Calculus of Variations

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Local Lagrangian Densities

Extremize over f ⁣:XRf \colon \mathcal{X} \to \mathbb{R}, where xXx \in \mathcal{X}:

S(f)=XL(x,f(x),f(x)) dV,\begin{equation} \label{eqn-1}\tag{1} S(f) = \int_{\mathcal{X}} \mathcal{L}\big(x, f(x), \nabla f(x)\big) {~\rm d}{V}, \end{equation}

where L\mathcal{L} is a real-valued Lagrangian density and VV is the measure for X\mathcal{X}. Specifically, we wish to solve for this extremization in terms of the functional derivative

δSδf=ddϵS(f+ϵg)ϵ=0=0,\begin{equation} \label{eqn-2}\tag{2} \frac{\delta S}{\delta f} = \frac{\rm d}{{\rm d}\epsilon} S(f + \epsilon g) \bigg \rvert_{\epsilon{=}0}= 0, \end{equation}

where the function g ⁣:XRg \colon \mathcal{X} \to \mathbb{R} is arbitrary up being zero on the boundary (denoted \partial) of X\mathcal{X}, and L\mathcal{L}, ff, and gg are differentiable. xX,g(x)=0.\forall x \in \partial \mathcal{X}, \quad g(x) = 0. Let us expand the induced perturbation dS{\rm d}S to first order in dϵ{\rm d}\epsilon about ϵ=0\epsilon=0:

dS=X[Lfdf+iLfidfi] dV=dϵX[Lfg+iLfigxi] dV,\begin{align*} {\rm d}S &= \int_{\mathcal{X}} \bigg[ \frac{\partial \mathcal{L}}{\partial f} {\rm d} f + \sum_{i} \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}} {\rm d} f^{\prime}_{i} \bigg] {~\rm d}{V} \\ &= {\rm d}\epsilon \int_{\mathcal{X}} \bigg[ \frac{\partial \mathcal{L}}{\partial f} g + \sum_{i} \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}} \frac{\partial g}{\partial x_{i}} \bigg] {~\rm d}{V}, \tag{3} \end{align*}

where fifxif^{\prime}_{i} \equiv \frac{\partial f}{\partial x_{i}}.

We next integrate by parts, where, by Generalized Stokes’s Theorem,

Xαydn^=X(αy) dV=X(α)y dV2nd term in (3)+Xα(y) dV.\begin{equation*} \int_{\partial \mathcal{X}} \alpha \mathbf{y} \cdot {\rm d}\hat{n} = \int_{\mathcal{X}} \nabla \cdot (\alpha \mathbf{y}) ~{\rm d}V = \underbrace{\int_{\mathcal{X}} (\nabla \alpha) \cdot \mathbf{y} ~{\rm d}V}_{\text{2nd term in (3)}} + \int_{\mathcal{X}} \alpha (\nabla \cdot \mathbf{y}) ~{\rm d}V. \end{equation*}

for scalar α\alpha, vector y\mathbf{y}, and unit vector n^\hat{n} normal to X\partial \mathcal{X}.

In our case,

dSdϵ=Xig(x)0 on XLfi dn^i+Xg(x)[Lfixi(Lfi)] dV.\begin{align*} \frac{{\rm d} S}{{\rm d} \epsilon} = \int_{\partial \mathcal{X}} \sum_{i} \underbrace{g(x)}_{0 \text{ on } \partial \mathcal{X}} \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}} {~\rm d}{\hat{n}_{i}} + \int_{\mathcal{X}} g(x) \bigg[ \frac{\partial \mathcal{L}}{\partial f} - \sum_{i} \frac{\partial }{\partial x_{i}} \bigg(\frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}}\bigg) \bigg] {~\rm d}{V}. \end{align*}

The only way to guarantee that this derivative is zero for all permissible gg is for the Euler-Lagrange equation to hold: Lfixi(Lfi)=0;fi=fxi.\frac{\partial \mathcal{L}}{\partial f} - \sum_{i} \frac{\partial }{\partial x_{i}} \bigg(\frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}}\bigg) =0 \quad ; \quad f^{\prime}_{i} = \frac{\partial f}{\partial x_{i}}.

Non-Local Lagrangian Densities

Consider now a situation in which the Lagrangian density L\mathcal{L} depends on the joint behavior of ff at nn disjoint points xaXx^{a} \in \mathcal{X} indexed by a[n]a \in [n], such that the relationship between these points is captured by [xa]MXn[x^{a}] \in \mathcal{M} \subseteq \mathcal{X}^{n}. That is, we seek to extremize S(f)=ML([xa],[f(xa)],[f(xa)]) dVn.S(f) = \int_{\mathcal{M}} \mathcal{L}\big([x^{a}], [f(x^{a})], [\nabla f(x^{a})] \big) {~\rm d}{V^{n}}. S(f)S(f) depends on non-local properties of the function ff, and M\mathcal{M} determines any specific constraints or symmetries of the nn points.

