Categories: Mathematics, Physics.

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:

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

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*Applied mathematics for physicists: lecture notes*, 2019, unpublished.

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