Single constraint

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

Let ${\displaystyle \ \operatorname {d} f\ }$ and ${\displaystyle \ \operatorname {d} g\ }$ be the exterior derivatives of ${\displaystyle \ f\ }$ and ${\displaystyle \ g\ }$. Stationarity for the restriction ${\displaystyle \ f|_{N}\ }$ at ${\displaystyle \ x\in N\ }$ means ${\displaystyle \ \operatorname {d} (f|_{N})_{x}=0~.}$ Equivalently, the kernel ${\displaystyle \ \ker(\operatorname {d} f_{x})\ }$ contains ${\displaystyle \ T_{x}N=\ker(\operatorname {d} g_{x})~.}$ In other words, ${\displaystyle \ \operatorname {d} f_{x}\ }$ and ${\displaystyle \ \operatorname {d} g_{x}\ }$ are proportional 1-forms. For this it is necessary and sufficient that the following system of ${\displaystyle \ {\tfrac {1}{2}}m(m-1)\ }$ equations holds:

${\displaystyle \operatorname {d} f_{x}\wedge \operatorname {d} g_{x}=0\in \Lambda ^{2}(T_{\star x}M)}$

where ${\displaystyle \ \wedge \ }$ denotes the exterior product. The stationary points ${\displaystyle \ x\ }$ are the solutions of the above system of equations plus the constraint ${\displaystyle \ g(x)=0~.}$ Note that the ${\displaystyle \ {\tfrac {1}{2}}m(m-1)\ }$ equations are not independent, since the left-hand side of the equation belongs to the subvariety of ${\displaystyle \ \Lambda ^{2}(T_{\star x}M)\ }$ consisting of decomposable elements.

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

Multiple constraints

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

${\displaystyle \ x\ }$ is a stationary point of ${\displaystyle f|_{N}}$ if and only if ${\displaystyle \ \ker(\operatorname {d} f_{x})\ }$ contains ${\displaystyle \ \ker(\operatorname {d} G_{x})~.}$ For convenience let ${\displaystyle \ L_{x}=\operatorname {d} f_{x}\ }$ and ${\displaystyle \ K_{x}=\operatorname {d} G_{x}\ ,}$ where ${\displaystyle \ \operatorname {d} G}$ denotes the tangent map or Jacobian ${\displaystyle \ TM\to T\mathbb {R} ^{p}~.}$ The subspace ${\displaystyle \ker(K_{x})}$ has dimension smaller than that of ${\displaystyle \ker(L_{x})}$, namely ${\displaystyle \ \dim(\ker(L_{x}))=n-1\ }$ and ${\displaystyle \ \dim(\ker(K_{x}))=n-p~.}$ ${\displaystyle \ker(K_{x})}$ belongs to ${\displaystyle \ \ker(L_{x})\ }$ if and only if ${\displaystyle L_{x}\in T_{\star x}M}$ belongs to the image of ${\displaystyle \ K_{\star x}:\mathbb {R} _{\star }^{p}\to T_{\star x}M~.}$ Computationally speaking, the condition is that ${\displaystyle L_{x}}$ belongs to the row space of the matrix of ${\displaystyle \ K_{x}\ ,}$ or equivalently the column space of the matrix of ${\displaystyle K_{\star x}}$ (the transpose). If ${\displaystyle \ \omega _{x}\in \Lambda ^{p}(T_{\star x}M)\ }$ denotes the exterior product of the columns of the matrix of ${\displaystyle \ K_{\star x}\ ,}$ the stationary condition for ${\displaystyle \ f|_{N}\ }$ at ${\displaystyle \ x\ }$ becomes

${\displaystyle L_{x}\wedge \omega _{x}=0\in \Lambda ^{p+1}\left(T_{\star x}M\right)}$

Once again, in this formulation it is not necessary to explicitly find the Lagrange multipliers, the numbers ${\displaystyle \ \lambda _{1},\ldots ,\lambda _{p}\ }$ such that

${\displaystyle \ \operatorname {d} f_{x}=\sum _{i=1}^{p}\lambda _{i}\operatorname {d} (g_{i})_{x}~.}$

