Functionals

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 2019/04/15   Share

Functionals

Just like functions take variables as arguments, functionals take functions as arguments.

In the following sections, the functional of the form

$G[y]=\int_\alpha^\beta f(y,y' ; x) \, dx$

is considered.

Functional derivatives

Definition:

$\begin{align} \delta G &= G[y+\delta y]-G[y]\\ &=\int_\alpha^\beta f(y+\delta y,y'+\delta y'; x)- f(y,y' ; x) \, dx\\ &=\int_\alpha^\beta [f(y,y' ; x)+\frac{\partial f}{\partial y}(\delta y)+\frac{\partial f}{\partial y'}(\delta y)']- f(y,y' ; x) \, dx && \mathrm{Taylor \ expansion}\\ &=\int_\alpha^\beta \frac{\partial f}{\partial y}(\delta y) -\frac{d}{dx}(\frac{\partial f}{\partial y'})(\delta y)\, dx +[\delta y \frac{\partial f}{\partial y'}]_\alpha^\beta \end{align}$

If y is fixed on the boundary, the B.T is 0, then

$\delta G=\int_\alpha^\beta [\frac{\partial f}{\partial y} -\frac{d}{dx}(\frac{\partial f}{\partial y'})](\delta y)\, dx \tag{1}\label{fcndif}$

and define the functional derivative to be

$\frac{\delta G}{\delta y(x)}=\frac{\partial f}{\partial y} -\frac{d}{dx}(\frac{\partial f}{\partial y'}) \tag{2}\label{fcnd}$

Stationary when $\frac{\delta G}{\delta y(x)}=0$ , i.e.

$\frac{\partial f}{\partial y} =\frac{d}{dx}(\frac{\partial f}{\partial y'})$

This is the Euler Lagrange equation.

Special cases

No dependence on $y$

If $\frac{\partial f}{\partial y}=0$ , then $\frac{\partial f}{\partial y'}$ is const.

No dependence on $x$ - First integral

If $\frac{\partial f}{\partial x}=0$ , use chain rule,

$\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial y'}\frac{d^2y}{dx^2}$

Use the EL equation

$\frac{df}{dx}=\frac{\partial f}{\partial x}+y'\frac{d}{dx}(\frac{\partial f}{\partial y'})+y''\frac{\partial f}{\partial y'}=\frac{d}{dx}(y'\frac{\partial f}{\partial y'})\\ \implies \frac{df}{dx}(y'\frac{\partial f}{\partial y'}-f)=0\\ \implies y'\frac{\partial f}{\partial y'}-f=\mathrm{const.}$

Relate to SL

Consider these two functionals

$\begin{align} F[y]&=\int_\alpha^\beta y\mathcal{L}y\, dx\\ &=\int_\alpha^\beta -y\frac{d}{dx}[\rho(x)y']+\sigma(x)y^2\, dx\\ &=\int_\alpha^\beta \rho(x)(y')^2+\sigma(x)y^2\, dx \end{align}$

where the last line used integration by parts, and the boundary term is 0; and

$G[y]=\int_\alpha^\beta w(x)y^2\, dx$

Then, use the expression $\ref{fcndif}$ , and we get

$\delta F=2\int_\alpha^\beta (\delta y)\mathcal{L}y\, dx\\ \delta G=2\int_\alpha^\beta (\delta y)w(x)y\, dx$

$\frac{\delta F}{\delta y(x)}=2\mathcal{L}y\\ \frac{\delta G}{\delta y(x)}=2wy$

Then consider

$\Lambda[y]=\frac{F[y]}{G[y]}\\ \implies \delta \Lambda=\frac{F[y]+\delta F[y]}{G[y]+\delta G[y]}=\frac{1}{G}(\delta F -\frac{F}{G}\delta G)\\ \implies \frac{\delta \Lambda}{\delta y(x)}=\frac{2}{G}(\mathcal{L}y-\Lambda wy)$

We have it. Extremizing $\Lambda$ is equivalent to $y$ being an eigenfunction, and the extremal values are eigenvalues.

