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Euler–Lagrange equation

From Wikipedia, the free encyclopedia
Second-order partial differential equation describing motion of mechanical system

In thecalculus of variations andclassical mechanics, theEuler–Lagrange equations[1] are a system of second-orderordinary differential equations whose solutions arestationary points of the givenaction functional. The equations were discovered in the 1750s by Swiss mathematicianLeonhard Euler and Italian mathematicianJoseph-Louis Lagrange.

Because a differentiable functional is stationary at its localextrema, the Euler–Lagrange equation is useful for solvingoptimization problems in which, given some functional, one seeks the function minimizing or maximizing it. This is analogous toFermat's theorem incalculus, stating that at any point where a differentiable function attains a local extremum itsderivative is zero. InLagrangian mechanics, according toHamilton's principle of stationary action, the evolution of a physical system is described by the solutions to the Euler equation for theaction of the system. In this context Euler equations are usually calledLagrange equations. Inclassical mechanics,[2] it is equivalent toNewton's laws of motion; indeed, the Euler-Lagrange equations will produce the same equations as Newton's Laws. This is particularly useful when analyzing systems whose force vectors are particularly complicated. It has the advantage that it takes the same form in any system ofgeneralized coordinates, and it is better suited to generalizations. Inclassical field theory there is ananalogous equation to calculate the dynamics of afield.

History

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The Euler–Lagrange equation was developed in connection with their studies of thetautochrone problem.

The Euler–Lagrange equation was developed in the 1750s by Euler and Lagrange in connection with their studies of thetautochrone problem. This is the problem of determining a curve on which a weighted particle will fall to a fixed point in a fixed amount of time, independent of the starting point.

Lagrange solved this problem in 1755 and sent the solution to Euler. Both further developed Lagrange's method and applied it tomechanics, which led to the formulation ofLagrangian mechanics. Their correspondence ultimately led to thecalculus of variations, a term coined by Euler himself in 1766.[3]

Statement

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Let(X,L){\displaystyle (X,L)} be areal dynamical system withn{\displaystyle n} degrees of freedom. HereX{\displaystyle X} is theconfiguration space andL=L(t,q(t),v(t)){\displaystyle L=L(t,{\boldsymbol {q}}(t),{\boldsymbol {v}}(t))} theLagrangian, i.e. a smoothreal-valued function such thatq(t)X,{\displaystyle {\boldsymbol {q}}(t)\in X,} andv(t){\displaystyle {\boldsymbol {v}}(t)} is ann{\displaystyle n}-dimensional "vector of speed". (For those familiar withdifferential geometry,X{\displaystyle X} is asmooth manifold, andL:Rt×TXR,{\displaystyle L\colon {\mathbb {R} }_{t}\times TX\to {\mathbb {R} },} whereTX{\displaystyle TX} is thetangent bundle ofX).{\displaystyle X).}[4]

LetP(a,b,xa,xb){\displaystyle {\cal {P}}(a,b,{\boldsymbol {x}}_{a},{\boldsymbol {x}}_{b})} be the set of smooth pathsq:[a,b]X{\displaystyle {\boldsymbol {q}}:[a,b]\to X} for whichq(a)=xa{\displaystyle {\boldsymbol {q}}(a)={\boldsymbol {x}}_{a}} andq(b)=xb.{\displaystyle {\boldsymbol {q}}(b)={\boldsymbol {x}}_{b}.}

Theaction functionalS:P(a,b,xa,xb)R{\displaystyle S\colon {\cal {P}}(a,b,{\boldsymbol {x}}_{a},{\boldsymbol {x}}_{b})\to \mathbb {R} } is defined viaS[q]=abL(t,q(t),q˙(t))dt.{\displaystyle S[{\boldsymbol {q}}]=\int _{a}^{b}L(t,{\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t))\,dt.}

A pathqP(a,b,xa,xb){\displaystyle {\boldsymbol {q}}\in {\cal {P}}(a,b,{\boldsymbol {x}}_{a},{\boldsymbol {x}}_{b})} is astationary point ofS{\displaystyle S} if and only if