Example 1

Illustration of the constrained optimization problem 1

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

For the method of Lagrange multipliers, the constraint is

${\displaystyle g(x,y)=x^{2}+y^{2}-1=0\ ,}$

hence the Lagrangian function,

${\displaystyle {\begin{aligned}{\mathcal {L}}(x,y,\lambda )&=f(x,y)+\lambda \cdot g(x,y)\\[4pt]&=x+y+\lambda (x^{2}+y^{2}-1)\ ,\end{aligned}}}$

is a function that is equivalent to ${\displaystyle \ f(x,y)\ }$ when ${\displaystyle \ g(x,y)\ }$ is set to 0.

Now we can calculate the gradient:

${\displaystyle {\begin{aligned}\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )&=\left({\frac {\partial {\mathcal {L}}}{\partial x}},{\frac {\partial {\mathcal {L}}}{\partial y}},{\frac {\partial {\mathcal {L}}}{\partial \lambda }}\right)\\[4pt]&=\left(1+2\lambda x,1+2\lambda y,x^{2}+y^{2}-1\right)\ \color {gray}{,}\end{aligned}}}$

and therefore:

${\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0\quad \Leftrightarrow \quad {\begin{cases}1+2\lambda x=0\\1+2\lambda y=0\\x^{2}+y^{2}-1=0\end{cases}}}$

Notice that the last equation is the original constraint.

The first two equations yield

${\displaystyle x=y=-{\frac {1}{2\lambda }},\qquad \lambda \neq 0~.}$

By substituting into the last equation we have:

${\displaystyle {\frac {1}{4\lambda ^{2}}}+{\frac {1}{4\lambda ^{2}}}-1=0\ ,}$

so

${\displaystyle \lambda =\pm {\frac {1}{\sqrt {2\ }}}\ ,}$

which implies that the stationary points of ${\displaystyle {\mathcal {L}}}$ are

${\displaystyle \left({\tfrac {\sqrt {2\ }}{2}},{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {1}{\sqrt {2\ }}}\right),\qquad \left(-{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {\sqrt {2\ }}{2}},{\tfrac {1}{\sqrt {2\ }}}\right)~.}$

Evaluating the objective function f at these points yields

${\displaystyle f\left({\tfrac {\sqrt {2\ }}{2}},{\tfrac {\sqrt {2\ }}{2}}\right)={\sqrt {2\ }}\ ,\qquad f\left(-{\tfrac {\sqrt {2\ }}{2}},-{\tfrac {\sqrt {2\ }}{2}}\right)=-{\sqrt {2\ }}~.}$

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

Example 2

Illustration of the constrained optimization problem 2

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

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

Example 3

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 of

${\displaystyle f(x,y)=x^{2}y}$

with the condition that the ${\displaystyle \ x\ }$- and ${\displaystyle \ y\ }$-coordinates lie on the circle around the origin with radius ${\displaystyle \ {\sqrt {3\ }}~.}$ That is, subject to the constraint

${\displaystyle g(x,y)=x^{2}+y^{2}-3=0~.}$

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

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

Applying the ordinary Lagrange multiplier method yields

${\displaystyle {\begin{aligned}{\mathcal {L}}(x,y,\lambda )&=f(x,y)+\lambda \cdot g(x,y)\\&=x^{2}y+\lambda (x^{2}+y^{2}-3)\ ,\end{aligned}}}$

from which the gradient can be calculated:

${\displaystyle {\begin{aligned}\nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )&=\left({\frac {\partial {\mathcal {L}}}{\partial x}},{\frac {\partial {\mathcal {L}}}{\partial y}},{\frac {\partial {\mathcal {L}}}{\partial \lambda }}\right)\\&=\left(2xy+2\lambda x,x^{2}+2\lambda y,x^{2}+y^{2}-3\right)~.\end{aligned}}}$

And therefore:

