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Lagrange multiplier

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Method to solve constrained optimization problems

"Lagrangian function" redirects here; not to be confused withLagrangian (physics).

Inmathematical optimization, themethod of Lagrange multipliers is a strategy for finding the localmaxima and minima of afunction subject toequation constraints (i.e., subject to the condition that one or moreequations have to be satisfied exactly by the chosen values of thevariables).[1] It is named after the mathematicianJoseph-Louis Lagrange.

Summary and rationale

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The basic idea is to convert a constrained problem into a form such that thederivative test of an unconstrained problem can still be applied. The relationship between the gradient of the function and gradients of the constraints rather naturally leads to a reformulation of the original problem, known as theLagrangian function or Lagrangian.[2] In the general case, the Lagrangian is defined as

L(x,λ)≡f(x)+⟨λ,g(x)⟩{\displaystyle {\mathcal {L}}(x,\lambda )\equiv f(x)+\langle \lambda ,g(x)\rangle }

for functionsf,g{\displaystyle f,g}; the notation⟨⋅,⋅⟩{\displaystyle \langle \cdot ,\cdot \rangle } denotes aninner product. The valueλ{\displaystyle \lambda } is called theLagrange multiplier.

In simple cases, where the inner product is defined as thedot product, the Lagrangian is

L(x,λ)≡f(x)+λ⋅g(x){\displaystyle {\mathcal {L}}(x,\lambda )\equiv f(x)+\lambda \cdot g(x)}

The method can be summarized as follows: in order to find the maximum or minimum of a functionf{\displaystyle f} subject to the equality constraintg(x)=0{\displaystyle g(x)=0}, find thestationary points ofL{\displaystyle {\mathcal {L}}} considered as a function ofx{\displaystyle x} and the Lagrange multiplierλ {\displaystyle \lambda ~}. This means that allpartial derivatives should be zero, including the partial derivative with respect toλ {\displaystyle \lambda ~}.[3]

∂L∂x=0{\displaystyle {\frac {\partial {\mathcal {L}}}{\partial x}}=0}  and  ∂L ∂λ=0 ;{\displaystyle {\frac {\ \partial {\mathcal {L}}\ }{\partial \lambda }}=0\ ;}

or equivalently

∂f(x)∂x+λ⋅∂g(x)∂x=0{\displaystyle {\frac {\partial f(x)}{\partial x}}+\lambda \cdot {\frac {\partial g(x)}{\partial x}}=0}  and g(x)=0 .{\displaystyle g(x)=0~.}

The solution corresponding to the originalconstrained optimization is always asaddle point of the Lagrangian function,[4][5] which can be identified among the stationary points from thedefiniteness of thebordered Hessian matrix.[6]

The great advantage of this method is that it allows the optimization to be solved without explicitparameterization in terms of the constraints. As a result, the method of Lagrange multipliers is widely used to solve challenging constrained optimization problems. Further, the method of Lagrange multipliers is generalized by theKarush–Kuhn–Tucker conditions, which can also take into account inequality constraints of the formh(x)≤c{\displaystyle h(\mathbf {x} )\leq c} for a given constantc{\displaystyle c}.

Statement

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The following is known as the Lagrange multiplier theorem.[7]

Letf:Rn→R{\displaystyle f\colon \mathbb {R} ^{n}\to \mathbb {R} } be theobjective function and letg:Rn→Rc{\displaystyle g\colon \mathbb {R} ^{n}\to \mathbb {R} ^{c}} be the constraints function, both belonging toC1{\displaystyle C^{1}} (that is, having continuous first derivatives). Letx⋆{\displaystyle x_{\star }} be an optimal solution to the following optimization problem such that, for the matrix of partial derivatives[D⁡g(x⋆)]j,k= ∂gj ∂xk{\displaystyle {\Bigl [}\operatorname {D} g(x_{\star }){\Bigr ]}_{j,k}={\frac {\ \partial g_{j}\ }{\partial x_{k}}}},rank⁡(D⁡g(x⋆))=c≤n{\displaystyle \operatorname {rank} (\operatorname {D} g(x_{\star }))=c\leq n}:

maximize f(x)subject to: g(x)=0{\displaystyle {\begin{aligned}&{\text{maximize }}f(x)\\&{\text{subject to: }}g(x)=0\end{aligned}}}

Then there exists a unique Lagrange multiplierλ⋆∈Rc{\displaystyle \lambda _{\star }\in \mathbb {R} ^{c}} such thatD⁡f(x⋆)=λ⋆TD⁡g(x⋆) .{\displaystyle \operatorname {D} f(x_{\star })=\lambda _{\star }^{\mathsf {T}}\operatorname {D} g(x_{\star })~.} (In this equation,λ⋆{\displaystyle \lambda _{\star }} is a column vector, so its transposeλ⋆T{\displaystyle \lambda _{\star }^{\mathsf {T}}} is a row vector. Alternatively, we can redefine the Lagrange multiplier directly as a row vector and thus avoid the transposition.)

The Lagrange multiplier theorem states that at any local maximum (or minimum) of the function evaluated under the equality constraints, if constraint qualification applies (explained below), then thegradient of the function (at that point) can be expressed as alinear combination of the gradients of the constraints (at that point), with the Lagrange multipliers acting ascoefficients.[8] This is equivalent to saying that any direction perpendicular to all gradients of the constraints is also perpendicular to the gradient of the function. Or still, saying that thedirectional derivative of the function is0 in every feasible direction.

Single constraint

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Figure 1: The red curve shows the constraintg(x,y) =c. The blue curves are contours off(x,y). The point where the red constraint tangentially touches a blue contour is the maximum off(x,y) along the constraint, sinced1 >d2.

For the case of only one constraint and only two choice variables (as exemplified in Figure 1), consider theoptimization problemmaximizex,yf(x,y)subject tog(x,y)=0.{\displaystyle {\begin{aligned}{\underset {x,y}{\text{maximize}}}\quad &f(x,y)\\{\text{subject to}}\quad &g(x,y)=0.\end{aligned}}}(Sometimes an additive constant is shown separately rather than being included ing{\displaystyle g}, in which case the constraint is writteng(x,y)=c,{\displaystyle g(x,y)=c,} as in Figure 1.) We assume that bothf{\displaystyle f} andg{\displaystyle g} have continuous firstpartial derivatives. We introduce a new variable (λ{\displaystyle \lambda }) called aLagrange multiplier (orLagrange undetermined multiplier) and study theLagrange function (orLagrangian orLagrangian expression) defined byL(x,y,λ)=f(x,y)+λ⋅g(x,y),{\displaystyle {\mathcal {L}}(x,y,\lambda )=f(x,y)+\lambda \cdot g(x,y),}where theλ{\displaystyle \lambda } term may be either added or subtracted. Iff(x0,y0){\displaystyle f(x_{0},y_{0})} is a maximum off(x,y){\displaystyle f(x,y)} for the original constrained problem and∇g(x0,y0)≠0,{\displaystyle \nabla g(x_{0},y_{0})\neq 0,} then there existsλ0{\displaystyle \lambda _{0}} such that (x0,y0,λ0{\displaystyle x_{0},y_{0},\lambda _{0}}) is astationary point for the Lagrange function (stationary points are those points where the first partial derivatives ofL{\displaystyle {\mathcal {L}}} are zero). The assumption∇g≠0{\displaystyle \nabla g\neq 0} is called constraint qualification. However, not all stationary points yield a solution of the original problem, as the method of Lagrange multipliers yields only anecessary condition for optimality in constrained problems.[9][10][11][12][13] Sufficient conditions for a minimum or maximumalso exist, but if a particularcandidate solution satisfies the sufficient conditions, it is only guaranteed that that solution is the best onelocally – that is, it is better than any permissible nearby points. Theglobal optimum can be found by comparing the values of the original objective function at the points satisfying the necessary and locally sufficient conditions.

