Dictionary:Calculus of variations

{{#category_index:C|calculus of variations}} The mathematics of finding a function that will maximize (or minimize) a definite integral.

Mathematical background

A 'function' is a mapping from a space of numbers to another space of numbers. A mathematical object known as a 'functional' is a mapping of a space of functions to a number. A definite integral is one such example of a functional.

Physics background

In mathematical physics there are problems which require that a path or surface represents the minimization of conserved physical quantities.

Four classic problems of variational calculus are the catenary, brachistochrone, geodesic, and the minimal surface problems.

Lagrangian formulation of the classical action

In general, all such problems are reduced to the minimization of the classical action $S$ is defined by the integral

$S=\int _{A}^{B}L(x,{\dot {x}},t)\;dt.$

in such a way that the optimal function $x(t)$ is produced. Here ${\dot {x}}\equiv dx/dt$. The function $L(x,{\dot {x}},t)$ is a scalar-valued function, but $x$ and ${\dot {x}}$ may be in $R^{n}$ where $n$ is greater than 1.

Application of Taylor's theorem

Regarding the application of Taylor's theorem, we may write $L(x+\delta x,{\dot {x}}+\delta {\dot {x}},t)$ as

$L(x+\delta x,{\dot {x}}+\delta {\dot {x}},t)=L(x,{\dot {x}},t)+\left[{\frac {\partial L}{\partial x}}\delta x+{\frac {\partial L}{\partial {\dot {x}}}}\delta {\dot {x}}\right]+{\frac {1}{2!}}\left\{{\frac {\partial ^{2}L}{\partial x^{2}}}\left(\delta x\right)^{2}+2{\frac {\partial ^{2}L}{\partial x\partial {\dot {x}}}}\delta x\delta {\dot {x}}+{\frac {\partial ^{2}L}{\partial {\dot {x}}^{2}}}\left(\delta {\dot {x}}\right)^{2}\right\}+O(\epsilon ^{3}).$

Here $\delta x=x(t)-h(t)$ such that $|x-h|<\epsilon <<1$. Terms in square brackets $[...]$ is commonly called the first variation and the part in braces $\{...\}$ is commonly called the second variation.

The process of solving a given problem is to minimize the first variation of the classical action which is given formally as

$\delta S=\int _{A}^{B}\left[L(x+\delta x,{\dot {x}}+\delta {\dot {x}},t)-L(x,{\dot {x}},t)\right]\;dt\sim \int _{A}^{b}\left[{\frac {\partial L}{\partial x}}\delta x+{\frac {\partial L}{\partial {\dot {x}}}}\delta {\dot {x}}\right]\;dt.$

The problem then is to find the curve $x(t)$ which satisfies the condition that $\delta S=0$ and that all of the potential solution curves have the same values at the endpoints of integration $A$ and $B$. This means that the variations of the solution $\delta x(A)=0$ and $\delta x(B)=0.$

The Euler-Lagrange equations

The integral representation of the first variation of the action

$\delta S=\int _{A}^{B}\left[{\frac {\partial L}{\partial x}}\delta x+{\frac {\partial L}{\partial {\dot {x}}}}\delta {\dot {x}}\right]\;dt$

may be simplified by performing integration by parts on the second term in the integrand, wherein the $\delta {\dot {x}}$ factor is integratied to yield

${\frac {\partial L}{\partial {\dot {x}}}}\delta x(t){\bigg |}_{A}^{B}-\int _{A}^{B}{\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}}}\right]\delta x(t)\;dt$

.

The endpoint evaluation vanishes because the endpoints of all of the solution curves are taken to be the same for all possible solutions, hence $\delta x(A)=\delta x(B)=0.$ Substituting the result back into the integral for $\delta S$ we have

$\delta S=\int _{A}^{B}\left\{{\frac {\partial L}{\partial x}}-{\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}}}\right]\right\}\delta x\;dt=0,$

implying that the integrand vanishes

${\frac {\partial L}{\partial x}}-{\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}}}\right]=0.$

Without loss of generality, we may consider $x$ and ${\dot {x}}$ to be in $n$-dimensions, permitting us to write

${\frac {\partial L}{\partial x_{i}}}-{\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}_{i}}}\right]=0,$

where $i=1,2,...,n$.

Solving the Euler-Lagrange equations is equivalent to finding the gradient of the Lagrangian in all of its variables, and setting this equal to zero. The solution curve $x(t)$ is the projection of the minimum path in phase space onto the spatial coordinates (the configuration space), depending on the problem.

Properties of the Lagrangian

The Lagrangian formulation is linear. The Lagrangian of a system is the sum of the Lagrangians of the subsystems that make up that system.

Case I

Case I: $L=L(x,{\dot {x}})$

In this case, the Lagrangian is not explicitly a function of the integration variable.