${\displaystyle \nabla _{x,y,\lambda }{\mathcal {L}}(x,y,\lambda )=0\quad \iff \quad {\begin{cases}2xy+2\lambda x=0\\x^{2}+2\lambda y=0\\x^{2}+y^{2}-3=0\end{cases}}\quad \iff \quad {\begin{cases}x(y+\lambda )=0&{\text{(i)}}\\x^{2}=-2\lambda y&{\text{(ii)}}\\x^{2}+y^{2}=3&{\text{(iii)}}\end{cases}}}$

(iii) is just the original constraint. (i) implies ${\displaystyle \ x=0\ }$ or ${\displaystyle \ \lambda =-y~.}$ If ${\displaystyle x=0}$ then ${\displaystyle \ y=\pm {\sqrt {3\ }}\ }$ by (iii) and consequently ${\displaystyle \ \lambda =0\ }$ from (ii). If ${\displaystyle \ \lambda =-y\ ,}$ substituting this into (ii) yields ${\displaystyle \ x^{2}=2y^{2}~.}$ Substituting this into (iii) and solving for ${\displaystyle \ y\ }$ gives ${\displaystyle \ y=\pm 1~.}$ Thus there are six critical points of ${\displaystyle \ {\mathcal {L}}\$ :}

${\displaystyle ({\sqrt {2\ }},1,-1);\quad (-{\sqrt {2\ }},1,-1);\quad ({\sqrt {2\ }},-1,1);\quad (-{\sqrt {2\ }},-1,1);\quad (0,{\sqrt {3\ }},0);\quad (0,-{\sqrt {3\ }},0)~.}$

Evaluating the objective at these points, one finds that

${\displaystyle f(\pm {\sqrt {2\ }},1)=2;\quad f(\pm {\sqrt {2\ }},-1)=-2;\quad f(0,\pm {\sqrt {3\ }})=0~.}$

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

Note that while ${\displaystyle \ ({\sqrt {2\ }},1,-1)\ }$ is a critical point of ${\displaystyle \ {\mathcal {L}}\ ,}$ it is not a local extremum of ${\displaystyle \ {\mathcal {L}}~.}$ We have

${\displaystyle {\mathcal {L}}\left({\sqrt {2\ }}+\varepsilon ,1,-1+\delta \right)=2+\delta \left(\varepsilon ^{2}+\left(2{\sqrt {2\ }}\right)\varepsilon \right)~.}$

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

Example 4

Entropy

Suppose we wish to find the discrete probability distribution on the points ${\displaystyle \ \{p_{1},p_{2},\ldots ,p_{n}\}\ }$ with maximal information entropy. This is the same as saying that we wish to find the least structured probability distribution on the points ${\displaystyle \ \{p_{1},p_{2},\cdots ,p_{n}\}~.}$ In other words, we wish to maximize the Shannon entropy equation:

${\displaystyle f(p_{1},p_{2},\ldots ,p_{n})=-\sum _{j=1}^{n}p_{j}\log _{2}p_{j}~.}$

For this to be a probability distribution the sum of the probabilities ${\displaystyle \ p_{i}\ }$ at each point ${\displaystyle \ x_{i}\ }$ must equal 1, so our constraint is:

${\displaystyle g(p_{1},p_{2},\ldots ,p_{n})=\sum _{j=1}^{n}p_{j}=1~.}$

We use Lagrange multipliers to find the point of maximum entropy, ${\displaystyle \ {\vec {p}}^{\,*}\ ,}$ across all discrete probability distributions ${\displaystyle \ {\vec {p}}\ }$ on ${\displaystyle \ \{x_{1},x_{2},\ldots ,x_{n}\}~.}$ We require that:

${\displaystyle \left.{\frac {\partial }{\partial {\vec {p}}}}(f+\lambda (g-1))\right|_{{\vec {p}}={\vec {p}}^{\,*}}=0\ ,}$

which gives a system of n equations, ${\displaystyle \ k=1,\ \ldots ,n\ ,}$ such that:

${\displaystyle \left.{\frac {\partial }{\partial p_{k}}}\left\{-\left(\sum _{j=1}^{n}p_{j}\log _{2}p_{j}\right)+\lambda \left(\sum _{j=1}^{n}p_{j}-1\right)\right\}\right|_{p_{k}=p_{\star k}}=0~.}$