The method of Lagrange multipliers relies on the intuition that at a maximum,f(x,y) cannot be increasing in the direction of any such neighboring point that also hasg = 0. If it were, we could walk alongg = 0 to get higher, meaning that the starting point wasn't actually the maximum. Viewed in this way, it is an exact analogue to testing if the derivative of an unconstrained function is0, that is, we are verifying that the directional derivative is 0 in any relevant (viable) direction.

We can visualizecontours off given byf(x,y) =d for various values ofd, and the contour ofg given byg(x,y) =c.

Suppose we walk along the contour line withg =c . We are interested in finding points wheref almost does not change as we walk, since these points might be maxima.

There are two ways this could happen:

  1. We could touch a contour line off, since by definitionf does not change as we walk along its contour lines. This would mean that the tangents to the contour lines off andg are parallel here.
  2. We have reached a "level" part off, meaning thatf does not change in any direction.

To check the first possibility (we touch a contour line off), notice that since thegradient of a function is perpendicular to the contour lines, the tangents to the contour lines off andg are parallel if and only if the gradients off andg are parallel. Thus we want points(x,y) whereg(x,y) =c and∇x,yf=λ∇x,yg,{\displaystyle \nabla _{x,y}f=\lambda \,\nabla _{x,y}g,}for someλ{\displaystyle \lambda } where∇x,yf=(∂f∂x,∂f∂y),∇x,yg=(∂g∂x,∂g∂y){\displaystyle \nabla _{x,y}f=\left({\frac {\partial f}{\partial x}},{\frac {\partial f}{\partial y}}\right),\qquad \nabla _{x,y}g=\left({\frac {\partial g}{\partial x}},{\frac {\partial g}{\partial y}}\right)}are the respective gradients. The constantλ{\displaystyle \lambda } is required because although the two gradient vectors are parallel, the magnitudes of the gradient vectors are generally not equal. This constant is called the Lagrange multiplier. (In some conventionsλ{\displaystyle \lambda } is preceded by a minus sign).

Notice that this method also solves the second possibility, thatf is level: iff is level, then its gradient is zero, and settingλ=0{\displaystyle \lambda =0} is a solution regardless of∇x,yg{\displaystyle \nabla _{x,y}g}.

To incorporate these conditions into one equation, we introduce an auxiliary functionL(x,y,λ)≡f(x,y)+λ⋅g(x,y),{\displaystyle {\mathcal {L}}(x,y,\lambda )\equiv f(x,y)+\lambda \cdot g(x,y)\,,}and solve∇x,y,λL(x,y,λ)=0 .{\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0~.}Note that this amounts to solving three equations in three unknowns. This is the method of Lagrange multipliers.

Note that ∇λL(x,y,λ)=0 {\displaystyle \ \nabla _{\lambda }{\mathcal {L}}(x,y,\lambda )=0\ } implies g(x,y)=0 ,{\displaystyle \ g(x,y)=0\ ,} as the partial derivative ofL{\displaystyle {\mathcal {L}}} with respect toλ{\displaystyle \lambda } is g(x,y) .{\displaystyle \ g(x,y)~.}

To summarize∇x,y,λL(x,y,λ)=0⟺{∇x,yf(x,y)=−λ∇x,yg(x,y)g(x,y)=0{\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0\iff {\begin{cases}\nabla _{x,y}f(x,y)=-\lambda \,\nabla _{x,y}g(x,y)\\g(x,y)=0\end{cases}}}The method generalizes readily to functions onn{\displaystyle n} variables∇x1,…,xn,λL(x1,…,xn,λ)=0{\displaystyle \nabla _{x_{1},\dots ,x_{n},\lambda }{\mathcal {L}}(x_{1},\dots ,x_{n},\lambda )=0}which amounts to solvingn + 1 equations inn + 1 unknowns.

The constrained extrema off arecritical points of the LagrangianL{\displaystyle {\mathcal {L}}}, but they are not necessarilylocal extrema ofL{\displaystyle {\mathcal {L}}} (see§ Example 2 below).

One mayreformulate the Lagrangian as aHamiltonian, in which case the solutions are local minima for the Hamiltonian. This is done inoptimal control theory, in the form ofPontryagin's maximum principle.

The fact that solutions of the method of Lagrange multipliers are not necessarily extrema of the Lagrangian, also poses difficulties for numerical optimization. This can be addressed by minimizing themagnitude of the gradient of the Lagrangian, as these minima are the same as the zeros of the magnitude, as illustrated inExample 5: Numerical optimization.

Multiple constraints

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Figure 2: A paraboloid constrained along two intersecting lines.
Figure 3: Contour map of Figure 2.

The method of Lagrange multipliers can be extended to solve problems with multiple constraints using a similar argument. Consider aparaboloid subject to two line constraints that intersect at a single point. As the only feasible solution, this point is obviously a constrained extremum. However, thelevel set off{\displaystyle f} is clearly not parallel to either constraint at the intersection point (see Figure 3); instead, it is a linear combination of the two constraints' gradients. In the case of multiple constraints, that will be what we seek in general: The method of Lagrange seeks points not at which the gradient off{\displaystyle f} is a multiple of any single constraint's gradient necessarily, but in which it is a linear combination of all the constraints' gradients.

Concretely, suppose we haveM{\displaystyle M} constraints and are walking along the set of points satisfyinggi(x)=0,i=1,…,M.{\displaystyle g_{i}(\mathbf {x} )=0,i=1,\dots ,M\,.} Every pointx{\displaystyle \mathbf {x} } on the contour of a given constraint functiongi{\displaystyle g_{i}} has a space of allowable directions: the space of vectors perpendicular to∇gi(x).{\displaystyle \nabla g_{i}(\mathbf {x} )\,.} The set of directions that are allowed by all constraints is thus the space of directions perpendicular to all of the constraints' gradients. Denote this space of allowable moves by A {\displaystyle \ A\ } and denote the span of the constraints' gradients byS.{\displaystyle S\,.} ThenA=S⊥,{\displaystyle A=S^{\perp }\,,} the space of vectors perpendicular to every element ofS.{\displaystyle S\,.}

We are still interested in finding points wheref{\displaystyle f} does not change as we walk, since these points might be (constrained) extrema. We therefore seekx{\displaystyle \mathbf {x} } such that any allowable direction of movement away fromx{\displaystyle \mathbf {x} } is perpendicular to∇f(x){\displaystyle \nabla f(\mathbf {x} )} (otherwise we could increasef{\displaystyle f} by moving along that allowable direction). In other words,∇f(x)∈A⊥=S.{\displaystyle \nabla f(\mathbf {x} )\in A^{\perp }=S\,.} Thus there are scalarsλ1,λ2, …,λM{\displaystyle \lambda _{1},\lambda _{2},\ \dots ,\lambda _{M}} such that∇f(x)=∑k=1Mλk∇gk(x)⟺∇f(x)−∑k=1Mλk∇gk(x)=0 .{\displaystyle \nabla f(\mathbf {x} )=\sum _{k=1}^{M}\lambda _{k}\,\nabla g_{k}(\mathbf {x} )\quad \iff \quad \nabla f(\mathbf {x} )-\sum _{k=1}^{M}{\lambda _{k}\nabla g_{k}(\mathbf {x} )}=0~.}

These scalars are the Lagrange multipliers. We now haveM{\displaystyle M} of them, one for every constraint.

