Paper 4, Section II, 16D16 \mathrm{D}

Variational Principles | Part IB, 2017

Consider the functional

F[y]=αβf(y,y,x)dxF[y]=\int_{\alpha}^{\beta} f\left(y, y^{\prime}, x\right) d x

of a function y(x)y(x) defined for x[α,β]x \in[\alpha, \beta], with yy having fixed values at x=αx=\alpha and x=βx=\beta.

By considering F[y+ϵξ]F[y+\epsilon \xi], where ξ(x)\xi(x) is an arbitrary function with ξ(α)=ξ(β)=0\xi(\alpha)=\xi(\beta)=0 and ϵ1\epsilon \ll 1, determine that the second variation of FF is

δ2F[y,ξ]=αβ{ξ2[2fy2ddx(2fyy)]+ξ22fy2}dx\delta^{2} F[y, \xi]=\int_{\alpha}^{\beta}\left\{\xi^{2}\left[\frac{\partial^{2} f}{\partial y^{2}}-\frac{d}{d x}\left(\frac{\partial^{2} f}{\partial y \partial y^{\prime}}\right)\right]+\xi^{\prime 2} \frac{\partial^{2} f}{\partial y^{\prime 2}}\right\} d x

The surface area of an axisymmetric soap film joining two parallel, co-axial, circular rings of radius a distance 2L2 L apart can be expressed by the functional

F[y]=LL2πy1+y2dxF[y]=\int_{-L}^{L} 2 \pi y \sqrt{1+y^{\prime 2}} d x

where xx is distance in the axial direction and yy is radial distance from the axis. Show that the surface area is stationary when

y=EcoshxE,y=E \cosh \frac{x}{E},

where EE is a constant that need not be determined, and that the stationary area is a local minimum if

L/EL/E(ξ2ξ2)sech2zdz>0\int_{-L / E}^{L / E}\left(\xi^{\prime 2}-\xi^{2}\right) \operatorname{sech}^{2} z d z>0

for all functions ξ(z)\xi(z) that vanish at z=±L/Ez=\pm L / E, where z=x/Ez=x / E.

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