Carrying out the differentiation of these n equations, we get

${\displaystyle -\left({\frac {1}{\ln 2}}+\log _{2}p_{\star k}\right)+\lambda =0~.}$

This shows that all ${\displaystyle \ p_{\star k}\ }$ are equal (because they depend on λ only). By using the constraint

${\displaystyle \sum _{j}p_{j}=1\ ,}$

we find

${\displaystyle p_{\star k}={\frac {1}{n}}~.}$

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

Example 5

Numerical optimization

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 at saddle points, rather than at local maxima (or minima).^[4][17] Unfortunately, many numerical optimization techniques, such as hill climbing, gradient descent, some of the quasi-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 the gradient of the Lagrangian as below), or else use an optimization technique that finds stationary points (such as Newton's method without an extremum seeking line search) and not necessarily extrema.

As a simple example, consider the problem of finding the value of x that minimizes ${\displaystyle \ f(x)=x^{2}\ ,}$ constrained such that ${\displaystyle \ x^{2}=1~.}$ (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:

${\displaystyle {\mathcal {L}}(x,\lambda )=x^{2}+\lambda (x^{2}-1)~.}$

The two critical points occur at saddle points where x = 1 and x = −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:

${\displaystyle {\begin{aligned}&{\frac {\partial {\mathcal {L}}}{\partial x}}=2x+2x\lambda \\[5pt]&{\frac {\partial {\mathcal {L}}}{\partial \lambda }}=x^{2}-1~.\end{aligned}}}$

If the target function is not easily differentiable, the differential with respect to each variable can be approximated as

${\displaystyle {\begin{aligned}{\frac {\ \partial {\mathcal {L}}\ }{\partial x}}\approx {\frac {{\mathcal {L}}(x+\varepsilon ,\lambda )-{\mathcal {L}}(x,\lambda )}{\varepsilon }},\\[5pt]{\frac {\ \partial {\mathcal {L}}\ }{\partial \lambda }}\approx {\frac {{\mathcal {L}}(x,\lambda +\varepsilon )-{\mathcal {L}}(x,\lambda )}{\varepsilon }},\end{aligned}}}$

where ${\displaystyle \ \varepsilon \ }$ 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:

${\displaystyle {\begin{aligned}h(x,\lambda )&={\sqrt {(2x+2x\lambda )^{2}+(x^{2}-1)^{2}\ }}\\[4pt]&\approx {\sqrt {\left({\frac {\ {\mathcal {L}}(x+\varepsilon ,\lambda )-{\mathcal {L}}(x,\lambda )\ }{\varepsilon }}\right)^{2}+\left({\frac {\ {\mathcal {L}}(x,\lambda +\varepsilon )-{\mathcal {L}}(x,\lambda )\ }{\varepsilon }}\right)^{2}\ }}~.\end{aligned}}}$

(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 of h occur at x = 1 and x = −1, just as in ${\displaystyle \ {\mathcal {L}}~.}$ Unlike the critical points in ${\displaystyle \ {\mathcal {L}}\ ,}$ however, the critical points in h occur at local minima, so numerical optimization techniques can be used to find them.

Control theory

In optimal control theory, the Lagrange multipliers are interpreted as costate variables, and Lagrange multipliers are reformulated as the minimization of the Hamiltonian, in Pontryagin's minimum principle.

Nonlinear programming

The Lagrange multiplier method has several generalizations. In nonlinear programming there are several multiplier rules, e.g. the Carathéodory–John Multiplier Rule and the Convex Multiplier Rule, for inequality constraints.^[18]

Power systems

Methods based on Lagrange multipliers have applications in power systems, e.g. in distributed-energy-resources (DER) placement and load shedding.^[19]

Safe Reinforcement Learning

The method of Lagrange multipliers applies to constrained Markov decision processes.^[20] It naturally produces gradient-based primal-dual algorithms in safe reinforcement learning.^[21]