As before, we introduce an auxiliary functionL(x1,…,xn,λ1,…,λM)=f(x1,…,xn)−∑k=1Mλkgk(x1,…,xn) {\displaystyle {\mathcal {L}}\left(x_{1},\ldots ,x_{n},\lambda _{1},\ldots ,\lambda _{M}\right)=f\left(x_{1},\ldots ,x_{n}\right)-\sum \limits _{k=1}^{M}{\lambda _{k}g_{k}\left(x_{1},\ldots ,x_{n}\right)}\ }and solve∇x1,…,xn,λ1,…,λML(x1,…,xn,λ1,…,λM)=0⟺{∇f(x)−∑k=1Mλk∇gk(x)=0g1(x)=⋯=gM(x)=0{\displaystyle \nabla _{x_{1},\ldots ,x_{n},\lambda _{1},\ldots ,\lambda _{M}}{\mathcal {L}}(x_{1},\ldots ,x_{n},\lambda _{1},\ldots ,\lambda _{M})=0\iff {\begin{cases}\nabla f(\mathbf {x} )-\sum _{k=1}^{M}{\lambda _{k}\,\nabla g_{k}(\mathbf {x} )}=0\\g_{1}(\mathbf {x} )=\cdots =g_{M}(\mathbf {x} )=0\end{cases}}}which amounts to solvingn+M{\displaystyle n+M} equations in n+M {\displaystyle \ n+M\ } unknowns.

The constraint qualification assumption when there are multiple constraints is that the constraint gradients at the relevant point are linearly independent.

Modern formulation via differentiable manifolds

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The problem of finding the local maxima and minima subject to constraints can be generalized to finding local maxima and minima on adifferentiable manifold M .{\displaystyle \ M~.}[14] In what follows, it is not necessary thatM{\displaystyle M} be a Euclidean space, or even aRiemannian manifold. All appearances of the gradient ∇ {\displaystyle \ \nabla \ } (which depends on a choice of Riemannian metric) can be replaced with theexterior derivative d⁡ .{\displaystyle \ \operatorname {d} ~.}

Single constraint

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Let M {\displaystyle \ M\ } be asmooth manifold of dimension m .{\displaystyle \ m~.} Suppose that we wish to find the stationary points x {\displaystyle \ x\ } of a smooth function f:M→R {\displaystyle \ f:M\to \mathbb {R} \ } when restricted to the submanifold N {\displaystyle \ N\ } defined by g(x)=0 ,{\displaystyle \ g(x)=0\ ,} where g:M→R {\displaystyle \ g:M\to \mathbb {R} \ } is a smooth function for which0 is aregular value.

Let d⁡f {\displaystyle \ \operatorname {d} f\ } and d⁡g {\displaystyle \ \operatorname {d} g\ } be theexterior derivatives of f {\displaystyle \ f\ } and g {\displaystyle \ g\ }. Stationarity for the restriction f|N {\displaystyle \ f|_{N}\ } at x∈N {\displaystyle \ x\in N\ } means d⁡(f|N)x=0 .{\displaystyle \ \operatorname {d} (f|_{N})_{x}=0~.} Equivalently, the kernel ker⁡(d⁡fx) {\displaystyle \ \ker(\operatorname {d} f_{x})\ } contains TxN=ker⁡(d⁡gx) .{\displaystyle \ T_{x}N=\ker(\operatorname {d} g_{x})~.} In other words, d⁡fx {\displaystyle \ \operatorname {d} f_{x}\ } and d⁡gx {\displaystyle \ \operatorname {d} g_{x}\ } are proportional 1-forms. For this it is necessary and sufficient that the following system of 12m(m−1) {\displaystyle \ {\tfrac {1}{2}}m(m-1)\ } equations holds:d⁡fx∧d⁡gx=0∈Λ2(Tx∗M){\displaystyle \operatorname {d} f_{x}\wedge \operatorname {d} g_{x}=0\in \Lambda ^{2}(T_{x}^{\ast }M)}where ∧ {\displaystyle \ \wedge \ } denotes theexterior product. The stationary points x {\displaystyle \ x\ } are the solutions of the above system of equations plus the constraint g(x)=0 .{\displaystyle \ g(x)=0~.} Note that the 12m(m−1) {\displaystyle \ {\tfrac {1}{2}}m(m-1)\ } equations are not independent, since the left-hand side of the equation belongs to the subvariety of Λ2(Tx∗M) {\displaystyle \ \Lambda ^{2}(T_{x}^{\ast }M)\ } consisting ofdecomposable elements.

In this formulation, it is not necessary to explicitly find the Lagrange multiplier, a number λ {\displaystyle \ \lambda \ } such that d⁡fx=λ⋅d⁡gx .{\displaystyle \ \operatorname {d} f_{x}=\lambda \cdot \operatorname {d} g_{x}~.}

Multiple constraints

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Let M {\displaystyle \ M\ } and f {\displaystyle \ f\ } be as in the above section regarding the case of a single constraint. Rather than the functiong{\displaystyle g} described there, now consider a smooth function G:M→Rp(p>1) ,{\displaystyle \ G:M\to \mathbb {R} ^{p}(p>1)\ ,} with component functions gi:M→R ,{\displaystyle \ g_{i}:M\to \mathbb {R} \ ,} for which0∈Rp{\displaystyle 0\in \mathbb {R} ^{p}} is aregular value. LetN{\displaystyle N} be the submanifold of M {\displaystyle \ M\ } defined by G(x)=0 .{\displaystyle \ G(x)=0~.}