Alternatively derived from Lagrange multipliers.

Variation principles

Fermat’s principle

Minimizes optical path length. (Or not so strictly minimizes the time taken.)

The OPD is defined with the refractive index $\mu(x)$ , through

$P=\int_A^B \mu(\mathbf{r})\,dl$

If no question involves a path that does not double back, use one of the coordinates to parametrize.
Requires the functional derivative w.r.t. the other coordinates to be 0.

Hamilton’s principle

The Lagrangian of a system is $L=T-V$ .

The action functional is

$S=\int_{t_i}^{t_f} L(\{q_i\},\{\dot{q_i}\}\dots,t)\,dt$

where $q$ s are general coordinates.

Hamilton’s principle is that the motion in configuration space extremizes the action functional $S$ .

The E-L equation becomes the Lagrange equation,

$\frac{d}{dt}(\frac{\partial L}{\partial \dot{q_i}})-\frac{\partial L}{\partial q_i}=0$

for all $i$ .

First integral

If $\frac{\partial L}{\partial t}=0$ , then

$\sum_{i=1}^N \dot{q}_i \frac{\partial L}{\partial \dot{q_i}}-L=\mathrm{const.}$

By considering $\frac{d L}{dt}$ .

Generally in this case, $T$ is a homogeneous quadratic in velocities, and $V$ does not depend on velocities, then,

$\sum_{i=1}^N \dot{q}_i \frac{\partial L}{\partial \dot{q_i}}-L=T+V=\mathrm{const.}$

i.e. Energy is conserved.

Constrained variation and Lagrange multipliers

Recall: $df=\nabla f\cdot d \mathbf{x}$ , stationary points at $\nabla f=0$ .

Now the path is not free, but on a given $d\mathbf{l}$ , such that $dp=d\mathbf{l}\cdot\nabla p=0$ (line of const. $p$ ). Therefore it has to be that $\nabla f=\lambda \nabla p$ .

The question becomes to extremize without constraint a function

$\phi(x,y,\lambda)=f(x,y)-\lambda p(x,y)$

where the $\lambda$ is a Lagrange multiplier. More generally, for functions,

$\phi(x,y,\dots,\lambda_1,\lambda_2,\dots)=f(x,y,\dots)-\sum_{i=1}^k\lambda_i p_i(x,y)$

And for functionals,

$\Phi_\lambda [y]=G[y]-\lambda P[y]$

Rayleigh-Ritz

For an SL problem, suppose everything ( $\rho,\ \sigma,\ w$ ) is positive, then $\Lambda \geq 0$ . There exists an absolute minimum in the extremal values $\lambda_0 \geq 0$ .

Therefore, we have

$\Lambda[y]\geq \lambda_0$

for all $y$ satisfying B.C.s, and only equal if $y=y_0$ .

Steps

Put into SL form
Come up with a trial function, with parameters
Compute $F$ and $G$ for the trail function
Minimize $\Lambda$ w.r.t parameters

Then we have the best estimate of the eigenvalue and corresponding values of parameters with this trail function.

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Author：2LalA猪

Permalink：https://butteraddict.fun/2019/04/Functionals/

Published on：15th April 2019, 2:12 pm

Updated on：9th June 2020, 3:24 pm

License：CC BY 4.0

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CATALOG

1. Functionals
2. Functional derivatives
1. 2.1. Special cases
  1. 2.1.1. No dependence on $y$
  2. 2.1.2. No dependence on $x$ - First integral
2. 2.2. Relate to SL
3. Variation principles
1. 3.1. Fermat’s principle
2. 3.2. Hamilton’s principle
  1. 3.2.1. First integral
4. Constrained variation and Lagrange multipliers
5. Rayleigh-Ritz
1. 5.1. Steps

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