Lqi(t,q(t),q˙(t))ddtLq˙i(t,q(t),q˙(t))=0,i=1,,n.{\displaystyle {\frac {\partial L}{\partial q^{i}}}(t,{\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t))-{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {q}}^{i}}}(t,{\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t))=0,\quad i=1,\dots ,n.}

Here,q˙(t){\displaystyle {\dot {\boldsymbol {q}}}(t)} is thetime derivative ofq(t).{\displaystyle {\boldsymbol {q}}(t).} When we say stationary point, we mean a stationary point ofS{\displaystyle S} with respect to any small perturbation inq{\displaystyle {\boldsymbol {q}}}. See proofs below for more rigorous detail.

Derivation of the one-dimensional Euler–Lagrange equation

The derivation of the one-dimensional Euler–Lagrange equation is one of the classic proofs inmathematics. It relies on thefundamental lemma of calculus of variations.

We wish to find a functionf{\displaystyle f} which satisfies the boundary conditionsf(a)=A{\displaystyle f(a)=A},f(b)=B{\displaystyle f(b)=B}, and which extremizes the functionalJ[f]=abL(x,f(x),f(x))dx .{\displaystyle J[f]=\int _{a}^{b}L(x,f(x),f'(x))\,\mathrm {d} x\ .}

We assume thatL{\displaystyle L} is twice continuously differentiable.[5] A weaker assumption can be used, but the proof becomes more difficult.[citation needed]

Iff{\displaystyle f} extremizes the functional subject to the boundary conditions, then any slight perturbation off{\displaystyle f} that preserves the boundary values must either increaseJ{\displaystyle J} (iff{\displaystyle f} is a minimizer) or decreaseJ{\displaystyle J} (iff{\displaystyle f} is a maximizer).

Letf+εη{\displaystyle f+\varepsilon \eta } be the result of such a perturbationεη{\displaystyle \varepsilon \eta } off{\displaystyle f}, whereε{\displaystyle \varepsilon } is small andη{\displaystyle \eta } is a differentiable function satisfyingη(a)=η(b)=0{\displaystyle \eta (a)=\eta (b)=0}. Then defineΦ(ε)=J[f+εη]=abL(x,f(x)+εη(x),f(x)+εη(x))dx .{\displaystyle \Phi (\varepsilon )=J[f+\varepsilon \eta ]=\int _{a}^{b}L(x,f(x)+\varepsilon \eta (x),f'(x)+\varepsilon \eta '(x))\,\mathrm {d} x\ .}

We now wish to calculate thetotal derivative ofΦ{\displaystyle \Phi } with respect toε.dΦdε=ddεabL(x,f(x)+εη(x),f(x)+εη(x))dx=abddεL(x,f(x)+εη(x),f(x)+εη(x))dx=ab[η(x)Lf(x,f(x)+εη(x),f(x)+εη(x))+η(x)Lf(x,f(x)+εη(x),f(x)+εη(x))]dx .{\displaystyle {\begin{aligned}{\frac {\mathrm {d} \Phi }{\mathrm {d} \varepsilon }}&={\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\int _{a}^{b}L(x,f(x)+\varepsilon \eta (x),f'(x)+\varepsilon \eta '(x))\,\mathrm {d} x\\&=\int _{a}^{b}{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}L(x,f(x)+\varepsilon \eta (x),f'(x)+\varepsilon \eta '(x))\,\mathrm {d} x\\&=\int _{a}^{b}\left[\eta (x){\frac {\partial L}{\partial {f}}}(x,f(x)+\varepsilon \eta (x),f'(x)+\varepsilon \eta '(x))+\eta '(x){\frac {\partial L}{\partial f'}}(x,f(x)+\varepsilon \eta (x),f'(x)+\varepsilon \eta '(x))\right]\mathrm {d} x\ .\end{aligned}}}

The third line follows from the fact thatx{\displaystyle x} does not depend onε{\displaystyle \varepsilon }, i.e.dxdε=0{\displaystyle {\frac {\mathrm {d} x}{\mathrm {d} \varepsilon }}=0}.