 x {\displaystyle \ x\ } is a stationary point off|N{\displaystyle f|_{N}} if and only if ker⁡(d⁡fx) {\displaystyle \ \ker(\operatorname {d} f_{x})\ } contains ker⁡(d⁡Gx) .{\displaystyle \ \ker(\operatorname {d} G_{x})~.} For convenience let Lx=d⁡fx {\displaystyle \ L_{x}=\operatorname {d} f_{x}\ } and Kx=d⁡Gx ,{\displaystyle \ K_{x}=\operatorname {d} G_{x}\ ,} where d⁡G{\displaystyle \ \operatorname {d} G} denotes the tangent map or Jacobian TM→TRp {\displaystyle \ TM\to T\mathbb {R} ^{p}~} ( TxRp{\displaystyle \ T_{x}\mathbb {R} ^{p}} can be canonically identified with Rp{\displaystyle \ \mathbb {R} ^{p}}). The subspaceker⁡(Kx){\displaystyle \ker(K_{x})} has dimension smaller than that ofker⁡(Lx){\displaystyle \ker(L_{x})}, namely dim⁡(ker⁡(Lx))=n−1 {\displaystyle \ \dim(\ker(L_{x}))=n-1\ } and dim⁡(ker⁡(Kx))=n−p .{\displaystyle \ \dim(\ker(K_{x}))=n-p~.}ker⁡(Kx){\displaystyle \ker(K_{x})} belongs to ker⁡(Lx) {\displaystyle \ \ker(L_{x})\ } if and only ifLx∈Tx∗M{\displaystyle L_{x}\in T_{x}^{\ast }M} belongs to the image of Kx∗:Rp∗→Tx∗M .{\displaystyle \ K_{x}^{\ast }:\mathbb {R} ^{p\ast }\to T_{x}^{\ast }M~.} Computationally speaking, the condition is thatLx{\displaystyle L_{x}} belongs to the row space of the matrix of Kx ,{\displaystyle \ K_{x}\ ,} or equivalently the column space of the matrix ofKx∗{\displaystyle K_{x}^{\ast }} (the transpose). If ωx∈Λp(Tx∗M) {\displaystyle \ \omega _{x}\in \Lambda ^{p}(T_{x}^{\ast }M)\ } denotes the exterior product of the columns of the matrix of Kx∗ ,{\displaystyle \ K_{x}^{\ast }\ ,} the stationary condition for f|N {\displaystyle \ f|_{N}\ } at x {\displaystyle \ x\ } becomesLx∧ωx=0∈Λp+1(Tx∗M){\displaystyle L_{x}\wedge \omega _{x}=0\in \Lambda ^{p+1}\left(T_{x}^{\ast }M\right)}Once again, in this formulation it is not necessary to explicitly find the Lagrange multipliers, the numbers λ1,…,λp {\displaystyle \ \lambda _{1},\ldots ,\lambda _{p}\ } such that d⁡fx=∑i=1pλid⁡(gi)x .{\displaystyle \ \operatorname {d} f_{x}=\sum _{i=1}^{p}\lambda _{i}\operatorname {d} (g_{i})_{x}~.}

Interpretation of the Lagrange multipliers

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In this section, we modify the constraint equations from the formgi(x)=0{\displaystyle g_{i}({\bf {x}})=0} to the form gi(x)=ci ,{\displaystyle \ g_{i}({\bf {x}})=c_{i}\ ,} where the ci {\displaystyle \ c_{i}\ } arem real constants that are considered to be additional arguments of the Lagrangian expressionL{\displaystyle {\mathcal {L}}}.

Often the Lagrange multipliers have an interpretation as some quantity of interest. For example, by parametrising the constraint's contour line, that is, if the Lagrangian expression isL(x1,x2,…;λ1,λ2,…;c1,c2,…)=f(x1,x2,…)+λ1(c1−g1(x1,x2,…))+λ2(c2−g2(x1,x2,…))+⋯{\displaystyle {\begin{aligned}&{\mathcal {L}}(x_{1},x_{2},\ldots ;\lambda _{1},\lambda _{2},\ldots ;c_{1},c_{2},\ldots )\\[4pt]={}&f(x_{1},x_{2},\ldots )+\lambda _{1}(c_{1}-g_{1}(x_{1},x_{2},\ldots ))+\lambda _{2}(c_{2}-g_{2}(x_{1},x_{2},\dots ))+\cdots \end{aligned}}}then ∂L∂ck=λk .{\displaystyle \ {\frac {\partial {\mathcal {L}}}{\partial c_{k}}}=\lambda _{k}~.}

So,λk is the rate of change of the quantity being optimized as a function of the constraint parameter.As examples, inLagrangian mechanics the equations of motion are derived by finding stationary points of theaction, the time integral of the difference between kinetic and potential energy. Thus, the force on a particle due to a scalar potential,F = −∇V, can be interpreted as a Lagrange multiplier determining the change in action (transfer of potential to kinetic energy) following a variation in the particle's constrained trajectory. In control theory this is formulated instead ascostate equations.

Moreover, by theenvelope theorem the optimal value of a Lagrange multiplier has an interpretation as the marginal effect of the corresponding constraint constant upon the optimal attainable value of the original objective function: If we denote values at the optimum with a star (⋆{\displaystyle \star }), then it can be shown that d⁡f( x1⋆(c1,c2,…), x2⋆(c1,c2,…), … ) d⁡ck=λ⋆k .{\displaystyle {\frac {\ \operatorname {d} f\left(\ x_{1\star }(c_{1},c_{2},\dots ),\ x_{2\star }(c_{1},c_{2},\dots ),\ \dots \ \right)\ }{\operatorname {d} c_{k}}}=\lambda _{\star k}~.}

For example, in economics the optimal profit to a player is calculated subject to a constrained space of actions, where a Lagrange multiplier is the change in the optimal value of the objective function (profit) due to the relaxation of a given constraint (e.g. through a change in income); in such a context λ⋆k {\displaystyle \ \lambda _{\star k}\ } is themarginal cost of the constraint, and is referred to as theshadow price.[15]

Sufficient conditions

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Main article:Bordered Hessian

Sufficient conditions for a constrained local maximum or minimum can be stated in terms of a sequence of principal minors (determinants of upper-left-justified sub-matrices) of the borderedHessian matrix of second derivatives of the Lagrangian expression.[6][16]

Examples

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Example 1

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Illustration of the constrained optimization problem 1

Suppose we wish to maximize f(x,y)=x+y {\displaystyle \ f(x,y)=x+y\ } subject to the constraint x2+y2=1 .{\displaystyle \ x^{2}+y^{2}=1~.} Thefeasible set is the unit circle, and thelevel sets off are diagonal lines (with slope −1), so we can see graphically that the maximum occurs at (12,12) ,{\displaystyle \ \left({\tfrac {1}{\sqrt {2}}},{\tfrac {1}{\sqrt {2}}}\right)\ ,} and that the minimum occurs at (−12,−12) .{\displaystyle \ \left(-{\tfrac {1}{\sqrt {2}}},-{\tfrac {1}{\sqrt {2}}}\right)~.}

For the method of Lagrange multipliers, the constraint isg(x,y)=x2+y2−1=0 ,{\displaystyle g(x,y)=x^{2}+y^{2}-1=0\ ,}hence the Lagrangian function,L(x,y,λ)=f(x,y)+λ⋅g(x,y)=x+y+λ(x2+y2−1) ,{\displaystyle {\begin{aligned}{\mathcal {L}}(x,y,\lambda )&=f(x,y)+\lambda \cdot g(x,y)\\[4pt]&=x+y+\lambda (x^{2}+y^{2}-1)\ ,\end{aligned}}}is a function that is equivalent to f(x,y) {\displaystyle \ f(x,y)\ } when g(x,y) {\displaystyle \ g(x,y)\ } is set to0.

