Categories: Mathematics, Physics.

Lagrange multiplier

The method of Lagrange multipliers or undetermined multipliers is a technique for optimizing (i.e. finding the extrema of) a function $$f(x, y, z)$$, subject to a given constraint $$\phi(x, y, z) = C$$, where $$C$$ is a constant.

If we ignore the constraint $$\phi$$, optimizing $$f$$ simply comes down to finding stationary points:

\begin{aligned} 0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z} \end{aligned}

This problem is easy: $$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$ are independent and arbitrary, so all we need to do is find the roots of the partial derivatives $$f_x$$, $$f_y$$ and $$f_z$$, which we respectively call $$x_0$$, $$y_0$$ and $$z_0$$, and then the extremum is simply $$(x_0, y_0, z_0)$$.

But the constraint $$\phi$$, over which we have no control, adds a relation between $$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$, so if two are known, the third is given by $$\phi = C$$. The problem is then a system of equations:

\begin{aligned} 0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z} \\ 0 &= \dd{\phi} = \phi_x \dd{x} + \phi_y \dd{y} + \phi_z \dd{z} \end{aligned}

Solving this directly would be a delicate balancing act of all the partial derivatives.

To help us solve this, we introduce a “dummy” parameter $$\lambda$$, the so-called Lagrange multiplier, and contruct a new function $$L$$ given by:

\begin{aligned} L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z) \end{aligned}

At the extremum, $$\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$$, so now the problem is a “single” equation again:

\begin{aligned} 0 = \dd{L} = (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z} \end{aligned}

Assuming $$\phi_z \neq 0$$, we now choose $$\lambda$$ such that $$f_z + \lambda \phi_z = 0$$. This choice represents satisfying the constraint, so now the remaining $$\dd{x}$$ and $$\dd{y}$$ are independent again, and we simply have to find the roots of $$f_x + \lambda \phi_x$$ and $$f_y + \lambda \phi_y$$.

In effect, after introducing $$\lambda$$, we have four unknowns $$(x, y, z, \lambda)$$, but also four equations:

\begin{aligned} L_x = L_y = L_z = 0 \qquad \quad \phi = C \end{aligned}

We are only really interested in the first three unknowns $$(x, y, z)$$, so $$\lambda$$ is sometimes called the undetermined multiplier, since it is just an algebraic helper whose value is irrelevant.

This method generalizes nicely to multiple constraints or more variables: suppose that we want to find the extrema of $$f(x_1, ..., x_N)$$ subject to $$M < N$$ conditions:

\begin{aligned} \phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M \end{aligned}

This once again turns into a delicate system of $$M+1$$ equations to solve:

\begin{aligned} 0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N} \\ 0 &= \dd{\phi_1} = \phi_{1, x_1} \dd{x_1} + ... + \phi_{1, x_N} \dd{x_N} \\ &\vdots \\ 0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N} \end{aligned}

Then we introduce $$M$$ Lagrange multipliers $$\lambda_1, ..., \lambda_M$$ and define $$L(x_1, ..., x_N)$$:

\begin{aligned} L = f + \sum_{m = 1}^M \lambda_m \phi_m \end{aligned}

As before, we set $$\dd{L} = 0$$ and choose the multipliers $$\lambda_1, ..., \lambda_M$$ to eliminate $$M$$ of its $$N$$ terms:

\begin{aligned} 0 = \dd{L} = \sum_{n = 1}^N \Big( f_{x_n} + \sum_{m = 1}^M \lambda_m \phi_{x_n} \Big) \dd{x_n} \end{aligned}

References

1. G.B. Arfken, H.J. Weber, Mathematical methods for physicists, 6th edition, 2005, Elsevier.
2. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.