${\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}}}\right]={\frac {\partial ^{2}L}{\partial t\partial x}}+{\frac {\partial ^{2}L}{\partial x\partial {\dot {x}}}}{\frac {dx}{dt}}+{\frac {\partial ^{2}L}{\partial {\dot {x}}^{2}}}{\frac {d^{2}x}{dt^{2}}}={\frac {\partial ^{2}L}{\partial x\partial {\dot {x}}}}{\frac {dx}{dt}}+{\frac {\partial ^{2}L}{\partial {\dot {x}}^{2}}}{\frac {d^{2}x}{dt^{2}}}$

Alternately

We may write the full time derivative of $L$

${\frac {dL}{dt}}={\frac {\partial L}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial L}{\partial {\dot {x}}}}{\frac {d^{2}x}{dt^{2}}}$

and

${\frac {d}{dt}}\left[{\dot {x}}{\frac {\partial L}{\partial {\dot {x}}}}\right]={\frac {\partial L}{\partial {\dot {x}}}}{\frac {d^{2}x}{dt^{2}}}+{\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}}}\right]{\frac {dx}{dt}}$

Combining the two identities we have

${\frac {d}{dt}}\left[L-{\dot {x}}{\frac {dL}{\partial x}}\right]={\frac {\partial L}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial L}{\partial {\dot {x}}}}{\frac {d^{2}x}{dt^{2}}}-{\frac {\partial L}{\partial {\dot {x}}}}{\frac {d^{2}x}{dt^{2}}}-{\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}}}\right]{\frac {dx}{dt}}=\left({\frac {\partial L}{\partial x}}-{\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}}}\right]\right){\frac {dx}{dt}}=0$

.

The term in parentheses $(...),$ is Euler-Lagrange.

Hence

${\frac {d}{dt}}\left[L-{\dot {x}}{\frac {dL}{\partial {\dot {x}}}}\right]=0$

is equivalent to the Euler-Lagrange equations when $L$ does not depend on the integration variable.

We may immediately write

$L-{\dot {x}}{\frac {dL}{\partial {\dot {x}}}}=c$

where $c$ is a constant.

Case II

Case II - $L=L({\dot {x}})$

The Lagrangian does not depend on $x$ or on the integration variable $t$.

With no explicit dependence of the Lagrangian on

${\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}_{j}}}\right]=0$

implying that

${\frac {\partial L}{\partial {\dot {x}}_{j}}}={\mbox{a constant vector}}.$

Also, because $L$ does not depend on space

${\frac {\partial L}{\partial x_{j}}}=0$

implying that the Lagrangian $L={\mbox{const.}}$.

Physics Ramifications

If we consider the problems of mechanics that different forms of the Lagrangian represent, we see important ramifications regarding the nature of symmetries that lead to the equations of classical mechanics.

Free space, no potential, $L=L({\dot {x}})$

As above this implies that

${\frac {\partial L}{\partial x}}=0$ meaning that $L$ is constant with respect to the spatial coordinat $x$.

We also have from the Euler-Lagrange equations that

${\frac {d}{dt}}\left[{\frac {\partial L}{\partial {\dot {x}}_{j}}}\right]=0$

implying that

${\frac {\partial L}{\partial {\dot {x}}_{j}}}=w_{j}$

,

where $w_{j}$ is a constant vector.

The last result follows because the partial derivative with respect to ${\dot {x}}$ is the gradient with respect to the velocity ${\dot {x}}_{j}=v_{j}$.

Thus, the Lagrangian cannot be a function of the direction of the velocity vector, meaning that the Lagrangian can only be a function of the magnitude of velocity, but not its direction. This is a statement of the isotropy of space.

The result implies that

$L=v_{j}w_{j}+c_{1}$

,

where $c_{1}$ is a constant of integration.

By the isotropy of space

$v_{j}w_{j}=|v||w|\cos \theta$

But because there can be no angular dependence, we may take $w_{j}=Kv_{j}$ and thus the expression for the Lagrangian becomes

$L=Kv_{j}v_{j}+c_{1}$

.

The simplest model is of a particle traveling at a constant velocity, in a constant direction, which is the definition of an inertial frame.

If we define $K=m/2$, where $m$ is mass, then the Lagrangian takes on the form of kinetic energy, to within a constant

$L=1/2mv_{j}v_{j}+c_{1}$

.

Conservation of linear momentum

Substituting this form of the Lagrangian into

${\frac {\partial L}{\partial v_{j}}}=mv_{j}\equiv p_{j}$

where $p_{j}$ is the momentum, which must be the same as the constant vector $w_{j}$ defined in the previous discussion. This is a statement of conservation of linear momentum.

Free Space with a potential field $U(x)$

By the linearity of the Lagrangian, we need only add the potential $-U(x)$ to the previous expression of the Lagrangian to obtain

$L={\frac {1}{2}}mv_{j}v_{j}-U(x_{j})$

.

Solving the Euler-Lagrange equations (here with ${\dot {x}}$ written as $v_{j}$ we have

${\frac {d}{dt}}\left[{\frac {\partial L}{\partial v_{j}}}\right]={\frac {\partial L}{\partial x_{j}}}$

.

Invoking the equivalence of the velocity gradient of L and momentum, we have

${\frac {d}{dt}}\left[mv_{j}\right]=-{\frac {\partial U}{\partial x_{j}}}=-\nabla _{j}U(x_{j}).$

.

Applying the time derivative to the velocity $v_{j}$ we obtain the acceleration $a_{j}$, we have Newton's equation of motion

$ma_{j}=-\nabla _{j}U\equiv F_{j}$

.

Thus, Newton's laws of mechanics follow from the consideration of the homogeneity and isotropy of space and time.

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