Now we can calculate the gradient:∇x,y,λL(x,y,λ)=(∂L∂x,∂L∂y,∂L∂λ)=(1+2λx,1+2λy,x2+y2−1) ,{\displaystyle {\begin{aligned}\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )&=\left({\frac {\partial {\mathcal {L}}}{\partial x}},{\frac {\partial {\mathcal {L}}}{\partial y}},{\frac {\partial {\mathcal {L}}}{\partial \lambda }}\right)\\[4pt]&=\left(1+2\lambda x,1+2\lambda y,x^{2}+y^{2}-1\right)\ \color {gray}{,}\end{aligned}}}and therefore:∇x,y,λL(x,y,λ)=0⇔{1+2λx=01+2λy=0x2+y2−1=0{\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0\quad \Leftrightarrow \quad {\begin{cases}1+2\lambda x=0\\1+2\lambda y=0\\x^{2}+y^{2}-1=0\end{cases}}}

Notice that the last equation is the original constraint.

The first two equations yieldx=y=−12λ,λ≠0 .{\displaystyle x=y=-{\frac {1}{2\lambda }},\qquad \lambda \neq 0~.}By substituting into the last equation we have:14λ2+14λ2−1=0 ,{\displaystyle {\frac {1}{4\lambda ^{2}}}+{\frac {1}{4\lambda ^{2}}}-1=0\ ,}soλ=±12  ,{\displaystyle \lambda =\pm {\frac {1}{\sqrt {2\ }}}\ ,}which implies that the stationary points ofL{\displaystyle {\mathcal {L}}} are(2 2,2 2,−12 ),(−2 2,−2 2,12 ) .{\displaystyle \left({\tfrac {\sqrt {2\ }}{2}},{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {1}{\sqrt {2\ }}}\right),\qquad \left(-{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {\sqrt {2\ }}{2}},{\tfrac {1}{\sqrt {2\ }}}\right)~.}

Evaluating the objective functionf at these points yieldsf(2 2,2 2)=2  ,f(−2 2,−2 2)=−2  .{\displaystyle f\left({\tfrac {\sqrt {2\ }}{2}},{\tfrac {\sqrt {2\ }}{2}}\right)={\sqrt {2\ }}\ ,\qquad f\left(-{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {\sqrt {2\ }}{2}}\right)=-{\sqrt {2\ }}~.}

Thus the constrained maximum is 2  {\displaystyle \ {\sqrt {2\ }}\ } and the constrained minimum is−2{\displaystyle -{\sqrt {2}}}.

Example 2

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Illustration of the constrained optimization problem 2

Now we modify the objective function of Example 1 so that we minimize f(x,y)=(x+y)2 {\displaystyle \ f(x,y)=(x+y)^{2}\ } instead of f(x,y)=x+y ,{\displaystyle \ f(x,y)=x+y\ ,} again along the circle g(x,y)=x2+y2−1=0 .{\displaystyle \ g(x,y)=x^{2}+y^{2}-1=0~.} Now the level sets off{\displaystyle f} are still lines of slope −1, and the points on the circle tangent to these level sets are again (2/2,2/2) {\displaystyle \ ({\sqrt {2}}/2,{\sqrt {2}}/2)\ } and (−2/2,−2/2) .{\displaystyle \ (-{\sqrt {2}}/2,-{\sqrt {2}}/2)~.} These tangency points are maxima of f .{\displaystyle \ f~.}

On the other hand, the minima occur on the level set for f=0 {\displaystyle \ f=0\ } (since by its construction f {\displaystyle \ f\ } cannot take negative values), at (2/2,−2/2) {\displaystyle \ ({\sqrt {2}}/2,-{\sqrt {2}}/2)\ } and (−2/2,2/2) ,{\displaystyle \ (-{\sqrt {2}}/2,{\sqrt {2}}/2)\ ,} where the level curves of f {\displaystyle \ f\ } are not tangent to the constraint. The condition that ∇x,y,λ(f(x,y)+λ⋅g(x,y))=0 {\displaystyle \ \nabla _{x,y,\lambda }\left(f(x,y)+\lambda \cdot g(x,y)\right)=0\ } correctly identifies all four points as extrema; the minima are characterized in by λ=0 {\displaystyle \ \lambda =0\ } and the maxima by λ=−2 .{\displaystyle \ \lambda =-2~.}

Example 3

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Illustration of constrained optimization problem 3.

This example deals with more strenuous calculations, but it is still a single constraint problem.

Suppose one wants to find the maximum values off(x,y)=x2y{\displaystyle f(x,y)=x^{2}y}with the condition that the x {\displaystyle \ x\ }- and y {\displaystyle \ y\ }-coordinates lie on the circle around the origin with radius 3  .{\displaystyle \ {\sqrt {3\ }}~.} That is, subject to the constraintg(x,y)=x2+y2−3=0 .{\displaystyle g(x,y)=x^{2}+y^{2}-3=0~.}

As there is just a single constraint, there is a single multiplier, say λ .{\displaystyle \ \lambda ~.}

The constraint g(x,y) {\displaystyle \ g(x,y)\ } is identically zero on the circle of radius 3  .{\displaystyle \ {\sqrt {3\ }}~.} Any multiple of g(x,y) {\displaystyle \ g(x,y)\ } may be added to g(x,y) {\displaystyle \ g(x,y)\ } leaving g(x,y) {\displaystyle \ g(x,y)\ } unchanged in the region of interest (on the circle where our original constraint is satisfied).

Applying the ordinary Lagrange multiplier method yieldsL(x,y,λ)=f(x,y)+λ⋅g(x,y)=x2y+λ(x2+y2−3) ,{\displaystyle {\begin{aligned}{\mathcal {L}}(x,y,\lambda )&=f(x,y)+\lambda \cdot g(x,y)\\&=x^{2}y+\lambda (x^{2}+y^{2}-3)\ ,\end{aligned}}}from which the gradient can be calculated:∇x,y,λL(x,y,λ)=(∂L∂x,∂L∂y,∂L∂λ)=(2xy+2λx,x2+2λy,x2+y2−3) .{\displaystyle {\begin{aligned}\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )&=\left({\frac {\partial {\mathcal {L}}}{\partial x}},{\frac {\partial {\mathcal {L}}}{\partial y}},{\frac {\partial {\mathcal {L}}}{\partial \lambda }}\right)\\&=\left(2xy+2\lambda x,x^{2}+2\lambda y,x^{2}+y^{2}-3\right)~.\end{aligned}}}And therefore:∇x,y,λL(x,y,λ)=0⟺{2xy+2λx=0x2+2λy=0x2+y2−3=0⟺{x(y+λ)=0(i)x2=−2λy(ii)x2+y2=3(iii){\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0\quad \iff \quad {\begin{cases}2xy+2\lambda x=0\\x^{2}+2\lambda y=0\\x^{2}+y^{2}-3=0\end{cases}}\quad \iff \quad {\begin{cases}x(y+\lambda )=0&{\text{(i)}}\\x^{2}=-2\lambda y&{\text{(ii)}}\\x^{2}+y^{2}=3&{\text{(iii)}}\end{cases}}}(iii) is just the original constraint. (i) implies x=0 {\displaystyle \ x=0\ } or λ=−y .{\displaystyle \ \lambda =-y~.} Ifx=0{\displaystyle x=0} then y=±3  {\displaystyle \ y=\pm {\sqrt {3\ }}\ } by (iii) and consequently λ=0 {\displaystyle \ \lambda =0\ } from (ii). If λ=−y ,{\displaystyle \ \lambda =-y\ ,} substituting this into (ii) yields x2=2y2 .{\displaystyle \ x^{2}=2y^{2}~.} Substituting this into (iii) and solving for y {\displaystyle \ y\ } gives y=±1 .{\displaystyle \ y=\pm 1~.} Thus there are six critical points of L :{\displaystyle \ {\mathcal {L}}\ :}(2 ,1,−1);(−2 ,1,−1);(2 ,−1,1);(−2 ,−1,1);(0,3 ,0);(0,−3 ,0) .{\displaystyle ({\sqrt {2\ }},1,-1);\quad (-{\sqrt {2\ }},1,-1);\quad ({\sqrt {2\ }},-1,1);\quad (-{\sqrt {2\ }},-1,1);\quad (0,{\sqrt {3\ }},0);\quad (0,-{\sqrt {3\ }},0)~.}

