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diff --git a/content/know/concept/lagrange-multiplier/index.pdc b/content/know/concept/lagrange-multiplier/index.pdc new file mode 100644 index 0000000..2b14897 --- /dev/null +++ b/content/know/concept/lagrange-multiplier/index.pdc @@ -0,0 +1,103 @@ +--- +title: "Lagrange multiplier" +firstLetter: "L" +publishDate: 2021-03-02 +categories: +- Mathematics +- Physics + +date: 2021-03-02T16:28:42+01:00 +draft: false +markup: pandoc +--- + +# 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: + +$$\begin{aligned} + L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z) +\end{aligned}$$ + +Clearly, $\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$, +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 $f_y + \lambda \phi_y$. + +This 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: + +$$\begin{aligned} + 0 = \dd{L} + = \sum_{n = 1}^N \Big( f_{x_n} + \sum_{m = 1}^M \lambda_m \phi_{x_n} \Big) \dd{x_n} +\end{aligned}$$ |