Whenε=0{\displaystyle \varepsilon =0},Φ{\displaystyle \Phi } has anextremum value, so thatdΦdε|ε=0=ab[η(x)Lf(x,f(x),f(x))+η(x)Lf(x,f(x),f(x))]dx=0 .{\displaystyle \left.{\frac {\mathrm {d} \Phi }{\mathrm {d} \varepsilon }}\right|_{\varepsilon =0}=\int _{a}^{b}\left[\eta (x){\frac {\partial L}{\partial f}}(x,f(x),f'(x))+\eta '(x){\frac {\partial L}{\partial f'}}(x,f(x),f'(x))\,\right]\,\mathrm {d} x=0\ .}

The next step is to useintegration by parts on the second term of the integrand, yieldingab[Lf(x,f(x),f(x))ddxLf(x,f(x),f(x))]η(x)dx+[η(x)Lf(x,f(x),f(x))]ab=0 .{\displaystyle \int _{a}^{b}\left[{\frac {\partial L}{\partial f}}(x,f(x),f'(x))-{\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {\partial L}{\partial f'}}(x,f(x),f'(x))\right]\eta (x)\,\mathrm {d} x+\left[\eta (x){\frac {\partial L}{\partial f'}}(x,f(x),f'(x))\right]_{a}^{b}=0\ .}

Using the boundary conditionsη(a)=η(b)=0{\displaystyle \eta (a)=\eta (b)=0},ab[Lf(x,f(x),f(x))ddxLf(x,f(x),f(x))]η(x)dx=0.{\displaystyle \int _{a}^{b}\left[{\frac {\partial L}{\partial f}}(x,f(x),f'(x))-{\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {\partial L}{\partial f'}}(x,f(x),f'(x))\right]\eta (x)\,\mathrm {d} x=0\,.}

Applying thefundamental lemma of calculus of variations now yields the Euler–Lagrange equationLf(x,f(x),f(x))ddxLf(x,f(x),f(x))=0.{\displaystyle {\frac {\partial L}{\partial f}}(x,f(x),f'(x))-{\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {\partial L}{\partial f'}}(x,f(x),f'(x))=0\,.}

Alternative derivation of the one-dimensional Euler–Lagrange equation

Given a functionalJ=abL(t,y(t),y(t))dt{\displaystyle J=\int _{a}^{b}L(t,y(t),y'(t))\,\mathrm {d} t}onC1([a,b]){\displaystyle C^{1}([a,b])} with the boundary conditionsy(a)=A{\displaystyle y(a)=A} andy(b)=B{\displaystyle y(b)=B}, we proceed by approximating the extremal curve by a polygonal line withn{\displaystyle n} segments and passing to the limit as the number of segments grows arbitrarily large.

Divide the interval[a,b]{\displaystyle [a,b]} inton{\displaystyle n} equal segments with endpointst0=a,t1,t2,,tn=b{\displaystyle t_{0}=a,t_{1},t_{2},\ldots ,t_{n}=b} and letΔt=tktk1{\displaystyle \Delta t=t_{k}-t_{k-1}}. Rather than a smooth functiony(t){\displaystyle y(t)} we consider the polygonal line with vertices(t0,y0),,(tn,yn){\displaystyle (t_{0},y_{0}),\ldots ,(t_{n},y_{n})}, wherey0=A{\displaystyle y_{0}=A} andyn=B{\displaystyle y_{n}=B}. Accordingly, our functional becomes a real function ofn1{\displaystyle n-1} variables given byJ(y1,,yn1)k=0n1L(tk,yk,yk+1ykΔt)Δt.{\displaystyle J(y_{1},\ldots ,y_{n-1})\approx \sum _{k=0}^{n-1}L\left(t_{k},y_{k},{\frac {y_{k+1}-y_{k}}{\Delta t}}\right)\Delta t.}