Evaluating the objective at these points, one finds thatf(±2 ,1)=2;f(±2 ,−1)=−2;f(0,±3 )=0 .{\displaystyle f(\pm {\sqrt {2\ }},1)=2;\quad f(\pm {\sqrt {2\ }},-1)=-2;\quad f(0,\pm {\sqrt {3\ }})=0~.}

Therefore, the objective function attains theglobal maximum (subject to the constraints) at (±2 ,1 ){\displaystyle \ (\pm {\sqrt {2\ }},1\ )} and theglobal minimum at (±2 ,−1) .{\displaystyle \ (\pm {\sqrt {2\ }},-1)~.} The point (0,3 ) {\displaystyle \ (0,{\sqrt {3\ }})\ } is alocal minimum of f {\displaystyle \ f\ } and (0,−3 ) {\displaystyle \ (0,-{\sqrt {3\ }})\ } is alocal maximum of f ,{\displaystyle \ f\ ,} as may be determined by consideration of theHessian matrix of L(x,y,0) .{\displaystyle \ {\mathcal {L}}(x,y,0)~.}

Note that while (2 ,1,−1) {\displaystyle \ ({\sqrt {2\ }},1,-1)\ } is a critical point of L ,{\displaystyle \ {\mathcal {L}}\ ,} it is not a local extremum of L .{\displaystyle \ {\mathcal {L}}~.} We haveL(2 +ε,1,−1+δ)=2+δ(ε2+(22 )ε) .{\displaystyle {\mathcal {L}}\left({\sqrt {2\ }}+\varepsilon ,1,-1+\delta \right)=2+\delta \left(\varepsilon ^{2}+\left(2{\sqrt {2\ }}\right)\varepsilon \right)~.}

Given any neighbourhood of (2 ,1,−1) ,{\displaystyle \ ({\sqrt {2\ }},1,-1)\ ,} one can choose a small positive ε {\displaystyle \ \varepsilon \ } and a small δ {\displaystyle \ \delta \ } of either sign to get L{\displaystyle \ {\mathcal {L}}} values both greater and less than 2 .{\displaystyle \ 2~.} This can also be seen from the Hessian matrix of L {\displaystyle \ {\mathcal {L}}\ } evaluated at this point (or indeed at any of the critical points) which is anindefinite matrix. Each of the critical points of L {\displaystyle \ {\mathcal {L}}\ } is asaddle point of L .{\displaystyle \ {\mathcal {L}}~.}[4]

Example 4 – Entropy

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Suppose we wish to find thediscrete probability distribution on the points {p1,p2,…,pn} {\displaystyle \ \{p_{1},p_{2},\ldots ,p_{n}\}\ } with maximalinformation entropy. This is the same as saying that we wish to find theleast structured probability distribution on the points {p1,p2,⋯,pn} .{\displaystyle \ \{p_{1},p_{2},\cdots ,p_{n}\}~.} In other words, we wish to maximize theShannon entropy equation:f(p1,p2,…,pn)=−∑j=1npjlog2⁡pj .{\displaystyle f(p_{1},p_{2},\ldots ,p_{n})=-\sum _{j=1}^{n}p_{j}\log _{2}p_{j}~.}

For this to be a probability distribution the sum of the probabilities pi {\displaystyle \ p_{i}\ } at each point xi {\displaystyle \ x_{i}\ } must equal 1, so our constraint is:g(p1,p2,…,pn)=∑j=1npj=1 .{\displaystyle g(p_{1},p_{2},\ldots ,p_{n})=\sum _{j=1}^{n}p_{j}=1~.}

We use Lagrange multipliers to find the point of maximum entropy, p→∗ ,{\displaystyle \ {\vec {p}}^{\,*}\ ,} across all discrete probability distributions p→ {\displaystyle \ {\vec {p}}\ } on {x1,x2,…,xn} .{\displaystyle \ \{x_{1},x_{2},\ldots ,x_{n}\}~.} We require that:∂∂p→(f+λ(g−1))|p→=p→∗=0 ,{\displaystyle \left.{\frac {\partial }{\partial {\vec {p}}}}(f+\lambda (g-1))\right|_{{\vec {p}}={\vec {p}}^{\,*}}=0\ ,}which gives a system ofn equations, k=1, …,n ,{\displaystyle \ k=1,\ \ldots ,n\ ,} such that:∂∂pk{−(∑j=1npjlog2⁡pj)+λ(∑j=1npj−1)}|pk=p⋆k=0 .{\displaystyle \left.{\frac {\partial }{\partial p_{k}}}\left\{-\left(\sum _{j=1}^{n}p_{j}\log _{2}p_{j}\right)+\lambda \left(\sum _{j=1}^{n}p_{j}-1\right)\right\}\right|_{p_{k}=p_{\star k}}=0~.}

Carrying out the differentiation of thesen equations, we get−(1ln⁡2+log2⁡p⋆k)+λ=0 .{\displaystyle -\left({\frac {1}{\ln 2}}+\log _{2}p_{\star k}\right)+\lambda =0~.}

This shows that all p⋆k {\displaystyle \ p_{\star k}\ } are equal (because they depend onλ only). By using the constraint∑jpj=1 ,{\displaystyle \sum _{j}p_{j}=1\ ,}we findp⋆k=1n .{\displaystyle p_{\star k}={\frac {1}{n}}~.}

Hence, the uniform distribution is the distribution with the greatest entropy, among distributions onn points.

Example 5 – Numerical optimization

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Lagrange multipliers cause the critical points to occur at saddle points (Example 5).
The magnitude of the gradient can be used to force the critical points to occur at local minima (Example 5).

The critical points of Lagrangians occur atsaddle points, rather than at local maxima (or minima).[4][17] Unfortunately, many numerical optimization techniques, such ashill climbing,gradient descent, some of thequasi-Newton methods, among others, are designed to find local maxima (or minima) and not saddle points. For this reason, one must either modify the formulation to ensure that it's a minimization problem (for example, by extremizing the square of thegradient of the Lagrangian as below), or else use an optimization technique that findsstationary points (such asNewton's method without an extremum seekingline search) and not necessarily extrema.

As a simple example, consider the problem of finding the value ofx that minimizes f(x)=x2 ,{\displaystyle \ f(x)=x^{2}\ ,} constrained such that x2=1 .{\displaystyle \ x^{2}=1~.} (This problem is somewhat untypical because there are only two values that satisfy this constraint, but it is useful for illustration purposes because the corresponding unconstrained function can be visualized in three dimensions.)

