Calculus of Variations

Wikipedia Article


Extremize over ff : S=XL(x,f(x),f(x)) dxS = \int_{\mathcal{X}} \mathcal{L}\big(x, f(x), \nabla f(x)\big) {~\rm d}{x} Specifically, we wish to solve for this extremization in terms of the functional derivative δSδf=ddϵS(f+ϵv)ϵ=0=0\frac{\delta S}{\delta f} = \frac{\rm d}{{\rm d}\epsilon} S(f + \epsilon v) \big \rvert_{\epsilon{=}0}= 0 where the function v(x)v(x) is arbitrary up being zero on the boundary of X\mathcal{X} and differentiable. xX,v(x)=0\forall x \in \partial \mathcal{X}, \quad v(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] dx;fi=fxi=dϵX[Lfv+iLfivxi] dx+O(dϵ2)\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}{x} \quad ; \quad f^{\prime}_{i} = \frac{\partial f}{\partial x_{i}} \\ &= {\rm d}\epsilon \int_{\mathcal{X}} \bigg[ \frac{\partial \mathcal{L}}{\partial f} v + \sum_{i} \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}} \frac{\partial v}{\partial x_{i}} \bigg] {~\rm d}{x} + \mathcal{O}({\rm d}\epsilon^{2}) \end{align}

Integrating by parts,

dSdϵ=XLfiv(x)=0 dx+Xv(x)[Lfixi(Lfi)] dx\begin{align} \frac{{\rm d} S}{{\rm d} \epsilon} = \int_{\partial \mathcal{X}} \frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}} \underbrace{v(x)}_{=0} {~\rm d}{x} + \int_{\mathcal{X}} v(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}{x} \end{align}

The only way to guarantee that this derivative is zero for all permissible vv 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}}


Consider now S=XYL(x,y,f(x),f(y),f(x),f(y)) dydxS = \int_{\mathcal{X}} \int_{\mathcal{Y}} \mathcal{L}\big(x, y, f(x), f(y), \nabla f(x), \nabla f(y) \big) {~\rm d}{y}{\rm d}{x} where X=Y\mathcal{X} = \mathcal{Y} . Using the same prescription to vary ff as above, df=vdϵ{\rm d} f = v {\rm d}\epsilon For extremal SS , we obtain

dSdϵ=XYv(x)[Lf(x)ixi(Lfi(x))]+v(y)[Lf(y)iyi(Lfi(y))]dydx=0\begin{align*} \frac{{\rm d} S}{{\rm d} \epsilon} = \int_{\mathcal{X}} \int_{\mathcal{Y}} v(x) &\bigg[ \frac{\partial \mathcal{L}}{\partial f(x)} - \sum_{i} \frac{\partial }{\partial x_{i}} \bigg(\frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}(x)}\bigg) \bigg] \\ \quad + v(y) &\bigg[ \frac{\partial \mathcal{L}}{\partial f(y)} - \sum_{i} \frac{\partial }{\partial y_{i}} \bigg(\frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}(y)}\bigg) \bigg] {\rm d}y {\rm d}x = 0 \end{align*}

Since vv is arbitrary up to boundary conditions, we may treat the values of vv at xx and yy as independent when xyx \neq y , but we may not do so when x=yx = y . When the event x=yx = y has measure 0, we arrive at the independent equations:

Y[Lf(x)ixi(Lfi(x))]dy=0X[Lf(y)iyi(Lfi(y))]dx=0\begin{align} \int_{\mathcal{Y}} \bigg[ \frac{\partial \mathcal{L}}{\partial f(x)} - \sum_{i} \frac{\partial }{\partial x_{i}} \bigg(\frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}(x)}\bigg) \bigg] {\rm d}y = 0 \\ \int_{\mathcal{X}} \bigg[ \frac{\partial \mathcal{L}}{\partial f(y)} - \sum_{i} \frac{\partial }{\partial y_{i}} \bigg(\frac{\partial \mathcal{L}}{\partial f^{\prime}_{i}(y)}\bigg) \bigg] {\rm d}x = 0 \end{align}