Extremals of this new functional defined on the discrete pointst0,,tn{\displaystyle t_{0},\ldots ,t_{n}} correspond to points whereJ(y1,,yn)ym=0.{\displaystyle {\frac {\partial J(y_{1},\ldots ,y_{n})}{\partial y_{m}}}=0.}

Note that change ofym{\displaystyle y_{m}} affects L not only at m but also at m-1 for the derivative of the 3rd argument.L(3rd argument)(ym+1(ym+Δym)Δt)=L(ym+1ymΔt)LyΔymΔt{\displaystyle L({\text{3rd argument}})\left({\frac {y_{m+1}-(y_{m}+\Delta y_{m})}{\Delta t}}\right)=L\left({\frac {y_{m+1}-y_{m}}{\Delta t}}\right)-{\frac {\partial L}{\partial y'}}{\frac {\Delta y_{m}}{\Delta t}}}L((ym+Δym)ym1Δt)=L(ymym1Δt)+LyΔymΔt{\displaystyle L\left({\frac {(y_{m}+\Delta y_{m})-y_{m-1}}{\Delta t}}\right)=L\left({\frac {y_{m}-y_{m-1}}{\Delta t}}\right)+{\frac {\partial L}{\partial y'}}{\frac {\Delta y_{m}}{\Delta t}}}

Evaluating the partial derivative givesJym=Ly(tm,ym,ym+1ymΔt)Δt+Ly(tm1,ym1,ymym1Δt)Ly(tm,ym,ym+1ymΔt).{\displaystyle {\frac {\partial J}{\partial y_{m}}}=L_{y}\left(t_{m},y_{m},{\frac {y_{m+1}-y_{m}}{\Delta t}}\right)\Delta t+L_{y'}\left(t_{m-1},y_{m-1},{\frac {y_{m}-y_{m-1}}{\Delta t}}\right)-L_{y'}\left(t_{m},y_{m},{\frac {y_{m+1}-y_{m}}{\Delta t}}\right).}

Dividing the above equation byΔt{\displaystyle \Delta t} givesJymΔt=Ly(tm,ym,ym+1ymΔt)1Δt[Ly(tm,ym,ym+1ymΔt)Ly(tm1,ym1,ymym1Δt)],{\displaystyle {\frac {\partial J}{\partial y_{m}\Delta t}}=L_{y}\left(t_{m},y_{m},{\frac {y_{m+1}-y_{m}}{\Delta t}}\right)-{\frac {1}{\Delta t}}\left[L_{y'}\left(t_{m},y_{m},{\frac {y_{m+1}-y_{m}}{\Delta t}}\right)-L_{y'}\left(t_{m-1},y_{m-1},{\frac {y_{m}-y_{m-1}}{\Delta t}}\right)\right],}and taking the limit asΔt0{\displaystyle \Delta t\to 0} of the right-hand side of this expression yieldsLyddtLy=0.{\displaystyle L_{y}-{\frac {\mathrm {d} }{\mathrm {d} t}}L_{y'}=0.}

The left hand side of the previous equation is thefunctional derivativeδJ/δy{\displaystyle \delta J/\delta y} of the functionalJ{\displaystyle J}. A necessary condition for a differentiable functional to have an extremum on some function is that its functional derivative at that function vanishes, which is granted by the last equation.

Example

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A standard example[citation needed] is finding the real-valued functiony(x) on the interval [a,b], such thaty(a) =c andy(b) =d, for which thepathlength along thecurve traced byy is as short as possible.

s=abdx2+dy2=ab1+y2dx,{\displaystyle {\text{s}}=\int _{a}^{b}{\sqrt {\mathrm {d} x^{2}+\mathrm {d} y^{2}}}=\int _{a}^{b}{\sqrt {1+y'^{2}}}\,\mathrm {d} x,}

the integrand function beingL(x,y,y)=1+y2{\textstyle L(x,y,y')={\sqrt {1+y'^{2}}}}.