Using Lagrange multipliers, this problem can be converted into an unconstrained optimization problem:L(x,λ)=x2+λ(x2−1) .{\displaystyle {\mathcal {L}}(x,\lambda )=x^{2}+\lambda (x^{2}-1)~.}

The two critical points occur at saddle points wherex = 1 andx = −1.

In order to solve this problem with a numerical optimization technique, we must first transform this problem such that the critical points occur at local minima. This is done by computing the magnitude of the gradient of the unconstrained optimization problem.

First, we compute the partial derivative of the unconstrained problem with respect to each variable:∂L∂x=2x+2xλ∂L∂λ=x2−1 .{\displaystyle {\begin{aligned}&{\frac {\partial {\mathcal {L}}}{\partial x}}=2x+2x\lambda \\[5pt]&{\frac {\partial {\mathcal {L}}}{\partial \lambda }}=x^{2}-1~.\end{aligned}}}

If the target function is not easily differentiable, the differential with respect to each variable can be approximated as ∂L ∂x≈L(x+ε,λ)−L(x,λ)ε, ∂L ∂λ≈L(x,λ+ε)−L(x,λ)ε,{\displaystyle {\begin{aligned}{\frac {\ \partial {\mathcal {L}}\ }{\partial x}}\approx {\frac {{\mathcal {L}}(x+\varepsilon ,\lambda )-{\mathcal {L}}(x,\lambda )}{\varepsilon }},\\[5pt]{\frac {\ \partial {\mathcal {L}}\ }{\partial \lambda }}\approx {\frac {{\mathcal {L}}(x,\lambda +\varepsilon )-{\mathcal {L}}(x,\lambda )}{\varepsilon }},\end{aligned}}}whereε{\displaystyle \varepsilon } is a small value.

Next, we compute the magnitude of the gradient, which is the square root of the sum of the squares of the partial derivatives:h(x,λ)=(2x+2xλ)2+(x2−1)2 ≈( L(x+ε,λ)−L(x,λ) ε)2+( L(x,λ+ε)−L(x,λ) ε)2  .{\displaystyle {\begin{aligned}h(x,\lambda )&={\sqrt {(2x+2x\lambda )^{2}+(x^{2}-1)^{2}\ }}\\[4pt]&\approx {\sqrt {\left({\frac {\ {\mathcal {L}}(x+\varepsilon ,\lambda )-{\mathcal {L}}(x,\lambda )\ }{\varepsilon }}\right)^{2}+\left({\frac {\ {\mathcal {L}}(x,\lambda +\varepsilon )-{\mathcal {L}}(x,\lambda )\ }{\varepsilon }}\right)^{2}\ }}~.\end{aligned}}}

(Since magnitude is always non-negative, optimizing over the squared-magnitude is equivalent to optimizing over the magnitude. Thus, the "square root" may be omitted from these equations with no expected difference in the results of optimization.)

The critical points ofh occur atx = 1 andx = −1, just as inL .{\displaystyle {\mathcal {L}}~.} Unlike the critical points inL,{\displaystyle {\mathcal {L}}\,,} however, the critical points inh occur at local minima, so numerical optimization techniques can be used to find them.

Applications

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Lagrangian Mechanics

[edit]

InLagrangian Mechanics, the Euler-Lagrange equations can be augmented with Lagrange multipliers as a method to impose physical constraints on systems.[18] This method is not required in general, because an alternative method is to choose a set of linearly independent generalised coordinates such that the constraints are implicitly imposed.

When Lagrange multipliers are used, the constraint equations need to be simultaneously solved with the Euler-Lagrange equations. Hence, the equations become a system ofdifferential algebraic equations (as opposed to a system ofordinary differential equations).[19]

The method of Lagrange multipliers is useful when it is difficult to write the Lagrangian in terms of a set of linearly independent generalised coordinates. For example, for use in programmatic dynamical systems modelling algorithms, or for use in modelling systems with closed kinematic chains.[20] They are also useful for imposing non-holonomic constraints.[18][20]

Given a set of holonomic constraint equationsfj(q,t)=0{\displaystyle f_{j}(\mathbf {q} ,t)=0}, the Euler-Lagrange equations with Lagrange multipliers can be written as[18][19]

ddt∂L∂q˙i−∂L∂qi+∑j=1Cλj∂fj∂qi⏟−τi,constraint=τi{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial L}{\partial {\dot {q}}_{i}}}-{\frac {\partial L}{\partial q_{i}}}+\underbrace {\sum _{j=1}^{C}\lambda _{j}{\frac {\partial f_{j}}{\partial q_{i}}}} _{-\tau _{i,{\text{constraint}}}}=\tau _{i}}

The meaning ofτi,constraint{\displaystyle \tau _{i,{\text{constraint}}}} can be interpreted by moving it to the other side of the equation and absorbing it into the generalised force termτi{\displaystyle \tau _{i}}. In this interpretation, the system hasC{\displaystyle C} number of additional degrees of freedom, and there are no additionally imposed constraints, but the constraint forcesτi,constraint{\displaystyle \tau _{i,{\text{constraint}}}} just happen to have the right values such that the constraints hold.[18][19]

Control theory

[edit]

Inoptimal control theory, the Lagrange multipliers are interpreted ascostate variables, and Lagrange multipliers are reformulated as the minimization of theHamiltonian, inPontryagin's maximum principle.

Nonlinear programming

[edit]

The Lagrange multiplier method has several generalizations. Innonlinear programming there are several multiplier rules, e.g. the Carathéodory–John Multiplier Rule and the Convex Multiplier Rule, for inequality constraints.[21]

Economics

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In many models inmathematical economics such asgeneral equilibrium models, consumer behavior is implemented asutility maximization and firm behavior asprofit maximization, both entities being subject to constraints such asbudget constraints andproduction constraints. The usual way to determine an optimal solution is achieved by maximizing some function, where the constraints are enforced using Lagrangian multipliers.[22][23][24][25]

Power systems

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Methods based on Lagrange multipliers have applications inpower systems, e.g. in distributed-energy-resources (DER) placement and load shedding.[26]

Safe Reinforcement Learning

[edit]

The method of Lagrange multipliers applies to constrainedMarkov decision processes.[27] It naturally produces gradient-based primal-dual algorithms in safe reinforcement learning.[28]

Normalized solutions

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Considering the PDE problems with constraints, i.e., the study of the properties of the normalized solutions, Lagrange multipliers play an important role.

