displaystyle x} {\displaystyle x}-retningen, svarer 4 til Newtons anden lov.

F = m x ¨ {\displaystyle F=m{\ddot {x}}} {\displaystyle F=m{\ddot {x}}}
  1. [1]
    Weisstein, Eric W., Beltrami Identity, Wolfram Alpha, arkiveret fra originalen 12. juli 2019, hentet 12. juli 2019
  2. [2]
    Weisstein, Eric W., Euler-Lagrange Differential Equation, Wolfram Alpha, arkiveret fra originalen 26. maj 2019, hentet 19. juli 2019
  3. [3]
    Goldstein, Herbert; Poole, Charles; Safko, John. "Variational Principles and Lagrange's Equations", Classical Mechanics (3. udgave), Addison Wesley, s. 34-35.
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