The partial derivatives ofL are:

L(x,y,y)y=y1+y2andL(x,y,y)y=0.{\displaystyle {\frac {\partial L(x,y,y')}{\partial y'}}={\frac {y'}{\sqrt {1+y'^{2}}}}\quad {\text{and}}\quad {\frac {\partial L(x,y,y')}{\partial y}}=0.}

By substituting these into the Euler–Lagrange equation, we obtain

ddxy(x)1+(y(x))2=0y(x)1+(y(x))2=C=constanty(x)=C1C2=:Ay(x)=Ax+B{\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {y'(x)}{\sqrt {1+(y'(x))^{2}}}}&=0\\{\frac {y'(x)}{\sqrt {1+(y'(x))^{2}}}}&=C={\text{constant}}\\\Rightarrow y'(x)&={\frac {C}{\sqrt {1-C^{2}}}}=:A\\\Rightarrow y(x)&=Ax+B\end{aligned}}}

that is, the function must have a constant first derivative, and thus itsgraph is astraight line.

Generalizations

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Single function of single variable with higher derivatives

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The stationary values of the functional

I[f]=x0x1L(x,f,f,f,,f(k)) dx ;  f:=dfdx, f:=d2fdx2, f(k):=dkfdxk{\displaystyle I[f]=\int _{x_{0}}^{x_{1}}{\mathcal {L}}(x,f,f',f'',\dots ,f^{(k)})~\mathrm {d} x~;~~f':={\cfrac {\mathrm {d} f}{\mathrm {d} x}},~f'':={\cfrac {\mathrm {d} ^{2}f}{\mathrm {d} x^{2}}},~f^{(k)}:={\cfrac {\mathrm {d} ^{k}f}{\mathrm {d} x^{k}}}}

can be obtained from the Euler–Lagrange equation[6]

Lfddx(Lf)+d2dx2(Lf)+(1)kdkdxk(Lf(k))=0{\displaystyle {\cfrac {\partial {\mathcal {L}}}{\partial f}}-{\cfrac {\mathrm {d} }{\mathrm {d} x}}\left({\cfrac {\partial {\mathcal {L}}}{\partial f'}}\right)+{\cfrac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\left({\cfrac {\partial {\mathcal {L}}}{\partial f''}}\right)-\dots +(-1)^{k}{\cfrac {\mathrm {d} ^{k}}{\mathrm {d} x^{k}}}\left({\cfrac {\partial {\mathcal {L}}}{\partial f^{(k)}}}\right)=0}

under fixed boundary conditions for the function itself as well as for the firstk1{\displaystyle k-1} derivatives (i.e. for allf(i),i{0,...,k1}{\displaystyle f^{(i)},i\in \{0,...,k-1\}}). The endpoint values of the highest derivativef(k){\displaystyle f^{(k)}} remain flexible.

Several functions of single variable with single derivative

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If the problem involves finding several functions (f1,f2,,fm{\displaystyle f_{1},f_{2},\dots ,f_{m}}) of a single independent variable (x{\displaystyle x}) that define an extremum of the functional

I[f1,f2,,fm]=x0x1L(x,f1,f2,,fm,f1,f2,,fm) dx ;  fi:=dfidx{\displaystyle I[f_{1},f_{2},\dots ,f_{m}]=\int _{x_{0}}^{x_{1}}{\mathcal {L}}(x,f_{1},f_{2},\dots ,f_{m},f_{1}',f_{2}',\dots ,f_{m}')~\mathrm {d} x~;~~f_{i}':={\cfrac {\mathrm {d} f_{i}}{\mathrm {d} x}}}

then the corresponding Euler–Lagrange equations are[7]

Lfiddx(Lfi)=0;i=1,2,...,m{\displaystyle {\begin{aligned}{\frac {\partial {\mathcal {L}}}{\partial f_{i}}}-{\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{i}'}}\right)=0;\quad i=1,2,...,m\end{aligned}}}

Single function of several variables with single derivative

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A multi-dimensional generalization comes from considering a function on n variables. IfΩ{\displaystyle \Omega } is some surface, then