See also

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References

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  1. ^Hoffmann, Laurence D.; Bradley, Gerald L. (2004).Calculus for Business, Economics, and the Social and Life Sciences (8th ed.). McGraw Hill Higher Education. pp. 575–588.ISBN 0-07-242432-X.
  2. ^Beavis, Brian; Dobbs, Ian M. (1990)."Static Optimization".Optimization and Stability Theory for Economic Analysis. New York: Cambridge University Press. p. 40.ISBN 0-521-33605-8.
  3. ^Protter, Murray H.;Morrey, Charles B. Jr. (1985).Intermediate Calculus (2nd ed.). New York, NY: Springer. p. 267.ISBN 0-387-96058-9.
  4. ^abcWalsh, G.R. (1975)."Saddle-point Property of Lagrangian Function".Methods of Optimization. New York, NY: John Wiley & Sons. pp. 39–44.ISBN 0-471-91922-5.
  5. ^Kalman, Dan (2009). "Leveling with Lagrange: An alternate view of constrained optimization".Mathematics Magazine.82 (3):186–196.doi:10.1080/0025570X.2009.11953617.JSTOR 27765899.S2CID 121070192.
  6. ^abSilberberg, Eugene; Suen, Wing (2001).The Structure of Economics: A Mathematical Analysis (Third ed.). Boston: Irwin McGraw-Hill. pp. 134–141.ISBN 0-07-234352-4.
  7. ^de la Fuente, Angel (2000).Mathematical Methods and Models for Economists. Cambridge: Cambridge University Press. p. 285.doi:10.1017/CBO9780511810756.ISBN 978-0-521-58512-5.
  8. ^Luenberger, David G. (1969).Optimization by Vector Space Methods. New York: John Wiley & Sons. pp. 188–189.
  9. ^Bertsekas, Dimitri P. (1999).Nonlinear Programming (Second ed.). Cambridge, MA: Athena Scientific.ISBN 1-886529-00-0.
  10. ^Vapnyarskii, I.B. (2001) [1994],"Lagrange multipliers",Encyclopedia of Mathematics,EMS Press.
  11. ^Lasdon, Leon S. (2002) [1970].Optimization Theory for Large Systems (reprint ed.). Mineola, New York, NY: Dover.ISBN 0-486-41999-1.MR 1888251.
  12. ^Hiriart-Urruty, Jean-Baptiste;Lemaréchal, Claude (1993). "Chapter XII: Abstract duality for practitioners".Convex analysis and minimization algorithms. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences]. Vol. 306. Berlin, DE: Springer-Verlag. pp. 136–193 (and Bibliographical comments pp. 334–335).ISBN 3-540-56852-2.MR 1295240. Volume II: Advanced theory and bundle methods.
  13. ^Lemaréchal, Claude (15–19 May 2000). "Lagrangian relaxation". In Jünger, Michael; Naddef, Denis (eds.).Computational combinatorial optimization: Papers from the Spring School held in Schloß Dagstuhl. Spring School held in Schloß Dagstuhl,May 15–19, 2000. Lecture Notes in Computer Science. Vol. 2241. Berlin, DE: Springer-Verlag (published 2001). pp. 112–156.doi:10.1007/3-540-45586-8_4.ISBN 3-540-42877-1.MR 1900016.S2CID 9048698.
  14. ^Lafontaine, Jacques (2015).An Introduction to Differential Manifolds. Springer. p. 70.ISBN 978-3-319-20735-3.
  15. ^Dixit, Avinash K. (1990)."Shadow Prices".Optimization in Economic Theory (2nd ed.). New York: Oxford University Press. pp. 40–54.ISBN 0-19-877210-6.
  16. ^Chiang, Alpha C. (1984).Fundamental Methods of Mathematical Economics (Third ed.). McGraw-Hill. p. 386.ISBN 0-07-010813-7.
  17. ^Heath, Michael T. (2005).Scientific Computing: An introductory survey. McGraw-Hill. p. 203.ISBN 978-0-07-124489-3.
  18. ^abcdGoldstein, Herbert;Poole, Charles P. Jr.; Safko, John L. (2002).Classical Mechanics (3rd ed.). San Francisco, CA: Addison Wesley. pp. 45–51.ISBN 0-201-65702-3.
  19. ^abcBrenan, K. E.; Campbell, S. L.; Petzold, L. R. (1995).Numerical Solution of Initial-Value Problems in Differential-Algebraic Equations. Society for Industrial and Applied Mathematics. pp. 4–5.doi:10.1137/1.9781611971224.
  20. ^abRoy, Featherstone (2008).Rigid Body Dynamics Algorithms. Springer New York. pp. 42–45,141–143.doi:10.1007/978-1-4899-7560-7.
  21. ^Pourciau, Bruce H. (1980)."Modern multiplier rules".American Mathematical Monthly.87 (6):433–452.doi:10.2307/2320250.JSTOR 2320250.
  22. ^Kamien, M. I.;Schwartz, N. L. (1991).Dynamic Optimization: The Calculus of Variations and Optimal Control in Economics and Management (Second ed.). New York: Elsevier.ISBN 0-444-01609-0.
  23. ^Glötzl, Erhard; Glötzl, Florentin; Richters, Oliver (2019). "From constrained optimization to constrained dynamics: extending analogies between economics and mechanics".Journal of Economic Interaction and Coordination.14 (3):623–642.doi:10.1007/s11403-019-00252-7.hdl:10419/171974.
  24. ^Baxley, John V.; Moorhouse, John C. (1984). "Lagrange Multiplier Problems in Economics".The American Mathematical Monthly.91 (7):404–412.doi:10.1080/00029890.1984.11971446..
  25. ^Janová, Jitka (2011). "Applications of a constrained mechanics methodology in economics".European Journal of Physics.32 (6):1443–1463.arXiv:1106.3455.Bibcode:2011EJPh...32.1443J.doi:10.1088/0143-0807/32/6/001.
  26. ^Gautam, Mukesh; Bhusal, Narayan; Benidris, Mohammed (2020).A sensitivity-based approach to adaptive under-frequency load shedding. 2020 IEEE Texas Power and Energy Conference (TPEC).Institute of Electronic and Electrical Engineers. pp. 1–5.doi:10.1109/TPEC48276.2020.9042569.
  27. ^Altman, Eitan (2021).Constrained Markov Decision Processes.Routledge.
  28. ^Ding, Dongsheng; Zhang, Kaiqing; Jovanovic, Mihailo; Basar, Tamer (2020).Natural policy gradient primal-dual method for constrained Markov decision processes. Advances in Neural Information Processing Systems.

Further reading

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  • Beavis, Brian; Dobbs, Ian M. (1990)."Static Optimization".Optimization and Stability Theory for Economic Analysis. New York, NY: Cambridge University Press. pp. 32–72.ISBN 0-521-33605-8.
  • Bertsekas, Dimitri P. (1982).Constrained optimization and Lagrange multiplier methods. New York, NY: Academic Press.ISBN 0-12-093480-9.
  • Beveridge, Gordon S.G.; Schechter, Robert S. (1970)."Lagrangian multipliers".Optimization: Theory and Practice. New York, NY: McGraw-Hill. pp. 244–259.ISBN 0-07-005128-3.
  • Binger, Brian R.; Hoffman, Elizabeth (1998). "Constrained optimization".Microeconomics with Calculus (2nd ed.). Reading: Addison-Wesley. pp. 56–91.ISBN 0-321-01225-9.
  • Carter, Michael (2001)."Equality constraints".Foundations of Mathematical Economics. Cambridge, MA: MIT Press. pp. 516–549.ISBN 0-262-53192-5.
  • Hestenes, Magnus R. (1966). "Minima of functions subject to equality constraints".Calculus of Variations and Optimal Control Theory. New York, NY: Wiley. pp. 29–34.
  • Wylie, C. Ray; Barrett, Louis C. (1995). "The extrema of integrals under constraint".Advanced Engineering Mathematics (Sixth ed.). New York, NY: McGraw-Hill. pp. 1096–1103.ISBN 0-07-072206-4.

External links

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The WikibookCalculus optimization methods has a page on the topic of:Lagrange multipliers

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