Using the same prescription to vary ff as above, let df=gdϵ.{\rm d} f = g {\rm d}\epsilon.

This time, however, rather than fixing the value of gg on X\partial \mathcal{X}, we will assume either a periodic structure for M\mathcal{M}, or a Lagrangian density L\mathcal{L} that does not depend on f(xa)\nabla f(x^{a}) for all aa, in order to guarantee

g,Ma[nG](g(xa)Lfi(xa)) dVn=0.\begin{equation} \label{eqn-3}\tag{3} \forall g, \quad \int_{\mathcal{M}} \sum_{a \in [nG]} \nabla \cdot \bigg(g(x^{a}) \frac{\partial \mathcal{L}}{f_{i}^{\prime}(x^{a})}\bigg) {~\rm d}{V^{n}} = 0. \end{equation}

Let us determine the perturbation dS{\rm d}S to first order in dϵ{\rm d}\epsilon by considering perturbations of ff and ff^{\prime} at each xax^{a} separately:

dSdϵ=Ma[n][Lf(xa)g(xa)+iLfi(xa)(xiag(xa))]dVn.\begin{equation*} \frac{{\rm d} S}{{\rm d} \epsilon} = \int_{\mathcal{M}} \sum_{a \in [n]}\bigg[ \frac{\partial \mathcal{L}}{\partial f(x^{a})} g(x^{a}) + \sum_{i} \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}(x^{a})} \bigg( \frac{\partial}{\partial x^{a}_{i}}g(x^{a})\bigg) \bigg] {\rm d}V^{n}. \end{equation*}

We again integrate by parts, discarding the boundary term by previous assumption (Eq. (3)), and set dS/dϵ=0{\rm d}S/{\rm d}\epsilon = 0 at the desired extremum. We obtain

Ma[n]g(xa)[Lf(xa)ixia(Lfi(xa))]dVn=0.\begin{equation*} \int_{\mathcal{M}} \sum_{a \in [n]} g(x^{a}) \bigg[ \frac{\partial \mathcal{L}}{\partial f(x^{a})} - \sum_i \frac{\partial}{\partial x^{a}_{i}} \bigg( \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}(x^a)} \bigg) \bigg] {\rm d}V^{n} = 0. \end{equation*}

Let us be mathematically explicit about how this equation must hold for all gg. Specifically, consider perturbations of the form g(x)=δ(xy)g(x) = \delta(x - y) where δ\delta is the Dirac delta, for any yy. It follows that

y,a[n]Mxa=y[Lf(xa)xa=yiyi(Lfi(xa)xa=y)]dVn1=0.\begin{equation} \label{eqn-4}\tag{4} \forall y, \quad \sum_{a \in [n]} \int_{\mathcal{M} | x^{a} = y} \bigg[ \frac{\partial \mathcal{L}}{\partial f(x^{a})}\Big|_{x^{a}=y} - \sum_i \frac{\partial}{\partial y_{i}} \bigg( \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}(x^{a})} \Big|_{x^{a}=y} \bigg) \bigg] {\rm d}V^{n-1} = 0. \end{equation}

A Sanity Check

Consider the problem min12(f(x)+f(xπ/2))2L(x,f,f)dx.\min \int_{-\infty}^{\infty} \underbrace{\frac{1}{2}\Big( f(x) + f^{\prime}(x-\pi/2) \Big)^{2}}_{\mathcal{L}(x, f, f^{\prime})} {\rm d}x. We know that translations of f=sinf = \sin minimize this objective, though we have not excluded other possibilities.

To solve this problem with variational calculus, we redefine the Lagrangian density as L=12(f(x)+f(z))2,\mathcal{L} = \frac{1}{2}\Big( f(x) + f^{\prime}(z) \Big)^{2}, and integrate over the manifold M={(x,z) ⁣:z=xπ/2}.\mathcal{M} = \big\{(x, z) \colon z = x - \pi / 2 \big\}.

Applying Eq. (4), we have

y,[f(y)+f(z)]z=yπ/2+[(f(x)+f(y))]y=xπ/2=0.\begin{equation*} \forall y, \quad \bigg[ f(y) + f^{\prime}(z) \bigg]_{z=y-\pi/2} + \bigg[ -\Big( f^{\prime}(x) + f^{\prime\prime}(y) \Big) \bigg]_{y=x-\pi/2} = 0. \end{equation*}


f(y)=f(y)+f(yπ/2)f(y+π/2).\begin{equation*} f^{\prime\prime}(y) = f(y) + f^{\prime}(y-\pi/2) - f^{\prime}(y+\pi/2). \end{equation*}

Indeed, translations of sin\sin solve this equation, as hoped. Without ruling out other solutions, Eq. (4) at least passes this test.