I[f]=ΩL(x1,,xn,f,f1,,fn)dx ;  fj:=fxj{\displaystyle I[f]=\int _{\Omega }{\mathcal {L}}(x_{1},\dots ,x_{n},f,f_{1},\dots ,f_{n})\,\mathrm {d} \mathbf {x} \,\!~;~~f_{j}:={\cfrac {\partial f}{\partial x_{j}}}}

is extremized only iff satisfies thepartial differential equation

Lfj=1nxj(Lfj)=0.{\displaystyle {\frac {\partial {\mathcal {L}}}{\partial f}}-\sum _{j=1}^{n}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{j}}}\right)=0.}

Whenn = 2 and functionalI{\displaystyle {\mathcal {I}}} is theenergy functional, this leads to the soap-filmminimal surface problem.

Several functions of several variables with single derivative

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If there are several unknown functions to be determined and several variables such that

I[f1,f2,,fm]=ΩL(x1,,xn,f1,,fm,f1,1,,f1,n,,fm,1,,fm,n)dx ;  fi,j:=fixj{\displaystyle I[f_{1},f_{2},\dots ,f_{m}]=\int _{\Omega }{\mathcal {L}}(x_{1},\dots ,x_{n},f_{1},\dots ,f_{m},f_{1,1},\dots ,f_{1,n},\dots ,f_{m,1},\dots ,f_{m,n})\,\mathrm {d} \mathbf {x} \,\!~;~~f_{i,j}:={\cfrac {\partial f_{i}}{\partial x_{j}}}}

the system of Euler–Lagrange equations is[6]

Lf1j=1nxj(Lf1,j)=01Lf2j=1nxj(Lf2,j)=02Lfmj=1nxj(Lfm,j)=0m.{\displaystyle {\begin{aligned}{\frac {\partial {\mathcal {L}}}{\partial f_{1}}}-\sum _{j=1}^{n}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{1,j}}}\right)&=0_{1}\\{\frac {\partial {\mathcal {L}}}{\partial f_{2}}}-\sum _{j=1}^{n}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{2,j}}}\right)&=0_{2}\\\vdots \qquad \vdots \qquad &\quad \vdots \\{\frac {\partial {\mathcal {L}}}{\partial f_{m}}}-\sum _{j=1}^{n}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{m,j}}}\right)&=0_{m}.\end{aligned}}}

Single function of two variables with higher derivatives

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If there is a single unknown functionf to be determined that is dependent on two variablesx1 andx2 and if the functional depends on higher derivatives off up ton-th order such that

I[f]=ΩL(x1,x2,f,f1,f2,f11,f12,f22,,f222)dxfi:=fxi,fij:=2fxixj,{\displaystyle {\begin{aligned}I[f]&=\int _{\Omega }{\mathcal {L}}(x_{1},x_{2},f,f_{1},f_{2},f_{11},f_{12},f_{22},\dots ,f_{22\dots 2})\,\mathrm {d} \mathbf {x} \\&\qquad \quad f_{i}:={\cfrac {\partial f}{\partial x_{i}}}\;,\quad f_{ij}:={\cfrac {\partial ^{2}f}{\partial x_{i}\partial x_{j}}}\;,\;\;\dots \end{aligned}}}

then the Euler–Lagrange equation is[6]

Lfx1(Lf1)x2(Lf2)+2x12(Lf11)+2x1x2(Lf12)+2x22(Lf22)+(1)nnx2n(Lf222)=0{\displaystyle {\begin{aligned}{\frac {\partial {\mathcal {L}}}{\partial f}}&-{\frac {\partial }{\partial x_{1}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{1}}}\right)-{\frac {\partial }{\partial x_{2}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{2}}}\right)+{\frac {\partial ^{2}}{\partial x_{1}^{2}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{11}}}\right)+{\frac {\partial ^{2}}{\partial x_{1}\partial x_{2}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{12}}}\right)+{\frac {\partial ^{2}}{\partial x_{2}^{2}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{22}}}\right)\\&-\dots +(-1)^{n}{\frac {\partial ^{n}}{\partial x_{2}^{n}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{22\dots 2}}}\right)=0\end{aligned}}}

which can be represented shortly as:

Lf+j=1nμ1μj(1)jjxμ1xμj(Lfμ1μj)=0{\displaystyle {\frac {\partial {\mathcal {L}}}{\partial f}}+\sum _{j=1}^{n}\sum _{\mu _{1}\leq \ldots \leq \mu _{j}}(-1)^{j}{\frac {\partial ^{j}}{\partial x_{\mu _{1}}\dots \partial x_{\mu _{j}}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{\mu _{1}\dots \mu _{j}}}}\right)=0}

whereinμ1μj{\displaystyle \mu _{1}\dots \mu _{j}} are indices that span the number of variables, that is, here they go from 1 to 2. Here summation over theμ1μj{\displaystyle \mu _{1}\dots \mu _{j}} indices is only overμ1μ2μj{\displaystyle \mu _{1}\leq \mu _{2}\leq \ldots \leq \mu _{j}} in order to avoid counting the samepartial derivative multiple times, for examplef12=f21{\displaystyle f_{12}=f_{21}} appears only once in the previous equation.

Several functions of several variables with higher derivatives

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If there arep unknown functionsfi to be determined that are dependent onm variablesx1 ...xm and if the functional depends on higher derivatives of thefi up ton-th order such that

I[f1,,fp]=ΩL(x1,,xm;f1,,fp;f1,1,,fp,m;f1,11,,fp,mm;;fp,11,,fp,mm)dxfi,μ:=fixμ,fi,μ1μ2:=2fixμ1xμ2,{\displaystyle {\begin{aligned}I[f_{1},\ldots ,f_{p}]&=\int _{\Omega }{\mathcal {L}}(x_{1},\ldots ,x_{m};f_{1},\ldots ,f_{p};f_{1,1},\ldots ,f_{p,m};f_{1,11},\ldots ,f_{p,mm};\ldots ;f_{p,1\ldots 1},\ldots ,f_{p,m\ldots m})\,\mathrm {d} \mathbf {x} \\&\qquad \quad f_{i,\mu }:={\cfrac {\partial f_{i}}{\partial x_{\mu }}}\;,\quad f_{i,\mu _{1}\mu _{2}}:={\cfrac {\partial ^{2}f_{i}}{\partial x_{\mu _{1}}\partial x_{\mu _{2}}}}\;,\;\;\dots \end{aligned}}}

whereμ1μj{\displaystyle \mu _{1}\dots \mu _{j}} are indices that span the number of variables, that is they go from 1 to m. Then the Euler–Lagrange equation is

Lfi+j=1nμ1μj(1)jjxμ1xμj(Lfi,μ1μj)=0{\displaystyle {\frac {\partial {\mathcal {L}}}{\partial f_{i}}}+\sum _{j=1}^{n}\sum _{\mu _{1}\leq \ldots \leq \mu _{j}}(-1)^{j}{\frac {\partial ^{j}}{\partial x_{\mu _{1}}\dots \partial x_{\mu _{j}}}}\left({\frac {\partial {\mathcal {L}}}{\partial f_{i,\mu _{1}\dots \mu _{j}}}}\right)=0}

where the summation over theμ1μj{\displaystyle \mu _{1}\dots \mu _{j}} is avoiding counting the same derivativefi,μ1μ2=fi,μ2μ1{\displaystyle f_{i,\mu _{1}\mu _{2}}=f_{i,\mu _{2}\mu _{1}}} several times, just as in the previous subsection. This can be expressed more compactly as

j=0nμ1μj(1)jμ1μjj(Lfi,μ1μj)=0{\displaystyle \sum _{j=0}^{n}\sum _{\mu _{1}\leq \ldots \leq \mu _{j}}(-1)^{j}\partial _{\mu _{1}\ldots \mu _{j}}^{j}\left({\frac {\partial {\mathcal {L}}}{\partial f_{i,\mu _{1}\dots \mu _{j}}}}\right)=0}

Field theories

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Main article:Lagrangian (field theory)

Generalization to manifolds

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LetM{\displaystyle M} be asmooth manifold, and letC([a,b]){\displaystyle C^{\infty }([a,b])} denote the space ofsmooth functionsf:[a,b]M{\displaystyle f\colon [a,b]\to M}. Then, for functionalsS:C([a,b])R{\displaystyle S\colon C^{\infty }([a,b])\to \mathbb {R} } of the form

S[f]=ab(Lf˙)(t)dt{\displaystyle S[f]=\int _{a}^{b}(L\circ {\dot {f}})(t)\,\mathrm {d} t}

whereL:TMR{\displaystyle L\colon TM\to \mathbb {R} } is the Lagrangian, the statementdSf=0{\displaystyle \mathrm {d} S_{f}=0} is equivalent to the statement that, for allt[a,b]{\displaystyle t\in [a,b]}, each coordinate frametrivialization(xi,Xi){\displaystyle (x^{i},X^{i})} of a neighborhood off˙(t){\displaystyle {\dot {f}}(t)} yields the followingdimM{\displaystyle \dim M} equations:

i:ddtLXi|f˙(t)=Lxi|f˙(t).{\displaystyle \forall i:{\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial X^{i}}}{\bigg |}_{{\dot {f}}(t)}={\frac {\partial L}{\partial x^{i}}}{\bigg |}_{{\dot {f}}(t)}.}

Euler-Lagrange equations can also be written in a coordinate-free form as[8]

LΔθL=dL{\displaystyle {\mathcal {L}}_{\Delta }\theta _{L}=dL}

whereθL{\displaystyle \theta _{L}} is the canonical momenta1-form corresponding to the LagrangianL{\displaystyle L}. The vector field generating time translations is denoted byΔ{\displaystyle \Delta } and theLie derivative is denoted byL{\displaystyle {\mathcal {L}}}. One can use local charts(qα,q˙α){\displaystyle (q^{\alpha },{\dot {q}}^{\alpha })} in whichθL=Lq˙αdqα{\displaystyle \theta _{L}={\frac {\partial L}{\partial {\dot {q}}^{\alpha }}}dq^{\alpha }} andΔ:=ddt=q˙αqα+q¨αq˙α{\displaystyle \Delta :={\frac {d}{dt}}={\dot {q}}^{\alpha }{\frac {\partial }{\partial q^{\alpha }}}+{\ddot {q}}^{\alpha }{\frac {\partial }{\partial {\dot {q}}^{\alpha }}}} and use coordinate expressions for the Lie derivative to see equivalence with coordinate expressions of the Euler Lagrange equation. The coordinate free form is particularly suitable for geometrical interpretation of the Euler Lagrange equations.

See also

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Look upEuler–Lagrange equation in Wiktionary, the free dictionary.

Notes

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  1. ^Fox, Charles (1987).An introduction to the calculus of variations. Courier Dover Publications.ISBN 978-0-486-65499-7.
  2. ^Goldstein, H.;Poole, C.P.; Safko, J. (2014).Classical Mechanics (3rd ed.). Addison Wesley.
  3. ^A short biography of LagrangeArchived 2007-07-14 at theWayback Machine
  4. ^"Elements of the Calculus of Variations",Lectures on the Geometry of Manifolds (3 ed.), WORLD SCIENTIFIC, pp. 201–226, November 2020,doi:10.1142/9789811214820_0005,ISBN 978-981-12-1481-3, retrieved2025-10-31
  5. ^Courant & Hilbert 1953, p. 184
  6. ^abcCourant, R;Hilbert, D (1953).Methods of Mathematical Physics. Vol. I (First English ed.). New York: Interscience Publishers, Inc.ISBN 978-0471504474.{{cite book}}:ISBN / Date incompatibility (help)
  7. ^Weinstock, R. (1952).Calculus of Variations with Applications to Physics and Engineering. New York: McGraw-Hill.
  8. ^José; Saletan (1998).Classical Dynamics: A contemporary approach. Cambridge University Press.ISBN 9780521636360. Retrieved2023-09-12.

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